Rotations in \(\mathbb{R}^3\).

1. We have \(\vec{x}\cdot\vec{y} = \vec{x}^T I \vec{y}\). Using \(\vec{x}' = Q\vec{x}\) and \(\vec{y}' = Q\vec{y}\) we find \[\vec{x}'\cdot\vec{y}' = (Q\vec{x})^T I Q \vec{y} = \vec{x}^T Q^T I Q \vec{y} ~.\] Setting this equal to \(\vec{x}^T I \vec{y}\) for all \(\vec{x}\) and \(\vec{y}\) it follows that \[\begin{equation} \label{eq:so3constraintsols} Q^T I Q = I ~. \end{equation}\]

2. Since determinant of a product of matrices is the product of the determinants we have \[\det(Q^T I Q) = \det Q^T \det I \det Q = (\det Q)^2 ~,\] where we have also used that \(\det Q^T = \det Q\) and \(\det I = 1\) since \(I\) is the identity matrix. Therefore, taking the determinant of eq. \(\eqref{eq:so3constraintsols}\) we find \[(\det Q)^2 = 1 \qquad \Rightarrow \qquad \det Q = \pm 1 ~,\] as required.

3. Taking \(Q = e^{\theta \mathcal{J}}\) and expanding the exponential for small \(\theta\) gives \[Q = e^{\theta \mathcal{J}} = I + \theta \mathcal{J} + \mathcal{O}(\theta^2) ~.\] Expanding the constraint \(\eqref{eq:so3constraintsols}\) and dropping terms of \(\mathcal{O}(\theta^2)\) we find \[I + \theta \mathcal{J}^T + \theta \mathcal{J} = I ~,\] which implies \[\begin{equation} \label{eq:so3algconstraintsols} \mathcal{J}^T + \mathcal{J} = 0 ~, \end{equation}\] hence \(\mathcal{J}\) is a \(3\times3\) antisymmetric matrix.

4. To see explicitly that there are three independent \(3\times3\) antisymmetric matrices we can consider the constraint equation \(\eqref{eq:so3algconstraintsols}\) in component form \[\mathcal{J}_{ij} + \mathcal{J}_{ji} = 0 ~,\] where \(i,j=1,2,3\). Setting \(j=i\) this immediately tells us that \(\mathcal{J}_{11} = \mathcal{J}_{22} = \mathcal{J}_{33} = 0\), while for \(j<i\) we have \(\mathcal{J}_{21} = - \mathcal{J}_{12}\), \(\mathcal{J}_{31} = -\mathcal{J}_{13}\) and \(\mathcal{J}_{32} = - \mathcal{J}_{23}\). For \(i<j\) we find the same set of equations as for \(j<i\). Therefore, the general solution to \(\eqref{eq:so3algconstraintsols}\) takes the form \[\begin{split} \mathcal{J} & = \begin{pmatrix} 0 & \mathcal{J}_{12} & \mathcal{J}_{13} \\ -\mathcal{J}_{12} & 0 & \mathcal{J}_{23} \\ -\mathcal{J}_{13} & -\mathcal{J}_{23} & 0 \end{pmatrix} \\ & = -\mathcal{J}_{23} \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} -\mathcal{J}_{13} \begin{pmatrix} 0 & 0 & -1 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{pmatrix} -\mathcal{J}_{12} \begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} ~, \end{split}\] where we have expanded in the basis \(\eqref{eq:so3basis}\). To see that these generate rotations we consider \[J_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} ~.\] Exponentiating this matrix we obtain the corresponding finite transformation \[\begin{equation} \label{eq:so3expansion} Q(\theta) = e^{\theta J_1} = \sum_{n=0}^{\infty} \frac{1}{n!} (\theta J_1)^n = I + \sum_{n=0}^\infty \frac{1}{(2n+2)!} (\theta J_1)^{2n+2} + \sum_{n=0}^\infty \frac{1}{(2n+1)!} (\theta J_1)^{2n+1} ~, \end{equation}\] where we have expanded the exponential and split the sum into two sums, one over even and the other over odd powers of \(J_1\). Noting that \[\begin{split} (J_1)^2 = - \begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \qquad & \Rightarrow \qquad (J_1)^{2n+2} = (-1)^{n} (J_1)^2 ~, \\ & \Rightarrow \qquad (J_1)^{2n+1} = (-1)^{n} J_1 ~, \end{split}\] for \(n\in\mathbb{N}_0\), the sums in eq. \(\eqref{eq:so3expansion}\) can be simplified as \[\begin{split} &\sum_{n=0}^\infty \frac{1}{(2n+2)!} (\theta J_1)^{2n+2} = (J_1)^2 \sum_{n=0}^\infty \frac{(-1)^n}{(2n+2)!} \theta^{2n+2} = (J_1)^2 (1-\cos\theta) ~, \\ & \sum_{n=0}^\infty \frac{1}{(2n+1)!} (\theta J_1)^{2n+1} = J_1 \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\theta^{2n+1} = J_1 \sin \theta ~. \end{split}\] Therefore, we find \[\begin{split} Q(\theta) = e^{\theta J_1} & = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} + (\cos\theta-1)\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} + \sin\theta \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -1 \\ 0 & 1 & 0 \end{pmatrix} \\ & = \begin{pmatrix} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{pmatrix} ~. \end{split}\] This is a rotation by angle \(\theta\) about the \(x^1\) axis. A similar analysis shows that \(J_2\) and \(J_3\) generate rotations about the \(x^2\) and \(x^3\) axes respectively.

Lorentz transformations.

1. Using \(x' = \Lambda x\) and \(y' = \Lambda y\) we have \[x'{}^T \eta y' = x^T \Lambda ^T \eta \Lambda y~.\] Setting this equal to \(x^T \eta y\) for all \(x\) and \(y\) it follows that \[\begin{equation} \label{eq:so13constraintsol} \Lambda^T \eta \Lambda = \eta ~. \end{equation}\]

2. Taking the determinant of eq. \(\eqref{eq:so13constraintsol}\) we find \[(\det \Lambda)^2 \det \eta = \det \eta ~, \qquad \det \Lambda = \pm 1 ~,\] where we have used that the determinant of a product of matrices is the product of the determinant and \(\det \Lambda^T = \det \Lambda\).

3. Taking \(\Lambda = e^{\omega \mathcal{M}}\) and expanding the exponential for small \(\omega\) gives \[\Lambda =e^{\omega \mathcal{M}} = I +\omega \mathcal{M} + \mathcal{O}(\omega^2) ~.\] Expanding the constraint \(\eqref{eq:so13constraintsol}\) and dropping terms of \(\mathcal{O}(\omega^2)\) we find \[\eta + \mathcal{M}^T \eta + \eta \mathcal{M} = \eta ~,\] which implies \[\mathcal{M}^T \eta + \eta \mathcal{M} = 0 ~.\] Multiplying on the right by \(\eta\) and using \(\eta^2 = I\) gives \[\begin{equation} \label{eq:so13algconstraintsol} \mathcal{M}^T = - \eta \mathcal{M} \eta ~, \end{equation}\] as required.

4. To show that there are six independent solutions to the constraint \(\eqref{eq:so13algconstraintsol}\) let us write it in component form \[\begin{equation} \label{eq:so13algconstraintcomp} \mathcal{M}^\nu{}_\mu = - \eta_{\mu\rho} \mathcal{M}^\rho{}_\sigma \eta^{\sigma\nu} ~. \end{equation}\] Setting \(\mu = \nu\) this immediately tells us that \(\mathcal{M}^0{}_0 = \mathcal{M}^1{}_1 = \mathcal{M}^2{}_2 = \mathcal{M}^3{}_3 = 0\) since \(\eta_{\mu\nu}\) and \(\eta^{\mu\nu}\) are only non-vanishing if \(\mu = \nu\) and \(\eta_{00}\eta^{00} = \eta_{11}\eta^{11} = \eta_{22}\eta^{22} = \eta_{33}\eta^{33} = 1\). For \(\mu < \nu\) we have \(\mathcal{M}^i{}_0 = \mathcal{M}^0{}_i\) and \(\mathcal{M}^j{}_i = - \mathcal{M}^i{}_j\) where \(i,j=1,2,3\) and \(i<j\). For \(\mu > \nu\) we find the same set of equations as for \(\mu < \nu\). It follows that the number of independent solutions is the same as the number of components of \(\mathcal{M}\) with \(\mu < \nu\), which for a \(4\times4\) matrix is six. One way to label a basis with six elements is by a pair of antisymmetrised indices running over four values since \(\binom{4}{2} = 6\). This is a convenient choice since we can take the indices to be \(\rho,\sigma = 0,1,2,3\) and write \[(M^{\rho\sigma})^\mu{}_\nu = \eta^{\sigma\mu}\delta^\rho_\nu - \eta^{\rho\mu}\delta^\sigma_\nu ~.\] The right-hand side is antisymmetric in \(\rho\) and \(\sigma\) as required. Moreover, we have \[-\eta_{\mu\tau}(M^{\rho\sigma})^\tau{}_\upsilon \eta^{\upsilon\nu} =- \eta_{\mu\tau} (\eta^{\sigma\tau}\delta^\rho_\upsilon - \eta^{\rho\tau}\delta^\sigma_\upsilon) \eta^{\upsilon\nu} = - \eta^{\rho\nu} \delta_\mu^\sigma + \eta^{\sigma\nu}\delta_\mu^\rho = - (M^{\rho\sigma})^\nu{}_\mu ~,\] hence the basis elements also satisfy \(\eqref{eq:so13algconstraintcomp}\) as required. Finally, we can explicitly check that there are six independent solutions by writing out the matrices as in eq. \(\eqref{eq:so13basis}\).

5. That the solutions parametrised by \(\omega_{12}\), \(\omega_{13}\) and \(\omega_{23}\) generate rotations around the \(x^3\), \(x^2\) and \(x^1\) axes follows from part iv of the solution [sol11] to problem [prob11].

6. To show that \(\omega_{01}\), \(\omega_{02}\) and \(\omega_{03}\) generate boosts along the \(x^1\), \(x^2\) and \(x^3\) directions let us consider \(M^{01}\) and take \(\omega_{01} = \frac12 \omega\) and \(\omega_{02} = \omega_{03} = \omega_{12} = \omega_{13} = \omega_{23} = 0\) \[\begin{equation} \label{eq:boostex} \Lambda(\omega) = e^{\omega M^{01}} =\sum_{n=0}^\infty \frac{1}{n!} (\omega M^{01})^n = I + \sum_{n=0}^\infty \frac{1}{(2n+2)!} (\omega M^{01})^{2n+2} + \sum_{n=0}^\infty \frac{1}{(2n+1)!} (\omega M^{01})^{2n+1} ~. \end{equation}\] Noting that \[\begin{split} (M^{01})^2 = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \qquad & (M^{01})^{2n+2} = (M^{01})^2 ~, \\ & (M^{01})^{2n+1} = (M^{01}) ~, \end{split}\] for \(n\in\mathbb{Z}_0\), the sums in eq. \(\eqref{eq:boostex}\) can be simplified as \[\begin{split} & \sum_{n=0}^\infty \frac{1}{(2n+2)!} (\omega M^{01})^{2n+2} = (M^{01})^2 \sum_{n=0}^\infty \frac{1}{(2n+2)!} \omega^{2n+2} = (M^{01})^2 (\cosh \omega -1) ~, \\ & \sum_{n=0}^\infty \frac{1}{(2n+1)!} (\omega M^{01})^{2n+1} = M^{01}\sum_{n=0}^\infty \frac{1}{(2n+1)!} \omega^{2n+1} = M^{01} \sinh\omega ~. \end{split}\] Therefore, we find \[\begin{split} \Lambda(\omega) = e^{\omega M^{01}} & = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} + (\cosh \omega -1) \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + \sinh \omega \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \\ & = \begin{pmatrix} \cosh\theta & \sinh\theta & 0 & 0 \\ \sinh\theta & \cosh\theta & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} ~. \end{split}\] This is a boost by rapidity \(\omega\) along the \(x^1\) direction. To recover the usual form, we can set \(\sinh\omega = -\gamma v\) and \(\cosh\omega = \gamma\) where \(\gamma = \frac{1}{\sqrt{1-v^2}}\) to give \[\Lambda(v) = \begin{pmatrix} \gamma & -\gamma v & 0 & 0 \\ -\gamma v & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} ~.\] A similar analysis shows that \(M^{02}\) and \(M^{03}\) generate boosts along the \(x^2\) and \(x^3\) axes.

7. Starting from \(\eqref{eq:so13algconstraintsol}\) and multiplying on the left by \(\eta^T\) we find \[\eta^T \mathcal{M}^T = - \mathcal{M} \eta ~,\] where we have used \(\eta\eta^T = I\). This can be rewritten as \[(\mathcal{M}\eta)^T = - (\mathcal{M}\eta) ~,\] hence \(\mathcal{M}\eta\) is antisymmetric.

8. Taking the general solution in eq. \(\eqref{eq:so13general}\) \[\omega_{\mu\nu}M^{\mu\nu} = \begin{pmatrix} 0 & 2\omega_{01} & 2\omega_{02} & 2\omega_{03} \\ 2\omega_{01} & 0 & -2\omega_{12} & -2\omega_{13} \\ 2\omega_{02} & 2\omega_{12} & 0 & -2\omega_{23} \\ 2\omega_{03} & 2\omega_{13} & 2\omega_{23} & 0 \end{pmatrix} ~,\] and multiplying on the left by \(\eta\) we find \[\begin{pmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 2\omega_{01} & 2\omega_{02} & 2\omega_{03} \\ 2\omega_{01} & 0 & -2\omega_{12} & -2\omega_{13} \\ 2\omega_{02} & 2\omega_{12} & 0 & -2\omega_{23} \\ 2\omega_{03} & 2\omega_{13} & 2\omega_{23} & 0 \end{pmatrix} = \begin{pmatrix} 0 & -2\omega_{01} & -2\omega_{02} & -2\omega_{03} \\ 2\omega_{01} & 0 & -2\omega_{12} & -2\omega_{13} \\ 2\omega_{02} & 2\omega_{12} & 0 & -2\omega_{23} \\ 2\omega_{03} & 2\omega_{13} & 2\omega_{23} & 0 \end{pmatrix} ~,\] which is explicitly antisymmetric as expected.

Lorentz transformations of fields.

1. Having defined \(\Lambda x = x + \delta x + \mathcal{O}(\omega^2)\), we can substitute in for \(\Lambda = I +\omega_{\rho\sigma}M^{\rho\sigma} + \mathcal{O}(\omega^2)\) to give \[x + \omega_{\rho\sigma}M^{\rho\sigma} x + \mathcal{O}(\omega^2) = x + \delta x + \mathcal{O}(\omega^2) ~,\] where we have used that \(Ix = x\). The leading term \(x\) cancels and we identify \(\delta x\) to be the term linear in the infinitesimal parameter \(\omega_{\rho\sigma}\) \[\delta x = \omega_{\rho\sigma}M^{\rho\sigma} x ~.\] In index notation we recall that the vector \(x\) is written \(x^\mu\). Therefore, \[\delta x^\mu = \omega_{\rho\sigma} (M^{\rho\sigma} x)^\mu ~.\] Moreover, while \(\omega_{\rho\sigma}\) does not carry additional indices, \(M^{\rho\sigma}\) is a \(4\times 4\) matrix, hence \[\delta x^\mu = \omega_{\rho\sigma} (M^{\rho\sigma})^\mu{}_\nu x^\nu ~.\]

2. Under Lorentz transformations \(\phi(x) \to \phi'(x)\) where \[\phi'(x') = \phi(x) ~, \qquad x' = \Lambda x ~.\] Expanding in powers of \(\omega_{\rho\sigma}\), we have \[\begin{split} \phi'(x') & = \phi'(\Lambda x) = \phi'(x + \omega_{\rho\sigma}M^{\rho\sigma} x + \mathcal{O}(\omega^2)) \\ & = \phi'(x) + \omega_{\rho\sigma}(M^{\rho\sigma} x) \cdot \frac{\partial}{\partial x} \phi'(x) + \mathcal{O}(\omega^2) ~, \\ & = \phi(x) + \delta\phi(x) + \omega_{\rho\sigma}(M^{\rho\sigma} x) \cdot \frac{\partial}{\partial x} \phi(x) + \mathcal{O}(\omega^2) ~. \end{split}\] Setting this equal to \(\phi(x)\) and requiring the \(\mathcal{O}(\omega)\) term to vanish we find \[\delta\phi(x) = \omega_{\rho\sigma} L^{\rho\sigma} ~,\] where \[L^{\rho\sigma} = - (M^{\rho\sigma} x) \cdot \frac{\partial}{\partial x} ~.\] In index notation we have \[L^{\rho\sigma} = - (M^{\rho\sigma} x)^\mu \frac{\partial}{\partial x^\mu} = -(M^{\rho\sigma})^\mu{}_\nu x^\nu \frac{\partial}{\partial x^\mu} ~,\] where we recall that \(\frac{\partial}{\partial x^\mu} = \partial_\mu\) transforms as an object with a lowered index under Lorentz transformations.

3. Substituting in for \((M^{\rho\sigma})^\mu{}_\nu\) in the expression for \(L^{\rho\sigma}\) gives \[L^{\rho\sigma} = -(M^{\rho\sigma})^\mu{}_\nu x^\nu \frac{\partial}{\partial x^\mu} = -(\eta^{\sigma\mu}\delta^\rho_\nu - \eta^{\rho\mu} \delta^\sigma_\nu) x^\nu \frac{\partial}{\partial x^\mu} = x^\sigma \frac{\partial}{\partial x_\rho} - x^\rho \frac{\partial}{\partial x_\sigma} ~.\]

4. To compute the commutator \([L^{\mu\nu},L^{\rho\sigma}]\) let us first compute \[\begin{split} \phantom{}[x^\nu \frac{\partial}{\partial x_\mu},x^\sigma \frac{\partial}{\partial x_\rho}] & = x^\nu \eta^{\mu\sigma} \frac{\partial}{\partial x_\rho} + x^\nu x^\sigma \frac{\partial^2}{\partial x_\mu \partial x_\rho} - x^\sigma \eta^{\rho\nu} \frac{\partial}{\partial x_\mu} - x^\sigma x^\nu \frac{\partial^2}{\partial x_\rho\partial x_\mu} \\ & = \eta^{\mu\sigma} x^\nu \frac{\partial}{\partial x_\rho} - \eta^{\rho\nu} x^\sigma \frac{\partial}{\partial x_\mu} ~, \end{split}\] where we have used that \[\frac{\partial}{\partial x_\mu} x^\nu = \eta^{\mu\rho} \frac{\partial}{\partial x^\rho} x^\nu = \eta^{\mu\rho} \delta_\rho^\nu = \eta^{\mu\nu} ~,\] and that partial derivatives commute. Antisymmetrising the indices \(\mu\) and \(\nu\), and the indices \(\rho\) and \(\sigma\), we find \[\begin{split} \phantom{}[L^{\mu\nu},L^{\rho\sigma}] & = \eta^{\mu\sigma} x^\nu \frac{\partial}{\partial x_\rho} - \eta^{\rho\nu} x^\sigma \frac{\partial}{\partial x_\mu} - \eta^{\nu\sigma} x^\mu \frac{\partial}{\partial x_\rho} + \eta^{\rho\mu} x^\sigma \frac{\partial}{\partial x_\nu} \\ & \quad - \eta^{\mu\rho} x^\nu \frac{\partial}{\partial x_\sigma} + \eta^{\sigma\nu} x^\rho \frac{\partial}{\partial x_\mu} + \eta^{\nu\rho} x^\mu \frac{\partial}{\partial x_\sigma} - \eta^{\sigma\mu} x^\rho \frac{\partial}{\partial x_\nu} \\ & = -\eta^{\nu\rho}(x^\sigma \frac{\partial}{\partial x_\mu}-x^\mu \frac{\partial}{\partial x_\sigma}) +\eta^{\mu\rho}(x^\sigma \frac{\partial}{\partial x_\nu}-x^\nu \frac{\partial}{\partial x_\sigma}) \\ & \quad + \eta^{\nu\sigma}(x^\rho \frac{\partial}{\partial x_\mu}-x^\mu \frac{\partial}{\partial x_\rho}) -\eta^{\mu\sigma}(x^\rho \frac{\partial}{\partial x_\nu}-x^\nu \frac{\partial}{\partial x_\rho}) \\ & = -\eta^{\nu\rho} L^{\mu\sigma} + \eta^{\mu\rho}L^{\nu\sigma} + \eta^{\nu\sigma} L^{\mu\rho} - \eta^{\mu\sigma} L^{\nu\rho} ~. \end{split}\]

5. Under translations \(\phi(x) \to \phi'(x)\) where \[\phi'(x') = \phi(x) ~, \qquad x' = x - a ~.\] Expanding in powers of \(a\), we have \[\begin{split} \phi'(x') & = \phi'(x-a) = \phi'(x) - a \cdot \frac{\partial}{\partial x} \phi'(x) + \mathcal{O}(a^2) \\ & = \phi(x) + \delta\phi(x) - a \cdot \frac{\partial}{\partial x} \phi(x) + \mathcal{O}(a^2) ~. \end{split}\] Setting this equal to \(\phi(x)\) and requiring the \(\mathcal{O}(a)\) term to vanish we find \[\delta\phi(x) = a \cdot \frac{\partial}{\partial x} \phi(x) = a^\mu \frac{\partial}{\partial x^\mu} \phi(x) ~.\]

6. We have \[\begin{split} \phantom{}[P^\mu,L^{\nu\rho}] & = [\frac{\partial}{\partial x_\mu},x^\rho \frac{\partial}{\partial x_\nu}-x^\nu \frac{\partial}{\partial x_\rho}] \\ & = \eta^{\mu\rho} \frac{\partial}{\partial x_\nu} + x^\rho \frac{\partial^2}{\partial x_\mu \partial x_\nu} - \eta^{\mu\nu} \frac{\partial}{\partial x_\rho} - x^\nu \frac{\partial^2}{\partial x_\mu \partial x_\rho} -x^\rho \frac{\partial^2}{\partial x_\mu \partial x_\nu} + x^\nu \frac{\partial^2}{\partial x_\mu \partial x_\rho} \\ & = - \eta^{\mu\nu} P^\rho + \eta^{\mu\rho} P^\nu ~, \end{split}\] and \[\phantom{}[P^\mu,P^\nu] = [\frac{\partial}{\partial x_\mu},\frac{\partial}{\partial x_\nu}] = 0 ~,\] where we have used that partial derivatives commute.

