# Supersymmetry – Exercises

These exercises accompany the Supersymmetry course held in the Epiphany term of the 2022-2023 academic year as part of the Particles, Strings and Cosmology MSc degree at Durham University.

Durham, 9 January 2023

last updated 2 February 2023

# 2 Poincaré and Lorentz groups, and their representations

Check that the transformations $x^\mu \to \Lambda^\mu{}_\nu x^\nu + c^\mu ~, \qquad \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma \eta_{\mu\nu} = \eta_{\rho\sigma} ~,$ are indeed isometries of the metric $ds^2 = \eta_{\mu\nu} dx^\mu dx^\nu ~, \qquad \eta_{\mu\nu} = \mathop{\mathrm{diag}}(+1,-1,-1,-1)_{\mu\nu} ~,$ and that they form a group.

We have that $\begin{split} x^\mu \to \Lambda^\mu{}_\nu x^\nu + c^\mu \qquad & \Rightarrow \qquad dx^\mu \to \Lambda^\mu{}_\nu dx^\nu \\ & \Rightarrow \qquad ds^2 = \eta_{\mu\nu} dx^\mu dx^\nu \to \eta_{\mu\nu} \Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma dx^\rho dx^\sigma ~. \end{split}$ Therefore, if $$\Lambda^\mu{}_\rho \Lambda^\nu{}_\sigma \eta_{\mu\nu} = \eta_{\rho\sigma}$$ we immediately see that the metric is invariant.

To check that the transformations form a group let us write $$(\Lambda,c) x^\mu = \Lambda^\mu{}_\nu x^\nu + c^\mu$$. We then have that $(\tilde\Lambda,\tilde c)(\Lambda,c) x^\mu = \tilde\Lambda^\mu{}_\nu (\Lambda^\nu{}_\rho x^\rho + c^\nu) + \tilde c^\mu = (\tilde\Lambda\Lambda, \tilde\Lambda c + \tilde c) x^\mu ~.$ Given that $$\Lambda^\mathrm{t}\eta \Lambda = \eta$$ and $$\tilde\Lambda^\mathrm{t}\eta\tilde\Lambda = \eta$$ implies that the same is true for $$\tilde\Lambda\Lambda$$, we find that the transformations close amongst themselves. The identity element is simply given by $$(\Lambda,c) = (\mathbf{1},0)$$ and the inverse of $$(\Lambda,c)$$ is $$(\Lambda,c)^{-1} = (\eta\Lambda^\mathrm{t}\eta,-\eta\Lambda^\mathrm{t}\eta c)$$. It is straightforward to check that these are of the required form using that $$\eta^{-1} = \eta^\mathrm{t}= \eta$$. Finally, we need to check associativity. We have $\begin{split} (\hat\Lambda,\hat c) \big((\tilde\Lambda,\tilde c)(\Lambda,c)\big) & = (\hat\Lambda,\hat c)(\tilde\Lambda\Lambda, \tilde\Lambda c + \tilde c) = (\hat\Lambda\tilde\Lambda\Lambda, \hat\Lambda\tilde\Lambda c + \hat\Lambda\tilde c + \hat c) ~, \\ \big((\hat\Lambda,\hat c)(\tilde\Lambda,\tilde c)\big)(\Lambda,c) & = (\hat\Lambda\tilde\Lambda, \hat\Lambda\tilde c + \hat c)(\Lambda,c) = (\hat\Lambda\tilde\Lambda\Lambda, \hat\Lambda\tilde\Lambda c + \hat\Lambda\tilde c + \hat c) ~, \end{split}$ showing that the transformations indeed form a group.

Show that $\Lambda^\mu{}_\nu = \frac12 \mathop{\mathrm{tr}}(N \sigma_\nu N^\dagger \bar \sigma^\mu) ~, \qquad N \in \mathrm{SL}(2,\mathbb{C}) ~,$ defines a group homomorphism $$\mathrm{SL}(2,\mathbb{C}) \to \mathrm{SO}^+(1,3)$$.

We would like to show that $$\Lambda^\mu{}_\nu(N)\Lambda^\nu{}_\rho(\tilde{N}) = \Lambda^\mu{}_\rho(N\tilde{N})$$. We have that $\begin{split} \Lambda^\mu{}_\rho(N\tilde{N}) & = \frac{1}{2}\mathop{\mathrm{tr}}(N \tilde{N} \sigma_\rho \tilde{N}^\dagger N^\dagger \bar \sigma^\mu) \\ & = \frac{1}{4} \mathop{\mathrm{tr}}(\tilde{N}\sigma_\rho \tilde{N}^\dagger \bar\sigma^\nu ) \mathop{\mathrm{tr}}( \sigma_\nu N^\dagger \bar \sigma^\mu N) \\ & = \frac{1}{2} \mathop{\mathrm{tr}}(N \sigma_\nu N^\dagger \bar \sigma^\mu) \frac{1}{2} \mathop{\mathrm{tr}}(\tilde{N}\sigma_\rho \tilde{N}^\dagger \bar\sigma^\nu ) \\ & = \Lambda^\mu{}_\nu(N) \Lambda^\nu{}_\rho(\tilde{N}) ~, \end{split}$ as required. Here we have used the cyclicity of the trace, and, to go from the first line to the second line, that $$\sigma_\mu = \bar \sigma^\mu$$ and that $$\{\tfrac{1}{\sqrt{2}}\sigma_\mu\}$$ is an orthonormal basis (over $$\mathbb{C}$$) for the set of $$2 \times 2$$ complex matrices; hence, $$\mathop{\mathrm{tr}}(M_1 M_2) = \frac12 \sum_\mu \mathop{\mathrm{tr}}(M_1 \sigma_\mu) \mathop{\mathrm{tr}}(\sigma_\mu M_2) = \frac12 \mathop{\mathrm{tr}}(M_1\bar\sigma^\mu) \mathop{\mathrm{tr}}(\sigma_\mu M_2)$$.

Show that $i(\sigma^{\mu\nu})_{\alpha}{}^\beta = \frac{i}{4}(\sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu)_{\alpha}{}^\beta ~, \qquad i(\bar\sigma^{\mu\nu})^{\dot\alpha}{}_{\dot\beta} = \frac{i}{4}(\bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu)^{\dot\alpha}{}_{\dot\beta} ~,$ satisfy the commutation relations of the Lorentz algebra.

Let us define $\begin{split} J_i & = \frac{i}{2}\epsilon_{ijk} \sigma_{jk} = - \frac{i}{8}\epsilon_{ijk}[\sigma_j,\sigma_k] = \frac{1}{4} \epsilon_{ijk} \epsilon_{jkl}\sigma_l = \frac{1}{2} \sigma_i ~, \\ K_i & = i\sigma_{0i} = \frac{i}{4}( -\sigma_i - \sigma_i) = - \frac{i}{2} \sigma_i ~. \end{split}$ It is now straightforward to compute $\begin{split} \phantom{}[J_i,J_j] & = \frac{1}{4}[\sigma_i,\sigma_j] = \frac{i}{2}\epsilon_{ijk}\sigma_k = i\epsilon_{ijk}J_k ~, \\ \phantom{}[K_i,K_j] & = -\frac{1}{4}[\sigma_i,\sigma_j] = -\frac{i}{2}\epsilon_{ijk}\sigma_k = -i\epsilon_{ijk}J_k ~, \\ \phantom{}[J_i,K_j] & = -\frac{i}{4}[\sigma_i,\sigma_j] = \frac{1}{2}\epsilon_{ijk}\sigma_k = i\epsilon_{ijk}K_k ~, \end{split}$ and we recover the commutation relations of the Lorentz algebra in one of its well-known forms as required. A similar computation follows for generators of the conjugate of the fundamental representation, with the only difference being that $$K_i = \frac{i}{2} \sigma_i$$.

Prove the following identities for anticommuting spinors: \begin{aligned} (i): & \quad \psi^\alpha\psi^\beta = -\frac{1}{2}\epsilon^{\alpha\beta} \psi\psi ~, & (ii): & \quad \bar\psi^{\dot\alpha}\bar\psi^{\dot\beta} = \frac{1}{2}\epsilon^{\dot\alpha\dot\beta} \bar\psi\bar\psi ~, \\ (iii): & \quad (\theta\phi)(\theta\psi) = -\frac{1}{2}(\theta\theta)(\phi\psi) ~, & (iv): & \quad (\bar\theta\bar\phi)(\bar\theta\bar\psi) = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\phi\bar\psi) ~, \\ (v): & \quad \chi\sigma^\mu \bar\psi = -\bar\psi \bar\sigma^\mu \chi ~, & (vi): & \quad \chi\sigma^\mu \bar\sigma^\nu \psi = \psi \sigma^\nu \bar\sigma^\mu \chi ~, \\ (vii): & \quad (\chi\sigma^\mu\bar\psi)^\dagger = \psi\sigma^\mu\bar\chi ~, & (viii): & \quad (\chi\sigma^\mu \bar\sigma^\nu \psi)^\dagger = \bar\psi\bar\sigma^\nu\sigma^\mu\bar\chi ~, \\ (ix): & \quad (\theta\psi)(\theta\sigma^\mu\bar\phi) = -\frac{1}{2}(\theta\theta)(\psi\sigma^\mu\bar\phi) ~, & (x): & \quad (\bar\theta\bar\psi)(\bar\theta\bar\sigma^\mu\phi) = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\psi\bar\sigma^\mu\phi) ~, \\ (xi): & \quad (\phi\psi)(\bar\chi\bar\theta) = \frac{1}{2}(\phi\sigma^\mu\bar\chi)(\psi\sigma_\mu\bar\theta) ~, & (xii): & \quad (\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) = \frac{1}{2}(\theta\theta)(\bar\theta\bar \theta)\eta^{\mu\nu} ~. \end{aligned}

$$(i)$$: $$\psi^\alpha\psi^\beta$$ is antisymmetric on interchange of $$\alpha$$ and $$\beta$$. Therefore, $$\psi^\alpha\psi^\beta \propto \epsilon^{\alpha\beta}$$. To fix the proportionality constant we contract with $$\epsilon_{\alpha\beta}$$. This gives $$\psi\psi = c \epsilon^{\alpha\beta}\epsilon_{\alpha\beta} = -2c$$; hence $$\psi^\alpha\psi^\beta = -\frac{1}{2}\epsilon^{\alpha\beta}\psi\psi$$.

$$(ii)$$: $$\bar\psi^{\dot\alpha}\bar\psi^{\dot\beta}$$ is antisymmetric on interchange of $$\dot\alpha$$ and $$\dot\beta$$. Therefore, $$\bar\psi^{\dot\alpha}\bar\psi^{\dot\beta} \propto \epsilon^{\dot\alpha\dot\beta}$$. To fix the proportionality constant we contract with $$\epsilon_{\dot\alpha\dot\beta}$$. This gives $$-\bar\psi\bar\psi = c \epsilon^{\dot\alpha\dot\beta}\epsilon_{\dot\alpha\dot\beta} = -2c$$; hence $$\bar\psi^{\dot\alpha}\bar\psi^{\dot\beta} = \frac{1}{2}\epsilon^{\dot\alpha\dot\beta}\bar\psi\bar\psi$$.

$$(iii)$$: We have $\begin{split} (\theta\phi)(\theta\psi) & = \epsilon_{\alpha\beta}\epsilon_{\gamma\delta} \theta^\alpha\phi^\beta\theta^\gamma\psi^\delta = - \epsilon_{\alpha\beta}\epsilon_{\gamma\delta}\theta^\alpha\theta^\gamma\phi^\beta\psi^\delta \\ & = \frac{1}{2} \epsilon_{\alpha\beta}\epsilon_{\gamma\delta}\epsilon^{\alpha\gamma} (\theta\theta)\phi^\beta\psi^\delta = - \frac{1}{2} \epsilon_{\beta\delta}(\theta\theta)\phi^\beta\psi^\delta = -\frac{1}{2}(\theta\theta)(\phi\psi) ~. \end{split}$

$$(iv)$$: We have $\begin{split} (\bar\theta\bar\phi)(\bar\theta\bar\psi) & = \epsilon_{\dot\alpha\dot\beta}\epsilon_{\dot\gamma\dot\delta} \bar\theta^{\dot\beta}\bar\phi^{\dot\alpha}\bar\theta^{\dot\delta}\bar\psi^{\dot\gamma} = -\epsilon_{\dot\alpha\dot\beta}\epsilon_{\dot\gamma\dot\delta} \bar\theta^{\dot\beta}\bar\theta^{\dot\delta}\bar\phi^{\dot\alpha}\bar\psi^{\dot\gamma} \\ & = -\frac{1}{2}\epsilon_{\dot\alpha\dot\beta}\epsilon_{\dot\gamma\dot\delta} \epsilon^{\dot\beta\dot\delta}(\bar\theta\bar\theta)\bar\phi^{\dot\alpha}\bar\psi^{\dot\gamma} = - \frac{1}{2} \epsilon_{\dot\gamma\dot\alpha}(\bar\theta\bar\theta)\bar\phi^{\dot\alpha}\bar\psi^{\dot\gamma} = -\frac{1}{2}(\bar\theta\bar\theta)(\bar\phi\bar\psi) ~. \end{split}$ Or alternatively, this is just the complex conjugate of identity $$(iii)$$.

$$(v)$$: We have $\begin{split} \chi\sigma^\mu \bar\psi & = \chi^\alpha(\sigma^\mu)_{\alpha\dot\alpha} \bar\psi^{\dot\alpha} = \epsilon^{\alpha\beta}\chi_\beta (\sigma^\mu)_{\alpha\dot\alpha} \epsilon^{\dot\alpha\dot\beta}\bar\psi_{\dot\beta} \\ & = \chi_\beta (\bar\sigma^\mu)^{\dot\beta\beta} \bar\psi_{\dot\beta} = - \bar\psi_{\dot\beta}(\bar\sigma^\mu)^{\dot\beta\beta}\chi_\beta = -\bar\psi \bar\sigma^\mu \chi ~. \end{split}$

$$(vi)$$: We have $\begin{split} \chi\sigma^\mu \bar\sigma^\nu \psi & = \chi^\alpha(\sigma^\mu)_{\alpha\dot\alpha}(\bar\sigma^\nu)^{\dot\alpha\beta}\psi_\beta = \epsilon^{\alpha\gamma}\chi_\gamma(\sigma^\mu)_{\alpha\dot\alpha}\epsilon^{\dot\alpha\dot\beta}\epsilon_{\dot\beta\dot\gamma} (\bar\sigma^\nu)^{\dot\gamma\beta} \epsilon_{\beta\delta}\psi^\delta \\ & = - \chi_\gamma(\bar\sigma^\mu)^{\dot\beta\gamma}(\sigma^\nu)_{\delta\dot\beta} \psi^\delta = \psi^\delta(\sigma^\nu)_{\delta\dot\beta}(\bar\sigma^\mu)^{\dot\beta\gamma}\chi_\gamma =\psi \sigma^\nu \bar\sigma^\mu \chi ~, \end{split}$

$$(vii)$$: We have $(\chi\sigma^\mu\bar\psi)^\dagger = \psi(\sigma^\mu)^\dagger \bar\chi = \psi\sigma^\mu\bar\chi ~.$

$$(viii)$$: We have $(\chi\sigma^\mu\bar\sigma^\nu\bar\psi)^\dagger = \psi(\bar\sigma^\nu)^\dagger(\sigma^\mu)^\dagger \bar\chi = \psi\bar\sigma^\nu\sigma^\mu\bar\chi ~.$

$$(ix)$$: Follows from identity $$(iii)$$ viewing $$\sigma^\mu \bar\phi$$ as a left-handed Weyl spinor with an undotted index.

$$(x)$$: Follows from identity $$(iv)$$ viewing $$\bar\sigma^\mu \phi$$ as a right-handed Weyl spinor with a dotted index. Or alternatively, this is just the complex conjugate of identity $$(ix)$$.