Energy-momentum tensor.

1. Under space-time translations the field \(\phi(x)\) transforms as \[\phi(x) \to \phi'(x) = \phi(x+a) ~.\] To show that the action is invariant under this transformation we note that \[\partial_\mu \phi(x) \to \partial_\mu\phi(x+a) = (\partial_\mu \phi)(x+a) ~,\] since \(a^\mu\) is a constant vector. Therefore, writing \[\mathcal{L}(x) = \mathcal{L}(\phi(x),\partial_\mu\phi(x)) = -\frac12 \partial_\mu\phi(x)\partial^\mu\phi(x) - \frac12 m^2 \phi(x)^2 ~,\] the Lagrangian density transforms as a scalar field, \(\mathcal{L}(x) \to \mathcal{L}(x+a)\), and the action transforms as \[S[\phi] = \int d^4x\, \mathcal{L}(x) \to \int d^4x \, \mathcal{L}(x+a) ~.\] Since the measure \(d^4x\) is invariant under translations, we have that \[\int d^4x \, \mathcal{L}(x+a) = \int d^4x\, \mathcal{L}(x) = S[\phi] ~,\] assuming sufficiently fast fall-off conditions at infinity, hence the action is invariant. We can also consider the infinitesimal version of this analysis starting from \[\delta_a\phi(x) = a^\mu\partial_\mu \phi(x) ~.\] The variation of the action is \[\begin{split} \delta_a S[\phi] & = \int d^4 x\, \Big(- \partial_\mu \phi(x) \partial^\mu (\delta_a\phi(x)) - m^2 \phi(x)\delta_a\phi(x) \Big) \\ & = \int d^4x\, \Big(- \partial_\mu \phi(x) \partial^\mu (a^\nu\partial_\nu \phi(x)) - m^2 \phi(x)a^\nu\partial_\nu \phi(x) \Big) \\ & = \int d^4x\, a^\nu\partial_\nu\Big(-\frac12 \partial_\mu\phi(x)\partial^\mu\phi(x) - \frac12 m^2 \phi(x)^2\Big) \\ & = \int d^4x \, a^\nu\partial_\nu \mathcal{L}(x) ~. \end{split}\] We again see that the Lagrangian density transforms as a scalar field, \(\delta_a \mathcal{L} = a^\mu\partial_\mu \mathcal{L}\), and the invariance of the action follows from demanding that the fields and their derivatives fall off sufficiently fast at infinity, which means the integral of a total derivative vanishes.

2. We have four parameters that we can vary, which are the four components of \(a^\mu\). Therefore, we expect to find four independent conserved currents associated with this symmetry.

3. To derive the conserved currents we again consider the infinitesimal variation of the action, but we now allow \(a^\mu\) to depend on \(x^\mu\). This gives \[\begin{split} \partial_{a(x)} S[\phi] & = \int d^4 x\, \Big(- \partial_\mu \phi(x) \partial^\mu (\delta_{a(x)}\phi(x)) - m^2 \phi(x)\delta_{a(x)}\phi(x) \Big) \\ & = \int d^4x\, \Big(- \partial_\mu \phi(x) \partial^\mu (a^\nu(x)\partial_\nu \phi(x)) - m^2 \phi(x)a^\nu(x)\partial_\nu \phi(x) \Big) \\ & = \int d^4x\, \Big(a^\nu(x)\partial_\nu \mathcal{L}(x) -\partial^\mu a^\nu(x) \partial_\mu \phi(x) \partial_\nu \phi(x) \Big) \\ & = \int d^4x\, \partial_\mu a^\nu(x) \Big(-\delta^{\mu}_{\nu}\mathcal{L}(x) - \partial^\mu \phi(x) \partial_\nu \phi(x) \Big) ~. \end{split}\] Therefore, we can read off the conserved currents \[T^\mu{}_\nu = \partial^\mu\phi\partial_\nu\phi + \delta^\mu_\nu \mathcal{L}~,\] where by convention we have redefined \(T^\mu{}_\nu \to - T^\mu{}_\nu\) compared to the definition of the conserved current in the lecture notes. Here \(\mu\) is the index of the conserved currents, while the index \(\nu\) labels the four conserved currents. This means that the expected conservation equation is \[\partial_\mu T^\mu{}_\nu = 0~.\] The tensor \(T^\mu{}_\nu\) is known as the energy-momentum tensor. To construct the conserved charges, by convention we raise the index \(\nu\) \[T^{\mu\nu} = \partial^\mu\phi\partial^\nu\phi + \eta^{\mu\nu}\Big(-\frac12 \partial_\rho\phi\partial^\rho\phi - \frac12 m^2 \phi^2\Big) ~,\] set \(\mu = 0\) and integrate over space. We find \[\begin{split} H & = \int d^3 x \, T^{00} = \int d^3x \, \Big(\frac12 (\partial_t \phi)^2 + \frac12 \vec{\nabla}\phi\cdot \vec{\nabla}\phi + \frac12 m^2\phi^2 \Big) ~, \\ P^i & = \int d^3x \, T^{0i} = \int d^3 x \, \big(-\partial_t \phi\partial^i \phi\big) ~. \end{split}\] The conserved charge \(H\) is the Hamiltonian or the energy of the system. This shows that the Hamiltonian is the conserved charge associated with invariance under time translations. The remaining three conserved charges \(P^i\) are momenta, which are conserved due to invariance under space translations.

4. To show the currents are conserved on-shell we recall the Euler-Lagrange equation for the field \(\phi\) \[\begin{equation} \label{eq:eomscalar} \partial_\mu\partial^\mu \phi - m^2\phi = 0 ~. \end{equation}\] We have \[\begin{split} \partial_\mu T^\mu{}_\nu &= \partial_\mu \partial^\mu \phi \partial_\nu \phi + \partial^\mu\phi \partial_\mu\partial_\nu\phi + \delta^{\mu}_{\nu}\Big(-\partial_\rho\phi\partial_\mu\partial^\rho \phi - m^2 \phi \partial_\mu\phi\Big) \\ & = m^2\phi\partial_\nu \phi + \partial^\mu\phi \partial_\mu\partial_\nu\phi -\partial_\rho\phi\partial_\nu\partial^\rho \phi - m^2 \phi \partial_\nu\phi \\ & = 0 ~, \end{split}\] where to go from the first to the second line we have used the Euler-Lagrange equation \(\eqref{eq:eomscalar}\), and to go from the second to the final line we have renamed dummy indices. Therefore, the currents \(T^\mu{}_\nu\) are conserved, which also implies that the Noether charges are constant on-shell \[\frac{d}{dt}H = \frac{d}{dt}P^i = 0 ~.\]

Hamiltonians in gauge theory.

1. To determine the Euler-Lagrange equations for the gauge potential \(A_\mu\) we can either use the standard result \[\partial_\mu\Big(\frac{\partial \mathcal{L}}{\partial(\partial_\mu A_\nu)}\Big) - \frac{\partial\mathcal{L}}{\partial A_\nu} = 0 ~,\] or vary the action directly. Since the former is straightforward, we will follow the latter approach. We have \[\begin{split} \delta_A S[A_\mu] & = \int d^4 x\, \Big(-\frac14 \delta_A F_{\mu\nu}F^{\mu\nu} -\frac14 F_{\mu\nu}\delta_A F^{\mu\nu} \Big) \\ & = \int d^4x \, \Big(-\frac12 F^{\mu\nu}\delta_A F_{\mu\nu} \Big) \\ & = \int d^4x \, \Big(-\frac12 F^{\mu\nu}(\partial_\mu \delta A_\nu - \partial_\nu \delta A_\mu)\Big) \\ & = \int d^4x \, \big(-F^{\mu\nu}\partial_\mu\delta A_\nu\big) \\ & = \int d^4x \, \big( \partial_\mu F^{\mu\nu} \delta A_\nu \big) ~, \end{split}\] where we have used the antisymmetry of \(F^{\mu\nu}\) to go from the third to the fourth line. From here we can read off the Euler-Lagrange equation, which requires that the coefficient of \(\delta A_\nu\) in the integrand vanishes \[\partial_\mu F^{\mu\nu} = 0 ~.\]

2. To find the Hamiltonian we need to split time and space such that the Lagrangian density is written as \[\mathcal{L}= -\frac14(\partial_i A_0 - \partial_0 A_i)(\partial^i A^0 - \partial^0 A^i) -\frac14(\partial_0 A_i - \partial_i A_0)(\partial^0 A^i - \partial^i A^0) - \frac14 F_{ij}F^{ij} ~.\] The first two terms are the same, hence we have \[\mathcal{L}= - \frac12 (\partial_0 A_i - \partial_i A_0)(\partial^0 A^i - \partial^i A^0) - \frac14 F_{ij}F^{ij} ~.\] To compute the canonical momentum we treat \(A_0\) and \(A_i\) separately. For the momentum conjugate to \(A_i\) we have \[\pi^i = \frac{\partial\mathcal{L}}{\partial(\partial_0 A_0)} = \partial^i A^0 - \partial^0 A^i ~,\] which are the components of the electric field. Note that this implies \[\pi_i = \partial_i A^0 - \partial^0 A_i = \partial_0 A_i - \partial_i A_0 ~.\] For the momenta conjugate to \(A_0\) we find \[\pi^0 = \frac{\partial\mathcal{L}}{\partial(\partial_0 A_i)} = 0 ~,\] i.e., the conjugate momenta vanishes identically. That no time derivatives of \(A_0\) appear in the action means that it is not a propagating degree of freedom. Its Euler-Lagrange equation is a constraint equation rather than a time evolution equation. This constraint is the statement of Gauss’ law, i.e. \(\vec{\nabla}.\vec{E} = 0\). We can now compute the Hamiltonian density \[\begin{split} \mathcal{H}& = \pi^i \partial_0 A_i - \mathcal{L} \\ & = \pi^i \partial_0 A_i + \frac12 (\partial_0 A_i - \partial_i A_0)(\partial^0 A^i - \partial^i A^0) + \frac14 F_{ij}F^{ij} \\ & = \pi^i\pi_i - \pi^i \partial_i A_0 - \frac12 \pi^i \pi_i + \frac14 F_{ij}F^{ij} \\ & = \frac12 \pi^i \pi^i - \pi^i \partial_i A_0 + \frac14 F_{ij}F^{ij} ~, \end{split}\] where we have substituted in for the time derivatives of the fields in terms of their conjugate momenta. Finally, we can write the integrated Hamiltonian as \[\begin{split} H & = \int d^3x\, \Big(\frac12 \pi^i \pi^i - \pi^i \partial_i A_0 + \frac14 F_{ij}F^{ij}\Big) \\ & = \int d^3x\, \Big(\frac12 \pi^i \pi^i + \partial_i \pi^i A_0 + \frac14 F_{ij}F^{ij}\Big) ~. \end{split}\] Note that Hamilton’s equation of motion for \(A_0\), which says that \(\frac{\partial\mathcal{H}}{\partial A_0}\) is equal to the conjugate momentum for \(A_0\), again tells us that \(\partial_i \pi^i = 0\), i.e. \(\vec{\nabla}.\vec{E} = 0\).

Noether’s theorem.

1. The Euler-Lagrange equations for a field theory for two real scalar fields \(\phi_1\) and \(\phi_2\) are \[\begin{equation} \label{eq:eulerlagrangegeneral} \frac{\partial\mathcal{L}}{\partial \phi_a} - \partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_a)}\Big) = 0 ~, \qquad a = 1,2 ~, \end{equation}\] where \(\mathcal{L}\) is the Lagrangian density, which in this case is \[-\frac12 \partial_\mu \phi_1 \partial^\mu \phi_1 -\frac12 \partial_\mu \phi_2 \partial^\mu \phi_2 -\frac12 m^2 (\phi_1^2 + \phi_2^2) - \lambda(\phi_1^2+\phi_2^2)^2 ~.\] Therefore, \[\frac{\partial\mathcal{L}}{\partial \phi_1} = -m^2\phi_1 - 4\lambda(\phi_1^2+\phi_2^2)\phi_1 ~, \qquad \frac{\partial\mathcal{L}}{\partial \phi_2} = -m^2\phi_2 - 4\lambda(\phi_1^2+\phi_2^2)\phi_2 ~.\] and \[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_1)} = - \partial^\mu\phi_1 ~, \qquad \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi_2)} = - \partial^\mu\phi_2 ~.\] Substituting into eq. \(\eqref{eq:eulerlagrangegeneral}\) we find the Euler-Lagrange equations \[\begin{equation} \begin{split} & \partial_\mu\partial^\mu \phi_1 - m^2 \phi_1 - 4 \lambda(\phi_1^2+\phi_2^2)\phi_1 = 0 ~, \\ & \partial_\mu\partial^\mu \phi_2 - m^2 \phi_2 - 4 \lambda(\phi_1^2+\phi_2^2)\phi_2 = 0 ~. \end{split}\label{eq:eulerlagrangeso2} \end{equation}\]

2. Under the continuous transformation \(\eqref{eq:so2symmetry}\) \[\begin{split} \phi_1^2 + \phi_2^2 \to \phi_1'{}^2 + \phi_2'{}^2 & = (\phi_1 \cos\alpha - \phi_2 \sin\alpha)^2 + (\phi_1\sin\alpha + \phi_2 \cos\alpha)^2 \\ & = \phi_1^2 (\cos^2 \alpha + \sin^2\alpha) + \phi_2^2 (\sin^2\alpha + \cos^2\alpha) \\ & \quad +\phi_1\phi_2(-2 \cos\alpha\sin\alpha + 2 \sin\alpha\cos\alpha) \\ & = \phi_1^2 + \phi_2^2 ~, \end{split}\] where we have used that \(\cos^2\alpha + \sin^2 \alpha = 1\). Similarly, we have \[\begin{split} \partial_\mu \phi_1 \partial^\mu \phi_1 + \partial_\mu \phi_2 \partial^\mu \phi_2 \to \, & \partial_\mu \phi'_1 \partial^\mu \phi'_1 + \partial_\mu \phi'_1 \partial^\mu \phi'_1 \\ & = (\partial_\mu\phi_1 \cos\alpha - \partial_\mu\phi_2 \sin\alpha)(\partial^\mu\phi_1 \cos\alpha - \partial^\mu\phi_2 \sin\alpha) \\ & \quad + (\partial_\mu\phi_1\sin\alpha + \partial_\mu\phi_2 \cos\alpha)(\partial^\mu\phi_1\sin\alpha + \partial^\mu\phi_2 \cos\alpha) \\ & =\partial_\mu \phi_1 \partial^\mu \phi_1 (\cos^2 \alpha + \sin^2\alpha) + \partial_\mu \phi_2 \partial^\mu \phi_2(\sin^2\alpha + \cos^2\alpha) \\ & \quad +\partial_\mu\phi_1\partial^\mu\phi_2(-\cos\alpha\sin\alpha + \sin\alpha\cos\alpha) \\ & \quad +\partial_\mu\phi_2\partial^\mu\phi_1(-\sin\alpha\cos\alpha + \cos\alpha\sin\alpha) \\ & = \partial_\mu \phi_1 \partial^\mu \phi_1 + \partial_\mu \phi_2 \partial^\mu \phi_2 ~, \end{split}\] where we have used that \(\alpha\) is a constant parameter. It immediately follows that the action \(\eqref{eq:acttwoscalar}\) is invariant under the continuous transformation \(\eqref{eq:so2symmetry}\) \[S[\phi_1',\phi_2'] = S[\phi_1,\phi_2] ~.\] Note that the continuous transformation \(\eqref{eq:so2symmetry}\) can be understood as an \(\mathrm{SO}(2)\) rotation of the two-component vector \[\phi = \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} ~,\] where \[\phi \to \phi' = \begin{pmatrix} \phi'_1 \\ \phi'_2 \end{pmatrix} =\begin{pmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha\end{pmatrix} \begin{pmatrix} \phi_1 \\ \phi_2 \end{pmatrix} ~.\] In terms of the vector \(\phi\), the action \(\eqref{eq:acttwoscalar}\) reads \[S[\phi] = \int d^4x\, \Big(-\frac12 \partial_\mu \phi \cdot \partial^\mu \phi - \frac12 m^2 |\phi|^2 - \lambda |\phi|^4\Big) ~.\]

3. To determine the conserved current we consider an infinitesimal transformation and let the parameter \(\alpha\) depend on space-time. Expanding \(\eqref{eq:so2symmetry}\) to first order in \(\alpha\) we find the infinitesimal transformation \[\delta_{\alpha} \phi_1 = -\alpha\phi_2 ~, \qquad \delta_{\alpha} \phi_2 = \alpha \phi_1 ~.\] The conserved current is defined to be the term in the variation of the action proportional to \(\partial_\mu \alpha\). Such a contribution can only come from the kinetic terms, while the potential terms will remain invariant even when \(\alpha\) is allowed to depend on space-time. Therefore, varying the action, we have \[\begin{split} \delta_{\alpha(x)} S[\phi_1,\phi_2] & = \int d^4x\,\big(- \partial_\mu \phi_1 \partial^\mu (\delta_{\alpha(x)} \phi_1) - \partial_\mu \phi_2 \partial^\mu (\delta_{\alpha(x)} \phi_2) \big) \\ & = \int d^4x\,\big(\partial_\mu \phi_1 \partial^\mu (\alpha\phi_2) - \partial_\mu \phi_2 \partial^\mu (\alpha \phi_1) \big) \\ & = \int d^4x\,\big(\phi_2 \partial^\mu \phi_1 - \phi_1 \partial^\mu \phi_2\big)\partial_\mu\alpha ~. \end{split}\] We can then read off the conserved current \[\begin{equation} \label{eq:so2conservedcurrent} j^\mu = \phi_2 \partial^\mu \phi_1 - \phi_1 \partial^\mu \phi_2 ~, \end{equation}\] and write down the conserved charge \[Q = \int d^3x\, j^0 = \int d^3x\, \big(\phi_1 \partial_t \phi_2 - \phi_2 \partial_t \phi_1\big) ~.\]

4. To show that the current is conserved on-shell we note that \[\partial_\mu j^\mu = \partial_\mu (\phi_2 \partial^\mu \phi_1 - \phi_1 \partial^\mu \phi_2) = \phi_2 \partial_\mu\partial^\mu \phi_1 - \phi_1 \partial_\mu\partial^\mu \phi_2 ~.\] We can now substitute in for \(\partial_\mu\partial^\mu \phi_1\) and \(\partial_\mu\partial^\mu \phi_2\) using the Euler-Lagrange equations \(\eqref{eq:eulerlagrangeso2}\) to give \[\begin{split} \partial_\mu j^\mu & = \phi_2(m^2 \phi_1 + 4 \lambda(\phi_1^2+\phi_2^2)\phi_1) - \phi_1(m^2 \phi_2 + 4 \lambda(\phi_1^2+\phi_2^2)\phi_2) \\ & = \phi_2\phi_1(m^2 +4 \lambda(\phi_1^2+\phi_2^2)) - \phi_1\phi_2 (m^2 +4 \lambda(\phi_1^2+\phi_2^2)) = 0 ~. \end{split}\]

5. To show that the two actions \(\eqref{eq:acttwoscalar}\) and \(\eqref{eq:actcomplexscalar}\) are the same, let us substitute \(\Phi = \frac{1}{\sqrt{2}} (\phi_1 + i \phi_2)\) into \(\eqref{eq:actcomplexscalar}\). We have \[\begin{split} \partial_\mu\Phi^*\partial^\mu\Phi & = \frac12 \partial_\mu(\phi_1 - i \phi_2) \partial^\mu (\phi_1 + i\phi_2) \\ & = \frac12 \partial_\mu\phi_1 \partial^\mu \phi_1 - \frac i2 \partial_\mu\phi_2 \partial^\mu \phi_1 + \frac i2 \partial_\mu\phi_1 \partial^\mu \phi_2 + \frac12 \partial_\mu\phi_2 \partial^\mu \phi_2 \\ & = \frac12 \partial_\mu\phi_1 \partial^\mu \phi_1 + \frac12 \partial_\mu\phi_2 \partial^\mu \phi_2 ~, \end{split}\] and \(\Phi^* \Phi = \frac12(\phi_1^2 + \phi_2^2)\). Therefore, \[- \partial_\mu\Phi^*\partial^\mu\Phi - m^2\Phi^* \Phi - 4\lambda(\Phi^* \Phi)^2 = - \frac12 \partial_\mu\phi_1 \partial^\mu \phi_1 - \frac12 \partial_\mu\phi_2 \partial^\mu \phi_2 - \frac12 m^2 (\phi_1^2 + \phi_2^2) - \lambda (\phi_1^2 + \phi_2^2)^2 ~,\] showing that the two Lagrangian densities, hence also the actions, are the same.

6. The Euler-Lagrange equations for a field theory of a complex scalar field \(\Phi\) and its conjugate \(\Phi^*\) are \[\begin{equation} \label{eq:eulerlagrangegeneralcomplex} \frac{\partial \mathcal{L}}{\partial \Phi} - \partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu \Phi)}\Big) = 0 ~,\qquad \frac{\partial \mathcal{L}}{\partial \Phi^*} - \partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu \Phi^*)}\Big) = 0 ~, \end{equation}\] where \(\mathcal{L}\) is the Lagrangian density, which in this case is \[\mathcal{L}= - \partial_\mu\Phi^*\partial^\mu\Phi - m^2\Phi^* \Phi - 4\lambda(\Phi^* \Phi)^2 ~.\] Therefore, \[\frac{\partial \mathcal{L}}{\partial \Phi} = -m^2 \Phi^* - 8 \lambda(\Phi^* \Phi) \Phi^* ~, \qquad \frac{\partial \mathcal{L}}{\partial \Phi^*} = -m^2 \Phi - 8 \lambda(\Phi^* \Phi) \Phi ~,\] and \[\partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu \Phi)}\Big) = - \partial_\mu \Phi^* ~,\qquad \partial_\mu\Big(\frac{\partial\mathcal{L}}{\partial(\partial_\mu \Phi^*)}\Big) = - \partial_\mu \Phi ~.\] Substituting into eq. \(\eqref{eq:eulerlagrangegeneralcomplex}\) we find the Euler-Lagrange equations \[\begin{equation} \begin{split} & \partial_\mu\partial^\mu \Phi^* -m^2 \Phi^* - 8 \lambda(\Phi^* \Phi) \Phi^* = 0 ~, \\ & \partial_\mu\partial^\mu \Phi -m^2 \Phi - 8 \lambda(\Phi^* \Phi) \Phi = 0 ~, \end{split}\label{eq:eulerlagrangecomplex} \end{equation}\]

7. Under the continuous transformation \(\eqref{eq:u1symmetry}\) \[\Phi^* \Phi \to e^{-i\alpha}\Phi^* e^{i\alpha}\Phi = \Phi^* \Phi ~,\] and \[\partial_\mu\Phi^* \partial^\mu\Phi \to \partial_\mu(e^{-i\alpha}\Phi^*)\partial^\mu( e^{i\alpha}\Phi) = e^{-i\alpha} \partial_\mu\Phi^* e^{i\alpha}\partial^\mu \Phi = \partial_\mu\Phi^* \partial^\mu\Phi ~,\] where we have used that \(\alpha\) is a constant parameter. It immediately follows that the action \(\eqref{eq:actcomplexscalar}\) is invariant under the continuous transformation \(\eqref{eq:u1symmetry}\). Substituting in for the complex field \(\Phi\) in terms of the two real scalar fields \(\phi_1\) and \(\phi_2\) in the continuous transformation \(\eqref{eq:u1symmetry}\) we find \[\begin{split} \frac1{\sqrt{2}} (\phi_1 + i \phi_2) \to \frac1{\sqrt{2}} (\phi_1' + i \phi_2') & = e^{i\alpha}\frac1{\sqrt{2}} (\phi_1 + i \phi_2) \\ & =\frac1{\sqrt{2}} (\cos\alpha + i\sin\alpha)(\phi_1 + i \phi_2) \\ & = \frac1{\sqrt{2}} ((\phi_1 \cos\alpha - \phi_2 \sin\alpha) + i (\phi_1 \sin\alpha + \phi_2 \cos\alpha)) ~. \end{split}\] Taking the real and imaginary parts of this transformation we recover the continuous transformation \(\eqref{eq:so2symmetry}\) demonstrating that the two transformations are equivalent.

8. To determine the conserved current we consider an infinitesimal transformation and let the parameter \(\alpha\) depend on space-time. Expanding \(\eqref{eq:u1symmetry}\) to first order in \(\alpha\) we find the transformation \[\delta_\alpha \Phi = i \alpha\Phi ~, \qquad \delta_\alpha \Phi^* = - i \alpha\Phi^* ~.\] The conserved current is defined to be the term in the variation of the action proportional to \(\partial_\mu\alpha\). Such a contribution can only come from the kinetic term, while the potential terms will remain invariant even when \(\alpha\) is allowed to depend on space-time. Therefore, varying the action, we have \[\begin{split} \delta_{\alpha(x)} S[\Phi,\Phi^*] & = \int d^4x\, \big( -\partial_\mu \Phi^* \partial^\mu (\delta_{\alpha(x)}\Phi) - \partial_\mu(\delta_{\alpha(x)} \Phi^*) \partial^\mu \Phi \big) \\ & = \int d^4x\,\big(- i \partial_\mu\Phi^* \partial^\mu(\alpha\Phi) + i \partial_\mu (\alpha\Phi^*) \partial^\mu \Phi\big) \\ & = \int d^4x\, i\big(\Phi^* \partial^\mu\Phi -\Phi \partial^\mu \Phi^*\big)\partial_\mu\alpha ~. \end{split}\] We can then read off the conserved current \[\begin{equation} ~\label{eq:u1conservedcurrent} j^\mu = i(\Phi^* \partial^\mu\Phi - \Phi \partial^\mu \Phi^*) ~, \end{equation}\] and write down the conserved charge \[Q = i \int d^3x \, \big(\Phi \partial_t \Phi^* - \Phi^* \partial_t\Phi ) ~.\]

9. Substituting in for the complex field \(\Phi\) in terms of the two real scalar fields \(\phi_1\) and \(\phi_2\) in the conserved current \(\eqref{eq:u1conservedcurrent}\) associated to the symmetry \(\eqref{eq:u1symmetry}\) gives \[\begin{split} j^\mu & = i(\Phi^* \partial^\mu\Phi - \Phi \partial^\mu \Phi^*) \\ & = \frac{i}{2} ((\phi_1-i\phi_2)\partial^\mu (\phi_1+i\phi_2) - (\phi_1+i\phi_2)\partial^\mu (\phi_1-i\phi_2)) \\ & = \frac{i}{2} (\phi_1\partial^\mu\phi_1 -i\phi_2 \partial^\mu\phi_1 + i \phi_1 \partial^\mu\phi_2 + \phi_2 \partial^\mu\phi_2 -\phi_1\partial^\mu\phi_1 -i\phi_2 \partial^\mu\phi_1 + i \phi_1 \partial^\mu\phi_2 - \phi_2 \partial^\mu\phi_2) \\ & = \phi_2 \partial^\mu\phi_1 - \phi_1 \partial^\mu\phi_2 ~, \end{split}\] which is the conserved current \(\eqref{eq:so2conservedcurrent}\) associated to the symmetry \(\eqref{eq:so2symmetry}\).

Many harmonic oscillators.

1. To derive the Hamiltonian we start by computing the conjugate momenta \[p_c = \frac{\partial L}{\partial \dot q_c} = \dot q_c^* ~, \qquad p_c^* = \frac{\partial L}{\partial \dot q_c^*} = \dot q_c ~,\] where the Lagrangian is \[L = \sum_{c=1}^N \big(\dot q_c^*(t) \dot q_c(t) - \omega_c^2 q_c^*(t) q_c(t) \big) ~.\] The Hamiltonian is then given by \[H = \sum_{c=1}^N \big(p_c(t) \dot q_c(t) + p_c^*(t) \dot q_c^*(t) \big) - L = \sum_{c=1}^N \big(p_c(t) p_c^*(t) + \omega_c^2 q_c^*(t) q_c(t) \big) ~,\] where we have substituted in for \(\dot q_c(t)\) and \(\dot q_c^*(t)\) in terms of \(p_c(t)\) and \(p_c^*(t)\).

2. Upon canonical quantization we have \[\begin{gathered} q_c \to \hat q_c ~, \qquad q_c^* \to \hat q_c^\dagger ~, \qquad p_c \to \hat p_c ~, \qquad p_c^* \to \hat p_c^\dagger ~, \\ a_c \to \hat a_c ~, \qquad b_c \to \hat b_c ~, \qquad a_c^* \to \hat a_c^\dagger ~, \qquad b_c^* \to \hat b_c^\dagger ~, \end{gathered}\] hence, \[\begin{equation} \begin{split} \hat q_c & = \frac{1}{\sqrt{2\omega_c}} \big(\hat a_c e^{-i\omega_c t} + \hat b_c^\dagger e^{i\omega_c t} \big) ~, \\ \hat q_c^\dagger & = \frac{1}{\sqrt{2\omega_c}} \big(\hat b_c e^{-i\omega_c t} + \hat a_c^\dagger e^{i\omega_c t} \big) ~. \\ \hat p_c & = \dot{\hat q}_c^\dagger =-i \sqrt{\frac{\omega_c}{2}} \big(\hat b_c e^{-i\omega_c t} - \hat a_c^\dagger e^{i\omega_c t} \big) ~, \\ \hat p_c^\dagger & = \dot{\hat q}_c = -i\sqrt{\frac{\omega_c}{2}} \big(\hat a_c e^{-i\omega_c t} - \hat b_c^\dagger e^{i\omega_c t} \big) ~, \end{split}\label{eq:qqpp} \end{equation}\] where we keep in mind that \(\hat q_c\), \(\hat q_c^\dagger\), \(\hat p_c\) and \(\hat p_c^\dagger\) depend on time, while \(\hat a_c\), \(\hat a_c^\dagger\), \(\hat b_c\) and \(\hat b_c^\dagger\) do not. Therefore, we find \[\begin{split} \hat q_c^\dagger \hat q_c & = \frac{1}{2\omega_c} \big(\hat b_c e^{-i\omega_c t} + \hat a_c^\dagger e^{i\omega_c t} \big) \big(\hat a_c e^{-i\omega_c t} + \hat b_c^\dagger e^{i\omega_c t} \big) \\ & = \frac{1}{2\omega_c} \big(\hat b_c a_c e^{-2i\omega_c t} + \hat b_c \hat b_c^\dagger + \hat a_c^\dagger a_c + \hat a_c^\dagger \hat b_c^\dagger e^{2i\omega_c t} \big) ~, \end{split}\] and \[\begin{split} \hat p_c \hat p_c^\dagger & = -\frac{\omega_c}{2} \big(\hat b_c e^{-i\omega_c t} - \hat a_c^\dagger e^{i\omega_c t} \big) \big(\hat a_c e^{-i\omega_c t} - \hat b_c^\dagger e^{i\omega_c t} \big) \\ & = \frac{\omega_c}{2} \big(-\hat b_c a_c e^{-2i\omega_c t} + \hat b_c \hat b_c^\dagger + \hat a_c^\dagger a_c - \hat a_c^\dagger \hat b_c^\dagger e^{2i\omega_c t} \big) ~. \end{split}\] Combining these expressions gives \[\begin{equation} \label{eq:shohamiltonian} \hat H = \sum_{c=1}^N \big(p_c p_c^\dagger + \omega_c^2 q_c^\dagger q_c \big) = \sum_{c=1}^N \omega_c\big(\hat a_c^\dagger a_c+ \hat b_c \hat b_c^\dagger\big) ~. \end{equation}\] To further simplify we need to know the commutators \([\hat a_c,\hat a_{c'}^\dagger]\) and \([\hat b_c,\hat b_{c'}^\dagger]\), which can be worked out from the canonical equal-time commutation relations \[\phantom{}[\hat q_c, \hat p_{c'}] = [\hat q_c^\dagger,\hat p_{c'}^\dagger] = i \delta_{cc'} ~,\] with the remaining equal-time commutators equal to zero. Solving the relations in eq. \(\eqref{eq:qqpp}\) for \(\hat a_c\), \(\hat a_c^\dagger\), \(\hat b_c\) and \(\hat b_c^\dagger\) we find \[\begin{split} \hat a_c & = e^{+i\omega_c t} \sqrt{\frac{\omega_c}{2}} \Big(\hat q_c + \frac{i}{\omega_c} \hat p_c^\dagger\Big) ~, \\ \hat a_c^\dagger & = e^{-i\omega_c t} \sqrt{\frac{\omega_c}{2}} \Big(\hat q_c^\dagger - \frac{i}{\omega_c} \hat p_c\Big) ~, \\ \hat b_c & = e^{+i\omega_c t} \sqrt{\frac{\omega_c}{2}} \Big(\hat q_c^\dagger + \frac{i}{\omega_c} \hat p_c\Big) ~, \\ \hat b_c^\dagger & = e^{-i\omega_c t} \sqrt{\frac{\omega_c}{2}} \Big(\hat q_c - \frac{i}{\omega_c} \hat p_c^\dagger\Big) ~. \end{split}\] Therefore, \[\begin{split} \phantom{}[\hat a_c,\hat a_{c'}^\dagger] & = e^{i (\omega_c-\omega_{c'}) t} \frac{\sqrt{\omega_c\omega_{c'}}}{2} [\hat q_c + \frac{i}{\omega_c} \hat p_c^\dagger,\hat q_{c'}^\dagger - \frac{i}{\omega_{c'}} \hat p_{c'}] \\ & = e^{i (\omega_c-\omega_{c'}) t} \frac{\sqrt{\omega_c\omega_{c'}}}{2} \Big([\hat q_c,\hat q_{c'}^\dagger] - \frac{i}{\omega_{c'}} [\hat q_c, \hat p_{c'}] + \frac{i}{\omega_c} [\hat p_c^\dagger,\hat q_{c'}^\dagger] + \frac{1}{\omega_c\omega_{c'}} [\hat p_c^\dagger,\hat p_{c'}]\Big) \\ & = e^{i (\omega_c-\omega_{c'}) t} \frac{\sqrt{\omega_c\omega_{c'}}}{2} \Big( 0 + \frac{1}{\omega_{c'}} \delta_{cc'} + \frac{1}{\omega_{c}} \delta_{cc'} + 0 \Big) \\ & = \delta_{cc'} ~, \end{split}\] and \[\begin{split} \phantom{}[\hat b_c,\hat b_{c'}^\dagger] & = e^{i (\omega_c-\omega_{c'}) t} \frac{\sqrt{\omega_c\omega_{c'}}}{2} [\hat q_c^\dagger + \frac{i}{\omega_c} \hat p_c,\hat q_{c'} - \frac{i}{\omega_{c'}} \hat p_{c'}^\dagger] \\ & = e^{i (\omega_c-\omega_{c'}) t} \frac{\sqrt{\omega_c\omega_{c'}}}{2} \Big([\hat q_c^\dagger,\hat q_{c'}] - \frac{i}{\omega_{c'}} [\hat q_c^\dagger, \hat p_{c'}^\dagger] + \frac{i}{\omega_c} [\hat p_c,\hat q_{c'}] + \frac{1}{\omega_c\omega_{c'}} [\hat p_c,\hat p_{c'}^\dagger]\Big) \\ & = e^{i (\omega_c-\omega_{c'}) t} \frac{\sqrt{\omega_c\omega_{c'}}}{2} \Big( 0 + \frac{1}{\omega_{c'}} \delta_{cc'} + \frac{1}{\omega_{c}} \delta_{cc'} + 0 \Big) \\ & = \delta_{cc'} ~. \end{split}\] We can similarly show that the remaining commutators vanish. The Hamiltonian \(\eqref{eq:shohamiltonian}\) can now be simplified \[\begin{split} \hat H & = \sum_{c=1}^N \omega_c\big(\hat a_c^\dagger a_c+ \hat b_c \hat b_c^\dagger\big) \\ & = \sum_{c=1}^N \omega_c\big(\hat a_c^\dagger a_c+ [\hat b_c, \hat b_c^\dagger] + \hat b_c^\dagger b_c \big) \\ & = \sum_{c=1}^N \omega_c\big(\hat a_c^\dagger a_c + \hat b_c^\dagger b_c + 1 \big) ~, \end{split}\] as required

Canonical quantization of a free massive complex scalar field.

1. Starting from \[\Phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}}\big(a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } + b_{\vec{k}}^* e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \big) ~,\] it is straightforward to take the complex conjugate to find \[\Phi^*(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}}\big(b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} + a^*_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) ~.\] To compute the derivatives we note that the only dependence on \(t\) and \(\vec{x}\) in \(\Phi(t,\vec{x})\) is contained in the exponentials. If we act with \(\vec{\nabla}\) on \(e^{\pm i\vec{k}\cdot\vec{x}}\) we bring down a factor of \(\pm i \vec{k}\). Similarly, if we act with \(\partial_t\) on \(e^{\pm i\omega_{\vec{k}}t}\) we bring down a factor of \(\pm i \omega_{\vec{k}}\). Therefore, we find \[\begin{split} \vec{\nabla}\Phi(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{i\vec{k}}{\sqrt{2\omega_{\vec{k}}}}\big(a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } - b_{\vec{k}}^* e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \big) ~, \\ \vec{\nabla}\Phi^*(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{i\vec{k}}{\sqrt{2\omega_{\vec{k}}}}\big(b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - a^*_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) ~, \\ \Pi^*(t,\vec{x}) & = \partial_t \Phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, (-i)\sqrt{\frac{\omega_{\vec{k}}}{2}}\big(a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } - b_{\vec{k}}^* e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \big) ~, \\ \Pi(t,\vec{x}) & = \partial_t \Phi^*(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, (-i)\sqrt{\frac{\omega_{\vec{k}}}{2}} \big(b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - a^*_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) ~. \end{split}\]

2. Upon canonical quantization we have \[\begin{gathered} \Phi(t,\vec{x}) \to \hat \Phi(t,\vec{x}) ~, \qquad \Phi^*(t,\vec{x}) \to \hat \Phi^\dagger(t,\vec{x}) ~, \qquad \Pi(t,\vec{x}) \to \hat \Pi(t,\vec{x}) ~, \qquad \Pi^*(t,\vec{x}) \to \hat \Pi^\dagger(t,\vec{x}) ~, \\ a_{\vec{k}} \to \hat a_{\vec{k}} ~, \qquad b_{\vec{k}} \to \hat b_{\vec{k}} ~, \qquad a_{\vec{k}}^* \to \hat a_{\vec{k}}^\dagger ~, \qquad b_{\vec{k}}^* \to \hat b_{\vec{k}}^\dagger ~, \end{gathered}\] hence \[\begin{split} \hat \Phi(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}}\big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } + \hat b_{\vec{k}}^\dagger e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \big) ~, \\ \hat\Phi^\dagger(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}}\big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} + \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) ~, \\ \vec{\nabla}\hat\Phi(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{i\vec{k}}{\sqrt{2\omega_{\vec{k}}}}\big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } - \hat b_{\vec{k}}^\dagger e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \big) ~, \\ \vec{\nabla}\hat\Phi^\dagger(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{i\vec{k}}{\sqrt{2\omega_{\vec{k}}}}\big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) ~, \\ \hat\Pi^\dagger(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, (-i)\sqrt{\frac{\omega_{\vec{k}}}{2}}\big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } - \hat b_{\vec{k}}^\dagger e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \big) ~, \\ \hat\Pi(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, (-i)\sqrt{\frac{\omega_{\vec{k}}}{2}} \big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) ~. \end{split}\] The Hamiltonian becomes \[\begin{equation} \label{eq:complexscalarhamiltonianquantum} \hat H = \int d^3x \, \big( \hat \Pi^\dagger \hat \Pi + \vec{\nabla}\hat \Phi^\dagger \cdot \vec{\nabla}\hat \Phi + m^2 \hat \Phi^\dagger \hat \Phi \big) ~. \end{equation}\]