$$(xi)$$: We have that $(\phi\psi)(\bar\chi\bar\theta) = \phi^\alpha \psi_\alpha \bar\chi_{\dot\alpha} \bar\theta^{\dot\alpha} = -\bar\chi^{\dot\alpha}\phi^\alpha\psi_\alpha \bar\theta_{\dot\alpha}= - \mathop{\mathrm{tr}}(M_1 M_2) ~,$ where $$(M_1)^{\dot\alpha\alpha} = \bar\chi^{\dot\alpha}\phi^\alpha$$ and $$(M_2)_{\alpha\dot\alpha} = \psi_\alpha\bar\theta_{\dot\alpha}$$. Since $$\sigma^\mu = \bar\sigma_\mu$$ and $$\{\tfrac{1}{\sqrt{2}}\sigma^\mu\}$$ is an orthonormal basis (over $$\mathbb{C}$$) for the set of $$2\times 2$$ complex matrices, we have that $$\mathop{\mathrm{tr}}(M_1M_2) = \frac{1}{2}\mathop{\mathrm{tr}}(M_1\sigma^\mu)\mathop{\mathrm{tr}}(M_2\bar\sigma_\mu)$$. Substituting back in for $$M_1$$ and $$M_2$$ gives $(\phi\psi)(\bar\chi\bar\theta) = - \frac{1}{2}(\phi\sigma^\mu\bar\chi)(\bar\theta\bar\sigma_\mu\psi) = \frac{1}{2}(\phi\sigma^\mu\bar\chi)(\psi\sigma_\mu\bar\theta) ~,$ where we have used identity $$(v)$$.

$$(xii)$$: We have $\begin{split} (\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) & = - \theta^\alpha \theta^\beta \bar\theta^{\dot\alpha}\bar\theta^{\dot\beta} (\sigma^\mu)_{\alpha\dot\alpha} (\sigma^\nu)_{\beta\dot\beta} = \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \epsilon^{\alpha\beta}\epsilon^{\dot\alpha\dot\beta} (\sigma^\mu)_{\alpha\dot\alpha} (\sigma^\nu)_{\beta\dot\beta} \\ & = \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu)_{\alpha\dot\alpha} (\bar\sigma^\nu)^{\dot\alpha\alpha} = \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \mathop{\mathrm{tr}}(\sigma^\mu\bar\sigma^\nu) =\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\eta^{\mu\nu} ~. \end{split}$

A Dirac spinor is made up of one left-handed and one right-handed Weyl spinor. In the Weyl basis we have $\Psi = \begin{pmatrix} \psi_\alpha \\ \bar\chi^{\dot\alpha} \end{pmatrix} ~, \qquad \gamma^\mu = \begin{pmatrix} \mathbf{0}& \sigma^\mu \\ \bar \sigma^\mu & \mathbf{0}\end{pmatrix} ~.$ Check that the gamma matrices $$\gamma^\mu$$ satisfy the Clifford algebra relations $$\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}\mathbf{1}$$. Express the Dirac Lagrangian $\mathcal{L}= -i\bar\Psi\gamma^\mu\partial_\mu \Psi - m \bar\Psi\Psi ~,$ in terms of the Weyl spinors $$\psi$$ and $$\bar\chi$$. What happens when $$\Psi$$ is taken the be a Majorana spinor?

Computing the anticommutator of the gamma matrices we find $\{\gamma^\mu,\gamma^\nu\} = \begin{pmatrix} \sigma^\mu\bar\sigma^\nu +\sigma^\nu\bar\sigma^\mu & \mathbf{0}\\ \mathbf{0}& \bar\sigma^\mu\sigma^\nu +\bar\sigma^\nu\sigma^\mu \end{pmatrix}~.$ We can then check explicitly that $\begin{split} \sigma^0\bar\sigma^0 +\sigma^0\bar\sigma^0 & = 2 \mathbf{1}~, \\ \sigma^0\bar\sigma^i +\sigma^i\bar\sigma^0 & = \sigma_i - \sigma_i = 0 ~, \\ \sigma^i\bar\sigma^j +\sigma^j\bar\sigma^i & = -\{\sigma_i,\sigma_j\}= - 2\delta_{ij}\mathbf{1}~, \end{split}$ so that we have $$\sigma^\mu\bar\sigma^\nu +\sigma^\nu\bar\sigma^\mu= 2 \eta^{\mu\nu}\mathbf{1}$$. Similarly, we find that $$\bar\sigma^\mu\sigma^\nu +\bar\sigma^\nu\sigma^\mu = 2 \eta^{\mu\nu}\mathbf{1}$$ and we have that the gamma matrices indeed satisfy the Clifford algebra relations.

To express the Dirac Lagrangian in terms of Weyl spinors, let us start by recalling that $$\bar\Psi = \Psi^\dagger \gamma^0$$. Therefore, $\bar\Psi = \begin{pmatrix} \bar\psi_{\dot\alpha} & \chi^{\alpha} \end{pmatrix} \begin{pmatrix} \mathbf{0}& \mathbf{1}\\ \mathbf{1}& \mathbf{0}\end{pmatrix} = \begin{pmatrix} \chi^{\alpha} & \bar\psi_{\dot\alpha} \end{pmatrix} ~.$ Now substituting into the Dirac Lagrangian gives $\begin{split} \mathcal{L}& = -i\bar\Psi\gamma^\mu\partial_\mu \Psi - m \bar\Psi\Psi \\ & = -i \begin{pmatrix} \chi^{\alpha} & \bar\psi_{\dot\alpha} \end{pmatrix} \begin{pmatrix} \mathbf{0}& (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu \\ (\bar \sigma^\mu)^{\dot\alpha\alpha} \partial_\mu & \mathbf{0}\end{pmatrix} \begin{pmatrix} \psi_\alpha \\ \bar\chi^{\dot\alpha} \end{pmatrix} - m\begin{pmatrix} \chi_{\alpha} & \bar\psi^{\dot\alpha} \end{pmatrix} \begin{pmatrix} \psi_\alpha \\ \bar\chi^{\dot\alpha} \end{pmatrix} \\ & = - i\chi^{\alpha}(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu\bar\chi^{\dot\alpha} -i\bar\psi_{\dot\alpha}(\bar\sigma^\mu)^{\dot\alpha\alpha} \partial_\mu \psi_\alpha -m \chi^{\alpha}\psi_\alpha -m \bar\psi_{\dot\alpha}\bar\chi^{\dot\alpha} \\ & = -i( \chi\sigma^\mu\partial_\mu\bar\chi +\bar\psi\bar\sigma^\mu \partial_\mu\psi) -m (\chi\psi + \bar\psi\bar\chi) \\ & = -i(\bar\chi\bar\sigma^\mu \partial_\mu\chi +\bar\psi\bar\sigma^\mu \partial_\mu\psi) -m(\chi\psi + \bar\psi\bar\chi) ~, \end{split}$ where to go to the final line we have integrated by parts and dropped a total derivative. For a Majorana spinor, we simply replace $$\chi$$ by $$\psi$$.

Can you think of any fields that you have encountered that transform in the real irreducible representations $$(1,0)\oplus(0,1)$$ and $$(1,1)$$?

The field strength $$F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$ of a $$\mathrm{U}(1)$$ gauge field transforms in the $$(1,0)\oplus(0,1)$$ representation. The two complex irreducible representations correspond to the self-dual and anti-self-dual part. As expected, we need to complexify the field strength to construct these since the Hodge dual squares to $$-1$$ on $$\mathbb{R}^{1,3}$$. The traceless part of a symmetric energy-momentum tensor $$T_{\mu\nu}$$ transforms in the $$(1,1)$$ representation.

A complex second rank antisymmetric tensor $$f^{\mu\nu} = - f^{\nu\mu}$$ can be decomposed into a self-dual part $$f_+^{\mu\nu}$$ and an anti-self-dual part $$f_-^{\mu\nu}$$ $f^{\mu\nu} = f_+^{\mu\nu} + f_-^{\mu\nu} ~, \qquad f_\pm^{\mu\nu} = \frac12 \Big(f^{\mu\nu} \pm \frac{i}{2}\epsilon^{\mu\nu\rho\sigma}f_{\rho\sigma}\Big) ~,$ where $$\epsilon^{\mu\nu\rho\sigma}$$ is the Levi-Civita symbol with $$\epsilon^{0123} = 1$$. The self-dual and anti-self-dual parts are therefore constrained such that $f_\pm^{\mu\nu} = \pm \frac{i}{2}\epsilon^{\mu\nu\rho\sigma}f_\pm{}_{\rho\sigma} ~.$ Check that $$\sigma^{\mu\nu} = \frac{1}{4}(\sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu)$$ and $$\bar\sigma^{\mu\nu} = \frac{1}{4}(\bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu)$$ are self-dual and anti-self-dual respectively. Raising and lowering indices with $$\epsilon_{\alpha\beta}$$, $$\epsilon_{\dot\alpha\dot\beta}$$ and their inverses, show that $$(\sigma^{\mu\nu})_{\alpha\beta}$$ and $$(\bar\sigma^{\mu\nu})^{\dot\alpha\dot\beta}$$ realise the isomorphism between self-dual and anti-self-dual second-rank antisymmetric tensors and the $$(1,0)$$ and $$(0,1)$$ representations of $$\mathrm{SL}(2,\mathbb{C})$$.

We first note that $$\sigma^{\mu\nu}$$ and $$\bar\sigma^{\mu\nu}$$ are antisymmetric on interchange of $$\mu$$ and $$\nu$$. Next we substitute in for $$\sigma^\mu$$ and $$\bar\sigma^\mu$$ to find $\sigma^{0i} = \frac{1}{2}\sigma_i ~, \qquad \bar\sigma^{0i} = -\frac{1}{2}\sigma_i ~, \qquad \sigma^{ij} = -\frac{i}{2}\epsilon_{ijk}\sigma_k ~, \qquad \bar\sigma^{ij} = -\frac{i}{2}\epsilon_{ijk}\sigma_k ~.$ Therefore, we have $\begin{split} \frac{i}{2}\epsilon^{0ijk} \sigma_{jk} & = \frac{1}{4}\epsilon_{ijk}\epsilon_{jkl} \sigma_l = \frac{1}{2}\sigma_i = \sigma^{0i} ~, \\ \frac{i}{2}\epsilon^{ij\mu\nu}\sigma_{\mu\nu} & = i\epsilon^{ij0k}\sigma_{0k} = -\frac{i}{2}\epsilon_{ijk} \sigma_k = \sigma^{ij} ~, \end{split}$ confirming that $$\sigma^{\mu\nu} = \frac{i}{2}\epsilon^{\mu\nu\rho\sigma} \sigma_{\mu\nu}$$. A similar computation holds for $$\bar\sigma^{\mu\nu}$$.

The matrices $$(\sigma^{\mu\nu})_{\alpha\beta} = \epsilon_{\beta\gamma} (\sigma^{\mu\nu})_{\alpha}{}^{\gamma}$$ and $$(\bar\sigma^{\mu\nu})^{\dot\alpha\dot\beta} = \epsilon^{\dot\beta\dot\gamma} (\bar\sigma^{\mu\nu})^{\dot\alpha}{}_{\dot\gamma}$$ are symmetric on interchange of $$\alpha$$ and $$\beta$$ or $$\dot\alpha$$ and $$\dot\beta$$. This follows since $$\sigma_i\epsilon$$ is symmetric for all the Pauli matrices $$\sigma_i$$. Therefore, contracting $$(\sigma^{\mu\nu})_{\alpha\beta}$$ with a self-dual second-rank antisymmetric tensor gives a symmetric bispinor with two undotted indices, which corresponds to the irreducible representation $$(1,0)$$. Similarly, contracting $$(\bar\sigma^{\mu\nu})^{\dot\alpha\dot\beta}$$ with an anti-self-dual second-rank antisymmetric tensor gives a symmetric bispinor with two dotted indices, which corresponds to the irreducible representation $$(0,1)$$.

Show that $$[W^2,P_\mu] = 0$$ where $$W^\mu$$ is the Pauli-Lubański vector $$W^\mu = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}P_\nu M_{\rho\sigma}$$.

We start by showing that $\begin{split} \phantom{}[W_\mu,P_\tau] & = \frac{1}{2}\epsilon^{\mu\nu\rho\sigma}P_\nu[M_{\rho\sigma},P_\tau] = -\frac{i}{2}\epsilon^{\mu\nu\rho\sigma}P_\nu(\eta_{\rho\tau} P_\sigma - \eta_{\sigma\tau}P_\rho) \\ & = i\epsilon^{\mu\nu\rho}{}_{\tau}P_\nu P_\rho = \frac{i}{2}\epsilon^{\mu\nu\rho}{}_{\tau}[P_\nu, P_\rho] = 0 ~. \end{split}$ It then immediately follows that $$[W^2,P_\mu] = 0$$.

# 3 Supersymmetry algebras and supermultiplets

Using commutators ($$[X,Y] = XY-YX$$) and anticommutators ($$\{X,Y\} = XY+YX$$), write down the four types of Jacobi identity.

The first type of Jacobi identity involves three even generators $\phantom{}[X_0,[Y_0,Z_0]] + [Y_0,[Z_0,X_0]] + [Z_0,[X_0,Y_0]] = 0 ~.$ The second type of Jacobi identity involves two even generators and one odd generator $\phantom{}[X_0,[Y_0,Z_1]] + [Y_0,[Z_1,X_0]] + [Z_1,[X_0,Y_0]] = 0 ~.$ The third type of Jacobi identity involves one even generator and two odd generators $\phantom{}[X_0,\{Y_1,Z_1\}] + \{Y_1,[Z_1,X_0]\} - \{Z_1,[X_0,Y_1]\} = 0 ~.$ The fourth type of Jacobi identity involves three odd generators $\phantom{}[X_1,\{Y_1,Z_1\}] + [Y_1,\{Z_1,X_1\}] + [Z_1,\{X_1,Y_1\}] = 0 ~.$

Supersymmetric quantum mechanics with two supercharges (sometimes called $$\mathcal{N}=2$$ or $$\mathcal{N}=(1,1)$$) has the superalgebra $\begin{gathered} \{Q,\bar Q\} = 2H ~, \qquad \{Q,Q\} = Z ~, \qquad \{\bar Q,\bar Q\} = \bar Z ~, \\ \phantom{}[R,Q] = -Q ~, \qquad [R,\bar Q] = \bar Q ~, \qquad [R,Z] = -2Z ~, \qquad [R,\bar Z] = 2\bar Z ~, \end{gathered}$ where $$H$$ is the Hamiltonian and all other commutators vanish. Check the graded Jacobi identity for this superalgebra.

We start by observing that $$H$$ is central, hence all Jacobi identities involving $$H$$ are trivially satisfied. It remains to check those Jacobi identities involving one of the seven non-vanishing commutators or anticommutators and one of $$Q$$, $$Z$$ or $$R$$: $\begin{split} & [Q,\{Q,\bar Q\}] + [Q,\{\bar Q, Q\}] + [\bar Q,\{Q,Q\}] = 4[Q,H] + [\bar Q,Z] = 0 ~, \\ & [Q,\{Q,Q\}] + [Q,\{Q, Q\}] + [Q,\{Q,Q\}] = 3[Q,Z] = 0 ~, %\\ %& [Q,\{\bar Q,\bar Q\}] + [\bar Q,\{\bar Q, Q\}] + [\bar Q,\{Q,\bar Q\}] = [Q,\bar Z] + 4[\bar Q,H] = 0 ~, \\ & \{Q,[R,Q]\} - [R,\{Q,Q\}] - \{Q,[Q,R]\} = -\{Q,Q\} - [R,Z] - \{Q,Q\} = -2Z + 2Z = 0 ~, \\ & \{Q,[R,\bar Q]\} - [R,\{\bar Q,Q\}] - \{\bar Q,[Q,R]\} = \{Q,\bar Q\} - 2[R,H] - \{\bar Q,Q\} = 0 ~, \\ & [Q,[R,Z]] + [R,[Z,Q]] + [Z,[Q,R]] = -2[Q,Z] + [Z,Q] = 0 ~, \\ & [Q,[R,\bar Z]] + [R,[\bar Z,Q]] + [\bar Z,[Q,R]] = 2[Q,\bar Z] + [\bar Z,Q] = 0 ~, \\ & [Z,\{Q,\bar Q\}] + \{Q,[\bar Q,Z]\} - \{\bar Q,[Z,Q]\} = 2[Z,H] = 0 ~, \\ & [Z,\{Q,Q\}] + \{Q,[Q,Z]\} - \{Q,[Z,Q]\} = [Z,Z] = 0 ~, \\ & [Z,\{\bar Q,\bar Q\}] + \{\bar Q,[\bar Q,Z]\} - \{\bar Q,[Z,\bar Q]\} = [Z,\bar Z] = 0 ~, %\\ %& [Z,[R,Q]] + [R,[Q,Z]] + [Q,[Z,R]] = -[Z,Q] + 2[Q,Z] = 0 ~, %\\ %& [Z,[R,\bar Q]] + [R,[\bar Q,Z]] + [\bar Q,[Z,R]] = [Z,\bar Q] + 2[\bar Q,Z] = 0 ~, \\ & [Z,[R,\bar Z]] + [R,[\bar Z,Z]] + [\bar Z,[Z,R]] = 2[Z,\bar Z] + 2[\bar Z,Z] = 0 ~. %\\ %& [R,\{Q,\bar Q\}] + \{Q,[\bar Q,R]\} - \{\bar Q,[R,Q]\} = 2[R,H] - \{Q,\bar Q\} + \{\bar Q,Q\} = 0 ~, %\\ %& [R,\{Q,Q\}] + \{Q,[Q,R]\} - \{Q,[R,Q]\} = [R,Z] + \{Q,Q\} + \{Q,Q\} = -2Z + Z + Z = 0 ~. \end{split}$ The remaining Jacobi identities are either trivially satisfied, e.g. $$[X_0,[X_0,Y_i]] + [X_0,[Y_i,X_0]] + [Y_i,[X_0,X_0]] = 0$$, or are related to those above by conjugation.