3. Substituting in the Fourier expansions gives \[\begin{split} \int d^3x \, m^2 \hat \Phi^\dagger \hat \Phi = \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{m^2}{2\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} + \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) \times \\ & \quad \big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}'}t + i\vec{k}'\cdot\vec{x} } + \hat b_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}'}t -i\vec{k}'\cdot\vec{x} }\big) ~. \end{split}\] Expanding and integrating over \(\vec{x}\) we find \[\begin{split} (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{m^2}{2\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \Big(\big(\hat b_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat b_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') + \\ & \quad \big( \hat b_{\vec{k}} \hat b_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~, \end{split}\] where we recall that \[\int d^3x \, e^{i\vec{p}\cdot\vec{x}} = (2\pi)^3 \delta^{(3)}(\vec{p}) ~.\] Also recalling that \(\omega_{-\vec{k}} = \omega_{\vec{k}}\), we can now use the delta function to integrate over \(\vec{k}'\) to give \[\begin{split} \int d^3x \, m^2 \hat \Phi^\dagger \hat \Phi = \int \frac{d^3k}{(2\pi)^3} \, \frac{m^2}{2\omega_{\vec{k}}} & \big(\hat b_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat b_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{split}\] Similarly, we have \[\begin{split} \int d^3x \, \vec{\nabla}\hat \Phi^\dagger \cdot \vec{\nabla} \hat \Phi = \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{-\vec{k}\cdot \vec{k}'}{2\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) \times \\ & \quad \big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}'}t + i\vec{k}'\cdot\vec{x} } - \hat b_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}'}t -i\vec{k}'\cdot\vec{x} }\big) ~. \end{split}\] Expanding and integrating over \(\vec{x}\) we find \[\begin{split} (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{-\vec{k}\cdot\vec{k}'}{2\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \Big(\big(\hat b_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat b_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') - \\ & \quad \big( \hat b_{\vec{k}} \hat b_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~. \end{split}\] We can now use the delta function to integrate over \(\vec{k}'\) to give \[\begin{split} \int d^3x \, \vec{\nabla}\hat \Phi^\dagger \cdot \vec{\nabla} \hat \Phi = \int \frac{d^3k}{(2\pi)^3} \, \frac{\vec{k}^2}{2\omega_{\vec{k}}} & \big(\hat b_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat b_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{split}\] Therefore, using \(\omega_{\vec{k}}^2 = \vec{k}^2 + m^2\), we find \[\begin{equation} \begin{split} & \int d^3x \, \big(\vec{\nabla}\hat \Phi^\dagger \cdot \vec{\nabla} \hat \Phi + m^2 \hat \Phi^\dagger \hat \Phi\big) \\ & \quad = \int \frac{d^3k}{(2\pi)^3} \, \frac{\omega_{\vec{k}}^2}{2\omega_{\vec{k}}} \big(\hat b_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat b_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) \\ & \quad = \int \frac{d^3k}{(2\pi)^3} \, \frac{\omega_{\vec{k}}}{2} \big(\hat b_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat b_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{split}\label{eq:secondpart} \end{equation}\] Finally, we have \[\begin{split} \int d^3x \, \hat \Pi \hat \Pi^\dagger = \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, (-1)\frac{\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}}{2} & \big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) \times \\ & \ \big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}'}t + i\vec{k}'\cdot\vec{x} } - \hat b_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}'}t -i\vec{k}'\cdot\vec{x} }\big) ~. \end{split}\] Expanding and integrating over \(\vec{x}\) we find \[\begin{split} (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, (-1)\frac{\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}}{2} & \Big(\big(\hat b_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat b_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') - \\ & \quad \big( \hat b_{\vec{k}} \hat b_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~. \end{split}\] We can now use the delta function to integrate over \(\vec{k}'\) to give \[\begin{equation} \label{eq:firstpart} \int d^3x \, \hat \Pi \hat \Pi^\dagger = \int \frac{d^3k}{(2\pi)^3} \, \frac{\omega_{\vec{k}}}{2} \big(-\hat b_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} - \hat a^\dagger_{\vec{k}}\hat b_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{equation}\] Now summing eq. \(\eqref{eq:secondpart}\) and eq. \(\eqref{eq:firstpart}\) we find \[\hat H = \int d^3x \, \big( \hat \Pi^\dagger \hat \Pi + \vec{\nabla}\hat \Phi^\dagger \cdot \vec{\nabla}\hat \Phi + m^2 \hat \Phi^\dagger \hat \Phi \big) = \int \frac{d^3k}{(2\pi)^3} \, \omega_{\vec{k}} \big(\hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger \big) ~.\]

4. Using \([\hat b_{\vec k}, \hat b^\dagger_{\vec k'}] = (2\pi)^3 \delta^{(3)}(\vec{k} - \vec{k}')\), we have \[\begin{split} \hat H & = \int \frac{d^3k}{(2\pi)^3} \, \omega_{\vec{k}} \big(\hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger \big) \\ & = \int \frac{d^3k}{(2\pi)^3} \, \omega_{\vec{k}} \big(\hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} + [\hat b_{\vec{k}}, \hat b_{\vec{k}}^\dagger] + \hat b^\dagger_{\vec{k}}\hat b_{\vec{k}} \big) \\ & = \int \frac{d^3k}{(2\pi)^3} \, \omega_{\vec{k}} \big(\hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} + \hat b^\dagger_{\vec{k}}\hat b_{\vec{k}} + (2\pi)^3 \delta^{(3)}(0)\big) ~, \end{split}\] as required.

The number operator.

1. Let us consider a general state with \(N_a\) particles and \(N_b\) antiparticles \[|N_a,N_b\rangle = \hat a_{\vec{p}_1}^\dagger \hat a_{\vec{p}_2}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle ~.\] We then have \[\hat{\mathcal{N}}_A|N_a,N_b\rangle = \int \frac{d^3k}{(2\pi)^3} \, \hat a_{\vec{k}}^\dagger \hat a_{\vec{k}} \hat a_{\vec{p}_1}^\dagger \hat a_{\vec{p}_2}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle ~.\] Using the fact that \(\hat a_{\vec{k}}\) commutes with \(\hat b_{\vec{p_i}}^\dagger\) and annihilates the vacuum, hence \[\begin{split} & \int \frac{d^3k}{(2\pi)^3} \, \hat a_{\vec{k}}^\dagger \hat a_{\vec{p}_1}^\dagger \hat a_{\vec{p}_2}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat a_{\vec{k}}\hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle \\ & = \int \frac{d^3k}{(2\pi)^3} \, \hat a_{\vec{k}}^\dagger \hat a_{\vec{p}_1}^\dagger \hat a_{\vec{p}_2}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger \hat a_{\vec{k}} |0\rangle = 0 ~, \end{split}\] we can write \[\hat{\mathcal{N}}_A|N_a,N_b\rangle = \int \frac{d^3k}{(2\pi)^3} \, \hat a_{\vec{k}}^\dagger [\hat a_{\vec{k}}, \hat a_{\vec{p}_1}^\dagger \hat a_{\vec{p}_2}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger] \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle ~.\] We can now use the identity \[\phantom{}[A,B_1 B_2 \dots B_N] = \sum_{i=1}^N B_1 \dots B_{i-1} [A,B_i] B_{i+1} \dots B_N ~,\] to show that \[\begin{split} \hat{\mathcal{N}}_A|N_a,N_b\rangle & = \sum_{i=1}^{N_a}\int \frac{d^3k}{(2\pi)^3} \, \hat a_{\vec{k}}^\dagger\hat a_{\vec{p}_1}^\dagger \dots \hat a_{\vec{p}_{i-1}}^\dagger [\hat a_{\vec{k}},\hat a_{\vec{p}_i}^\dagger] \hat a_{\vec{p}_{i+1}}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle \\ & = \sum_{i=1}^{N_a}\int \frac{d^3k}{(2\pi)^3} \, \hat a_{\vec{k}}^\dagger\hat a_{\vec{p}_1}^\dagger \dots \hat a_{\vec{p}_{i-1}}^\dagger (2\pi)^3 \delta^{(3)} (\vec{k} - \vec{p}_i)\hat a_{\vec{p}_{i+1}}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle \\ & = \sum_{i=1}^{N_a} \hat a_{\vec{p}_i}^\dagger\hat a_{\vec{p}_1}^\dagger \dots \hat a_{\vec{p}_{i-1}}^\dagger \hat a_{\vec{p}_{i+1}}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle \\ & = \sum_{i=1}^{N_a} \hat a_{\vec{p}_1}^\dagger \dots \hat a_{\vec{p}_{i-1}}^\dagger \hat a_{\vec{p}_i}^\dagger \hat a_{\vec{p}_{i+1}}^\dagger \dots \hat a_{\vec{p}_{N_a}}^\dagger \hat b_{\vec{p}_1}^\dagger \hat b_{\vec{p}_2}^\dagger \hat b_{\vec{p}_{N_b}}^\dagger |0\rangle \\ & = \sum_{i=1}^{N_a} |N_a,N_b\rangle = N_a |N_a,N_b\rangle ~, \end{split}\] where we have used the commutation relation \([\hat a_{\vec{k}},\hat a_{\vec{p}_i}^\dagger] = (2\pi)^3 \delta^{(3)} (\vec{k} - \vec{p}_i)\) and the resulting delta function to integrate over \(\vec{k}\).

2. By analogy, the operator that counts the number of antiparticles in a multiparticle state is \[\hat{\mathcal{N}}_B = \int \frac{d^3k}{(2\pi)^3} \, \hat b_{\vec{k}}^\dagger \hat b_{\vec{k}} ~.\]

The charge operator and normal ordering.

1. Taking the hermitian conjugate of \(\hat Q\) we have \[\begin{split} \hat Q^\dagger &= i \int d^3x \, \big(\hat\Pi \hat\Phi - \hat\Phi^\dagger \hat\Pi^\dagger \big)^\dagger = i \int d^3x \, \big( \hat\Phi^\dagger \hat \Pi^\dagger - \hat \Pi \hat\Phi \big) = - i \int d^3x \, \big(\hat\Pi \hat\Phi - \hat\Phi^\dagger \hat\Pi^\dagger \big) = \hat Q ~, \end{split}\] hence \(\hat Q\) is hermitian.

2. We recall the Fourier expansions \[\begin{split} \hat \Phi(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}}\Big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } + \hat b_{\vec{k}}^\dagger e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \Big) ~, \\ \hat\Phi^\dagger(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}}\Big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} + \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \Big) ~, \\ \hat\Pi^\dagger(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, (-i)\sqrt{\frac{\omega_{\vec{k}}}{2}}\Big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x} } - \hat b_{\vec{k}}^\dagger e^{+ i\omega_{\vec{k}}t -i\vec{k}\cdot\vec{x} } \Big) ~, \\ \hat\Pi(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, (-i)\sqrt{\frac{\omega_{\vec{k}}}{2}} \Big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \Big) ~. \end{split}\] Therefore, \[\begin{split} -i\int d^3x \, \hat\Pi \hat\Phi = \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{(-1)}{2}\sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}} & \Big(\hat b_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \Big) \times \\ & \Big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}}t + i\vec{k}'\cdot\vec{x} } + \hat b_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}}t -i\vec{k}'\cdot\vec{x} } \Big) ~. \end{split}\] Expanding and doing the integral over \(\vec{x}\) we find \[\begin{split} (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \,\frac{(-1)}{2} \sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}} & \Big(\big(\hat b_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} - \hat a^\dagger_{\vec{k}}\hat b_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') + \\ & \quad \big( \hat b_{\vec{k}} \hat b_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} - \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~. \end{split}\] We can now use the delta function to integrate over \(\vec{k}'\) to give \[\begin{equation} \label{eq:firsttermq} -i\int d^3x \, \hat\Pi \hat\Phi = \int \frac{d^3k}{(2\pi)^3} \, \frac{(-1)}{2} \big(\hat b_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} - \hat a^\dagger_{\vec{k}}\hat b_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger - \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{equation}\] The hermitian conjugate of this expression is \[\begin{equation} \begin{split} i \int d^3x \, \hat\Phi^\dagger \hat\Pi^\dagger &= \int \frac{d^3k}{(2\pi)^3} \, \frac{(-1)}{2} \big(\hat a_{-\vec{k}}^\dagger \hat b_{\vec{k}}^\dagger e^{2 i\omega_{\vec{k}} t} - \hat b_{-\vec{k}} \hat a_{\vec{k}} e^{- 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}}^\dagger \hat b_{\vec{k}} - \hat a_{\vec{k}}\hat a_{\vec{k}}^\dagger \big) \\ & = \int \frac{d^3k}{(2\pi)^3} \, \frac{(-1)}{2} \big(\hat a_{\vec{k}}^\dagger \hat b_{-\vec{k}}^\dagger e^{2 i\omega_{\vec{k}} t} - \hat b_{\vec{k}} \hat a_{-\vec{k}} e^{- 2i\omega_{\vec{k}}t} + \hat b_{\vec{k}}^\dagger \hat b_{\vec{k}} - \hat a_{\vec{k}}\hat a_{\vec{k}}^\dagger \big) ~, \end{split}\label{eq:secondtermq} \end{equation}\] where we made the change of variables \(\vec{k} \to -\vec{k}\) in the first two terms. Summing eq. \(\eqref{eq:firsttermq}\) and eq. \(\eqref{eq:secondtermq}\) we find \[\hat Q = \int \frac{d^3k}{(2\pi)^3} \,\big(\hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} - \hat b_{\vec{k}} \hat b_{\vec{k}}^\dagger \big) ~.\] Normal ordering moves all the creation operators to the left and all the annihilation operators to the right to give \[\,:\!\hat Q\!:\, = \int \frac{d^3k}{(2\pi)^3} \,\big(\hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} - \hat b_{\vec{k}}^\dagger \hat b_{\vec{k}} \big) ~.\] Note that \[\,:\!\hat Q\!:\, = \hat{\mathcal{N}}_A - \hat{\mathcal{N}}_B ~,\] i.e., the difference between the number of particles and antiparticles.

Canonical quantization of a free massive real scalar field.

1. The Euler-Lagrange equation for the field \(\phi\) in position space is \[\begin{equation} \label{eq:el51} \partial_\mu\partial^\mu\phi - m^2 \phi = 0 ~, \qquad \partial_\mu\partial^\mu = -\partial_t^2 + \vec{\nabla}^2 ~. \end{equation}\] In momentum space, this equation becomes \[(\partial_t^2 + \omega_{\vec{k}}^2) \tilde\phi(t,\vec{k}) ~, \qquad \omega_{\vec{k}} = \sqrt{\vec{k}^2 + m^2} ~.\] Therefore, the Euler-Lagrange equation at fixed \(\vec{k}\) is that of a simple harmonic oscillator. It follows that the general solution is given by \[\tilde\phi(t,\vec{k}) = \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(a_{\vec{k}} e^{-i\omega_{\vec{k}}t} + b^*_{-\vec{k}} e^{+i\omega_{\vec{k}}t}\big) ~,\] where \(a_{\vec{k}}\) and \(b_{\vec{k}}\) are momentum-dependent integration constants and the overall normalisation is for convenience. Returning to position space using the Fourier transform gives \[\phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(a_{\vec{k}} e^{-i\omega_{\vec{k}}t} + b^*_{-\vec{k}} e^{+i\omega_{\vec{k}}t}\big) e^{+i\vec{k}\cdot \vec{x}} ~.\] Changing variables \(\vec{k} \to - \vec{k}\) in the second term we find \[\phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} + b^*_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~.\] Finally, we impose that the field is real, i.e. \(\phi^* = \phi\). This implies that \[\begin{split} & \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big((a_{\vec{k}}-b_{\vec{k}}) e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} + (b^*_{\vec{k}} - a^*_{\vec{k}}) e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) = 0 \\ \Rightarrow \quad & \frac{1}{\sqrt{2\omega_{\vec{k}'}}} \big((a_{\vec{k}'}-b_{\vec{k}'}) e^{-i\omega_{\vec{k}'}t} + (b^*_{-\vec{k}'} - a^*_{-\vec{k}'}) e^{+i\omega_{\vec{k}'}t} \big) = 0 ~, \end{split}\] where we have multiplied by \(e^{-i\vec{k}'\cdot \vec{x}}\) and integrated over \(\vec{x}\). Since \(e^{-i\omega_{\vec{k}'}t}\) and \(e^{+i\omega_{\vec{k}'}t}\) are linearly independent functions, it follows that \(b_{\vec{k}} = a_{\vec{k}}\). Therefore, we have \[\phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} + a^*_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~.\] Note it is also possible to directly check that this satisfies the Euler-Lagrange equation \(\eqref{eq:el51}\) as follows \[(-\partial_t^2 + \vec{\nabla}^2 + m^2) \phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} (\omega_{\vec{k}}^2 - \vec{k}^2 - m^2) \big(a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} + a^*_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) = 0 ~,\] since \(\omega^2_{\vec{k}} = \vec{k}^2 + m^2\). Now quantizing, we let the \(\phi\) and \(a_{\vec{k}}\) become operators, with the complex conjugate of \(a_{\vec{k}}\) becoming the hermitian conjugate \[\hat \phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} + \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~.\] That solving the Euler-Lagrange equations and quantizing the resulting classical solution yields the quantum field in the Heisenberg picture is special to free theories. In interacting theories, the expression for the classical field will receive non-trivial quantum corrections.

2. The scalar field \(\hat\phi\) and its conjugate momentum \(\hat\pi\) satisfy the canonical equal time commutation relation \[\phantom{}[\hat\phi(t,\vec{x}),\hat\pi(t,\vec{y})] = i \delta^{(3)}(\vec{x}- \vec{y}) ~.\] The expression for the conjugate momentum in a free theory can be found from its classical definition \[\pi = \frac{\partial\mathcal{L}}{\partial(\partial_t\phi)} = \partial_t\phi ~.\] Therefore, we have \[\begin{equation} \begin{split} \hat \phi(t,\vec{x}) & = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} + \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~, \\ \hat \pi(t,\vec{x}) & = - i \int \frac{d^3k}{(2\pi)^3} \, \sqrt{\frac{\omega_{\vec{k}}}{2}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} - \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~. \end{split}\label{eq:fourier51} \end{equation}\] Taking the expression for \(\hat\phi(t,\vec{x})\), multiplying by \(e^{i\vec{k}'\cdot\vec{x}}\) and integrating over \(\vec{x}\), we find \[\begin{split} \int d^3x \, e^{i\vec{k}'\cdot\vec{x}} \hat\phi(t,\vec{x}) & = \int d^3x \, e^{i\vec{k}'\cdot\vec{x}} \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i \vec{k}\cdot \vec{x}} + \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t -i \vec{k}\cdot \vec{x}} \big) \\ & = \int d^3x \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i (\vec{k} + \vec{k}')\cdot \vec{x}} + \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t -i (\vec{k} -\vec{k}')\cdot \vec{x}} \big) \\ & = \int d^3k \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t}\delta^{(3)}(\vec{k} + \vec{k}') +\hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t}\delta^{(3)}(\vec{k} - \vec{k}')\big) \\ & =\frac{1}{\sqrt{2\omega_{\vec{k}'}}} \big(\hat a_{-\vec{k}'} e^{-i\omega_{\vec{k}'}t}+\hat a^\dagger_{\vec{k}'} e^{+i\omega_{\vec{k}'}t}\big)~, \end{split}\] where we recall that \(\omega_{-\vec{k}} = \omega_{\vec{k}}\). Similarly, starting from the expression for \(\hat \pi(t,\vec{x})\), we find \[\int d^3x\,e^{i\vec{k}'\cdot\vec{x}} \hat\pi(t,\vec{x}) =-i\sqrt{\frac{\omega_{\vec{k}'}}{2}} \big(\hat a_{-\vec{k}'} e^{-i\omega_{\vec{k}'}t}-\hat a^\dagger_{\vec{k}'} e^{+i\omega_{\vec{k}'}t}\big)~.\] Taking linear combinations of these two expressions we find \[\begin{split} \hat a_{\vec{k}} &= \int d^3x\, e^{+i\omega_{\vec{k}}t -i \vec{k}\cdot \vec{x}} \Big( \sqrt{\frac{\omega_{\vec{k}}}{2}}\hat\phi(t,\vec{x}) + \frac{i}{\sqrt{2\omega_{\vec{k}}}}\hat\pi(t,\vec{x})\Big) ~, \\ \hat a^\dagger_{\vec{k}} &= \int d^3x\, e^{-i\omega_{\vec{k}}t +i \vec{k}\cdot \vec{x}} \Big( \sqrt{\frac{\omega_{\vec{k}}}{2}}\hat\phi(t,\vec{x}) - \frac{i}{\sqrt{2\omega_{\vec{k}}}}\hat\pi(t,\vec{x})\Big) ~, \end{split}\] where we have renamed \(\vec{k}' \to -\vec{k}\) in the expression for \(\hat a_{\vec{k}}\) and \(\vec{k}' \to \vec{k}\) in the expression for \(\hat a^\dagger_{\vec{k}}\). We can now directly compute the commutator \[\begin{split} \phantom{}[\hat a_{\vec{k}},\hat a_{\vec{k}'}^\dagger] = \frac12 \int d^3x \int d^3x' \, e^{it(\omega_{\vec{k}}-\omega_{\vec{k}'}) - i \vec{k}\cdot\vec{x} + i \vec{k}'\cdot\vec{x}'} \Big( & \sqrt{\omega_{\vec{k}}\omega_{\vec{k}'}}[\hat\phi(t,\vec{x}),\hat\phi(t,\vec{x}')] \\& -i\sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}}[\hat\phi(t,\vec{x}),\hat\pi(t,\vec{x}')] \\& + i\sqrt{\frac{\omega_{\vec{k}'}}{\omega_{\vec{k}}}}[\hat\pi(t,\vec{x}),\hat\phi(t,\vec{x}')] \\& + \frac{1}{\sqrt{\omega_{\vec{k}}\omega_{\vec{k}'}}}[\hat\pi(t,\vec{x}),\hat\pi(t,\vec{x}')]\Big) ~. \end{split}\] The commutators of \(\hat\phi(t,\vec{x})\) with itself and \(\hat\pi(t,\vec{x}')\) with itself vanish, while the other two commutators give delta functions and we find \[\phantom{}[\hat a_{\vec{k}},\hat a_{\vec{k}'}^\dagger] = \frac12 \int d^3x \int d^3x' \, e^{it(\omega_{\vec{k}}-\omega_{\vec{k}'}) - i \vec{k}\cdot\vec{x} + i \vec{k}'\cdot\vec{x}'}\Big( \sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}} + \sqrt{\frac{\omega_{\vec{k}'}}{\omega_{\vec{k}}}}\Big)\delta^{(3)}(\vec{x} - \vec{x}') ~.\] Now we can use the delta function to integrate over \(\vec{x}'\), which amounts to setting \(\vec{x}'\) to \(\vec{x}\), yielding \[\phantom{}[\hat a_{\vec{k}},\hat a_{\vec{k}'}^\dagger] = \frac12 \int d^3x \, e^{it(\omega_{\vec{k}}-\omega_{\vec{k}'}) - i (\vec{k}-\vec{k'})\cdot\vec{x} }\Big( \sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}} + \sqrt{\frac{\omega_{\vec{k}'}}{\omega_{\vec{k}}}}\Big)~.\] Using the identity \[\begin{equation} \label{eq:identity51} \int d^3x \, e^{i\vec{p}\cdot\vec{x}} = (2\pi)^3 \delta^{(3)} (\vec{p}) ~, \end{equation}\] the remaining integral over \(\vec{x}\) gives a delta function \[\phantom{}[\hat a_{\vec{k}},\hat a_{\vec{k}'}^\dagger] = \frac12 (2\pi)^3 \delta^{(3)}(\vec{k}-\vec{k}')e^{it(\omega_{\vec{k}}-\omega_{\vec{k}'})}\Big( \sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}} + \sqrt{\frac{\omega_{\vec{k}'}}{\omega_{\vec{k}}}}\Big)~.\] Finally, since \(\delta^{(3)}(\vec{k}-\vec{k}')\) only has support when \(\vec{k} = \vec{k}'\), i.e., it is zero if \(\vec{k} \neq \vec{k}'\), we can freely set \(\vec{k} = \vec{k}'\). Doing so the commutator simplifies to \[\phantom{}[\hat a_{\vec{k}},\hat a_{\vec{k}'}^\dagger] = (2\pi)^3 \delta^{(3)}(\vec{k}-\vec{k}') ~.\] We can similarly show that \[\phantom{}[\hat a_{\vec{k}},\hat a_{\vec{k}'}] = [\hat a^\dagger_{\vec{k}},\hat a^\dagger_{\vec{k}'}] = 0 ~.\]

3. The classical Hamiltonian for a free massive real scalar field \[\begin{equation} \label{eq:realscalarhamiltonian} H = \int d^3x \, \Big( \frac12\pi^2 + \frac12\vec{\nabla}\phi \cdot \vec{\nabla}\phi + \frac12 m^2 \phi^2 \Big) ~, \end{equation}\] becomes \[\begin{equation} \label{eq:realscalarhamiltonianquantum} \hat H = \int d^3x \, \Big( \frac12\hat\pi^2 + \frac12\vec{\nabla}\hat\phi \cdot \vec{\nabla}\hat\phi + \frac12 m^2 \hat\phi^2 \Big) ~, \end{equation}\] upon quantization. Substituting in the Fourier expansions \(\eqref{eq:fourier51}\) gives \[\begin{split} \int d^3x \, \frac12 m^2 \hat \phi^2 = \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{m^2}{4\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} + \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) \times \\ & \quad \big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}'}t + i\vec{k}'\cdot\vec{x} } + \hat a_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}'}t -i\vec{k}'\cdot\vec{x} }\big) ~. \end{split}\] Expanding and integrating over \(\vec{x}\) we find \[\begin{split} (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{m^2}{4\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \Big(\big(\hat a_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') + \\ & \quad \big( \hat a_{\vec{k}} \hat a_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~, \end{split}\] where we recall the identity \(\eqref{eq:identity51}\). Also recalling that \(\omega_{-\vec{k}} = \omega_{\vec{k}}\), we can now use the delta function to integrate over \(\vec{k}'\) to give \[\begin{split} \int d^3x \, \frac12 m^2 \hat \phi^2 = \int \frac{d^3k}{(2\pi)^3} \, \frac{m^2}{4\omega_{\vec{k}}} & \big(\hat a_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat a_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{split}\] Similarly, we have \[\begin{split} \int d^3x \, \frac12\vec{\nabla}\hat \phi \cdot \vec{\nabla} \hat \phi = \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{-\vec{k}\cdot \vec{k}'}{4\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) \times \\ & \quad \big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}'}t + i\vec{k}'\cdot\vec{x} } - \hat a_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}'}t -i\vec{k}'\cdot\vec{x} }\big) ~. \end{split}\] Expanding and integrating over \(\vec{x}\) we find \[\begin{split} (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{-\vec{k}\cdot\vec{k}'}{4\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}} & \Big(\big(\hat a_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') - \\ & \quad \big( \hat a_{\vec{k}} \hat a_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~. \end{split}\] We can now use the delta function to integrate over \(\vec{k}'\) to give \[\begin{split} \int d^3x \, \frac12 \vec{\nabla}\hat \phi \cdot \vec{\nabla} \hat \phi = \int \frac{d^3k}{(2\pi)^3} \, \frac{\vec{k}^2}{4\omega_{\vec{k}}} & \big(\hat a_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat a_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{split}\] Therefore, using \(\omega_{\vec{k}}^2 = \vec{k}^2 + m^2\), we find \[\begin{equation} \begin{split} & \int d^3x \, \Big(\frac12\vec{\nabla}\hat \phi \cdot \vec{\nabla} \hat \phi +\frac12 m^2 \hat \phi^2 \Big) \\ \quad & = \int \frac{d^3k}{(2\pi)^3} \, \frac{\omega_{\vec{k}}^2}{4\omega_{\vec{k}}} \big(\hat a_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat a_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) \\ \quad & = \int \frac{d^3k}{(2\pi)^3} \, \frac{\omega_{\vec{k}}}{4} \big(\hat a_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat a_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{split} \label{eq:secondpart51} \end{equation}\] Finally, we have \[\begin{split} \int d^3x \, \frac12 \hat \pi^2 = \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, (-1)\frac{\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}}{4} & \big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) \times \\ & \ \big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}'}t + i\vec{k}'\cdot\vec{x} } - \hat a_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}'}t -i\vec{k}'\cdot\vec{x} }\big) ~. \end{split}\] Expanding and integrating over \(\vec{x}\) we find \[\begin{split} (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, (-1)\frac{\sqrt{\omega_{\vec{k}}\omega_{\vec{k'}}}}{4} & \Big(\big(\hat a_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') - \\ & \quad \big( \hat a_{\vec{k}} \hat a_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~. \end{split}\] We can now use the delta function to integrate over \(\vec{k}'\) to give \[\begin{equation} \label{eq:firstpart51} \int d^3x \, \frac12\hat \pi^2 = \int \frac{d^3k}{(2\pi)^3} \, \frac{\omega_{\vec{k}}}{4} \big(-\hat a_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} - \hat a^\dagger_{\vec{k}}\hat a_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~. \end{equation}\] Now summing eq. \(\eqref{eq:secondpart51}\) and eq. \(\eqref{eq:firstpart51}\) we find \[\hat H = \int d^3x \, \Big( \frac12\hat\pi^2 + \frac12\vec{\nabla}\hat\phi \cdot \vec{\nabla}\hat\phi + \frac12 m^2 \hat\phi^2 \Big) = \int \frac{d^3k}{(2\pi)^3} \, \frac{\omega_{\vec{k}}}{2} \big(\hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} + \hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger \big) ~.\] To normal order, we move all creation operators to the left and all annihilation operators to the right. Therefore, \[\,:\!\hat H\!:\, = \int \frac{d^3k}{(2\pi)^3} \, \omega_{\vec{k}} \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} ~,\] as required.

Momentum of single particle states.

1. To quantize we promote the fields and the Noether charge to operators \[\hat{\vec{P}} = - \int d^3x \, \partial_t \hat{\phi} \vec{\nabla}\hat{\phi} ~.\] The quantum field written in terms of annihilation and creation operators is \[\hat \phi(t,\vec{x}) = \int \frac{d^3k}{(2\pi)^3} \, \frac{1}{\sqrt{2\omega_{\vec{k}}}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} + \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~.\] Taking derivatives we have \[\begin{split} \partial_t\hat \phi(t,\vec{x}) & = -i\int \frac{d^3k}{(2\pi)^3} \, \sqrt{\frac{\omega_{\vec{k}}}{2}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} - \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~, \\ \vec{\nabla} \hat \phi(t,\vec{x}) & = i \int \frac{d^3k}{(2\pi)^3} \, \frac{\vec{k}}{\sqrt{2\omega_{\vec{k}}}} \big(\hat a_{\vec{k}} e^{-i\omega_{\vec{k}}t + i\vec{k}\cdot \vec{x}} - \hat a^\dagger_{\vec{k}} e^{+i\omega_{\vec{k}}t-i\vec{k}\cdot \vec{x}} \big) ~. \end{split}\] Substituting in these Fourier expansions, we have \[\begin{split} - \int d^3x \, \partial_t \hat{\phi} \vec{\nabla}\hat{\phi} = - \int d^3x \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{\vec{k}'}{2} \sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}} & \big(\hat a_{\vec{k}} e^{- i\omega_{\vec{k}}t + i\vec{k}\cdot\vec{x}} - \hat a^\dagger_{\vec{k}} e^{+ i\omega_{\vec{k}}t-i\vec{k}\cdot\vec{x}} \big) \times \\ & \ \big(\hat a_{\vec{k}'} e^{- i\omega_{\vec{k}'}t + i\vec{k}'\cdot\vec{x} } - \hat a_{\vec{k}'}^\dagger e^{+ i\omega_{\vec{k}'}t -i\vec{k}'\cdot\vec{x} }\big) ~. \end{split}\] Expanding and integrating over \(\vec{x}\) we find \[\begin{split} - (2\pi)^3 \int \frac{d^3k}{(2\pi)^3} \int \frac{d^3k'}{(2\pi)^3} \, \frac{\vec{k}'}{2} \sqrt{\frac{\omega_{\vec{k}}}{\omega_{\vec{k}'}}} & \Big(\big(\hat a_{\vec{k}} \hat a_{\vec{k}'} e^{- i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'}^\dagger e^{+ i(\omega_{\vec{k}}+ \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} + \vec{k}') - \\ & \quad \big( \hat a_{\vec{k}} \hat a_{\vec{k}'}^\dagger e^{- i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}'} e^{+ i(\omega_{\vec{k}} - \omega_{\vec{k}'}) t} \big) \delta^{(3)}(\vec{k} - \vec{k}') \Big) ~. \end{split}\] We can now use the delta function to integrate over \(\vec{k}'\) to give \[- \int d^3x \, \partial_t \hat{\phi} \vec{\nabla}\hat{\phi} = \int \frac{d^3k}{(2\pi)^3} \, \frac{\vec{k}}{2} \big(\hat a_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} + \hat a^\dagger_{\vec{k}}\hat a_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t} + \hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~,\] where we have used that \(\omega_{\vec{k}} = \omega_{-\vec{k}}\). Noting that \(\vec{k} \hat a_{\vec{k}} \hat a_{-\vec{k}} e^{-2 i\omega_{\vec{k}} t} = - (-\vec{k} \hat a_{-\vec{k}} \hat a_{\vec{k}} e^{-2 i\omega_{-\vec{k}} t})\) since the annihilation operators \(\hat a_{\vec{k}}\) and \(\hat a_{-\vec{k}}\) commute, i.e., the first term in the integrand is an odd function of \(\vec{k}\), it follows that when we integrate over all \(\vec{k}\) the contribution of this term vanishes. The same holds for the second term in the integrand \(\vec{k} \hat a^\dagger_{\vec{k}}\hat a_{-\vec{k}}^\dagger e^{+ 2i\omega_{\vec{k}}t}\). Therefore, we are left with \[\hat{\vec{P}} = - \int d^3x \, \partial_t \hat{\phi} \vec{\nabla}\hat{\phi} = \int \frac{d^3k}{(2\pi)^3} \, \frac{\vec{k}}{2} \big(\hat a_{\vec{k}} \hat a_{\vec{k}}^\dagger + \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} \big) ~.\] Finally, to normal order, we move all creation operators to the left and all annihilation operators to the right. Therefore, \[\,:\!\hat{\vec{P}}\!:\, = \int \frac{d^3k}{(2\pi)^3} \, \vec{k} \hat a^\dagger_{\vec{k}}\hat a_{\vec{k}} ~,\] as required.

2. Since normal ordering moves all annihilation operators to the right, \(\,:\!\hat{\vec{P}}\!:\,\) annihilates the vacuum state \[\,:\!\hat{\vec{P}}\!:\, |0\rangle = 0 ~.\] To compute the action of \(\hat{\vec{P}}\) on excited states it is useful to first compute the commutator \[\begin{split} \phantom{}[\,:\!\hat{\vec{P}}\!:\,,\hat a_{\vec{k}}^\dagger] & = \Big[ \int \frac{d^3k'}{(2\pi)^3} \, \vec{k}' \hat a_{\vec{k}'}^\dagger \hat a_{\vec{k}'} , \hat a_{\vec{k}}^\dagger \Big] \\ & = \int \frac{d^3k'}{(2\pi)^3} \, \vec{k}' \hat a_{\vec{k}'}^\dagger [\hat a_{\vec{k}'},\hat a_{\vec{k}}^\dagger] \\ & = \int \frac{d^3k'}{(2\pi)^3} \, \vec{k}' \hat a_{\vec{k}'}^\dagger (2\pi)^3 \delta^{(3)}(\vec{k}-\vec{k}') \\ & = \vec{k} \hat a_{\vec{k}}^\dagger ~. \end{split}\] Similarly, we have \[\phantom{}[\,:\!\hat{\vec{P}}\!:\,,\hat a_{\vec{k}}] = - \vec{k} \hat a_{\vec{k}} ~.\] Now we consider the excited state \(\hat a_{\vec{k}}^\dagger|0\rangle\) and compute the action of \(\,:\!\hat{\vec{P}}\!:\,\) \[\,:\!\hat{\vec{P}}\!:\, \hat a_{\vec{k}}^\dagger|0\rangle = [\,:\!\hat{\vec{P}}\!:\, ,\hat a_{\vec{k}}^\dagger]|0\rangle - \hat a_{\vec{k}}^\dagger\,:\!\hat{\vec{P}}\!:\,|0\rangle = \vec{k} \hat a_{\vec{k}}^\dagger |0\rangle ~.\] Therefore, the excited state is an eigenstate of \(\,:\!\hat{\vec{P}}\!:\,\) with eigenvalue \(\vec{k}\). We say that the state created by \(\hat a_{\vec{k}}^\dagger\) has momentum \(\vec{k}\).

Propagators in 1+1d.

1. Using the Lorentz covariance of the integral allows us to choose a Lorentz frame in which \(y^1 - x^1 = 0\) for timelike separated \(x\) and \(y\) and \(y^0 - x^0 = 0\) for spacelike separated \(x\) and \(y\). Therefore, for timelike separated \(x\) and \(y\) we compute \[G(y^0-x^0,0) = \int_{-\infty}^{+\infty} \frac{dk}{(2\pi)2\omega_k} \, e^{-i\omega_{k} (y^0-x^0)} ~.\] We now make the substitution \(k = m \sinh\eta\), which implies that \(dk = m\cosh\eta \, d\eta\) and \(\omega_k = \sqrt{k^2+ m^2} = m \sqrt{\sinh^2\eta + 1} = m \cosh\eta\). This leaves us with \[G(y^0-x^0,0) = \int_{-\infty}^{+\infty} \frac{d\eta}{4\pi} \, e^{-im(y^0-x^0)\cosh\eta } ~.\] Using the given integral identity we find \[G(y^0-x^0,0) = -\frac{i}{4}H_0^{(2)}(m(y^0-x^0)) ~.\] On the other hand, for spacelike separated \(x\) and \(y\) we compute \[G(0,y^1-x^1) = \int_{-\infty}^{+\infty} \frac{dk}{(2\pi)2\omega_k} \, e^{+ik (y^1-x^1)} ~.\] We again make the substitution \(k = m \sinh\eta\) leaving us with \[\begin{split} G(0,y^1-x^1) & = \int_{-\infty}^{+\infty} \frac{d\eta}{4\pi} \, e^{+i m (y^1-x^1)\sinh\eta} \\ & =\int_{-\infty}^{+\infty} \frac{d\eta}{4\pi} \, \big(\cos(m (y^1-x^1)\sinh\eta) + i\sin(m(y^1-x^1)\sinh\eta )\big) ~. \end{split}\] The second term in the integrand is an odd function of \(\eta\), hence it does not contribute when we integrate over all \(\eta\). The first term in the integrand is an even function of \(\eta\), hence integrating over all \(\eta\) is equivalent to integrating twice over positive \(\eta\). Therefore, we are left with \[G(0,y^1-x^1) = \int_{0}^{+\infty} \frac{d\eta}{2\pi} \, \cos(m(y^1-x^1)\sinh\eta)~,\] and using the given integral identity we find \[G(0,y^1-x^1) = \frac{1}{2\pi} K_0(m(y^1-x^1)) ~.\]

2. From the known asymptotic behaviour of the Hankel and Bessel functions we have \[G(y^0-x^0,0) = -\frac{i}{4}H_0^{(2)}(m(y^0-x^0)) \sim e^{-im(y^0-x^0)} \ \text{ as $y^0-x^0 \to \infty$}~,\] and \[G(0,y^1-x^1) = \frac{1}{2\pi} K_0(m(y^1-x^1)) \sim \frac{1}{\sqrt{m(y^1-x^1)}} e^{-m(y^1-x^1)} \ \text{ as $y^1-x^1 \to \infty$}~.\] Therefore, for large timelike separations the Feynman propagator oscillates with only the positive-frequency wave propagating into the future, while for large spacelike separations it decays exponentially.

Feynman, retarded and advanced propagators.

1. The integrand is singular, i.e. it diverges, at \(k^0 = \pm \omega_{\vec{k}}\). Close to the singularities the dominant behaviour of the integrand is \[\begin{split} -i \frac{e^{-ik^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{-(k^0)^2 + \omega_{\vec{k}}^2} & = i\frac{e^{-ik^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{(k^0 + \omega_{\vec{k}})(k^0 - \omega_{\vec{k}})} \\ & \sim \pm i\frac{e^{\mp i \omega_{\vec{k}} (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}} (k^0 \mp \omega_{\vec{k}})} \ \text{ as $k^0 \to \pm \omega_{\vec{k}}$} ~. \end{split}\] Therefore, up to an overall finite normalisation that depends on \(\vec{k}\), we see that close to the singularities the integral behaves as \[\int_a^b dx \, \frac{1}{x} ~, \quad a < 0 \text{ and } b > 0 ~,\] which is an improper integral with a singularity at \(x = 0\). Splitting the integral into two, one from \(a\) to \(-\epsilon_a\) and the other from \(\epsilon_b\) to \(b\), and taking the limits \(\epsilon_a \to 0^+\) and \(\epsilon_b \to 0^+\), we see that both integrals diverge, hence the integral is not well-defined. Note that the contribution from large \(|k^0|\) is well-defined since the integrand behaves as \[\begin{split} -i \frac{e^{-ik^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{-(k^0)^2 + \omega_{\vec{k}}^2} & \sim i\frac{e^{-i k^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{(k^0)^2} \ \text{ as $k^0 \to \pm \infty$} ~. \end{split}\] Moreover, we have that \[\mathop{\mathrm{Re}}\Big(i\frac{e^{-i k^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{(k^0)^2}\Big) \leq \frac{1}{(k^0)^2} ~, \qquad \mathop{\mathrm{Im}}\Big(i\frac{e^{-i k^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{(k^0)^2}\Big) \leq \frac{1}{(k^0)^2} ~.\] Therefore, we can use that \[\begin{split} & \lim_{R_a\to+\infty} \int_a^{R_a} dx\, \frac{1}{x^2} = \frac1a ~, \quad a> 0 ~, \\ & \lim_{R_b\to+\infty} \int_{-R_b}^{b} dx\, \frac{1}{x^2} = \frac1a ~, \quad b <0 ~, \end{split}\] to infer that the contribution from large \(|k^0|\) is well-defined.