Write down the content of the $$\mathcal{N}=2$$ gravitino ($$\lambda_0 = \frac{1}{2}$$) and graviton ($$\lambda_0 = 1$$) multiplets. Decompose these supermultiplets in terms of $$\mathcal{N}=1$$ massless supermultiplets.

The content of the gravitino multiplet is $$\big(\frac{1}{2},2\times(1),\frac{3}{2}\big)$$ plus its CPT-conjugate $$\big(-\frac{3}{2},2\times(-1),-\frac{1}{2}\big)$$. Combined this gives $$\big(-\frac{3}{2},2\times(-1),-\frac{1}{2},\frac{1}{2},2\times(1),\frac{3}{2}\big)$$ and contains one $$\mathcal{N}=1$$ gravitino multiplet and one $$\mathcal{N}=1$$ vector multiplet. The content of the graviton multiplet is $$\big(1,2\times(\frac{3}{2}),2\big)$$ plus its CPT-conjugate $$\big(-2,2\times(-\frac{3}{2}),-1\big)$$. Combined this gives $$\big(-2,2\times(-\frac{3}{2}),-1,1,2\times(\frac{3}{2}),2\big)$$ and contains one $$\mathcal{N}=1$$ graviton multiplet and one $$\mathcal{N}=1$$ gravitino multiplet.

Explicitly construct the most general massless supermultiplet in a SQFT with $$\mathcal{N}=3$$ supersymmetry. What is the content of this supermultiplet? Show that it coincides with the content of the $$\mathcal{N} =4$$ vector multiplet.

Starting with a Clifford vacuum of helicity $$\lambda_0$$ we construct the supermultiplet as follows: $|\lambda_0\rangle ~, \qquad a_I^\dagger |\lambda_0\rangle ~, \qquad a_{[J}^\dagger a_{K]}^\dagger |\lambda_0\rangle ~, \qquad a_1^\dagger a_2^\dagger a_3^\dagger |\lambda_0\rangle ~, \qquad I=1,2,3~,$ which contains $$1+3+3+1=8$$ states. Since each creation operator raises the helicity of the state by $$\frac{1}{2}$$, the content of this supermultiplet is $$\big(\lambda_0,3\times(\lambda_0+\frac{1}{2}),3\times(\lambda_0+1),\lambda_0+\frac{3}{2}\big)$$. Recalling that under CPT the helicity of a state changes sign, this supermultiplet cannot be self-conjugate with integer or half-integer $$\lambda_0$$. Therefore, we also need to include its CPT-conjugate, which has content $$\big(-\lambda_0-\frac{3}{2},3\times(-\lambda_0-1),3\times(-\lambda_0-\frac{1}{2}),-\lambda_0\big)$$. In SQFTs supermultiplets cannot have states with $$|\lambda| > 1$$. This immediately implies that $$\lambda_0 = -\frac{1}{2}$$, and the total content of the $$\mathcal{N}=3$$ vector multiplet is $$\big(-1,4\times(-\frac{1}{2}),6\times (0),4\times(\frac{1}{2}),1\big)$$, hence it contains one massless vector, four massless Weyl fermions and six massless scalars. This is the same content as the $$\mathcal{N}=4$$ vector multiplet.

$$\mathcal{N}=8$$ is special because there is a single massless supermultiplet for which all states have helicity $$|\lambda| \leq 2$$. This is the $$\mathcal{N}=8$$ graviton multiplet. Find the content of the $$\mathcal{N}=8$$ graviton multiplet. Decompose the $$\mathcal{N}=8$$ graviton multiplet in terms of $$\mathcal{N}=1$$ massless supermultiplets.

Starting with a Clifford vacuum of helicity $$\lambda_0$$ the supermultiplet has content $$\big(\lambda_0,8\times(\lambda_0+\frac{1}{2}),28\times(\lambda_0+1),56\times(\lambda_0+\frac{3}{2}),70\times(\lambda_0+2),56\times(\lambda_0+\frac{5}{2}),28\times(\lambda_0+3),8\times(\lambda_0+\frac{7}{2}),\lambda_0+4\big)$$. Requiring that there are no states with $$|\lambda|>2$$ implies that $$\lambda_0 = -2$$, which also means that we have a self-conjugate supermultiplet. Therefore, the content of the $$\mathcal{N}=8$$ graviton multiplet is $$\big(-2,8\times(-\frac{3}{2}),28\times(-1),56\times(-\frac{1}{2}),70\times(0),56\times(\frac{1}{2}),28\times(1),8\times(\frac{3}{2}),2\big)$$, hence it contains 1 graviton, 8 gravitinos, 28 vectors, 56 Weyl fermions and 70 scalars. In terms of $$\mathcal{N}=1$$ massless supermultiplets this is 1 graviton multiplet, 7 ($$= 8-1$$) gravitino multiplets, 21 ($$=28-7$$) vector multiplets and 35 ($$=56-21$$) chiral multiplets. As a consistency check, the 35 chiral multiplets indeed contain 70 scalars.

Starting from the super-Poincaré algebra with $$\mathcal{N}=2$$, fixing the Lorentz frame $$p_\mu = k_\mu = (m,0,0,0)_\mu$$ and setting $$Z^{IJ} = \big(\begin{smallmatrix} 0 & z \\ -z & 0 \end{smallmatrix}\big)^{IJ}$$, $$z\in\mathbb{R}$$ and $$z\geq0$$, check that the operators $a_\alpha = \frac{1}{\sqrt{2}}\big(Q_\alpha^{1} + \epsilon_{\alpha}{}^{\beta} \bar Q_{2\beta}\big) ~,\qquad b_\alpha = \frac{1}{\sqrt{2}}\big(Q_\alpha^{1} - \epsilon_{\alpha}{}^{\beta} \bar Q_{2\beta}\big) ~, \qquad$ satisfy the anticommutation relations $\{a_\alpha,a_\beta^\dagger\} = (2m+z)\delta_{\alpha\beta} ~, \qquad \{b_\alpha,b_\beta^\dagger\} = (2m-z)\delta_{\alpha\beta} ~,$ with all other anticommutators vanishing. Note that $$\epsilon_{1}{}^2 = - \epsilon_{2}{}^1 = - 1$$, $$\epsilon_1{}^1 = \epsilon_2{}^2 = 0$$ and we have dropped the distinction between dotted and undotted indices. Working in an eigenbasis of $$J_3 = M_{12}$$, determine the effect of the fermionic annihilation and creation operators on the $$z$$-component of the spin (the eigenvalue of $$J_3$$).

Since $$\{Q_\alpha^1,Q_\beta^1\} = \{\bar Q_{2\alpha},\bar Q_{2\beta}\} = 0$$ and $$\{Q_\alpha^1,\bar Q_{2\beta}\} = 0$$ we have $$\{a_\alpha,a_\beta\} = \{b_\alpha,b_\beta\} = \{a_\alpha,b_\beta\} = 0$$. By conjugation this also implies that $$\{a_\alpha^\dagger,a_\beta^\dagger\} = \{b_\alpha^\dagger,b_\beta^\dagger\} = \{a_\alpha^\dagger,b_\beta^\dagger\} = 0$$. For the remaining anticommutators we observe that $$(\sigma^\mu p_\mu)_{\alpha\beta} = m \delta_{\alpha\beta}$$. We then have $\begin{split} \{a_\alpha,a_\beta^\dagger\} & = \frac{1}{2} \big(\{Q^1_\alpha,\bar Q_{1\beta}\} + \epsilon_{\beta}{}^{\gamma}\{Q^1_\alpha,Q^2_\gamma\} + \epsilon_{\alpha}{}^{\gamma}\{\bar Q_{2\gamma},\bar Q_{1\beta}\} + \epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\{\bar Q_{2\gamma},Q^2_\delta\} \\ & = \frac{1}{2} \big(2m\delta_{\alpha\beta} + z\epsilon_{\beta}{}^{\gamma}\epsilon_{\alpha\gamma} - z \epsilon_{\alpha}{}^{\gamma}\epsilon_{\gamma\beta} + 2m\epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\delta_{\gamma\delta}\big) = (2m + z) \delta_{\alpha\beta} ~, \\ \{b_\alpha,b_\beta^\dagger\} & = \frac{1}{2} \big(\{Q^1_\alpha,\bar Q_{1\beta}\} - \epsilon_{\beta}{}^{\gamma}\{Q^1_\alpha,Q^2_\gamma\} - \epsilon_{\alpha}{}^{\gamma}\{\bar Q_{2\gamma},\bar Q_{1\beta}\} + \epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\{\bar Q_{2\gamma},Q^2_\delta\} \\ & = \frac{1}{2} \big(2m\delta_{\alpha\beta} - z\epsilon_{\beta}{}^{\gamma}\epsilon_{\alpha\gamma} + z \epsilon_{\alpha}{}^{\gamma}\epsilon_{\gamma\beta} + 2m\epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\delta_{\gamma\delta}\big) = (2m - z) \delta_{\alpha\beta} ~, \\ \{a_\alpha,b_\beta^\dagger\} & = \frac{1}{2} \big(\{Q^1_\alpha,\bar Q_{1\beta}\} - \epsilon_{\beta}{}^{\gamma}\{Q^1_\alpha,Q^2_\gamma\} + \epsilon_{\alpha}{}^{\gamma}\{\bar Q_{2\gamma},\bar Q_{1\beta}\} - \epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\{\bar Q_{2\gamma},Q^2_\delta\} \\ & = \frac{1}{2} \big(2m\delta_{\alpha\beta} - z\epsilon_{\beta}{}^{\gamma}\epsilon_{\alpha\gamma} - z \epsilon_{\alpha}{}^{\gamma}\epsilon_{\gamma\beta} - 2m\epsilon_{\alpha}{}^{\gamma}\epsilon_{\beta}{}^{\delta}\delta_{\gamma\delta}\big) = 0 ~. \end{split}$ To determine the effect of the fermionic annihilation and creation operators on the $$z$$-component of the spin we compute their commutators with $$M_{12}$$. We have \begin{aligned} \phantom{}[M_{12},Q_1^I] & = i(\sigma_{12})_1{}^1 Q_1^I = \frac{1}{2} Q_1^I ~, \qquad & \phantom{}[M_{12},Q_2^I] & = i(\sigma_{12})_2{}^2 Q_2^I = -\frac{1}{2} Q_2^I ~, \\ \phantom{}[M_{12},\bar Q_{I1}] & = i\epsilon_{12} (\bar\sigma_{12})^2{}_2 \epsilon^{21}\bar Q_{I1} = - \frac{1}{2} \bar Q_{I1} ~, \qquad & \phantom{}[M_{12},\bar Q_{I2}] & = i\epsilon_{21} (\bar\sigma_{12})^1{}_1 \epsilon^{12}\bar Q_{I2} = \frac{1}{2} \bar Q_{I2} ~. \end{aligned} Therefore, $$a_1$$, $$a_2^\dagger$$, $$b_1$$ and $$b_2^\dagger$$, which are linear combinations of $$Q^I_1$$ and $$\bar Q_{I2}$$, raise the $$z$$-component of the spin by $$\frac{1}{2}$$, while $$a_2$$, $$a_1^\dagger$$, $$b_2$$ and $$b_1^\dagger$$, which are linear combinations of $$Q^I_2$$ and $$\bar Q_{I1}$$, lower the $$z$$-component of the spin by $$\frac{1}{2}$$.

Construct the physical states of the long and short $$\mathcal{N}=2$$ massive vector multiplets in terms of oscillators.

For $$\mathcal{N}=2$$ we have four fermionic creation operators ($$a_1^\dagger$$, $$b_1^\dagger$$, $$a_2^\dagger$$ and $$b_2^\dagger$$) for the long multiplet and two ($$a_1^\dagger$$ and $$a_2^\dagger$$) for the short multiplet. $$a_1^\dagger$$ and $$b_1^\dagger$$ lower the $$z$$-component of the spin by $$\frac{1}{2}$$, while $$a_2^\dagger$$ and $$b_2^\dagger$$ raise it by $$\frac{1}{2}$$.

The physical states of the long $$\mathcal{N}=2$$ massive vector multiplet are The content of the supermultiplet is $$\big(-1,4\times(-\frac{1}{2}),6\times (0),4\times(\frac{1}{2}),1\big)$$, hence it contains one massive vector, two massive Dirac fermions and five massive real scalars.

The physical states of the short $$\mathcal{N}=2$$ massive vector multiplet are The content of the supermultiplet is $$\big(-1,2\times(-\frac{1}{2}), 2\times (0), 2\times(\frac{1}{2}),1\big)$$, hence it contains one massive vector, one massive Dirac fermion and one massive real scalar.

# 4 Superspace and superfields

Starting from the differential operator realisation of $$P_\mu$$, $$Q_\alpha$$ and $$\bar Q_{\dot \alpha}$$ $P_\mu = -i\partial_\mu ~, \qquad Q_\alpha = - i \big(\partial_\alpha -i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu\big) ~, \qquad \bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} - i \bar c (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big) ~,$ where $$c$$ and $$\bar c$$ are free real constants, show that the anticommutation relations $\{Q_\alpha,\bar Q_{\dot\alpha}\} = 2 (\sigma^{\mu})_{\alpha\dot\alpha} P_\mu ~, \qquad \{Q_\alpha, Q_{\beta}\} = \{\bar Q_{\dot\alpha}, \bar Q_{\dot\beta}\} = 0 ~,$ together with $$(Q_\alpha)^\dagger = \bar Q_{\dot\alpha}$$, imply that $$c=\bar c=1$$. Show that the anticommutation relations are equivalent to $$[\varepsilon_1 Q,\bar \varepsilon_2 \bar Q] = 2 \varepsilon_1\sigma^\mu\bar\varepsilon_2 P_\mu$$ and $$[\varepsilon_1 Q,\varepsilon_2 Q] = [\bar\varepsilon_1\bar Q,\bar\varepsilon_2\bar Q] = 0$$.

Since $$Q_\alpha$$ only depends on $$\partial_\alpha$$ and $$\bar \theta^{\dot\alpha}$$, while $$\bar Q_{\dot\alpha}$$ only depends on $$\bar\partial_{\dot\alpha}$$ and $$\theta^\alpha$$, it immediately follows that $$\{Q_\alpha, Q_{\beta}\} = \{\bar Q_{\dot\alpha}, \bar Q_{\dot\beta}\} = 0$$. For the final anticommutation relation we have $\begin{split} \{Q_\alpha,\bar Q_{\dot\alpha}\} & = \{- i \big(\partial_\alpha -i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu\big), i \big(\bar\partial_{\dot\alpha} - i \bar c (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big)\} \\ & = \{\partial_\alpha,-i\bar c(\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\} + \{-i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu,\bar\partial_{\dot\alpha}\} \\ & = -i(\bar c+c) (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu = (\bar c+c) (\sigma^\mu)_{\alpha\dot\alpha} P_\mu ~, \end{split}$ hence we require that $$\bar c+c = 2$$. On the other hand $\begin{split} (Q_\alpha)^\dagger & = i\big(\partial_\alpha -i c (\sigma^\mu\bar\theta)_\alpha\partial_\mu\big)^\dagger = i \big(\partial_{\dot\alpha} - i c (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu\big) ~, \end{split}$ where we have used that $$c$$ is real. Therefore, $$(Q_\alpha)^\dagger$$ is equal to $$\bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} - i \bar c (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big)$$ if $$c = \bar c$$. It immediately follows that $$c=\bar c= 1$$.

Expanding out the commutator $$[\varepsilon_1 Q,\bar \varepsilon_2 \bar Q]$$ we find $\begin{split} [\varepsilon_1 Q,\bar \varepsilon_2 \bar Q] & = - \varepsilon_1^\alpha Q_\alpha \varepsilon_2^{\dot \alpha} \bar Q_{\dot\alpha} + \varepsilon_2^{\dot\alpha} \bar Q_{\dot\alpha} \varepsilon_1^\alpha Q_\alpha = \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} Q_\alpha \bar Q_{\dot\alpha} + \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} \bar Q_{\dot\alpha} Q_\alpha \\ & = \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} \{Q_\alpha, \bar Q_{\dot\alpha} \} = 2 \varepsilon_1^\alpha\varepsilon_2^{\dot \alpha} (\sigma^\mu)_{\alpha\dot\alpha} P_\mu = 2 \varepsilon_1\sigma^\mu\bar\varepsilon_2 P_\mu ~. \end{split}$ Similarly, $\begin{split} [\varepsilon_1 Q,\varepsilon_2 Q] & = - \varepsilon_1^\alpha Q_\alpha \varepsilon_2^{\beta} Q_{\beta} + \varepsilon_2^{\beta} Q_{\beta} \varepsilon_1^\alpha Q_\alpha = \varepsilon_1^\alpha\varepsilon_2^{\beta} Q_\alpha Q_{\beta} + \varepsilon_1^\alpha\varepsilon_2^{\beta} Q_{\beta} Q_\alpha = \varepsilon_1^\alpha\varepsilon_2^{\beta} \{Q_\alpha, Q_{\beta} \} = 0 ~, \\ [\bar \varepsilon_1 \bar Q,\bar \varepsilon_2 \bar Q] & = - \varepsilon_1^{\dot\alpha} \bar Q_{\dot\alpha} \varepsilon_2^{\dot \beta} \bar Q_{\dot\beta} + \varepsilon_2^{\dot\beta} \bar Q_{\dot\beta}\bar \varepsilon_1^{\dot\alpha} Q_{\dot\alpha} = \varepsilon_1^{\dot\alpha}\varepsilon_2^{\dot \beta} \bar Q_{\dot\alpha} \bar Q_{\dot\beta} + \varepsilon_1^{\dot\alpha}\varepsilon_2^{\dot \beta} \bar Q_{\dot\beta} \bar Q_{\dot\alpha} = \varepsilon_1^{\dot\alpha}\varepsilon_2^{\dot \beta} \{\bar Q_{\dot\alpha}, \bar Q_{\dot\beta} \} = 0 ~. \end{split}$

Given that $$\{\partial_\alpha, \theta^\beta\} = \delta_\alpha^\beta$$ and $$\{\bar\partial_{\dot\alpha},\bar\theta^{\dot\beta}\} = \delta_{\dot\alpha}^{\dot\beta}$$, show that $$\{\partial^\alpha, \theta_\beta\} = -\delta^\alpha_\beta$$ and $$\{\bar\partial^{\dot\alpha},\bar\theta_{\dot\beta}\} = -\delta^{\dot\alpha}_{\dot\beta}$$, where we define $$\theta_\alpha = \epsilon_{\alpha\beta}\theta^\beta$$, $$\partial^\alpha = \epsilon^{\alpha\beta}\partial_\beta$$, $$\bar\theta_{\dot\alpha}=\epsilon_{\dot\alpha\dot\beta}\bar\theta^{\dot\beta}$$ and $$\bar\partial^{\dot\alpha} = \epsilon^{\dot\alpha\dot\beta} \bar\partial_{\dot\beta}$$, i.e. in the usual way.

We have $\{\partial^\alpha, \theta_\beta\} = \epsilon^{\alpha\gamma}\epsilon_{\beta\delta}\{\partial_\gamma,\theta^\delta\} = \epsilon^{\alpha\gamma}\epsilon_{\beta\delta} \delta_\gamma^\delta = \epsilon^{\alpha\gamma}\epsilon_{\beta\gamma} = -\epsilon^{\alpha\gamma}\epsilon_{\gamma\beta} = -\delta^\alpha_\beta ~.$ Similarly $\{\bar\partial^{\dot\alpha}, \bar\theta_{\dot\beta}\} = \epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\beta\dot\delta}\{\bar\partial_{\dot\gamma},\bar\theta^{\dot\delta}\} = \epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\beta\dot\delta} \delta_{\dot\gamma}^{\dot\delta} = \epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\beta\dot\gamma} = -\epsilon^{\dot\alpha\dot\gamma}\epsilon_{\dot\gamma\dot\beta} = -\delta^{\dot\alpha}_{\dot\beta} ~.$

Count the number of bosonic and fermionic degrees of freedom of a general superfield $\begin{split} Y(x,\theta,\bar\theta) & = y(x) + \theta\psi(x) + \bar\theta \bar\chi(x) + (\theta\theta) m(x) + (\bar\theta\bar\theta)\bar n(x) \\ & \quad + (\theta\sigma^\mu\bar\theta) v_\mu(x) + (\theta\theta) (\bar\theta\bar\lambda(x)) + (\bar\theta\bar\theta) (\theta \rho(x)) + (\theta\theta)(\bar\theta\bar\theta) D(x) ~. \end{split}$ Derive the supersymmetry transformations of the component fields by equating $\delta_{\varepsilon,\bar\varepsilon} Y(x,\theta,\bar\theta) = i [\varepsilon Q + \bar\varepsilon \bar Q,Y(x,\theta,\bar\theta)] ~,$ and $\delta_{\varepsilon,\bar\varepsilon} Y(x,\theta,\bar\theta) = \delta_{\varepsilon,\bar\varepsilon} y(x) + \theta \delta_{\varepsilon,\bar\varepsilon} \psi(x) + \bar\theta \delta_{\varepsilon,\bar\varepsilon} \bar\chi(x) + \dots + (\theta\theta)(\bar\theta\bar\theta) \delta_{\varepsilon,\bar\varepsilon} D(x) ~,$ where $Q_\alpha = - i \big(\partial_\alpha -i(\sigma^\mu\bar\theta)_\alpha\partial_\mu\big) ~, \qquad \bar Q_{\dot\alpha} = i \big(\bar\partial_{\dot\alpha} - i (\theta \sigma^\mu)_{\dot\alpha}\partial_\mu\big) ~.$ What is special about the variation of $$D$$?

The bosonic component fields are the complex scalars $$y$$, $$m$$, $$\bar n$$ and $$D$$ each with 2 real degrees of freedom, and the complex vector $$v_\mu$$ with $$2\times 4 = 8$$ real degrees of freedom. Therefore, the total number of real bosonic degrees of freedom is 16. The fermionic component fields are the Weyl fermions $$\psi$$, $$\bar\chi$$, $$\bar\lambda$$ and $$\rho$$ each with 4 real degrees of freedom. Therefore, the total number of real fermionic degrees of freedom is also 16.

To derive the supersymmetry transformations of the component fields, we start by noting that $i (\varepsilon Q + \bar\varepsilon \bar Q) = \varepsilon^{\alpha}\partial_\alpha - i(\varepsilon\sigma^\mu\bar\theta)\partial_\mu + \bar\varepsilon^{\dot\alpha} \bar\partial_{\dot\alpha} + i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu ~,$ and observe that $$\varepsilon^{\alpha}\partial_\alpha$$ lowers the power of $$\theta$$ by one, while $$i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu$$ raises it by one. Similarly, $$\bar\varepsilon^{\dot\alpha} \bar\partial_{\dot\alpha}$$ lowers the power of $$\bar \theta$$ by one, while $$- i(\varepsilon\sigma^\mu\bar\theta)\partial_\mu$$ raises it by one.

To derive the supersymmetry transformation of $$y$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon} y$$, we need to extract those terms in $$i [\varepsilon Q + \bar\varepsilon \bar Q, Y(x,\theta,\bar\theta)]$$ that are independent of $$\theta$$ and $$\bar\theta$$. These terms can only come from lowering the power of $$\theta$$ in $$\theta \psi$$ by one, i.e. applying $$\varepsilon^{\alpha}\partial_\alpha$$, or lowering the power of $$\bar\theta$$ in $$\bar\theta \bar\chi$$ by one, i.e. applying $$\bar\varepsilon^{\dot\alpha} \bar\partial_{\dot\alpha}$$. Therefore, it follows that $\delta_{\varepsilon,\bar\varepsilon} y = \varepsilon \psi + \bar\varepsilon \bar\chi ~.$

To derive the supersymmetry transformation of $$\psi$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon} \psi$$, we need to extract those terms in $$i [\varepsilon Q + \bar\varepsilon \bar Q, Y(x,\theta,\bar\theta)]$$ that are linear in $$\theta$$ and independent of $$\bar\theta$$. There are three such terms, which come from raising the power of $$\theta$$ in $$y$$, lowering the power of $$\theta$$ in $$(\theta\theta)m$$ by one, and lowering the power of $$\bar\theta$$ in $$(\theta\sigma^\mu\bar\theta)v_\mu$$ by one. Therefore, we have $\begin{split} \theta \delta_{\varepsilon,\bar\varepsilon} \psi & = i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu y + 2 (\varepsilon\theta) m + (\theta\sigma^\mu\bar\varepsilon) v_\mu = i (\theta \sigma^\mu\bar\varepsilon)\partial_\mu y + 2 (\theta\varepsilon) m + (\theta\sigma^\mu\bar\varepsilon) v_\mu ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon} \psi = i (\sigma^\mu\bar\varepsilon)\partial_\mu y + 2 \varepsilon m + (\sigma^\mu\bar\varepsilon) v_\mu ~.$

For the supersymmetry transformation of $$\bar\chi$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon} \bar\chi$$, we have $\begin{split} \bar \theta \delta_{\varepsilon,\bar\varepsilon} \bar\chi & = - i(\varepsilon\sigma^\mu\bar\theta)\partial_\mu y + 2(\bar\varepsilon\bar\theta)\bar n + (\varepsilon\sigma^\mu\bar\theta)v_\mu = i (\bar\theta\bar\sigma^\mu\varepsilon)\partial_\mu y + 2(\bar\theta\bar\varepsilon)\bar n -(\bar\theta\bar\sigma^\mu\varepsilon)v_\mu ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon} \bar\chi = i (\bar\sigma^\mu\varepsilon)\partial_\mu y + 2\bar\varepsilon\bar n -(\bar\sigma^\mu\varepsilon)v_\mu ~.$

For the supersymmetry transformation of $$m$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon}m$$, we have $\begin{split} (\theta\theta) \delta_{\varepsilon,\bar\varepsilon}m & = i (\theta \sigma^\mu\bar\varepsilon)(\theta\partial_\mu\psi) + (\theta\theta)(\bar\varepsilon\bar\lambda) = - \frac{i}{2}(\theta\theta)(\partial_\mu\psi\sigma^\mu\bar\varepsilon) + (\theta\theta)(\bar\varepsilon\bar\lambda) \\ & = \frac{i}{2}(\theta\theta)(\bar\varepsilon\bar\sigma^\mu\partial_\mu\psi) + (\theta\theta)(\bar\varepsilon\bar\lambda) ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon}m = \frac{i}{2}\bar\varepsilon\bar\sigma^\mu\partial_\mu\psi + \bar\varepsilon\bar\lambda ~.$

For the supersymmetry transformation of $$\bar n$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon}\bar n$$, we have $\begin{split} (\bar\theta\bar\theta) \delta_{\varepsilon,\bar\varepsilon}\bar n & = - i(\varepsilon\sigma^\mu\bar\theta) (\bar\theta\partial_\mu\bar\chi) + (\bar\theta\bar\theta)(\varepsilon\rho) = i(\bar\theta\bar\sigma^\mu\varepsilon) (\bar\theta\partial_\mu\bar\chi) + (\bar\theta\bar\theta)(\varepsilon\rho) \\ & = -\frac{i}{2}(\bar\theta\bar\theta)(\partial_\mu\bar\chi\bar\sigma^\mu\varepsilon) + (\bar\theta\bar\theta)(\varepsilon\rho) = \frac{i}{2}(\bar\theta\bar\theta)(\varepsilon\sigma^\mu\partial_\mu\bar\chi) + (\bar\theta\bar\theta)(\varepsilon\rho) ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon}\bar n = \frac{i}{2}\varepsilon\sigma^\mu\partial_\mu\bar\chi + \varepsilon\rho ~.$

For the supersymmetry transformation of $$v_\mu$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon}v_\mu$$, we have $\begin{split} (\theta\sigma^\mu\bar\theta)\delta_{\varepsilon,\bar\varepsilon}v_\mu & = - i(\varepsilon\sigma^\nu\bar\theta)(\theta\partial_\nu\psi) + i (\theta \sigma^\nu\bar\varepsilon)(\bar\theta\partial_\nu\bar\chi) +2 (\varepsilon\theta)(\bar\theta\bar\lambda) +2 (\bar\varepsilon\bar\theta)(\theta\rho) \\ & = i(\bar\theta\bar\sigma^\nu\varepsilon)(\theta\partial_\nu\psi) + i (\theta \sigma^\nu\bar\varepsilon)(\bar\theta\partial_\nu\bar\chi) +2 (\theta\varepsilon)(\bar\theta\bar\lambda) +2 (\bar\theta\bar\varepsilon)(\theta\rho) \\ & = \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\partial_\nu\psi\sigma_\mu\bar\sigma^\nu\varepsilon) - \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\partial_\nu\bar\chi\bar\sigma_\mu\sigma^\nu\bar\varepsilon) +(\theta\sigma^\mu\bar\theta) (\varepsilon\sigma_\mu \bar\lambda) +(\theta\sigma^\mu\bar\theta)(\rho\sigma_\mu\bar\varepsilon) \\ & = \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\varepsilon\sigma^\nu\bar\sigma_\mu\partial_\nu\psi) - \frac{i}{2}(\theta\sigma^\mu\bar\theta)(\bar\varepsilon\bar\sigma^\nu\sigma_\mu\partial_\nu\bar\chi) +(\theta\sigma^\mu\bar\theta) (\varepsilon\sigma_\mu \bar\lambda) -(\theta\sigma^\mu\bar\theta)(\bar\varepsilon\bar\sigma_\mu\rho) ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon}v_\mu = \frac{i}{2}\varepsilon\sigma^\nu\bar\sigma_\mu\partial_\nu\psi - \frac{i}{2}\bar\varepsilon\bar\sigma^\nu\sigma_\mu\partial_\nu\bar\chi +\varepsilon\sigma_\mu \bar\lambda -\bar\varepsilon\bar\sigma_\mu\rho ~.$

For the supersymmetry transformation of $$\bar\lambda$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon}\bar\lambda$$, we have $\begin{split} (\theta\theta)(\bar\theta\delta_{\varepsilon,\bar\varepsilon}\bar\lambda) & = -i(\varepsilon\sigma^\mu\bar\theta)(\theta\theta)\partial_\mu m +i (\theta \sigma^\mu\bar\varepsilon)(\theta\sigma^\nu\bar\theta)\partial_\mu v_\nu +2(\theta\theta)(\bar\varepsilon\bar\theta)D \\ & = i(\theta\theta)(\bar\theta\bar\sigma^\mu\varepsilon)\partial_\mu m +\frac{i}{2}(\theta\theta)(\bar\theta\bar\sigma^\nu\sigma^\mu\bar\varepsilon)\partial_\mu v_\nu +2(\theta\theta)(\bar\theta\bar\varepsilon)D ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon}\bar\lambda = i(\bar\sigma^\mu\varepsilon)\partial_\mu m +\frac{i}{2}(\bar\sigma^\nu\sigma^\mu\bar\varepsilon)\partial_\mu v_\nu +2\bar\varepsilon D ~.$

For the supersymmetry transformation of $$\rho$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon}\rho$$, we have $\begin{split} (\bar\theta\bar\theta)(\theta\delta_{\varepsilon,\bar\varepsilon}\rho) & = i (\theta \sigma^\mu\bar\varepsilon)(\bar\theta\bar\theta)\partial_\mu \bar n - i(\varepsilon\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta)\partial_\mu v_\nu +2(\varepsilon\theta)(\bar\theta\bar\theta)D \\ & = i (\bar\theta\bar\theta)(\theta \sigma^\mu\bar\varepsilon)\partial_\mu \bar n -i(\bar\theta\bar\sigma^\mu\varepsilon)(\bar\theta\bar\sigma^\nu\theta)\partial_\mu v_\nu +2(\bar\theta\bar\theta)(\theta\varepsilon)D \\ & = i (\bar\theta\bar\theta)(\theta \sigma^\mu\bar\varepsilon)\partial_\mu \bar n -\frac{i}{2}(\bar\theta\bar\theta)(\theta\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu v_\nu +2(\bar\theta\bar\theta)(\theta\varepsilon)D ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon}\rho = i (\sigma^\mu\bar\varepsilon)\partial_\mu \bar n -\frac{i}{2}(\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu v_\nu +2\varepsilon D ~,$

For the supersymmetry transformation of $$D$$, i.e. $$\delta_{\varepsilon,\bar\varepsilon} D$$, we have $\begin{split} (\theta\theta)(\bar\theta\bar\theta)\delta_{\varepsilon,\bar\varepsilon} D & = - i(\varepsilon\sigma^\mu\bar\theta)(\theta\theta)(\bar\theta\partial_\mu\bar\lambda) + i (\theta \sigma^\mu\bar\varepsilon)(\bar\theta\bar\theta)(\theta\partial_\mu\rho) \\ & = i(\theta\theta) (\bar\theta\bar\sigma^\mu\varepsilon)(\bar\theta\partial_\mu\bar\lambda) + i (\bar\theta\bar\theta)(\theta \sigma^\mu\bar\varepsilon)(\theta\partial_\mu\rho) \\ & = -\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\partial_\mu\bar\lambda\bar\sigma^\mu\varepsilon) - \frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\partial_\mu\rho\sigma^\mu\bar\varepsilon) \\ & = \frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\varepsilon\sigma^\mu\partial_\mu\bar\lambda) +\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\bar\varepsilon\bar\sigma^\mu\partial_\mu\rho) ~, \end{split}$ which implies $\delta_{\varepsilon,\bar\varepsilon} D = \frac{i}{2}\varepsilon\sigma^\mu\partial_\mu\bar\lambda +\frac{i}{2}\bar\varepsilon\bar\sigma^\mu\partial_\mu\rho ~,$ which is a total derivative.