2. Let us start by defining \[\begin{split} f(\vec{k}) & = \frac{e^{-ik^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{-(k^0)^2 + \omega_{\vec{k}}^2 - i\epsilon} ~, \\ f_R(\vec{k}) & = \frac{e^{-ik^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{-(k^0 + i\epsilon)^2 + \omega_{\vec{k}}^2} ~, \\ f_A(\vec{k}) & = \frac{e^{-ik^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{-(k^0 - i\epsilon)^2 + \omega_{\vec{k}}^2} ~, \end{split}\] and computing the positions of the poles to first order in \(\epsilon\). For \(f(\vec{k})\) the poles are located at \[(k^0)^2 = \omega_{\vec{k}}^2 - i\epsilon \quad \Rightarrow \quad k^0 = \pm \omega_{\vec{k}} \sqrt{1-\frac{i\epsilon}{\omega_{\vec{k}}^2}} = \pm \omega_{\vec{k}} \mp \frac{i\epsilon}{2\omega_{\vec{k}}} + \mathcal{O}(\epsilon^2) ~.\] The poles of \(f_R(\vec{k})\) are located at \[(k^0 + i\epsilon)^2 = \omega_{\vec{k}}^2 \quad \Rightarrow \quad k^0 = \pm \omega_{\vec{k}} - i \epsilon ~,\] and the poles of \(f_A(\vec{k})\) are located at \[(k^0 - i\epsilon)^2 = \omega_{\vec{k}}^2 \quad \Rightarrow \quad k^0 = \pm \omega_{\vec{k}} + i \epsilon ~.\] We can also compute the residues to leading order in \(\epsilon\). These are equal to the residues we find setting \(\epsilon = 0\), i.e., \[\begin{split} & \mathop{\mathrm{res}}_{k^0 = \pm \omega_{\vec{k}} \mp \frac{i\epsilon}{2\omega_{\vec{k}}} + \mathcal{O}(\epsilon^2)}f(\vec{k}) =\mathop{\mathrm{res}}_{k^0 = \pm\omega_{\vec{k}} - i \epsilon } f_R(\vec{k}) =\mathop{\mathrm{res}}_{k^0 = \pm\omega_{\vec{k}} + i \epsilon } f_A(\vec{k}) \\ & \quad =\mathop{\mathrm{res}}_{k^0 = \pm\omega_{\vec{k}}} \frac{e^{-ik^0 (y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{-(k^0)^2 + \omega_{\vec{k}}^2} + \mathcal{O}(\epsilon) = \mp \frac{e^{\mp i\omega_{\vec{k}}(y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}}} +\mathcal{O}(\epsilon) ~. \end{split}\] To compute \(I(\vec{k})\) for \(y^0 > x^0\) we consider the following contour

   

   which goes clockwise around the pole at \(k^0 = + \omega_{\vec{k}} - \frac{i\epsilon}{2\omega_{\vec{k}}}\). To compute the contribution from the semicircle we parametrise \(k^0 = - i R e^{-i\theta}\) with \(\theta \in (-\frac\pi2,\frac\pi2)\) and take \(R \to \infty\). On the semicircle we have \[e^{-i k^0(y^0-x^0)} = e^{-R e^{-i\theta} (y^0-x^0)} = e^{-R (y^0-x^0)\cos\theta} e^{+i R(y^0-x^0)\sin \theta} \to 0 \ \text{ as $R\to\infty$} ~,\] since \(y^0 > x^0\) and \(\cos\theta > 0\) for \(\theta \in (-\frac\pi2,\frac\pi2)\). Therefore, the contribution from the semicircle vanishes. Recalling that for a clockwise contour the contour integral is given by \(-2\pi i\) times the sum of residues, we find \[I(\vec{k}) = -i \oint \frac{dk^0}{2\pi} \, f(\vec{k}) = - \mathop{\mathrm{res}}_{k^0 = + \omega_{\vec{k}} - \frac{i\epsilon}{2\omega_{\vec{k}}} + \mathcal{O}(\epsilon^2)}f(\vec{k}) = \frac{e^{- i\omega_{\vec{k}}(y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}}} ~, \qquad y^0 > x^0 ~.\] To compute \(I_R(\vec{k})\) for \(y^0 > x^0\) we consider the following contour

   

   which is a clockwise contour encircling both poles. By the same reasoning as for \(I(\vec{k})\) the contribution from the semicircle vanishes and we find \[\begin{split} I_R(\vec{k}) & = -i \oint \frac{dk^0}{2\pi} \, f_R(\vec{k}) = - \mathop{\mathrm{res}}_{k^0 = + \omega_{\vec{k}} - i\epsilon}f_R(\vec{k}) - \mathop{\mathrm{res}}_{k^0 = - \omega_{\vec{k}} - i\epsilon}f_R(\vec{k}) \\ & = \frac{e^{- i\omega_{\vec{k}}(y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}}} - \frac{e^{+ i\omega_{\vec{k}}(y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}}} ~, \qquad y^0 > x^0 ~. \end{split}\] To compute \(I_A(\vec{k})\) for \(y^0 > x^0\) we consider the following contour

   

   which is a clockwise contour encircling neither pole. Again by the same reasoning as for \(I(\vec{k})\) the contribution from the semicircle vanishes and we find \[\begin{split} I_A(\vec{k}) & = -i \oint \frac{dk^0}{2\pi} \, f_A(\vec{k}) = 0 ~, \qquad y^0 > x^0 ~. \end{split}\]

3. To compute \(I(\vec{k})\) for \(x^0 > y^0\) we consider the following contour

   

   which goes anticlockwise around the pole at \(k^0 = - \omega_{\vec{k}} + \frac{i\epsilon}{2\omega_{\vec{k}}}\). To compute the contribution from the semicircle we parametrise \(k^0 = i R e^{i\theta}\) with \(\theta \in (-\frac\pi2,\frac\pi2)\) and take \(R \to \infty\). On the semicircle we have \[e^{-i k^0(y^0-x^0)} = e^{R e^{i\theta} (y^0-x^0)} = e^{-R e^{i\theta} (x^0-y^0)} = e^{-R (x^0-y^0) \cos\theta} e^{+i R (x^0 - y^0) \sin \theta} \to 0 \ \text{ as $R\to\infty$} ~,\] since \(x^0 > y^0\) and \(\cos\theta > 0\) for \(\theta \in (-\frac\pi2,\frac\pi2)\). Therefore, the contribution from the semicircle vanishes. Recalling that for an anticlockwise contour the contour integral is given by \(2\pi i\) times the sum of residues, we find \[I(\vec{k}) = -i \oint \frac{dk^0}{2\pi} \, f(\vec{k}) = \mathop{\mathrm{res}}_{k^0 = - \omega_{\vec{k}} + \frac{i\epsilon}{2\omega_{\vec{k}}} + \mathcal{O}(\epsilon^2)}f(\vec{k}) = \frac{e^{+ i\omega_{\vec{k}}(y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}}} ~, \qquad x^0 > y^0 ~.\] To compute \(I_R(\vec{k})\) for \(x^0 > y^0\) we consider the following contour

   

   which is an anticlockwise contour encircling neither pole. By the same reasoning as for \(I(\vec{k})\) the contribution from the semicircle vanishes and we find \[I_R(\vec{k}) = -i \oint \frac{dk^0}{2\pi} \, f_R(\vec{k}) = 0 ~, \qquad x^0 > y^0 ~.\] To compute \(I_A(\vec{k})\) for \(x^0 > y^0\) we consider the following contour

   

   which is an anticlockwise contour encircling both poles. Again by the same reasoning as for \(I(\vec{k})\) the contribution from the semicircle vanishes and we find \[\begin{split} I_A(\vec{k}) & = -i \oint \frac{dk^0}{2\pi} \, f_A(\vec{k}) = \mathop{\mathrm{res}}_{k^0 = + \omega_{\vec{k}} + i\epsilon}f_A(\vec{k}) + \mathop{\mathrm{res}}_{k^0 = - \omega_{\vec{k}} + i\epsilon}f_A(\vec{k}) \\ & = - \frac{e^{- i\omega_{\vec{k}}(y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}}} + \frac{e^{+ i\omega_{\vec{k}}(y^0-x^0) + i \vec{k}\cdot(\vec{y}-\vec{x})}}{2\omega_{\vec{k}}} ~, \qquad x^0 > y^0 ~. \end{split}\]

Dyson’s formula.

1. Expanding to \(\mathcal{O}(\epsilon^2)\) we have \[\exp\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) = 1 -i \epsilon \int_{t_0}^t dt'\, H_I(t') - \frac12 \epsilon^2 \int_{t_0}^t dt'\, H_I(t')\int_{t_0}^t dt''\, H_I(t'') + \mathcal{O}(\epsilon^3) ~.\] It follows that \[i\frac{d}{dt}\exp\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) = \epsilon H_I(t) - \frac{i}{2} \epsilon^2 H_I(t) \int_{t_0}^t dt'\, H_I(t') - \frac{i}{2} \epsilon^2 \int_{t_0}^t dt'\, H_I(t') H_I(t) + \mathcal{O}(\epsilon^3)~.\] On the other hand, we have \[\epsilon H_I(t)\exp\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) = \epsilon H_I(t) -i \epsilon^2 H_I(t) \int_{t_0}^t dt'\, H_I(t') + \mathcal{O}(\epsilon^3) ~.\] Therefore, we find \[i\frac{d}{dt}\exp\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) - \epsilon H_I(t)\exp\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) =\frac{i}{2} \epsilon^2\Big[ H_I(t) ,\int_{t_0}^t dt'\, H_I(t')\Big] + \mathcal{O}(\epsilon^3)~,\] i.e., the differential equation \(\eqref{eq:timeevolution}\) is solved up to \(\mathcal{O}(\epsilon)\), but this is no longer the case at \(\mathcal{O}(\epsilon^2)\) since the operator \(H_I(t)\) does not necessarily commute with \(H_I(t')\) for \(t\neq t'\).

2. To see that this is resolved if the exponential is replaced by a time ordered exponential we can simply observe that \[\begin{split} \overset{\leftarrow}{\mathrm{T}}\Big(\Big[ H_I(t) ,\int_{t_0}^t dt'\, H_I(t')\Big]\Big) & = \overset{\leftarrow}{\mathrm{T}}\Big(H_I(t) \int_{t_0}^t dt'\, H_I(t')\Big) - \overset{\leftarrow}{\mathrm{T}}\Big(\int_{t_0}^t dt'\, H_I(t')H_I(t) \Big) \\ & =H_I(t) \int_{t_0}^t dt'\, H_I(t') - H_I(t) \int_{t_0}^t dt'\, H_I(t') = 0 ~, \end{split}\] since time ordering places operators evaluated at later times to the left and we have \(t>t_0\). More explicitly, expanding to \(\mathcal{O}(\epsilon^2)\) we have \[\begin{split} & \mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) \\ & = 1 -i\epsilon \overset{\leftarrow}{\mathrm{T}}\Big(\int_{t_0}^t dt'\, H_I(t')\Big) - \frac12 \epsilon^2 \overset{\leftarrow}{\mathrm{T}}\Big(\int_{t_0}^t dt'\, H_I(t')\int_{t_0}^t dt''\, H_I(t'')\Big) + \mathcal{O}(\epsilon^3) \\ & = 1 -i\epsilon \int_{t_0}^t dt'\, H_I(t') \\ & \quad -\frac12 \epsilon^2 \int_{t_0}^t dt' \int_{t_0}^{t'} dt'' \, H_I(t') H_I(t'') -\frac12 \epsilon^2 \int_{t_0}^t dt' \int_{t'}^t dt'' \, H_I(t'') H_I(t') + \mathcal{O}(\epsilon^3) ~. \end{split}\] It follows that \[\begin{split} i\frac{d}{dt}\mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) & = \epsilon \int_{t_0}^t dt'\, H_I(t') -\frac i2 \epsilon^2 \int_{t_0}^{t} dt' \, H_I(t) H_I(t') \\ & \quad -\frac i2 \epsilon^2 \int_{t}^t dt' \, H_I(t') H_I(t) - \frac i2 \epsilon^2 \int_{t_0}^t dt' H_I(t) H_I(t') + \mathcal{O}(\epsilon^3) \\ & = \epsilon \int_{t_0}^t dt'\, H_I(t') - i \epsilon^2 H_I(t) \int_{t_0}^{t} dt' \, H_I(t') + \mathcal{O}(\epsilon^3) ~, \end{split}\] where we have used that the integral between a point and itself vanishes. On the other hand, we have \[\epsilon H_I(t)\mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) = \epsilon H_I(t) -i \epsilon^2 H_I(t) \int_{t_0}^t dt'\, H_I(t') + \mathcal{O}(\epsilon^3) ~.\] Therefore, we find \[i\frac{d}{dt}\mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) - \epsilon H_I(t)\mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\epsilon\int_{t_0}^t dt' \, H_I (t')\big) =\mathcal{O}(\epsilon^3)~,\] i.e., the differential equation \(\eqref{eq:timeevolution}\) is solved up to \(\mathcal{O}(\epsilon^2)\) as required.

3. We have \[\begin{split} U(t_3,t_2)U(t_2,t_1) & = \mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\epsilon\int_{t_2}^{t_3} dt' \, H_I (t')\big) \mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\epsilon\int_{t_1}^{t_2} dt'' \, H_I (t'')\big) \\ & = \overset{\leftarrow}{\mathrm{T}}\big(\exp\big(-i\epsilon\int_{t_2}^{t_3} dt' \, H_I (t')\big) \exp\big(-i\epsilon\int_{t_1}^{t_2} dt'' \, H_I (t'')\big) ~. \end{split}\] This follows since the first time ordered exponential depends on operators evaluated between times \(t_2\) and \(t_3\), while the second time ordered exponential depends on operators evaluated between times \(t_1\) and \(t_2\). Since any time between \(t_2\) and \(t_3\) is later than any time between \(t_1\) and \(t_2\) by assumption we can combine the two time orderings into one. Now, \(H_I(t)\) depends on a single variable, time, since we have integrated over space and all other parameters are kept fixed. Given that the ordering in time is fixed by the time ordering, we can treat the operators \(H_I(t)\) as if they commute within a time ordered expression. Therefore, we can combine the two exponentials into one to give \[\begin{split} U(t_3,t_2)U(t_2,t_1) & =\overset{\leftarrow}{\mathrm{T}}\big(\exp\big(-i\epsilon\int_{t_2}^{t_3} dt' \, H_I (t') -i\epsilon\int_{t_1}^{t_2} dt'' \, H_I (t'')\big) \\ & = \overset{\leftarrow}{\mathrm{T}}\big(\exp\big(-i\epsilon\int_{t_1}^{t_3} dt' \, H_I (t') \big) = U(t_3,t_1) ~, \end{split}\] as required.

Real scalar fields.

1. The definition of the time ordered product is \[\overset{\leftarrow}{\mathrm{T}}\big(\phi(y)\phi(x)\big) = \phi(y)\phi(x)\theta(y^0 - x^0) + \phi(x)\phi(y) \theta(x^0 - y^0) ~,\] which is explicitly symmetric under interchanging \(x\) and \(y\), hence \[\overset{\leftarrow}{\mathrm{T}}\big(\phi(y)\phi(x)\big) = \overset{\leftarrow}{\mathrm{T}}\big(\phi(x)\phi(y)\big) ~.\] The show the normal ordered product is also symmetric let us decompose the field \(\phi\) into its creation \(\phi^-\) and annihilation \(\phi^+\) parts \[\phi(x) = \phi^+(x) +\phi^-(x) ~.\] We then have \[\begin{split} \,:\!\phi(y)\phi(x)\!:\, & = \,:\!(\phi^+(y) +\phi^-(y))(\phi^+(x) +\phi^-(x))\!:\, \\ & = \,:\!\phi^+(y)\phi^+(y) + \phi^+(y)\phi^-(x) + \phi^-(y)\phi^+(x) + \phi^-(y)\phi^-(x)\!:\, \\ & = \phi^+(y)\phi^+(y) + \phi^-(x)\phi^+(y) + \phi^-(y)\phi^+(x) + \phi^-(y)\phi^-(x) ~, \end{split}\] since the normal ordering moves all creation operators to the left and all annihilation operators to the right. This is again explicitly symmetric under interchanging \(x\) and \(y\), hence \[\,:\!\phi(y)\phi(x)\!:\, = \,:\!\phi(x)\phi(y)\!:\, ~.\]

2. The Feynman propagator is defined as \[G(y-x) = \langle 0| \overset{\leftarrow}{\mathrm{T}}\big(\phi(y)\phi(x)\big) |0\rangle ~,\] hence \(G(y-x) = G(x-y)\) follows from the symmetry of the time ordered product.

3. We have \[\begin{split} & \int d^4z \, G_1(y-z) G_1(z-x) \\ &= (-i)^2 \int d^4z \int \frac{d^4 k}{(2\pi)^4} \int \frac{d^4 k'}{(2\pi)^4} \, \frac{e^{+i k\cdot(y-z)}}{k^2 + m_1^2 - i\epsilon} \frac{e^{+i k'\cdot(z-x)}}{k'^2 + m_1^2 - i\epsilon} \\ & = -\int \frac{d^4 k}{(2\pi)^4} \int \frac{d^4 k'}{(2\pi)^4} \,\frac{e^{+i k\cdot y - i k'\cdot x}}{(k^2 + m_1^2 - i\epsilon)(k'{}^2 + m_1^2 - i\epsilon)} \int d^4z \, e^{-i(k-k') \cdot z} ~, \end{split}\] where we have used two different integration variables, \(k\) and \(k'\), for the two Feynman propagators. The integral over \(z\) gives the delta function \((2\pi)^4 \delta^{(4)}(k-k')\), which can be used to do the integral over \(k'\), setting \(k' = k\) and yielding \[\begin{equation} \label{eq:productoffeynman} \int d^4z \, G_1(y-z) G_1(z-x) = -\int \frac{d^4 k}{(2\pi)^4} \,\frac{e^{+i k \cdot (y- x)}}{(k^2 + m_1^2 - i\epsilon)^2} ~, \end{equation}\] as required. This is an example of a convolution. Changing variables \(z \to z+x\) we have \[\int d^4z \, G_1(y-z) G_1(z-x) = \int d^4z \, G_1(y-x-z) G_1(z) ~,\] which is the convolution of two Feynman propagators. The Fourier transform of the convolution will be the product of the two Fourier transformed Feynman propagators \[\tilde G_1(k)^2 = -\frac{1}{(k^2 + m_1^2 - i\epsilon)^2} ~.\] Inverting the Fourier transform we recover the right-hand side of \(\eqref{eq:productoffeynman}\).

4. To prove this identity we proceed by induction. We have already shown it holds for \(N=2\). Therefore, we need to show that it holds for \(N=n\) assuming it holds for \(N=n-1\). Using the identity for \(N=n-1\) we have \[\begin{split} &\int d^4z_1 \int d^4z_2 \hspace{1ex} \dots \int d^4z_{n-1} \int d^4z_n \, G_1(y-z_1) G_1(z_1-z_2) \dots G_1(z_{n-1}-z_{n}) G_1(z_n - x) \\& = (-i)^{n} \int d^4z_n \int \frac{d^4k}{(2\pi)^4} \, \frac{e^{+i k\cdot(y-z_n)}}{(k^2 + m_1^2 - i\epsilon)^{n}} G_1(z_n - x) \\& = (-i)^{n+1} \int d^4z_n \int \frac{d^4k}{(2\pi)^4} \int \frac{d^4k'}{(2\pi)^4} \, \frac{e^{+i k\cdot(y-z_n)}}{(k^2 + m_1^2 - i\epsilon)^{n}} \frac{e^{+ik\cdot(z_n-x)}}{(k'{}^2 + m_1^2 - i\epsilon)} \\& = (-i)^{n+1} \int \frac{d^4k}{(2\pi)^4} \int \frac{d^4k'}{(2\pi)^4} \, \frac{e^{+i k\cdot y - ik'\cdot x}}{(k^2 + m_1^2 - i\epsilon)^{n} (k'{}^2 + m_1^2 - i\epsilon)} \int d^4z_n \, e^{-i(k-k')z_n} ~. \end{split}\] As before, the integral over \(z_n\) gives the delta function \((2\pi)^4 \delta^{(4)}(k-k')\), which can be used to do the integral over \(k'\), setting \(k' = k\) and yielding \[\begin{equation} \begin{split} &\int d^4z_1 \int d^4z_2 \hspace{1ex} \dots \int d^4z_{n-1} \int d^4z_n \, G_1(y-z_1) G_1(z_1-z_2) \dots G_1(z_{n-1}-z_{n}) G_1(z_n - x) \\& = (-i)^{n+1} \int \frac{d^4k}{(2\pi)^4} \, \frac{e^{+i k\cdot (y - x)}}{(k^2 + m_1^2 - i\epsilon)^{n+1}} ~, \end{split}\label{eq:productoffeynman2} \end{equation}\] which is the identity for \(N=n\) as required. This can also be shown using convolutions. Let us define \[\begin{split} &g_N(y-x)\\ & = \int d^4z_1 \int d^4z_2 \hspace{1ex} \dots \int d^4z_{N-1} \int d^4z_N \, G_1(y-z_1) G_1(z_1-z_2) \dots G_1(z_{N-1}-z_{N}) G_1(z_N - x) ~. \end{split}\] It then follows that \[g_n(y-x) = \int d^4z_n \, g_{n-1}(y-z_n) G_1(z_n-x) = \int d^4z_n \, g_{n-1}(y-x-z_n) G_1(z_n) ~,\] where we have changed variables \(z_n \to z_n + x\). Fourier transforming, this equation becomes \[\tilde g_n(k) = \tilde g_{n-1}(k) \tilde G_1(k) ~.\] Assuming the identity holds for \(N=n-1\) we have \[\tilde g_{n-1}(k) = \frac{(-i)^{n}}{(k^2 + m_1^2 - i\epsilon)^{n}} ~, \qquad \tilde G_1(k) = \frac{-i}{k^2 + m_1^2 - i\epsilon} ~,\] hence \[\tilde g_n(k) = \tilde g_{n-1}(k) \tilde G_1(k) = \frac{(-i)^{n+1}}{(k^2 + m_1^2 - i\epsilon)^{n+1}} ~.\] Inverting the Fourier transform we recover the right-hand side of \(\eqref{eq:productoffeynman2}\).