Note that we have made repeated use of the spinor identities derived in exercise 2.4.

Starting from the differential operators \begin{aligned} Q_\alpha & = -i\big( \partial_\alpha - i (\sigma^\mu\bar\theta)_\alpha \partial_\mu \big)~, \qquad & \bar Q_{\dot\alpha} & = i\big(\bar\partial_{\dot\alpha} - i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu \big) ~, \\ D_\alpha & = \partial_\alpha + i (\sigma^\mu\bar\theta)_\alpha \partial_\mu ~, \qquad & \bar D_{\dot\alpha} & = \bar\partial_{\dot\alpha} + i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu ~, \end{aligned} show that $\begin{gathered} \{D_\alpha,Q_\beta\} = \{D_\alpha,\bar Q_{\dot\alpha}\} = \{\bar D_{\dot\alpha},Q_\alpha\} = \{\bar D_{\dot\alpha},\bar Q_{\dot\beta} \} = 0 ~, \\ \{D_\alpha,\bar D_{\dot\alpha}\} = 2 i (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu ~, \qquad \{D_\alpha,D_\beta\} = \{\bar D_{\dot\alpha},\bar D_{\dot\beta} \} = 0 ~. \end{gathered}$

Since $$Q_\alpha$$ and $$D_\alpha$$ only depend on $$\partial_\alpha$$ and $$\bar \theta^{\dot\alpha}$$, while $$\bar Q_{\dot\alpha}$$ and $$\bar D_{\dot\alpha}$$ only depend on $$\bar\partial_{\dot\alpha}$$ and $$\theta^\alpha$$, it immediately follows that $$\{D_\alpha, Q_{\beta}\} = \{\bar D_{\dot\alpha}, \bar Q_{\dot\beta}\} = \{D_\alpha, D_{\beta}\} = \{\bar D_{\dot\alpha}, \bar D_{\dot\beta}\} = 0$$. For the remaining three anticommutation relations we have $\begin{split} \{D_\alpha,\bar Q_{\dot\alpha}\} & = \{\partial_\alpha,(\theta\sigma^\mu)_{\dot\alpha}\partial_\mu\} + \{i(\sigma^\mu\bar\theta)_\alpha\partial_\mu,i\bar\partial_{\dot\alpha}\} =(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu - (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu = 0 ~, \\ \{\bar D_{\dot\alpha},Q_{\alpha}\} & = \{\bar\partial_{\dot\alpha},-(\sigma^\mu\bar\theta)_{\alpha}\partial_\mu\} + \{i(\theta\sigma^\mu)_{\dot\alpha}\partial_\mu,-i\partial_{\alpha}\} =-(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu + (\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu = 0 ~, \\ \{D_\alpha,\bar D_{\dot\alpha}\} & = \{\partial_\alpha,i(\theta\sigma^\mu)_{\dot\alpha}\partial_\mu\} + \{i(\sigma^\mu\bar\theta)_\alpha\partial_\mu,\bar\partial_{\dot\alpha}\} =2i(\sigma^\mu)_{\alpha\dot\alpha} \partial_\mu ~. \end{split}$

Show that $= [D_{\alpha}, \bar y^\mu] = 0 ~,$ where \begin{aligned} y^\mu & = x^\mu + i \theta \sigma^\mu\bar \theta ~, & \qquad D_\alpha & = \partial_\alpha + i (\sigma^\mu\bar\theta)_\alpha \partial_\mu ~, \\ \bar y^\mu & = x^\mu - i \theta \sigma^\mu \bar \theta ~, & \qquad \bar D_{\dot\alpha} & = \bar\partial_{\dot\alpha} + i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu ~. \end{aligned}

We have $= [\bar\partial_{\dot\alpha} + i (\theta\sigma^\nu)_{\dot\alpha} \partial_\nu,x^\mu + i \theta \sigma^\mu \bar \theta] = - i (\theta\sigma^\mu)_{\dot\alpha} + i(\theta\sigma^\mu)_{\dot\alpha} = 0 ~,$ and $= [\partial_{\alpha} + i (\sigma^\nu\bar\theta)_{\alpha} \partial_\nu,x^\mu - i \theta \sigma^\mu \bar \theta] = - i (\sigma^\mu\bar\theta)_{\alpha} + i(\sigma^\mu\bar\theta)_{\alpha} = 0 ~.$

Express the supercovariant derivatives $$D_\alpha$$ and $$\bar D_{\dot\alpha}$$ and the supercharges $$Q_\alpha$$ and $$\bar Q_{\dot\alpha}$$, realised as differential operators, \begin{aligned} D_\alpha & = \partial_\alpha + i (\sigma^\mu\bar\theta)_\alpha \partial_\mu ~, \qquad & \bar D_{\dot\alpha} & = \bar\partial_{\dot\alpha} + i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu ~, \\ Q_\alpha & = -i\big( \partial_\alpha - i (\sigma^\mu\bar\theta)_\alpha \partial_\mu \big)~, \qquad & \bar Q_{\dot\alpha} & = i\big(\bar\partial_{\dot\alpha} - i (\theta\sigma^\mu)_{\dot\alpha} \partial_\mu \big) ~, \end{aligned} in terms of derivatives with respect to $$(y,\theta,\bar\theta)$$ and $$(\bar y,\theta,\bar\theta)$$ where $y^\mu = x^\mu + i \theta \sigma^\mu\bar \theta ~, \qquad \bar y^\mu = x^\mu - i \theta \sigma^\mu \bar \theta ~.$

We introduce the notation $$(\hat \partial_\mu,\hat\partial_\alpha,\hat{\bar\partial}_{\dot\alpha})$$ to denote partial derivatives with respect to $$(y,\theta,\bar\theta)$$ and continue to use $$(\partial_\mu,\partial_\alpha,\bar\partial_{\dot\alpha})$$ to denote partial derivatives with respect to $$(x,\theta,\bar\theta)$$. We then have $\begin{split} \partial_\mu & = \frac{\partial y^\nu}{\partial x^\mu} \hat\partial_{\nu} + \frac{\partial \theta^\alpha}{\partial x^\mu} \hat\partial_\alpha + \frac{\partial \bar\theta^{\dot\alpha}}{\partial x^\mu} \hat{\bar\partial}_{\dot\alpha} = \delta_\mu^\nu \hat\partial_{\nu} = \hat\partial_{\mu} ~, \\ \partial_\alpha & = \frac{\partial y^\mu}{\partial \theta^\alpha} \hat\partial_{\mu} + \frac{\partial \theta^\beta}{\partial \theta^\alpha} \hat\partial_\beta + \frac{\partial \bar\theta^{\dot\alpha}}{\partial \theta^\alpha} \hat{\bar\partial}_{\dot\alpha} = i (\sigma^\mu\bar\theta)_{\alpha} \hat\partial_{\mu} + \delta_{\alpha}^\beta \hat\partial_\beta = \hat\partial_\alpha + i (\sigma^\mu\bar\theta)_{\alpha} \hat\partial_{\mu} ~, \\ \bar\partial_{\dot\alpha} & = \frac{\partial y^\mu}{\partial \bar\theta^{\dot\alpha}} \hat\partial_{\mu} + \frac{\partial \theta^\alpha}{\partial \bar\theta^{\dot\alpha}} \hat\partial_\alpha + \frac{\partial \bar\theta^{\dot\beta}}{\partial \bar \theta^{\dot\alpha}} \hat{\bar\partial}_{\dot\beta} = - i (\theta\sigma^\mu)_{\dot\alpha} \hat\partial_{\mu} + \delta_{\dot\alpha}^{\dot\beta} \hat{\bar\partial}_{\dot\beta} = \hat{\bar\partial}_{\dot\alpha} - i (\theta\sigma^\mu)_{\dot\alpha} \hat\partial_{\mu} ~. \end{split}$ Substituting into the expressions for the supercovariant derivatives and supercharges realised as differential operators we find \begin{aligned} D_\alpha & = \hat\partial_\alpha + 2i (\sigma^\mu\bar\theta)_{\alpha} \hat\partial_{\mu} ~, \qquad & \bar D_{\dot\alpha} & = \hat{\bar\partial}_{\dot\alpha} ~, \\ Q_\alpha & = -i \hat\partial_\alpha ~, & \qquad \bar Q_{\dot\alpha} & = i\big(\hat{\bar\partial}_{\dot\alpha} - 2 i (\theta\sigma^\mu)_{\dot\alpha} \hat\partial_{\mu}\big) ~. \end{aligned}

A similar derivation shows that \begin{aligned} D_\alpha & = \tilde\partial_\alpha ~, \qquad & \bar D_{\dot\alpha} & = \tilde{\bar\partial}_{\dot\alpha} + 2 i (\theta\sigma^\mu)_{\dot\alpha} \tilde\partial_{\mu} ~, \\ Q_\alpha & = -i \big(\tilde\partial_\alpha - 2i (\sigma^\mu\bar\theta)_{\alpha} \tilde\partial_{\mu}\big) ~, & \qquad \bar Q_{\dot\alpha} & = i\tilde{\bar\partial}_{\dot\alpha} ~, \end{aligned} where $$(\tilde \partial_\mu,\tilde\partial_\alpha,\tilde{\bar\partial}_{\dot\alpha})$$ denote partial derivatives with respect to $$(\bar y,\theta,\bar\theta)$$.

Starting from a chiral superfield expressed in terms of components as $\Phi(y,\theta) = \phi(y) + \sqrt{2} \theta \psi(y) - (\theta\theta) F(y) ~,$ derive the component field expansion in terms of the coordinates $$(x,\theta,\bar\theta)$$ $\begin{split} \Phi(x,\theta,\bar\theta) & = \phi(x) + i (\theta\sigma^\mu\bar\theta)\partial_\mu\phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \phi(x) \\ & \quad + \sqrt{2}\theta\psi(x) -\frac{i}{\sqrt{2}}(\theta\theta)(\partial_\mu\psi(x) \sigma^\mu\bar\theta) -(\theta\theta) F(x) ~, \end{split}$ where $$y^\mu = x^\mu + i \theta \sigma^\mu\bar \theta$$.

We start by Taylor expanding $$\Phi(y,\theta)$$ around $$y^\mu = x^\mu$$ to give $\Phi(x,\theta,\bar\theta) = \Phi(x,\theta) + i (\theta \sigma^\mu\bar \theta) \partial_\mu \Phi(x,\theta) - \frac{1}{2} (\theta \sigma^\mu\bar \theta)(\theta \sigma^\nu\bar \theta) \partial_\mu\partial_\nu \Phi(x,\theta) ~.$ The expansion terminates at quadratic order in either $$\theta_{\alpha}$$ or $$\bar\theta_{\dot\alpha}$$ since they are anticommuting coordinates and the indices $$\alpha=1,2$$ and $$\dot\alpha=1,2$$. Now substituting in for $$\Phi(x,\theta)$$ we find $\begin{split} \Phi(x,\theta,\bar\theta) & = \phi(x) + \sqrt{2} \theta \psi(x) - (\theta\theta) F(x) \\ & \quad + i (\theta \sigma^\mu\bar \theta) \partial_\mu \phi(x) + \sqrt{2} i (\theta \sigma^\mu\bar \theta) (\theta \partial_\mu\psi(x)) - \frac{1}{2} (\theta \sigma^\mu\bar \theta)(\theta \sigma^\nu\bar \theta) \partial_\mu\partial_\nu \phi(x) ~. \end{split}$ Finally, using the following spinor identities $(\theta\phi) (\theta\psi) = -\frac{1}{2} (\theta\theta)(\phi\psi) ~, \qquad (\theta \sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) = \frac{1}{2}(\theta\theta)(\bar\theta\bar\theta) \eta^{\mu\nu} ~,$ which were proven in exercise 2.4, we can rearrange the final two terms to give $\begin{split} \Phi(x,\theta,\bar\theta) & = \phi(x) + i (\theta \sigma^\mu\bar \theta) \partial_\mu \phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \phi(x) \\ & \quad + \sqrt{2} \theta \psi(x) -\frac{i}{\sqrt{2}} (\theta \theta)( \partial_\mu\psi(x) \sigma^\mu\bar \theta) - (\theta\theta) F(x) ~, \end{split}$ as required.

# 5 SQFTs of chiral multiplets

Compute the expansion in component fields of the superfield $$W(\Phi)$$, where $$\Phi$$ denotes a set of superfields $$\Phi^i$$, and show that the coefficient of $$-\theta\theta$$ is $F_W = \partial_i W(\phi) F^i + \frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j ~,$ where $$\partial_i = \frac{\partial}{\partial \Phi^i}$$.

Show that the component of the canonical Kähler potential $$\bar\Phi_i\Phi^i$$ proportional to $$(\theta\theta)(\bar\theta\bar\theta)$$ is $D_K = \partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i + \text{total derivative} ~.$ Determine the total derivative.

Expanding $$W(\Phi) = W\big(\phi(y) + \sqrt{2}\theta\psi(y) - \theta\theta F(y)\big)$$ around $$\Phi = \phi(y)$$ we find $\begin{split} W(\Phi) & = W(\phi(y)) + \big(\sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big) \partial_i W(\phi(y)) \\ & \quad +\frac{1}{2}\big(\sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big)\big(\sqrt{2}\theta\psi^j(y) - \theta\theta F^j(y)\big)\partial_i\partial_jW(\phi(y)) \\ & = W(\phi(y)) + \big(\sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big) \partial_i W(\phi(y)) + (\theta\psi^i(y))(\theta\psi^j(y))\partial_i\partial_jW(\phi(y)) \\ & = W(\phi(y)) + \sqrt{2}\theta\psi^i(y) \partial_i W(\phi(y)) - \theta\theta F^i(y) \partial_i W(\phi(y)) -\frac{1}{2} (\theta\theta) (\psi^i(y)\psi^j(y))\partial_i\partial_jW(\phi(y)) ~. \end{split}$ Therefore, the coefficient of $$-\theta\theta$$ is $F_W = \partial_i W(\phi) F^i + \frac{1}{2}\partial_i\partial_jW(\phi)\psi^i\psi^j ~,$ as claimed.

The canonical Kähler potential is $\bar\Phi_i\Phi^i = \big(\bar\phi_i(\bar y) + \sqrt{2}\bar\theta\bar\psi_i(\bar y) - \bar\theta\bar\theta\bar F_i(\bar y)\big)\big(\phi^i(y) + \sqrt{2}\theta\psi^i(y) - \theta\theta F^i(y)\big) ~.$ We now recall that $$y^\mu = x^\mu + i \theta\sigma^\mu\bar\theta$$ and $$\bar y^\mu = x^\mu - i \theta\sigma^\mu\bar\theta$$. Therefore, expanding around $$y^\mu = x^\mu$$ and $$\bar y^\mu = x^\mu$$, and keeping only those terms proportional to $$(\theta\theta)(\bar\theta\bar\theta)$$, we find $\begin{split} \bar\Phi_i\Phi^i & = - \frac{1}{2}(\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) \partial_\mu\partial_\nu \bar\phi_i \phi^i - \frac{1}{2}\bar\phi_i (\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta) \partial_\mu\partial_\nu \phi^i +(\theta\sigma^\mu\bar\theta)\partial_\mu\bar\phi_i(\theta\sigma^\nu\bar\theta)\partial^\mu\phi^i \\ & \quad -2i(\theta\sigma^\mu\bar\theta) (\bar\theta \partial_\mu \bar\psi_i)(\theta\psi^i) +2i(\bar\theta\bar\psi_i)(\theta\sigma^\mu\bar\theta) (\theta\partial_\mu\psi^i) + (\bar\theta\bar\theta)(\theta\theta)\bar F_i F^i + \dots \\ & = - \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \bar\phi_i \phi^i - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\bar\phi_i \partial_\mu\partial^\mu \phi^i +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\bar\phi_i\partial^\mu\phi^i \\ & \quad +i(\theta\theta)(\psi^i\sigma^\mu\bar\theta) (\bar\theta \partial_\mu \bar\psi_i) -i(\theta\theta) (\bar\theta\bar\psi_i)(\partial_\mu\psi^i\sigma^\mu\bar\theta) + (\theta\theta) (\bar\theta\bar\theta)\bar F_i F^i + \dots \\ & = - \frac{1}{4} (\theta\theta)(\bar\theta\bar\theta) \partial_\mu\partial^\mu \bar\phi_i \phi^i - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\bar\phi_i \partial_\mu\partial^\mu \phi^i +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\bar\phi_i\partial^\mu\phi^i \\ & \quad +\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)( \partial_\mu \bar\psi_i\bar\sigma^\mu\psi^i) -\frac{i}{2}(\theta\theta)(\bar\theta\bar\theta)(\bar\psi_i\bar\sigma^\mu\partial_\mu\psi^i) + (\theta\theta) (\bar\theta\bar\theta)\bar F_i F^i + \dots \\ & = (\theta\theta) (\bar\theta\bar\theta) \Big( -\frac{1}{4}\partial_\mu\big(\partial^\mu \bar\phi_i \phi^i + \bar\phi_i\partial^\mu\phi^i -2i\bar\psi_i\bar\sigma^\mu\psi^i\big) +\partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i \Big) + \dots ~, \end{split}$ where all fields are evaluated at $$x^\mu$$. Therefore, as claimed, we have $D_K = \partial_\mu\bar\phi_i \partial^\mu\phi^i - i\bar\psi_i \bar\sigma^\mu \partial_\mu \psi^i + \bar F_i F^i + \text{total derivative} ~,$ where the total derivative is given by $-\frac{1}{4}\partial_\mu\big(\partial^\mu \bar\phi_i \phi^i + \bar\phi_i\partial^\mu\phi^i -2i\bar\psi_i\bar\sigma^\mu\psi^i\big) ~.$

Consider the Wess-Zumino model of a single chiral superfield $$\Phi$$ with $K(\Phi,\bar\Phi) = \bar\Phi\Phi ~, \qquad W(\Phi) = \frac{m}{2}\Phi^2 + \frac{\lambda}{3}\Phi^3 ~.$ Argue that this $$W(\Phi)$$ is the most general renormalisable superpotential and find the supersymmetric vacua of the theory.

Express the action of the Wess-Zumino model in terms of component fields both before and after integrating out the auxiliary fields. Show that the effective physical mass of $$\phi$$ is equal to that of $$\psi$$ and is given by $$m_{\text{eff}}(\langle\phi\rangle) = m +2 \lambda\langle\phi\rangle$$ when the action is expanded around one of the supersymmetric vacua. How is the quartic coupling in the scalar potential related to the Yukawa coupling? Interpret these two results.

Derive the equations of motion for the component fields $$\phi$$, $$\psi$$ and $$F$$ directly from the action expressed in terms of component fields. Finally, expand the equation of motion for the chiral superfield $$\Phi$$ and show it is equivalent to the equations of motion for the component fields.

The dimension of the chiral superfield is $$[\Phi] = 1$$. The action should have dimension 0, which means that $$[W]=3$$. To ensure renormalisability we assume that there are no irrelevant couplings. This rules out terms in the superpotential of the form $$\lambda_n \Phi^n$$ for $$n>3$$ since $$[\lambda_n\Phi^n] = [\lambda_n] + n[\Phi] = 3$$ implies that $$[\lambda_n] = 3-n$$. Therefore, $$W(\Phi)$$ can at most be a cubic polynomial in $$\Phi$$. We can then set the linear term in $$W(\Phi)$$ to zero by shifting $$\Phi$$, while the constant term can be ignored since $$\int d^2\theta \, 1 = 0$$. Therefore, the most general renormalisable superpotential is $W(\Phi) = \frac{m}{2}\Phi^2 + \frac{\lambda}{3}\Phi^3 ~.$

The supersymmetric vacua of the theory are the solutions to $$W'(\phi)\big|_{\langle\phi\rangle} = 0$$. We have $W'(\phi) = m\phi + \lambda \phi^2 = \phi(m+\lambda\phi) ~,$ hence the two supersymmetric vacua are $\langle\phi\rangle = 0 ~, \qquad \langle\phi\rangle = -\frac{m}{\lambda} ~.$

The action of the Wess-Zumino model in terms of component fields is given by $\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi \partial^\mu\phi - i\bar\psi \bar\sigma^\mu \partial_\mu \psi + \bar F F - W'(\phi) F -\bar W'(\bar\phi) \bar F -\frac{1}{2}W''(\phi)\psi\psi -\frac{1}{2}\bar W''(\bar\phi)\bar\psi\bar\psi \big) ~, \end{split}$ where \begin{aligned} W'(\phi) & = m\phi + \lambda \phi^2 ~, \qquad & W''(\phi) & = m + 2\lambda \phi ~, \\ \bar W'(\bar \phi) & = \bar m\bar \phi + \bar \lambda \bar \phi^2 ~, \qquad & \bar W''(\bar\phi) & = \bar m + 2\bar \lambda \bar \phi ~. \end{aligned} Varying with respect to the auxiliary fields $$F$$ and $$\bar F$$ we find $\bar F = W'(\phi) = m\phi + \lambda \phi^2 ~,\qquad F = \bar W'(\bar \phi) = \bar m\bar \phi + \bar \lambda \bar \phi^2 ~.$ Substituting back into the action we find that the action of the Wess-Zumino model in terms of component fields after integrating out the auxiliary fields is given by $\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi \partial^\mu\phi - i\bar\psi \bar\sigma^\mu \partial_\mu \psi - (\bar m\bar \phi + \bar \lambda \bar \phi^2 )(m\phi + \lambda \phi^2) -\frac{1}{2}(m + 2\lambda \phi)\psi\psi -\frac{1}{2} (\bar m + 2\bar \lambda \bar \phi )\bar\psi\bar\psi \big) ~. \end{split}$

To compute the effective masses we expand the action around $$\phi = \langle\phi\rangle$$ and $$\psi = 0$$ to quadratic order in fluctuations, which we denote by $$\varphi$$ and $$\vartheta$$, we find $\begin{split} \mathcal{S}& = \int d^4x \, \big(-(\bar m\langle\bar \phi\rangle + \bar \lambda\langle\bar \phi\rangle^2)(m\langle\phi\rangle + \lambda\langle \phi\rangle^2) \\ & \qquad\qquad -(\bar m\langle\bar \phi\rangle + \bar \lambda\langle\bar \phi\rangle^2)(m +2\lambda)\varphi -(m\langle\phi\rangle + \lambda\langle \phi\rangle^2)(\bar m+2\bar\lambda)\bar\varphi \\ & \qquad \qquad + \partial_\mu\bar{\varphi} \partial^\mu\varphi - (\bar m\langle\bar \phi\rangle + \bar \lambda\langle\bar \phi\rangle^2) \lambda \varphi^2 - (m\langle\phi\rangle + \lambda\langle \phi\rangle^2)\bar\lambda \bar{\varphi}^2 \\ & \qquad\qquad - (\bar m + 2 \bar\lambda\langle\bar \phi\rangle)(m + 2\lambda\langle \phi\rangle)\bar\varphi \varphi \\ & \qquad\qquad - i\bar{\vartheta} \bar\sigma^\mu \partial_\mu \vartheta -\frac{1}{2}(m + 2\lambda \langle\phi\rangle)\vartheta\vartheta -\frac{1}{2} (\bar m + 2\bar \lambda\langle \bar \phi\rangle )\bar{\vartheta}\bar{\vartheta} \big) + \dots ~. \end{split}$ The supersymmetric vacua are $$\langle \phi \rangle = 0$$ or $$\langle \phi \rangle = -\frac{m}{\lambda}$$, for which the constant term and the terms proportional to $$\varphi$$, $$\bar\varphi$$, $$\varphi^2$$ and $$\bar\varphi^2$$ vanish. Therefore, the action becomes $\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar{\varphi} \partial^\mu\varphi - (\bar m + 2 \bar\lambda\langle\bar \phi\rangle)(m + 2\lambda\langle \phi\rangle)\bar\varphi \varphi \\ &\qquad\qquad - i\bar{\vartheta} \bar\sigma^\mu \partial_\mu \vartheta -\frac{1}{2}(m + 2\lambda \langle\phi\rangle)\vartheta\vartheta -\frac{1}{2} (\bar m + 2\bar \lambda\langle \bar \phi\rangle )\bar{\vartheta}\bar{\vartheta} \big) + \dots ~, \end{split}$ and we can read off the effective mass of $$\phi$$ and $$\psi$$ $m^{\phi}_{\text{eff}} = m^{\psi}_{\text{eff}} = m_{\text{eff}} = |m + 2\lambda\langle \phi\rangle| ~.$ Note that the effective mass is nothing but $$m_{\text{eff}} = |W''(\phi)|$$. Expanding to quartic in order in fluctuations we see that the quartic coupling is $$\bar\lambda\lambda\bar\varphi^2\varphi^2$$, while the Yukawa couplings are $$\lambda\varphi\vartheta\vartheta$$ and $$\bar\lambda\bar\varphi\bar\vartheta\bar\theta$$. Therefore, schematically we have that $$(\text{quartic}) = (\text{Yukawa})^2$$. This is a consequence of supersymmetry and, together with the equal effective masses, leads to cancellations between quantum corrections due to bosons and those due to fermions around supersymmetric vacua.

The equations of motion for the component fields are \begin{aligned} \phi: \quad & \partial_\mu\partial^\mu \bar\phi + W''(\phi)F + \frac{1}{2}W'''(\phi) \psi\psi = 0 ~, & \qquad \bar\phi: \quad & \partial_\mu\partial^\mu \phi + \bar W''(\bar \phi)\bar F + \frac{1}{2}\bar W'''(\bar \phi) \bar \psi\bar \psi = 0 ~, \\ \psi: \quad & i\sigma^\mu\partial_\mu\bar\psi + W''(\phi)\psi = 0 ~, & \qquad \bar\psi: \quad & i\bar \sigma^\mu\partial_\mu\psi + \bar W''(\bar\phi)\bar\psi = 0 ~, \\ F: \quad & \bar F = W'(\phi) ~, & \bar F: \quad & F = \bar W'(\bar\phi) ~. \end{aligned} The equations of motion for the chiral and antichiral superfield are $\bar D^2 \bar\Phi = 4 W'( \Phi) ~, \qquad D^2 \Phi = 4\bar W'(\bar \Phi) ~.$ We start by writing the antichiral superfield $$\bar\Phi(\bar y,\bar\theta)=\bar\phi(\bar y) + \sqrt{2}\bar\theta\bar\psi(\bar y) - \bar\theta\bar\theta \bar F(\bar y)$$ in terms of the coordinates $$(y,\theta,\bar\theta)$$ $\begin{split} \bar\Phi(y,\theta,\bar\theta) & = \bar\phi(y) + \sqrt{2}\bar\theta\bar\psi(y) - \bar\theta\bar\theta \bar F(y) \\ & \quad - 2i(\theta\sigma^\mu\bar\theta)\hat{\partial}_\mu\bar\phi(y) - 2\sqrt{2}i(\theta\sigma^\mu\bar\theta)(\bar\theta\hat{\partial}_\mu\bar\psi(y)) - 2(\theta\sigma^\mu\bar\theta)(\theta\sigma^\nu\bar\theta)\hat{\partial}_\mu\hat{\partial}_\nu \bar\phi(y) \\ & = \bar\phi(y) + \sqrt{2}\bar\theta\bar\psi(y) - \bar\theta\bar\theta \bar F(y) \\ & \quad - 2i(\theta\sigma^\mu\bar\theta)\hat{\partial}_\mu\bar\phi(y) + \sqrt{2}i(\bar\theta\bar\theta)(\theta\sigma^\mu\hat{\partial}_\mu\bar\psi(y)) - (\theta\theta)(\bar\theta\bar\theta)\hat{\partial}_\mu\hat{\partial}^\mu \bar\phi(y) ~, \end{split}$ where we have used that $$\bar y^\mu = y^\mu - 2i\theta\sigma^\mu\bar\theta$$. In terms of derivatives with respect to $$(y,\theta,\bar\theta)$$, $$\bar D_{\dot\alpha} = \hat{\bar\partial}_{\dot\alpha}$$. Using that $$\hat{\bar\partial}^2(\bar\theta\bar\theta) = -4$$ we find $\bar D^2 \bar\Phi (y,\theta) = 4\big(\bar F(y) - \sqrt{2}i\theta\sigma^\mu\hat{\partial}_\mu\bar\psi(y) + \theta\theta\hat{\partial}_\mu\hat{\partial}^\mu \bar\phi(y)\big) ~.$ Next we observe that $\begin{split} 4W'(\Phi(y,\theta)) & = 4 W'(\phi(y)+\sqrt{2}\theta \psi(y) - \theta\theta F(y)) \\ & = 4 \big(W'(\phi(y)) + (\sqrt{2}\theta \psi(y) - \theta\theta F(y)) W''(\phi(y)) + (\theta\psi(y))(\theta\psi(y)) W'''(\phi(y))\big) \\ & = 4 \big(W'(\phi(y)) + \sqrt{2}\theta \psi(y)W''(\phi(y)) - \theta\theta (F(y) W''(\phi(y)) +\frac{1}{2} \psi(y)\psi(y) W'''(\phi(y)))\big) ~. \end{split}$ Equating $$\bar D^2 \bar\Phi(y,\theta)$$ with $$4W'(\Phi(y,\theta))$$, we recover the equations of motion for the component fields $\bar F = W'(\phi) ~, \qquad i\sigma^\mu\partial_\mu\bar\psi + W''(\phi) \psi = 0 ~, \qquad \partial_\mu\partial^\mu \bar\phi + W''(\phi) F + \frac{1}{2}W'''(\phi) \psi\psi = 0 ~,$ as the coefficients of $$1$$, $$\theta$$ and $$\theta\theta$$ respectively, where we have set $$y^\mu = x^\mu$$. A similar derivation holds for the conjugate equation of motion $$D^2\Phi = 4\bar W'(\bar\Phi)$$.

Consider the Wess-Zumino model of three chiral superfields $$X$$, $$Y$$ and $$Z$$ with the homogeneous trilinear superpotential $$W(X,Y,Z) = \lambda XYZ$$. Write down the scalar potential of the theory. Find the moduli space of supersymmetric vacua.