5. There is no contraction between \(\phi_1\) and \(\phi_2\). This follows since they are decoupled in the action, i.e., the action can be written as the sum of an action for \(\phi_1\) and an action for \(\phi_2\). This means that, as operators, \(\phi_1\) and \(\phi_2\) commute and \[\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_2(x)\big)|0\rangle = 0 ~.\] Since \(\phi_1\) and \(\phi_2\) commute, we can freely move the creation operators to the left and the annihilation operators to the right, yielding the normal ordered product, which has a vanishing vacuum expectation value. Since there is no contraction between \(\phi_1\) and \(\phi_2\), the time ordered vacuum expectation value factorises into a part that only contains the fields \(\phi_1\) and a part that only contains the fields \(\phi_2\) \[\begin{split} & \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(x_6)\phi_2(x_5)\phi_2(x_4)\phi_2(x_3)\phi_1(x_2)\phi_1(x_1)\big)|0\rangle \\ & = \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(x_6)\phi_2(x_5)\phi_2(x_4)\phi_2(x_3)\big)|0\rangle\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x_2)\phi_1(x_1)\big)|0\rangle ~. \end{split}\] We have \[\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x_2)\phi_1(x_1)\big)|0\rangle = G_1(x_2-x_1) ~,\] and \[\begin{split} & \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(x_6)\phi_2(x_5)\phi_2(x_4)\phi_2(x_3)\big)|0\rangle \\ & = G_2(x_6-x_5)G_2(x_4-x_3) + G_2(x_6-x_4)G_2(x_5-x_3) + G_2(x_6-x_3)G_2(x_5-x_4) ~, \end{split}\] since there are three ways to fully contract four fields. Therefore, \[\begin{split} & \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(x_6)\phi_2(x_5)\phi_2(x_4)\phi_2(x_3)\phi_1(x_2)\phi_1(x_1)\big)|0\rangle \\ & = G_1(x_2-x_1)\big(G_2(x_6-x_5)G_2(x_4-x_3) + G_2(x_6-x_4)G_2(x_5-x_3) + G_2(x_6-x_3)G_2(x_5-x_4)\big) ~. \end{split}\]

6. Again the time ordered vacuum expectation value factorises since there is no contraction between \(\phi_1\) and \(\phi_2\) \[\begin{split} & \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(x_{103})\phi_2(x_{102})\phi_2(x_{101})\phi_2(x_{100})\phi_1(x_{99})\dots\phi_1(x_2)\phi_1(x_1\big))|0\rangle \\ & = \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(x_{103})\phi_2(x_{102})\phi_2(x_{101})\phi_2(x_{100})\big)|0\rangle\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x_{99})\dots\phi_1(x_2)\phi_1(x_1\big))|0\rangle ~. \end{split}\] By Wick’s theorem \(\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x_{99})\dots\phi_1(x_2)\phi_1(x_1\big))|0\rangle\) is the sum of all possible contractions with all fields contracted. However, it is impossible to fully contract an odd number of fields. Therefore, this time ordered vacuum expectation value vanishes \[\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(x_{103})\phi_2(x_{102})\phi_2(x_{101})\phi_2(x_{100})\phi_1(x_{99})\dots\phi_1(x_2)\phi_1(x_1\big))|0\rangle = 0 ~.\]

Feynman diagrams.

1. To compute the symmetry factor, we use the following rules:

   1. Let \(g\) be the number of permutations of fully internal vertices, i.e., not connected to an external point, that give an identical diagram. When computing \(g\) it is important to explicitly label the internal vertices to check that the permutation really gives an identical diagram. In particular, the permutation should give the same set of propagators between the labelled vertices.

   2. Let \(d\) be the number of double bubble diagrams.

   3. Let \(\beta\) be the number of lines connecting a vertex to itself.

   4. Let \(\alpha_n\), \(n=1,2,3,4\), be the number of pairs of vertices connected by \(n\) lines.

   5. The symmetry factor is then given by \[S = g 2^\beta 2^d \prod_{n=1}^4 (n!)^{\alpha_n} ~.\]

   The first diagram we consider is

   

   where we have labelled the internal vertices. There is only one fully internal vertex labelled by \(v\). As such, the only permutation that gives an identical diagram is the identity permutation, hence \(g=1\). There are no double bubbles, hence \(d=0\). There is one line connecting the vertex at \(v\) to itself and no lines connecting the other vertices to themselves, hence \(\beta = 1\). There is one pair of vertices (\(\{u,v\}\)) connected by two lines, hence \(\alpha_2 = 1\). There are no pairs of vertices connected by one, three or four lines, hence \(\alpha_1 = \alpha_3 = \alpha_4 = 0\). Therefore, the symmetry factor is \(S = 2 \times 2 = 4\). The second diagram we consider is

   

   where we have labelled the internal vertices. There are no fully internal vertices. As such, the only permutation that gives an identical diagram is the identity permutation, hence \(g=1\). There are no double bubbles, hence \(d=0\). There are no lines connecting a vertex to itself, hence \(\beta = 0\). There is one pair of vertices (\(\{u,v\}\)) connected by three lines, hence \(\alpha_3 = 1\). There are no pairs of vertices connected by one, two or four lines, hence \(\alpha_1 = \alpha_2 = \alpha_4 = 0\). Therefore, the symmetry factor is \(S = 3! = 6\).

   The third diagram we consider is

   

   where we have labelled the internal vertices. Note that this diagram involves cubic vertices, but the rules for computing the symmetry factor are the same as for \(\phi^4\) theory. There are two fully internal vertices at \(\tilde u\) and \(\tilde v\). However, permuting these does not give an identical diagram since they would no longer be connected to \(u\) and \(v\) respectively, but to \(v\) and \(u\) instead. As such, the only permutation that gives an identical diagram is the identity permutation, hence \(g=1\). There are no double bubbles, hence \(d=0\). There are no lines connecting a vertex to itself, hence \(\beta = 0\). There are two pairs of vertices (\(\{u,\tilde u\}\) and \(\{v,\tilde v\}\)) connected by one line, hence \(\alpha_1 = 2\). There are two pairs of vertices (\(\{u,v\}\) and \(\{\tilde u,\tilde v\}\)) connected by two lines, hence \(\alpha_2 = 2\). There are no pairs of vertices connected by three or four lines, hence \(\alpha_3 = \alpha_4 = 0\). Therefore, the symmetry factor is \(S = 2^2 = 4\). The fourth diagram we consider is

   

   where we have labelled the internal vertices. Note that this diagram again involves cubic vertices, but the rules for computing the symmetry factor are the same as for \(\phi^4\) theory. There are no fully internal vertices. As such, the only permutation that gives an identical diagram is the identity permutation, hence \(g=1\). There are no double bubbles, hence \(d=0\). There are no lines connecting a vertex to itself, hence \(\beta = 0\). There are three pairs of vertices (\(\{\tilde x,\tilde y\}\), \(\{\tilde y,\tilde z\}\) and \(\{\tilde z,\tilde x\}\)) connected by one line, hence \(\alpha_1 = 3\). There are no pairs of vertices connected by two, three or four lines, hence \(\alpha_2 = \alpha_3 = \alpha_4 = 0\). Therefore, the symmetry factor is \(S = 1\). The fifth diagram we consider is

   

   where we have labelled the internal vertices. There are no fully internal vertices. As such, the only permutation that gives an identical diagram is the identity permutation, hence \(g=1\). There are no double bubbles, hence \(d=0\). There are no lines connecting a vertex to itself, hence \(\beta = 0\). There is one pair of vertices (\(\{u,v\}\)) connected by two lines, hence \(\alpha_2 = 1\). There are no pairs of vertices connected by one, three or four lines, hence \(\alpha_1 = \alpha_3 = \alpha_4 = 0\). Therefore, the symmetry factor is \(S = 2\). The sixth diagram we consider is

   

   where we have labelled the internal vertices. There are two fully internal vertices at \(\tilde u\) and \(\tilde v\). Permuting these gives an identical diagram since, in addition to being connected to each other, they are both connected to the vertices at \(u\) and \(v\) by one line. As such, there are two permutations that give an identical diagram, hence \(g=2\). There are no double bubbles, hence \(d=0\). There are no lines connecting a vertex to itself, hence \(\beta = 0\). There are four pairs of vertices (\(\{u,\tilde u\}\), \(\{u,\tilde v\}\), \(\{v,\tilde u\}\) and \(\{v,\tilde v\}\)) connected by one line, hence \(\alpha_1 = 4\). There is one pair of vertices (\(\{\tilde u,\tilde v\}\)) connected by two lines, hence \(\alpha_2 = 1\). There are no pairs of vertices connected by three or four lines, hence \(\alpha_3 = \alpha_4 = 0\). Therefore, the symmetry factor is \(S = 2 \times 2 = 4\). The final diagram we consider is

   

   where we have labelled the internal vertices. There is only one fully internal vertex labelled by \(\tilde u\). As such, the only permutation that gives an identical diagram is the identity permutation, hence \(g=1\). There are no double bubbles, hence \(d=0\). There is one line connecting the vertex at \(\tilde u\) to itself and no lines connecting the other vertices to themselves, hence \(\beta = 1\). There are three pairs of vertices (\(\{u,\tilde u\}\), \(\{v,\tilde u\}\) and \(\{u,v\}\)) connected by one line, hence \(\alpha_1 = 3\). There are no pairs of vertices connected by two, three or four lines, hence \(\alpha_2 = \alpha_3 = \alpha_4 = 0\). Therefore, the symmetry factor is \(S = 2\).

2. Expanding the vacuum-to-vacuum amplitude up to and including \(\mathcal{O}(\lambda^2)\) gives \[\begin{split} & \langle 0|\mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-\frac{i\lambda}{4!} \int d^4z \, \phi_I(z)^4 \big) |0\rangle \\ & = \langle 0|0\rangle - \frac{i\lambda}{4!} \int d^4z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_I(z)\phi_I(z)\phi_I(z)\phi_I(z)\big)|0\rangle \\ & \quad + \frac12 \Big(- \frac{i\lambda}{4!}\Big)^2 \int d^4z_1 \int d^4z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_I(z_1)\phi_I(z_1)\phi_I(z_1)\phi_I(z_1)\phi_I(z_2)\phi_I(z_2)\phi_I(z_2)\phi_I(z_2)\big)|0\rangle \\ & \quad + \mathcal{O}(\lambda^3) ~. \end{split}\] At leading order we have \[\langle 0|0\rangle = 1 ~.\] At \(\mathcal{O}(\lambda)\) we have \[- \frac{i\lambda}{4!} \int d^4z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_I(z)\phi_I(z)\phi_I(z)\phi_I(z)\big)|0\rangle ~.\] Applying Wick’s theorem there are three ways to fully contract the four fields, all of which give the same analytic expression, hence \[\begin{split} & - \frac{i\lambda}{4!} \int d^4z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_I(z)\phi_I(z)\phi_I(z)\phi_I(z)\big)|0\rangle = 3 \times - \frac{i\lambda}{4!} \int d^4z \, G(0)^2 = -\frac{i\lambda}{8}\int d^4z \, G(0)^2 ~. \end{split}\] At \(\mathcal{O}(\lambda^2)\) we have \[-\frac{\lambda^2}{2\times 4! \times 4!} \int d^4z_1 \int d^4z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_I(z_1)\phi_I(z_1)\phi_I(z_1)\phi_I(z_1)\phi_I(z_2)\phi_I(z_2)\phi_I(z_2)\phi_I(z_2)\big)|0\rangle ~.\] Applying Wick’s theorem there are \(7 \times 5 \times 3 \times 1 = 105\) ways to fully contract the eight fields giving three different analytic expressions. The first comes from contracting each field at point \(z_1\) with a field at point \(z_2\) giving \[-\frac{\lambda^2}{2\times 4! \times 4!} \int d^4z_1 \int d^4z_2 \, G(z_1-z_2)^4 ~.\] There are \(4 \times 3 \times 2 \times 1\) = 24 ways to do this. The second comes from contracting two fields at point \(z_1\) with fields at point \(z_2\), the remaining two fields at point \(z_1\) with each other, and the remaining two fields at point \(z_2\) with each other giving \[-\frac{\lambda^2}{2\times 4! \times 4!} \int d^4z_1 \int d^4z_2 \, G(z_1-z_2)^2G(0)^2 ~.\] There are \(\binom{4}{2}^2 \times 2 = 72\) ways to do this, \(\binom{4}{2}\) ways to contract \(\phi_I(z_1)\) with \(\phi_I(z_1)\), \(\binom{4}{2}\) ways to contract \(\phi_I(z_2)\) with \(\phi_I(z_2)\) and 2 ways to contract the remaining \(\phi_I(z_1)\) with \(\phi_I(z_2)\). The third comes from contracting fields at point \(z_1\) with fields at point \(z_1\), and fields at point \(z_2\) with fields at point \(z_2\) giving \[-\frac{\lambda^2}{2\times 4! \times 4!} \int d^4z_1 \int d^4z_2 \, G(0)^4 ~.\] There are \(3 \times 3 = 9\) ways to do this, 3 ways to contract \(\phi_I(z_1)\) with each other and 3 ways to contract \(\phi_I(z_2)\) with each other. Note that if we try to contract three fields at point \(z_1\) with fields at point \(z_2\) we would be forced to also contract the fourth field at point \(z_1\) with the fourth field at point \(z_2\) leading us back to the first case. Similarly, we cannot only contract one field at point \(z_1\) with one field at point \(z_2\). Furthermore, we have \(24+72+9=105\), which confirms we have found all the ways of contracting the fields. Therefore, we have \[\begin{split} & -\frac{\lambda^2}{2\times 4! \times 4!} \int d^4z_1 \int d^4z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_I(z_1)\phi_I(z_1)\phi_I(z_1)\phi_I(z_1)\phi_I(z_2)\phi_I(z_2)\phi_I(z_2)\phi_I(z_2)\big)|0\rangle \\ & = -\lambda^2\int d^4z_1 \int d^4z_2 \,\frac{1}{48} G(z_1-z_2)^4 + \frac{1}{16} G(z_1-z_2)^2G(0)^2 + \frac{1}{128} G(0)^4 ~. \end{split}\]

More real scalar fields.

1. For each propagator associated to the scalar field \(\phi_1\)

   

   we write down a factor of \(G_1(y-x)\). For each propagator associated to the scalar field \(\phi_2\)

   

   we write down a factor of \(G_2(y-x)\). For each vertex

   

   we write down a factor of \(\displaystyle{-i\lambda \int d^4 z}\). Note that the interaction term is \(\displaystyle{-\int d^4 x\, \lambda \phi_1\phi_2}\) so each vertex has two legs. Finally we divide by the symmetry factor of the diagram.

2. Expanding the vacuum-to-vacuum amplitude to \(\mathcal{O}(\lambda^4)\) gives \[\begin{split} & \langle 0| \mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)|0\rangle \\ & = \langle 0|0\rangle -i\lambda \int d^4 z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z)\phi_2(z)\big)|0\rangle - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\big)|0\rangle \\ & +i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\big)|0\rangle \\ & + \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\phi_1(z_4)\phi_2(z_4)\big)|0\rangle \\ & + \mathcal{O}(\lambda^5) ~. \end{split}\] At leading order we have \[\langle 0|0\rangle = 1 ~.\] At \(\mathcal{O}(\lambda)\) we have \[-i\lambda \int d^4 z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z)\phi_2(z)\big)|0\rangle = - i\lambda \int d^4z \, \langle 0|\phi_1(z)|0\rangle\langle 0|\phi_2(z) |0\rangle = 0 ~,\] since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields. At \(\mathcal{O}(\lambda^2)\) we have \[\begin{split} & - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\big)|0\rangle \\ & = - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_1(z_2)\big) |0\rangle\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(z_1)\phi_2(z_2)\big)|0\rangle \\ &= -\frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, G_1(z_1-z_2)G_2(z_1-z_2) ~, \end{split}\] where we have used that there is no contraction between \(\phi_1\) and \(\phi_2\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 2 (\(g = 2\) since we can interchange the two vertices, which are fully internal, to give an identical diagram). This gives the correct overall factor. At \(\mathcal{O}(\lambda^3)\) we have \[\begin{split} & i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\big)|0\rangle \\ & = i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_1(z_2)\phi_1(z_3)\big)|0\rangle\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(z_1)\phi_2(z_2)\phi_2(z_3)\big)|0\rangle = 0 ~, \end{split}\] since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields. At \(\mathcal{O}(\lambda^4)\) we have \[\begin{split} & \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\phi_1(z_4)\phi_2(z_4)\big)|0\rangle \\ & = \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(z_1)\phi_1(z_2)\phi_1(z_3)\phi_1(z_4)\big)|0\rangle \\&\hspace{35ex}\times\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(z_1)\phi_2(z_2)\phi_2(z_3)\phi_2(z_4)\big)|0\rangle ~, \end{split}\] where we have used that there is no contraction between \(\phi_1\) and \(\phi_2\). Applying Wick’s theorem there are 3 ways to contract the fields \(\phi_1\) and 3 ways to contract the fields \(\phi_2\). Therefore, we have \(3\times 3 = 9\) possible contractions in total giving two different analytic expressions. The first arises when for any contraction between \(\phi_1(z_i)\) and \(\phi_1(z_j)\) we also have a contraction between \(\phi_2(z_i)\) and \(\phi_2(z_j)\) giving \[\frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(z_1-z_2)G_2(z_1-z_2)G_1(z_3-z_4)G_2(z_3-z_4) ~.\] There are 3 ways to do this since once the contractions of \(\phi_1\) are fixed so are the contractions of \(\phi_2\). Note that all contractions of this type give the same analytic expression since we can permute the integration variables \(z_1\), \(z_2\), \(z_3\) and \(z_4\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 8 (\(g = 8\) since we can permute \((z_1,z_2,z_3,z_4)\) to \((z_2,z_1,z_3,z_4)\), \((z_1,z_2,z_4,z_3)\), \((z_3,z_4,z_1,z_2)\) or any composition of these). This gives the correct overall factor since \(\displaystyle{\frac{3}{4!} = \frac{1}{8}}\). The second arises when for any contraction between \(\phi_1(z_i)\) and \(\phi_1(z_j)\) we have that \(\phi_2(z_i)\) and \(\phi_2(z_j)\) are not contracted giving \[\frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(z_1-z_2)G_2(z_2-z_3)G_1(z_3-z_4)G_2(z_1-z_4) ~.\] There are 6 ways of doing this since once the contractions of \(\phi_1\) are fixed there are two choices for contracting \(\phi_2\). Again all contractions of this type give the same analytic expression by permuting the integration variables \(z_1\), \(z_2\), \(z_3\) and \(z_4\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 4 (\(g=4\) since we can permute \((z_1,z_2,z_3,z_4)\) to \((z_3,z_2,z_1,z_4)\), \((z_1,z_4,z_3,z_2)\) or the composition of these). This gives the correct overall factor since \(\displaystyle{\frac{6}{4!} = \frac14}\). Therefore, we have \[\begin{split} & \langle 0| \mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)|0\rangle \\ & = 1 -\frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, G_1(z_1-z_2)G_2(z_1-z_2) \\ & \quad + \frac{\lambda^4}{8} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(z_1-z_2)G_1(z_3-z_4)G_2(z_1-z_2)G_2(z_3-z_4) \\ & \quad + \frac{\lambda^4}{4} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(z_1-z_2)G_1(z_3-z_4)G_2(z_1-z_3)G_2(z_2-z_4) \\ & \quad + \mathcal{O}(\lambda^5) ~. \end{split}\]

3. Expanding \(\displaystyle{\langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_2(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big)|0\rangle}\) to \(\mathcal{O}(\lambda^4)\) gives \[\begin{split} & \langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_2(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big)|0\rangle \\ & = \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\big)|0\rangle -i\lambda \int d^4 z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z)\phi_2(z)\big)|0\rangle \\ & \quad - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\big)|0\rangle \\ & \quad +i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\big)|0\rangle \\ & \quad + \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2) \\ & \hspace{47.5ex} \times \phi_1(z_3)\phi_2(z_3)\phi_1(z_4)\phi_2(z_4)\big)|0\rangle + \mathcal{O}(\lambda^5) ~. \end{split}\] At leading order we have \[\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\big)|0\rangle = \langle 0|\phi_1(x)|0\rangle \langle 0|\phi_2(y)|0\rangle = 0 ~,\] since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields. At \(\mathcal{O}(\lambda)\) we have \[\begin{split} & -i\lambda \int d^4 z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z)\phi_2(z)\big)|0\rangle \\ & = - i\lambda \int d^4z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x)\phi_1(z)\big)|0\rangle\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_2(z)\big) |0\rangle \\ & = - i\lambda \int d^4z \, G_2(y-z)G_1(z-x) ~, \end{split}\] where we have used that there is no contraction between \(\phi_1\) and \(\phi_2\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 1. This gives the correct overall factor. At \(\mathcal{O}(\lambda^2)\) we have \[\begin{split} & - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\big)|0\rangle \\ & = - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x)\phi_1(z_1)\phi_1(z_2)\big)|0\rangle \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_2(z_1)\phi_2(z_2)\big)|0\rangle = 0 ~, \end{split}\] since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields. At \(\mathcal{O}(\lambda^3)\) we have \[\begin{split} & i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\big)|0\rangle \\ & = i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x)\phi_1(z_1)\phi_1(z_2)\phi_1(z_3)\big)|0\rangle \\ & \hspace{35ex} \times \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_2(z_1)\phi_2(z_2)\phi_2(z_3)\big)|0\rangle ~, \end{split}\] where we have used that there is no contraction between \(\phi_1\) and \(\phi_2\). Applying Wick’s theorem there are 3 ways to contract the fields \(\phi_1\) and 3 ways to contract the fields \(\phi_2\). Therefore, we have \(3\times 3 = 9\) possible contractions in total giving two different analytic expressions. The first arises when for a contraction between \(\phi_1(z_i)\) and \(\phi_1(x)\) we also have a contraction between \(\phi_2(z_i)\) and \(\phi_2(y)\) giving \[i\frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, G_2(y-z_1)G_1(z_1-x)G_2(z_2-z_3)G_1(z_2-z_3) ~.\] There are 3 ways to do this since once the contractions of \(\phi_1\) are fixed so are the contractions of \(\phi_2\). Note that all contractions of this type give the same analytic expression since we can permute the integration variables \(z_1\), \(z_2\) and \(z_3\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 2 (\(g = 2\) since we can permute \((z_1,z_2,z_3)\) to \((z_1,z_3,z_2)\)). This gives the correct overall factor since \(\displaystyle{\frac{3}{3!} = \frac{1}{2}}\). The second arises when for any contraction between \(\phi_1(z_i)\) and \(\phi_1(x)\) we have that \(\phi_2(z_i)\) and \(\phi_2(y)\) are not contracted giving \[i\frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, G_2(y-z_1)G_1(z_1-z_2)G_2(z_2-z_3)G_1(z_3-x) ~.\] There are 6 ways of doing this since once the contractions of \(\phi_1\) are fixed there are two choices for contracting \(\phi_2\). Again all contractions of this type give the same analytic expression by permuting the integration variables \(z_1\), \(z_2\) and \(z_3\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 1. This gives the correct overall factor since \(\displaystyle{\frac{6}{3!} = 1}\). At \(\mathcal{O}(\lambda^4)\) we have \[\begin{split} & \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2) \\ & \hspace{47.5ex} \times \phi_1(z_3)\phi_2(z_3)\phi_1(z_4)\phi_2(z_4)\big)|0\rangle \\ & = \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \,\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(x)\phi_1(z_1)\phi_1(z_2)\phi_1(z_3)\phi_1(z_4)\big)|0\rangle \\ & \hspace{47.5ex}\times\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_2(z_1)\phi_2(z_2)\phi_2(z_3)\phi_2(z_4)\big)|0\rangle \\ & = 0 ~, \end{split}\] since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields. Therefore, we have \[\begin{split} & \langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_2(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big)|0\rangle \\ & = - i\lambda \int d^4z \, G_2(y-z)G_1(z-x) \\ & \quad + i \frac{\lambda^3}{2} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, G_2(y-z_1)G_1(z_1-x)G_2(z_2-z_3)G_1(z_2-z_3) \\ & \quad + i \lambda^3 \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, G_2(y-z_1)G_1(z_1-z_2)G_2(z_2-z_3)G_1(z_3-x) \\ & \quad + \mathcal{O}(\lambda^5) ~. \end{split}\] Expanding \(\displaystyle{\langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_1(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big)|0\rangle}\) to \(\mathcal{O}(\lambda^4)\) gives \[\begin{split} & \langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_1(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big)|0\rangle \\ & = \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\big)|0\rangle -i\lambda \int d^4 z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z)\phi_2(z)\big)|0\rangle \\ & \quad - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\big)|0\rangle \\ & \quad +i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\big)|0\rangle \\ & \quad + \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2) \\ & \hspace{47.5ex} \times \phi_1(z_3)\phi_2(z_3)\phi_1(z_4)\phi_2(z_4)\big)|0\rangle + \mathcal{O}(\lambda^5) ~. \end{split}\] At leading order we have \[\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\big)|0\rangle = G_1(y-x) ~.\] At \(\mathcal{O}(\lambda)\) we have \[\begin{split} & -i\lambda \int d^4 z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z)\phi_2(z)\big)|0\rangle \\ & = - i\lambda \int d^4z \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z)\big)|0\rangle\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(z)\big) |0\rangle = 0 ~, \end{split}\] since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields. At \(\mathcal{O}(\lambda^2)\) we have \[\begin{split} & - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\big)|0\rangle \\ & = - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_1(z_2)\big)|0\rangle \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(z_1)\phi_2(z_2)\big)|0\rangle ~, \end{split}\] where we have used that there is no contraction between \(\phi_1\) and \(\phi_2\). Applying Wick’s theorem there are 3 ways to contract the fields \(\phi_1\) and 1 way to contract the fields \(\phi_2\). Therefore, we have \(3\times 1 = 3\) possible contractions in total giving two different analytic expressions. The first arises when \(\phi_1(y)\) is contracted with \(\phi_1(x)\) giving \[- \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, G_1(y-x)G_1(z_1-z_2)G_2(z_1-z_2) ~.\] There is 1 way to do this since once \(\phi_1(y)\) is contracted with \(\phi_1(x)\) the other contractions are fixed. The corresponding Feynman diagram is

   

   which has a symmetry factor of 2 (\(g=2\) since we can interchange the two vertices to give an identical diagram). This gives the correct overall factor. The second arises when \(\phi_1(y)\) is contracted with \(\phi_1(z_i)\) giving \[- \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, G_1(y-z_1)G_2(z_1-z_2)G_1(z_2-x) ~.\] There are 2 ways to do this since there are two choices for contracting \(\phi_1(y)\), which once made fixes the other contractions. Note that both contractions give the same analytic expression since we can permute the integration variables \(z_1\) and \(z_2\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 1. This gives the correct overall factor since \(\displaystyle{\frac{2}{2!} = 1}\). At \(\mathcal{O}(\lambda^3)\) we have \[\begin{split} & i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2)\phi_1(z_3)\phi_2(z_3)\big)|0\rangle \\ & = i \frac{\lambda^3}{3!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_1(z_2)\phi_1(z_3)\big)|0\rangle \\ & \hspace{40ex} \times \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(z_1)\phi_2(z_2)\phi_2(z_3)\big)|0\rangle \\ & =0 ~, \end{split}\] since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields. At \(\mathcal{O}(\lambda^4)\) we have \[\begin{split} & \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, \langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_2(z_1)\phi_1(z_2)\phi_2(z_2) \\ & \hspace{47.5ex} \times \phi_1(z_3)\phi_2(z_3)\phi_1(z_4)\phi_2(z_4)\big)|0\rangle \\ & = \frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \,\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\phi_1(z_1)\phi_1(z_2)\phi_1(z_3)\phi_1(z_4)\big)|0\rangle \\ & \hspace{47.5ex}\times\langle 0|\overset{\leftarrow}{\mathrm{T}}\big(\phi_2(z_1)\phi_2(z_2)\phi_2(z_3)\phi_2(z_4)\big)|0\rangle ~, \end{split}\] where we have used that there is no contraction between \(\phi_1\) and \(\phi_2\). Applying Wick’s theorem there are 15 ways to contract the fields \(\phi_1\) and 3 ways to contract the fields \(\phi_2\). Therefore, we have \(15\times 3 = 45\) possible contractions in total giving four different analytic expressions. The first arises when \(\phi_1(y)\) is contracted with \(\phi_1(x)\) and when for any contraction between \(\phi_1(z_i)\) and \(\phi_1(z_j)\) we also have a contraction between \(\phi_2(z_i)\) and \(\phi_2(z_j)\) giving \[\frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-x)G_1(z_1-z_2)G_2(z_1-z_2)G_1(z_3-z_4)G_2(z_3-z_4) ~.\] There are 3 ways to do this since once \(\phi_1(y)\) is contracted with \(\phi_1(x)\) there are three ways to contract the remaining fields \(\phi_1\), and once the contractions of \(\phi_1\) are fixed so are the contractions of \(\phi_2\). Note that all contractions of this type give the same analytic expression since we can permute the integration variables \(z_1\), \(z_2\), \(z_3\) and \(z_4\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 8 (\(g = 8\) since we can permute \((z_1,z_2,z_3,z_4)\) to \((z_2,z_1,z_3,z_4)\), \((z_1,z_2,z_4,z_3)\), \((z_3,z_4,z_1,z_2)\) or any composition of these). This gives the correct overall factor since \(\displaystyle{\frac{3}{4!} = \frac{1}{8}}\). The second arises when \(\phi_1(y)\) is contracted with \(\phi_1(x)\) and when for any contraction between \(\phi_1(z_i)\) and \(\phi_1(z_j)\) we have that \(\phi_2(z_i)\) and \(\phi_2(z_j)\) are not contracted giving \[\frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-x)G_1(z_1-z_2)G_2(z_2-z_3)G_1(z_3-z_4)G_2(z_1-z_4) ~.\] There are 6 ways of doing this since once \(\phi_1(y)\) is contracted with \(\phi_1(x)\) there are three ways to contract the remaining fields \(\phi_1\), and once the contractions of \(\phi_1\) are fixed there are two choices for contracting \(\phi_2\). Again all contractions of this type give the same analytic expression by permuting the integration variables \(z_1\), \(z_2\), \(z_3\) and \(z_4\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 4 (\(g=4\) since we can permute \((z_1,z_2,z_3,z_4)\) to \((z_3,z_2,z_1,z_4)\), \((z_1,z_4,z_3,z_2)\) or the composition of these). This gives the correct overall factor since \(\displaystyle{\frac{6}{4!} = \frac14}\). The third arises when \(\phi_1(y)\) is contracted with \(\phi_1(z_i)\), \(\phi_1(x)\) is contracted with \(\phi_1(z_j)\), and \(\phi_2(z_i)\) is contracted with \(\phi_2(z_j)\) giving \[\frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-z_1)G_2(z_1-z_2)G_1(z_2-x)G_1(z_3-z_4)G_2(z_3-z_4) ~.\] There are 12 ways of doing this since there are four choices for contracting \(\phi_1(y)\) and subsequently three choices for contracting \(\phi_1(x)\), which once made fixes the other contractions. Again all contractions of this type give the same analytic expression by permuting the integration variables \(z_1\), \(z_2\), \(z_3\) and \(z_4\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 2 (\(g=2\) since we can permute \((z_1,z_2,z_3,z_4)\) to \((z_1,z_2,z_4,z_3)\)). This gives the correct overall factor since \(\displaystyle{\frac{12}{4!} = \frac12}\). The fourth arises when \(\phi_1(y)\) is contracted with \(\phi_1(z_i)\), \(\phi_1(x)\) is contracted with \(\phi_1(z_j)\), and \(\phi_2(z_i)\) is not contracted with \(\phi_2(z_j)\) giving \[\frac{\lambda^4}{4!} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-z_1)G_2(z_1-z_2)G_1(z_2-z_3)G_2(z_3-z_4)G_1(z_4-x) ~.\] There are 24 ways of doing this since there are four choices for contracting \(\phi_1(y)\) with \(\phi_1(z_i)\) and subsequently three choices for contracting \(\phi_1(x)\) with \(\phi_1(z_j)\). Once these are made there are two choices for contracting \(\phi_2(z_i)\) and \(\phi_2(z_j)\) with \(\phi_2(z_k)\) and \(\phi_2(z_l)\). Again all contractions of this type give the same analytic expression by permuting the integration variables \(z_1\), \(z_2\), \(z_3\) and \(z_4\). The corresponding Feynman diagram is

   

   which has a symmetry factor of 1. This gives the correct overall factor since \(\displaystyle{\frac{24}{4!} = 1}\). Note that \(3+6+12+24 = 45\) as anticipated. Therefore, we have \[\begin{split} & \langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_1(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big)|0\rangle \\ & = G_1(y-x) \\ & \quad - \frac{\lambda^2}{2} \int d^4 z_1 \int d^4 z_2 \, G_1(y-x)G_1(z_1-z_2)G_2(z_1-z_2) \\ & \quad - \lambda^2 \int d^4 z_1 \int d^4 z_2 \, G_1(y-z_1)G_2(z_1-z_2)G_1(z_2-x) \\ & \quad + \frac{\lambda^4}{8} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-x)G_1(z_1-z_2)G_2(z_1-z_2)G_1(z_3-z_4)G_2(z_3-z_4) \\ & \quad + \frac{\lambda^4}{4} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-x)G_1(z_1-z_2)G_2(z_2-z_3)G_1(z_3-z_4)G_2(z_1-z_4) \\ & \quad + \frac{\lambda^4}{2} \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-z_1)G_2(z_1-z_2)G_1(z_2-x)G_1(z_3-z_4)G_2(z_3-z_4) \\ & \quad + \lambda^4 \int d^4 z_1 \int d^4 z_2 \int d^4 z_3 \int d^4 z_4 \, G_1(y-z_1)G_2(z_1-z_2)G_1(z_2-z_3)G_2(z_3-z_4)G_1(z_4-x) \\ & \quad + \mathcal{O}(\lambda^5) ~. \end{split}\]

4. Let us start by considering the time ordered two-point correlation function \[\langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\big) |\Omega\rangle = \frac{\langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_2(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big) |0\rangle}{\langle 0| \mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big) |0\rangle} ~.\] Only Feynman diagrams without vacuum bubble diagrams contribute to \(\langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\big) |\Omega\rangle\). At \(\mathcal{O}(\lambda^{2k})\) in the expansion there are an odd number of fields \(\phi_1\) and \(\phi_2\). Since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields, by Wick’s theorem the contribution at \(\mathcal{O}(\lambda^{2k})\) will vanish. At \(\mathcal{O}(\lambda^{2k+1})\) there is a unique Feynman diagram given by

   

   with \(2k+1\) vertices. This diagram has a symmetry factor of 1 so the corresponding analytic expression is \[\begin{split} & (-i\lambda)^{2k+1} \int d^4 z_1 \hspace{1ex} \dots \int d^4 z_{2k+1} \, G_2(y-z_1)G_1(z_1-z_2) \dots G_2(z_{2k}-z_{2k+1})G_1(z_{2k+1} - x) \\ & = i \lambda^{2k+1} \int \frac{d^4p}{(2\pi)^4} \, \frac{e^{+ip\cdot(y-x)}}{(p^2 + m_1^2 - i\epsilon)^{k+1}(p^2 + m_2^2 - i\epsilon)^{k+1}} ~, \end{split}\] where we recall \[G_1(y-x)= -i \int \frac{d^4p}{(2\pi)^4} \, \frac{e^{+ip\cdot(y-x)}}{p^2 + m_1^2 - i\epsilon} ~, \qquad G_2(y-x)= -i \int \frac{d^4p}{(2\pi)^4} \, \frac{e^{+ip\cdot(y-x)}}{p^2 + m_2^2 - i\epsilon} ~.\] Therefore, we have \[\begin{split} \langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\phi_2(y)\phi_1(x)\big) |\Omega\rangle & = \sum_{k=0}^\infty i \lambda^{2k+1} \int \frac{d^4p}{(2\pi)^4} \, \frac{e^{+ip\cdot(y-x)}}{(p^2 + m_1^2 - i\epsilon)^{k+1}(p^2 + m_2^2 - i\epsilon)^{k+1}} \\ & = i\lambda \int \frac{d^4p}{(2\pi)^4} \,e^{+ip\cdot(y-x)}\frac{1}{(p^2 + m_1^2 - i\epsilon)(p^2 + m_2^2 - i\epsilon)-\lambda^2}~. \end{split}\] Now let us consider the time ordered two-point correlation function \[\langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\big) |\Omega\rangle = \frac{\langle 0| \overset{\leftarrow}{\mathrm{T}}\Big(\phi_2(y)\phi_1(x)\exp\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big)\Big) |0\rangle}{\langle 0| \mathrm{T}\hspace{-1pt}\overset{\longleftarrow}{\exp}\big(-i\lambda\int d^4z \, \phi_1(z)\phi_2(z)\big) |0\rangle} ~.\] Only Feynman diagrams without vacuum bubble diagrams contribute to \(\langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\big) |\Omega\rangle\). At \(\mathcal{O}(\lambda^{2k+1})\) in the expansion there are an odd number of fields \(\phi_1\) and \(\phi_2\). Since there is no contraction between \(\phi_1\) and \(\phi_2\) and we cannot fully contract an odd number of fields, by Wick’s theorem the contribution at \(\mathcal{O}(\lambda^{2k+1})\) will vanish. At \(\mathcal{O}(\lambda^{2k})\) there is a unique Feynman diagram given by

   

   with \(2k\) vertices. This diagram has a symmetry factor of 1 so the corresponding analytic expression is \[\begin{split} & (-i\lambda)^{2k} \int d^4 z_1 \hspace{1ex} \dots \int d^4 z_{2k} \, G_1(y-z_1)G_2(z_1-z_2) \dots G_2(z_{2k-1}-z_{2k})G_1(z_{2k} - x) ~. \\ & = -i \lambda^{2k} \int \frac{d^4p}{(2\pi)^4} \, \frac{e^{+ip\cdot(y-x)}}{(p^2 + m_1^2 - i\epsilon)^{k+1}(p^2 + m_2^2 - i\epsilon)^{k}} ~. \end{split}\] Therefore, we have \[\begin{split} \langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\phi_1(y)\phi_1(x)\big) |\Omega\rangle & = \sum_{k=0}^\infty -i\lambda^{2k} \int \frac{d^4p}{(2\pi)^4} \, \frac{e^{+ip\cdot(y-x)}}{(p^2 + m_1^2 - i\epsilon)^{k+1}(p^2 + m_2^2 - i\epsilon)^{k}} \\ & = -i \int \frac{d^4p}{(2\pi)^4} \,e^{+ip\cdot(y-x)}\frac{p^2 + m_2^2 - i\epsilon}{(p^2 + m_1^2 - i\epsilon)(p^2 + m_2^2 - i\epsilon)-\lambda^2} ~. \end{split}\] By symmetry, we can infer that \[\begin{split} &\langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\phi_i(y)\phi_j(x)\big) |\Omega\rangle \\ & = -i \int \frac{d^4p}{(2\pi)^4} \,e^{+ip\cdot(y-x)}\frac{1}{(p^2 + m_1^2 - i\epsilon)(p^2 + m_2^2 - i\epsilon)-\lambda^2} \begin{pmatrix} p^2 + m_2^2 - i\epsilon & -\lambda \\ -\lambda & p^2 + m_1^2 - i\epsilon \end{pmatrix}_{ij} ~. \end{split}\] Implementing the transformation \[\begin{equation} \label{eq:rottotilde} \begin{pmatrix} \tilde\phi_1\\\tilde\phi_2\end{pmatrix} = \begin{pmatrix} \frac{\sqrt{M^2 +m_1^2 - m_2^2}}{\sqrt{2}M} & \frac{\sqrt{M^2 - m_1^2 + m_2^2}}{\sqrt{2}M} \\ -\frac{\sqrt{M^2 - m_1^2 + m_2^2}}{\sqrt{2}M} & \frac{\sqrt{M^2 +m_1^2 - m_2^2}}{\sqrt{2}M} \end{pmatrix} \begin{pmatrix} \phi_1\\\phi_2\end{pmatrix} ~, \qquad M = \sqrt[4]{(m_1^2-m_2^2)^2+4\lambda^2} ~, \end{equation}\] we find that the time ordered two point correlation functions of the rotated fields are \[\begin{equation} \begin{split}\label{eq:feynmanrot} &\langle\Omega| \overset{\leftarrow}{\mathrm{T}}\big(\tilde\phi_i(y)\tilde\phi_j(x)\big) |\Omega\rangle = -i \int \frac{d^4p}{(2\pi)^4} \,e^{+ip\cdot(y-x)} \begin{pmatrix} \frac{1}{p^2 + \tilde m_1^2 - i\epsilon} & 0 \\ 0 & \frac{1}{p^2 + \tilde{m}_2^2 - i\epsilon} \end{pmatrix}_{ij} ~, \end{split} \end{equation}\] where \[\tilde m_1^2 = \frac12\big(m_1^2+m_2^2+M^2\big) ~, \qquad \tilde m_2^2 = \frac12\big(m_1^2+m_2^2-M^2\big) ~.\] Note that if we implement the transformation \(\eqref{eq:rottotilde}\) in the action \[S[\phi_1,\phi_2] = \int d^4x \, \Big(-\frac12 \partial_\mu\phi_1\partial^\mu\phi_1 - \frac12 m_1^2 \phi_1^2 -\frac12 \partial_\mu\phi_2\partial^\mu\phi_2 -\frac12 m_2^2 \phi_2^2 - \lambda \phi_1 \phi_2 \Big) ~,\] we find \[S[\tilde\phi_1,\tilde\phi_2] = \int d^4x \, \Big(-\frac12 \partial_\mu\tilde\phi_1\partial^\mu\tilde\phi_1 - \frac12 \tilde m_1^2 \tilde\phi_1^2 -\frac12 \partial_\mu\tilde\phi_2\partial^\mu\tilde\phi_2 -\frac12 \tilde m_2^2 \tilde\phi_2^2 \Big) ~,\] i.e., a theory of two free massive real scalar fields with no interaction whose Feynman propagator is given by \(\eqref{eq:feynmanrot}\) as expected. Therefore, we have used perturbation theory in \(\lambda\) to recover an exact result that we could have found much faster by first transforming the fields. Nevertheless, this demonstrates that perturbation theory works.