The scalar potential of the theory is given by $\begin{split} V(x,y,z,\bar x,\bar y,\bar z) & = \sum_{i = x,y,z} \bar\partial^i \bar W(\bar x,\bar y,\bar z) \partial_i W(x,y,z) \\ & = \bar\lambda\lambda (\bar y y \bar z z + \bar x x \bar z z + \bar x x \bar y y) ~, \end{split}$ where the component field expansions of the chiral superfields $$X$$, $$Y$$ and $$Z$$ are $X = x + \mathcal{O}(\theta) ~, \qquad Y = y + \mathcal{O}(\theta) ~, \qquad Z = z + \mathcal{O}(\theta) ~.$ The moduli space of supersymmetric vacua is the space of solutions to $$\partial_i W(x,y,z)\big|_{\langle x\rangle,\langle y\rangle,\langle z\rangle} = 0$$. This gives us the following three equations $\langle y \rangle\langle z \rangle = \langle x\rangle\langle z \rangle = \langle x\rangle\langle y \rangle = 0 ~.$ These are solved by setting any two of $$\langle x\rangle$$, $$\langle y\rangle$$ and $$\langle z\rangle$$ equal to zero. There are three ways to do this, each of which gives a branch or component of the moduli space of vacua. The moduli space of supersymmetric vacua is therefore given by $\begin{split} \mathcal{M} & = \mathcal{M}_x \cup \mathcal{M}_y \cup \mathcal{M}_z ~, \\ \mathcal{M}_x & = \{x \in \mathbb{C}, ~ y=z = 0\} \cong \mathbb{C}~, \\ \mathcal{M}_y & = \{y \in \mathbb{C}, ~ x=z = 0\} \cong \mathbb{C}~, \\ \mathcal{M}_z & = \{z \in \mathbb{C}, ~ x=y = 0\} \cong \mathbb{C}~. \end{split}$ The three branches intersect at the origin.

Argue that the superpotentials $$W(\Phi) = \sum_{n=2}^N \lambda^n \Phi^n$$ and $$W(\Phi) = \frac{1}{2} m_{ij} \Phi^i\Phi^j + \frac{1}{3} \lambda_{ijk}\Phi^i\Phi^j\Phi^k$$ are not renormalised in an SQFT of chiral multiplets.

We start with the superpotential $W(\Phi) = \sum_{n=2}^N \lambda^n \Phi^n ~.$ Holomorphy implies that $$W_{\text{eff},\mu}$$ is a holomorphic function of $$\Phi$$ and $$\lambda_n$$. There are two broken symmetries: $$\mathrm{U}(1)_{_R}$$ with $$[W]_{_R} = 2$$ and $$[\Phi]_{_R} = 1$$, and $$\mathrm{U}(1)_{_F}$$ with $$[W]_{_F} = 0$$ and $$[\Phi]_{_F} = 1$$. The coupling constants have charges $$[\lambda_n]_{_R} = -n+2$$ and $$[\lambda_n]_{_F} = -n$$. Therefore, the effective superpotential is constrained to take the form $W_{\text{eff},\mu} = \lambda_2\Phi^2 f(t_i,\mu) ~, \qquad t_i = \frac{\lambda_i\Phi^{i-2}}{\lambda_2} ~,$ where $$t_i = \frac{\lambda_i\Phi^{i-2}}{\lambda_2}$$, $$i=3,\dots,N$$, are a maximal set of independent combinations of the superfield and the coupling constants with zero R-charge and zero flavour charge. Now taking the limit $$\lambda_n \to 0$$ with $$t_i$$ fixed the theory becomes free and we should have that $$W_{\text{eff},\mu} \to W$$. Therefore, $$f(t_i,\mu) \to 1 + \sum_{n=3}^N t_i$$, which immediately implies that $$f(t_i,\mu) = 1 + \sum_{n=3}^N t_i$$. As a result, we find that $W_{\text{eff},\mu} = \sum_{n=2}^N \lambda^n \Phi^n = W ~.$

For the superpotential $W(\Phi) = \frac{1}{2} m_{ij} \Phi^i\Phi^j + \frac{1}{3} \lambda_{ijk}\Phi^i\Phi^j\Phi^k ~, \qquad i=1,\dots,n ~,$ holomorphy implies that $$W_{\text{eff},\mu}$$ is a holomorphic function of $$\Phi^i$$, $$m_{ij}$$ and $$\lambda_{ijk}$$. The broken symmetries are now $$\mathrm{U}(1)_{_R}$$ with $$[W]_{_R} = 2$$, $$[\Phi]_{_R} = 1$$, $$[m_{ij}]_{_R} = 0$$ and $$[\lambda_{ijk}]_{_R} = -1$$, and $$\mathrm{U}(n)_{_F}$$, where $$\Phi^i$$ transforms in the fundamental representation, and $$m_{ij}$$ and $$\lambda_{ijk}$$ transform in the double and triple symmetric tensor products of the antifundamental representation respectively. Let us now choose one chiral superfield $$\Phi^{\hat\imath}$$. The superpotential is a cubic polynomial in $$\Phi^{\hat\imath}$$ with effective couplings that depend on $$m_{ij}$$, $$\lambda_{ijk}$$ and $$\Phi^{i\neq\hat{\imath}}$$. Therefore, the non-renormalisation argument applies. Since this holds for all $$\Phi^i$$, it follows that the superpotential $W(\Phi) = \frac{1}{2} m_{ij} \Phi^i\Phi^j + \frac{1}{3} \lambda_{ijk}\Phi^i\Phi^j\Phi^k ~,$ is not renormalised.

# 6 Supersymmetric gauge theories

Starting from an abelian vector superfield $$V$$ and its supersymmetric gauge transformation $$V \to V + \Lambda + \bar\Lambda$$, where $$\Lambda$$ is a chiral superfield, determine the gauge transformations of the component fields of $$V$$.

Fixing the Wess-Zumino (WZ) gauge $$V = V_{\text{WZ}}$$, compute the supersymmetry transformation of $$V_{\text{WZ}}$$ and determine the compensating gauge transformation that is needed to restore the WZ gauge.

Derive the component field expansion of the gaugino superfield $$W_\alpha = -\frac{1}{4} \bar D^2 D_\alpha V$$ in terms of the coordinates $$(y,\theta,\bar\theta)$$ using that $$W_\alpha$$ is gauge invariant.

A general abelian vector superfield $$V$$ takes the form $\begin{split} V(x,\theta,\bar\theta) &= C(x) + \theta\chi(x) + \bar\theta\bar\chi(x) +\theta\theta M(x) +\bar\theta\bar\theta\bar M(x) + (\theta\sigma^\mu\bar\theta) A_\mu(x) \\ & \quad + i(\theta\theta) \bar\theta\big(\bar\lambda(x) + \frac{1}{2}\bar\sigma^\mu\partial_\mu\chi(x)\big) - i(\bar\theta\bar\theta) \theta\big(\lambda(x) - \frac{1}{2}\sigma^\mu\partial_\mu\bar\chi(x)\big) \\ & \quad +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\big(D(x) - \frac{1}{2}\partial_\mu\partial^\mu C(x)\big) ~, \end{split}$ and we take the chiral superfield $$\Lambda$$ and its conjugate $$\bar\Lambda$$ to be given by $\Lambda(y,\theta) = \phi(y,\theta) + \sqrt{2}\theta\psi(y) - \theta\theta F(y) ~, \qquad \bar\Lambda(\bar y,\bar \theta) = \bar \phi(\bar y,\bar \theta) + \sqrt{2}\bar\theta\bar\psi(y) - \bar\theta\bar\theta \bar F(\bar y) ~,$ or, in terms of the coordinates $$(x,\theta,\bar\theta)$$, $\begin{split} \Lambda(x,\theta,\bar\theta) & = \phi(x) + i(\theta\sigma^\mu\bar\theta)\partial_\mu\phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\partial^\mu \phi(x) \\ & \quad + \sqrt{2}\theta\psi(x) - \frac{i}{\sqrt{2}}(\theta\theta)(\partial_\mu\psi(x)\sigma^\mu\bar\theta)-\theta\theta F(x) ~, \\ \bar \Lambda(x,\theta,\bar\theta) & = \bar \phi(x) - i(\theta\sigma^\mu\bar\theta)\partial_\mu\bar\phi(x) - \frac{1}{4}(\theta\theta)(\bar\theta\bar\theta)\partial_\mu\partial^\mu \bar\phi(x) \\ & \quad + \sqrt{2}\bar\theta\bar\psi(x) + \frac{i}{\sqrt{2}}(\bar\theta\bar\theta)(\theta\sigma^\mu\partial_\mu\bar\psi(x))-\bar\theta\bar\theta \bar F(x) ~. \end{split}$ Therefore, we have that $\begin{split} V(x,\theta,\bar\theta) + \Lambda(x,\theta,\bar\theta) + \bar\Lambda(x,\theta,\bar\theta) & = C(x) + \phi(x) + \bar\phi(x) +\theta\big(\chi(x) + \sqrt{2}\psi(x)\big) +\bar\theta\big(\bar\chi(x) + \sqrt{2}\bar\psi(x)\big) \\ & \quad + \theta\theta \big(M(x) - F(x)\big) + \bar\theta\bar\theta\big(\bar M(x)- \bar F(x)\big) \\ & \quad + (\theta\sigma^\mu\bar\theta) \big(A_\mu(x) + i\partial_\mu\phi(x) - i\partial_\mu\bar\phi(x)\big) \\ & \quad + i(\theta\theta) \bar\theta\big(\bar\lambda(x) + \frac{1}{2}\bar\sigma^\mu\partial_\mu\chi(x) + \frac{1}{\sqrt{2}}\bar\sigma^\mu\partial_\mu\psi(x)\big) \\ & \quad - i(\bar\theta\bar\theta) \theta\big(\lambda(x) - \frac{1}{2}\sigma^\mu\partial_\mu\bar\chi(x) - \frac{1}{\sqrt{2}}\sigma^\mu\partial_\mu\bar\psi(x)\big) \\ & \quad +\frac{1}{2}(\theta\theta)(\bar\theta\bar\theta)\big(D(x) - \frac{1}{2}\partial_\mu\partial^\mu C(x) \\ & \hspace{100pt} - \frac{1}{2}\partial_\mu\partial^\mu \phi(x) - \frac{1}{2}\partial_\mu\partial^\mu \bar\phi(x)\big) ~, \end{split}$ hence the gauge transformations of the component fields are \begin{aligned} C & \to C + 2\mathop{\mathrm{Re}}\phi ~, & \qquad \chi & \to \chi + \sqrt{2}\psi ~, & \qquad M & \to M - F ~, \\ A_\mu & \to A_\mu - 2\partial_\mu \mathop{\mathrm{Im}}\phi ~, & \qquad \lambda & \to \lambda ~, & \qquad D & \to D ~. \end{aligned}

From the results of exercise 4.3, we have that the supersymmetry transformations of the component fields of an abelian vector superfield are \begin{aligned} \delta_{\varepsilon,\bar\varepsilon} C & = \varepsilon \chi + \bar\varepsilon\bar\chi ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \chi = i(\sigma^\mu\bar\varepsilon)\big(\partial_\mu C -i A_\mu\big) + 2\varepsilon M ~, \qquad \delta_{\varepsilon,\bar\varepsilon} M = i\bar\varepsilon\big(\bar\lambda + \bar\sigma^\mu\partial_\mu\chi\big) ~, \\ \delta_{\varepsilon,\bar\varepsilon} A_\mu & = i\varepsilon\big(\sigma_\mu\bar\lambda +\frac{1}{2}(\sigma_\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma_\mu)\partial_\nu\chi\big) +i\bar\varepsilon\big(\bar\sigma_\mu\lambda - \frac{1}{2}(\bar\sigma_\mu\sigma^\nu+\bar\sigma^\nu\sigma_\mu)\partial_\nu\bar\chi\big) \\ & = i\varepsilon\big(\sigma_\mu\bar\lambda +\partial_\mu\chi\big) +i\bar\varepsilon\big(\bar\sigma_\mu\lambda - \partial_\mu\bar\chi\big) ~, \\ \delta_{\varepsilon,\bar\varepsilon} \lambda & = \frac{1}{2}\sigma^\mu\partial_\mu\delta_{\varepsilon,\bar\varepsilon} \bar\chi - (\sigma^\mu\bar\varepsilon)\partial_\mu \bar M + \frac{1}{2}(\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu A_\nu +i \varepsilon \big(D-\frac{1}{2}\partial_\mu\partial^\mu C\big) \\ & = \frac{1}{2}\sigma^\mu\partial_\mu(i\bar\sigma^\nu\varepsilon(\partial_\nu C + i A_\nu) + 2\bar\varepsilon \bar M) - (\sigma^\mu\bar\varepsilon)\partial_\mu \bar M + \frac{1}{2}(\sigma^\nu\bar\sigma^\mu\varepsilon)\partial_\mu A_\nu +i \varepsilon \big(D-\frac{1}{2}\partial_\mu\partial^\mu C\big) \\ & = \frac{i}{4}(\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu)\varepsilon\partial_\mu\partial_\nu C - \frac{i}{2}\varepsilon \partial_\mu\partial^\mu C - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D \\ & = - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D ~, \\ \delta_{\varepsilon,\bar\varepsilon} D & = \frac{1}{2}\partial_\mu\partial^\mu\delta_{\varepsilon,\bar\varepsilon} C + i\varepsilon\sigma^\mu\partial_\mu\big(i\bar\lambda + \frac{i}{2}\bar\sigma^\nu\partial_\nu\chi\big) + i\bar\varepsilon\bar\sigma^\mu\partial_\mu\big(-i \lambda+\frac{i}{2}\sigma^\nu\partial_\nu\bar\chi\big) \\ & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda + \frac{1}{2}\varepsilon\partial_\mu\partial^\mu\chi + \frac{1}{2}\bar\varepsilon\partial_\mu\partial^\mu\bar\chi \\ & \quad - \frac{1}{4}\varepsilon(\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu)\partial_\mu\partial_\nu\chi - \frac{1}{4}\bar\varepsilon(\bar\sigma^\mu\sigma^\nu+\bar\sigma^\nu\sigma^\mu)\partial_\mu\partial_\nu\bar\chi \\ & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda ~, \end{aligned} where we have used that $$\sigma^\mu\bar\sigma^\nu+\sigma^\nu\bar\sigma^\mu = \bar\sigma^\mu\sigma^\nu+\bar\sigma^\nu\sigma^\mu = 2\eta^{\mu\nu}\mathbf{1}$$. In the WZ gauge, $$C = \chi = M = 0$$, the supersymmetry transformations simplify to \begin{aligned} \delta_{\varepsilon,\bar\varepsilon} C & = 0 ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \chi = (\sigma^\mu\bar\varepsilon) A_\mu ~, \qquad \delta_{\varepsilon,\bar\varepsilon} M = i\bar\varepsilon\bar\lambda ~, \\ \delta_{\varepsilon,\bar\varepsilon} A_\mu & = i(\varepsilon\sigma_\mu\bar\lambda) +i(\bar\varepsilon\bar\sigma_\mu\lambda) ~, \\ \delta_{\varepsilon,\bar\varepsilon} \lambda & = - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D ~, \\ \delta_{\varepsilon,\bar\varepsilon} D & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda ~, \end{aligned} and we see explicitly that the WZ gauge is not supersymmetric. Recalling that the gauge transformations of the component fields are \begin{aligned} \delta_{\Lambda,\bar\Lambda} C & = 2\mathop{\mathrm{Re}}\phi ~, & \qquad \delta_{\Lambda,\bar\Lambda} \chi & = \sqrt{2}\psi ~, & \qquad \delta_{\Lambda,\bar\Lambda} M & = -F ~, \\ \delta_{\Lambda,\bar\Lambda} A_\mu & = -2\partial_\mu\mathop{\mathrm{Im}}\phi ~, & \qquad \delta_{\Lambda,\bar\Lambda} \lambda & = 0 ~, & \qquad \delta_{\Lambda,\bar\Lambda} D & = 0~, \end{aligned} it follows that to restore the WZ gauge we should set $\phi = 0 ~, \qquad \psi = - \frac{1}{\sqrt{2}}(\sigma^\mu\bar\varepsilon)A_\mu ~, \qquad F = i \bar\varepsilon \bar\lambda ~.$ Therefore, the compensating gauge transformation needed to restore the WZ gauge is $\begin{split} V \to V + \Lambda + \bar\Lambda ~, \qquad \Lambda(y,\theta) & = - (\theta\sigma^\mu\bar\varepsilon)A_\mu(y) - i(\theta\theta)(\bar\varepsilon \bar\lambda(y)) ~, \\ \Lambda(x,\theta,\bar\theta) & = - (\theta\sigma^\mu\bar\varepsilon)A_\mu(x) - i(\theta\sigma^\mu\bar\varepsilon)(\theta\sigma^\nu\bar\theta)\partial_\nu A_\mu(x) - i(\theta\theta)(\bar\varepsilon \bar\lambda(x)) \\ & = - (\theta\sigma^\mu\bar\varepsilon)A_\mu(x) - \frac{i}{2}(\theta\theta)(\bar\theta\bar\sigma^\mu\sigma^\nu \bar\varepsilon)\partial_\mu A_\nu(x) - i(\theta\theta)(\bar\varepsilon \bar\lambda(x)) ~, \end{split}$ such that, in the WZ gauge, we have \begin{aligned} (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) C & = 0 ~, \qquad (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) \chi = 0 ~, \qquad (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) M = 0 ~, \\ (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) A_\mu & = i(\varepsilon\sigma_\mu\bar\lambda) +i(\bar\varepsilon\bar\sigma_\mu\lambda) ~, \\ (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) \lambda & = - \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\varepsilon)(\partial_\mu A_\nu - \partial_\nu A_\mu) + i\varepsilon D ~, \\ (\delta_{\varepsilon,\bar\varepsilon}+\delta_{\Lambda,\bar\Lambda}) D & = - \varepsilon\sigma^\mu\partial_\mu\bar\lambda + \bar\varepsilon\bar\sigma^\mu\partial_\mu\lambda ~. \end{aligned}

Since $$W_\alpha$$ is gauge invariant, its general form is equal to its form in the WZ gauge, i.e. $$W_\alpha = - \frac{1}{4} \bar D^2 D_\alpha V = - \frac{1}{4} \bar D^2 D_\alpha V_{\text{WZ}}$$. In terms of the coordinates $$(y,\theta,\bar\theta)$$ we have $$D_\alpha = \hat\partial_\alpha + 2i(\sigma^\mu\bar\theta)_\alpha \hat\partial_\mu$$, $$\bar D_{\dot\alpha} = \hat{\bar\partial}_{\dot\alpha}$$ and $\begin{split} V_{\text{WZ}}(y,\theta,\bar\theta) & = (\theta\sigma^\mu\bar\theta) \big(A_\mu(y)-i (\theta\sigma^\nu\bar\theta)\hat\partial_\nu A_\mu\big) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) D(y) \\ & = (\theta\sigma^\mu\bar\theta) A_\mu(y) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) ~. \end{split}$ Therefore, $\begin{split} D_\alpha V_{\text{WZ}}(y,\theta,\bar\theta) & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + 2i (\sigma^\nu\bar\theta)_\alpha(\theta\sigma^\mu\bar\theta) \hat\partial_\nu A_\mu(y) + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) - 2 (\sigma^\mu\bar\theta)_\alpha (\theta\theta) (\bar\theta\hat\partial_\mu \bar\lambda(y)) \\ & \quad - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^\nu\bar\sigma^\mu\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + \frac{i}{2} ((2\eta^{\mu\nu}\mathbf{1}- 4\sigma^{\mu\nu})\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} (\bar\theta\bar\theta) F_{\mu\nu}(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) ~, \end{split}$ where $$F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)$$ is the abelian field strength and we recall that \begin{aligned} \sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu & = 4\sigma^{\mu\nu} ~, \qquad & \bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu & = 4\bar\sigma^{\mu\nu} ~, \\ \sigma^\mu\bar\sigma^\nu + \sigma^\nu\bar\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~, \qquad & \bar\sigma^\mu\sigma^\nu + \bar\sigma^\nu\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~. \end{aligned} Now using that $$\hat{\bar\partial}^2(\bar\theta\bar\theta) = -4$$ we find $\begin{split} W_\alpha(y,\theta,\bar\theta) & = -\frac{1}{4} \bar D^2 D_\alpha V_{\text{WZ}}(y,\theta,\bar\theta) \\ & = - i \lambda_\alpha(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} F_{\mu\nu}(y) + \theta_\alpha D(y) + (\theta\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} ~. \end{split}$

The field content of an abelian supersymmetric gauge theory consists of $$r$$ abelian vector superfields $$V_a$$, $$a=1,\dots,r$$, and $$N$$ chiral superfields $$\Phi^i$$, $$i=1,\dots,N$$, with charges $$[\Phi^i]_{_{Q_a}} = q^i_a$$ under the $$\mathrm{U}(1)^r$$ gauge group.

Write down the component field expansion of the most general supersymmetric action $\begin{split} \mathcal{S}& = \mathcal{S}_{\text{Maxwell}} + \mathcal{S}_{\text{FI}} + \mathcal{S}_{\text{matter}} + \mathcal{S}_{\text{W}} ~, \\ \mathcal{S}_{\text{Maxwell}} & = \sum_{a=1}^r \mathop{\mathrm{Im}}\Big(\int d^4x d^2\theta \, \frac{\tau_a}{8\pi} W_a^{\alpha} W_{a\alpha} \Big) ~, \qquad \tau_a = \frac{\theta_a}{2\pi} + \frac{4\pi i}{g_a^2} ~, \\ \mathcal{S}_{\text{FI}} & = -2 \sum_{a=1}^r \xi_a \int d^4x d^2\theta d^2\bar\theta \, V_a ~, \\ \mathcal{S}_{\text{matter}} & = \sum_{i=1}^N \int d^4x d^2\theta d^2\bar\theta \, \bar \Phi_i e^{2\sum_{a=1}^r q_a^i V_a} \Phi^i ~, \\ \mathcal{S}_{\text{W}} & = \int d^4x d^2\theta \, W(\Phi) + \int d^4x d^2\bar \theta \, \bar W(\bar\Phi) ~, \end{split}$ where the superpotential $$W(\Phi)$$ is a gauge invariant holomorphic function of $$\Phi^i$$, i.e. $$[W]_{_{Q_a}} = 0$$ for all $$a=1,\dots r$$.

Integrate out the auxiliary fields $$F^i$$, $$\bar F_i$$ and $$D_a$$ in the chiral and abelian vector superfields, and derive the scalar potential $\begin{split} V(\phi,\bar\phi) & = \sum_{i=1}^N \bar F_i F^i + \sum_{a=1}^r \frac{1}{2g_a^2} (D_a)^2 \\ & = \sum_{i=1}^N \partial_i W(\phi) \bar\partial^i \bar W(\bar\phi) + \sum_{a=1}^r \frac{g_a^2}{2}(\mu_a(\phi,\bar\phi)-\xi_a)^2 ~, \qquad \mu_a(\phi,\bar\phi) = \sum_{i=1}^N q_a^i \bar\phi_i\phi^i ~. \end{split}$

Up total derivatives, the component field expansion of the most general supersymmetric action for $$r$$ vector superfields and $$N$$ chiral superfields is $\begin{split} \mathcal{S}& = \sum_{a=1}^r \int d^4x\, \big(\frac{1}{g_a^2}(-\frac{1}{4} F_a^{\mu\nu}F_{a\mu\nu} - i\bar \lambda_a\bar\sigma^\mu\partial_\mu\lambda_a + \frac{1}{2} (D_a)^2) + \frac{\theta_a}{32\pi^2} \tilde F_a^{\mu\nu}F_{a\mu\nu} - \xi_a D_a \big) \\ & \quad + \sum_{i=1}^N \int d^4x\,\big(\bar D_\mu \bar\phi_i D^\mu \phi^i - i \bar \psi_i \bar\sigma^\mu D_\mu \psi^i + \bar F_i F^i + \sum_{a=1}^r q_a^i (i \sqrt{2} \bar\phi_i\lambda_a\psi^i - i \sqrt{2} \bar\psi_i\bar\lambda_a\phi^i + \bar\phi_i D_a \phi^i)\big) \\ & \quad - \sum_{i=1}^N \int d^4 x \, \big(\partial_i W(\phi) F^i - \bar\partial^i \bar W(\bar \phi) \bar F_i\big) -\frac{1}{2} \sum_{i=1}^N \sum_{j=1}^N \int d^4 x \, \big(\partial_i\partial_j W(\phi) \psi^i\psi^j + \bar\partial^i\bar\partial^j \bar W(\bar \phi) \bar\psi_i\bar\psi_j\big) ~, \end{split}$ where $D_\mu\phi^i = \partial_\mu\phi^i - i \sum_{a=1}^r q_a^i A_{a\mu} \phi^i ~, \\ \qquad D_\mu\psi^i = \partial_\mu\psi^i - i \sum_{a=1}^r q_a^i A_{a\mu} \psi^i ~.$ The terms involving the auxiliary fields are $\mathcal{S}_{\text{aux}} = \int d^4x \,\big( \sum_{a=1}^r (\frac{1}{2g_a^2} (D_a)^2 - \xi_a D_a + D_a \sum_{i=1}^Nq_a^i \bar\phi_i \phi^i) + \sum_{i=1}^N(\bar F_i F^i - \partial_i W(\phi) F^i - \bar\partial^i \bar W(\bar\phi) \bar F_i) \big) ~.$ Varying with respect to $$F^i$$, $$\bar F_i$$ and $$D_a$$ we find $\bar F_i = \partial_i W(\phi) ~, \qquad F^i = \bar\partial^i \bar W(\bar\phi) ~, \qquad D_a = - g_a^2 \big(\sum_{i=1}^N q_a^i \bar\phi_i \phi^i - \xi_a\big) ~.$ Substituting back into $$\mathcal{S}_{\text{aux}}$$ we find $\mathcal{S}_{\text{aux}} = - \int d^4x \,\big( \sum_{a=1}^r \frac{1}{2g_a^2} (D_a)^2 + \sum_{i=1}^N \bar F_i F^i \big) ~,$ where the auxiliary fields are understood to solve their equations of motion, hence are functions of the scalar fields. These are all the terms in the action that involve only scalar fields without derivatives. Therefore, the scalar potential is indeed given by $\begin{split} V(\phi,\bar\phi) & = \sum_{i=1}^N \bar F_i F^i + \sum_{a=1}^r \frac{1}{2g_a^2} (D_a)^2 \\ & = \sum_{i=1}^N \partial_i W(\phi) \bar\partial^i \bar W(\bar\phi) + \sum_{a=1}^r \frac{g_a^2}{2}(\mu_a(\phi,\bar\phi)-\xi_a)^2 ~, \qquad \mu_a(\phi,\bar\phi) = \sum_{i=1}^N q_a^i \bar\phi_i\phi^i ~. \end{split}$

Supersymmetric quantum electrodynamics (SQED) with one flavour is a supersymmetric $$\mathrm{U}(1)$$ gauge theory with matter fields, i.e. chiral superfields, $$Q$$ of charge $$1$$ and $$\tilde Q$$ of charge $$-1$$. Its action is given by $\begin{split} \mathcal{S}& = \mathop{\mathrm{Im}}\Big(\int d^4 x d^2\theta \, \frac{\tau}{8\pi} W^\alpha W_\alpha \Big) + \int d^4x d^2\theta d^2\bar\theta \, \big(\bar Q e^{2V} Q + \bar{\tilde{Q}} e^{-2V} \tilde{Q} - 2\xi V\big) \\ & \quad + \int d^4 x d^2\theta \, m \tilde{Q} Q + \int d^4x d^2\bar\theta \, \bar m \bar{\tilde{Q}} \bar{Q} ~, \end{split}$ where $$m$$ is a complex mass parameter, $$\tau = \frac{4\pi i}{g^2} + \frac{\theta}{2\pi}$$ is the complexified gauge coupling, $$\xi$$ is the real Fayet-Iliopoulos (FI) parameter, and $$W_\alpha = - \frac{1}{4} \bar D^2 D_\alpha V$$ is the gaugino (or photino) superfield.

Derive the scalar potential of this theory in terms of the component fields of the chiral superfields $$Q$$ and $$\tilde{Q}$$. Determine the moduli space of supersymmetric vacua for

• $$m = 0$$, $$\xi = 0$$;

• $$m = 0$$, $$\xi \neq 0$$;

• $$m \neq 0$$, $$\xi = 0$$;

• $$m \neq 0$$, $$\xi \neq 0$$.

Determine the vacuum expectation values of the gauge invariant chiral operator $$M = \tilde{Q} Q$$ in these supersymmetric vacua. You may use that unbroken supersymmetry implies that $$\langle XY \rangle = \langle X\rangle\langle Y \rangle$$ for any two chiral superfields $$X$$ and $$Y$$.

The component field expansion of the chiral superfields $$Q$$ and $$\tilde{Q}$$ is $Q = \phi^Q + \sqrt{2} \theta\psi^Q - \theta\theta F^Q ~, \qquad \tilde Q = \phi^{\tilde Q} + \sqrt{2} \theta\psi^{\tilde Q} - \theta\theta F^{\tilde Q} ~,$ The scalar potential is then given by $V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = \bar F_Q F^Q + \bar F_{\tilde{Q}} F^{\tilde{Q}} + \frac{1}{2g^2} D^2 ~,$ where \begin{aligned} F^Q & = \partial_{\bar {\phi}_Q}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{\tilde{Q}} ~, \qquad & \bar F_Q & = \partial_{\phi^Q}(m \phi^{\tilde Q}\phi^{Q}) = m \phi^{\tilde{Q}} ~, \\ F^{\tilde{Q}} & = \partial_{\bar{\phi}_{\tilde{Q}}}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{Q} ~, \qquad & \bar F_{\tilde{Q}} & = \partial_{\phi^{\tilde{Q}}}(m \phi^{\tilde Q} \phi^{Q}) = m \phi^Q ~, \\ D & = -g^2 (\bar \phi_Q \phi^Q - \bar\phi_{\tilde{Q}} \phi^{\tilde Q} - \xi) ~. && \end{aligned} Therefore, $V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = \bar m m(|\phi^Q|^2 + |\phi^{\tilde Q})|^2) + \frac{g^2}{2}(|\phi^Q|^2 - |\phi^{\tilde Q}|^2 - \xi)^2 ~.$

To determine the moduli space of supersymmetric vacua we solve $$\bar F_Q = \bar F_{\tilde{Q}} = D = 0$$ and quotient out by the $$\mathrm{U}(1)$$ gauge symmetry \begin{aligned} Q & \to e^{i\alpha(x)} Q ~, \qquad & \bar Q & \to e^{-i\alpha(x)} \bar Q ~, \\ \tilde{Q} & \to e^{-i\alpha(x)} \tilde{Q} ~, \qquad & \bar{\tilde{Q}} & \to e^{i\alpha(x)} \bar{\tilde{Q}} ~. \end{aligned}

• For $$m = 0$$, $$\xi = 0$$, we have $$\bar F_Q = \bar F_{\tilde{Q}} = 0$$ and $$D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2)$$. Solving $$D = 0$$, we find $$|\phi^Q|^2 = |\phi^{\tilde Q}|^2$$. We can use the $$\mathrm{U}(1)$$ gauge symmetry to fix $$\arg \phi^Q = \arg \phi^{\tilde{Q}}$$, hence $$\phi^Q = \phi^{\tilde{Q}}$$. Therefore, $\mathcal{M} = \Big\{\phi^Q, \phi^{\tilde{Q}} ~:~ \phi^Q = \phi^{\tilde{Q}}\Big\} = \mathbb{C}~.$ The vacuum expectation value of $$M = \tilde{Q} Q$$ is $$\langle M \rangle = \langle \tilde{Q}\rangle \langle Q\rangle = \langle \phi^{\tilde{Q}}\rangle\langle \phi^Q \rangle$$, hence takes all complex values on $$\mathcal{M}$$.

• For $$m = 0$$, $$\xi \neq 0$$, we have $$\bar F_Q = \bar F_{\tilde{Q}} = 0$$ and $$D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2 - \xi)$$. Solving $$D = 0$$, we find $$|\phi^Q|^2 = |\phi^{\tilde Q}|^2 + \xi$$. We can again use the $$\mathrm{U}(1)$$ gauge symmetry to fix $$\arg \phi^Q = \arg \phi^{\tilde{Q}}$$. Then for $$\xi > 0$$, $$\phi^Q$$ is determined in terms of $$\tilde\phi^Q$$, while for $$\xi < 0$$, $$\tilde\phi^Q$$ is determined in terms of $$\phi^Q$$. In particular, we have $\begin{split} \xi > 0 ~:~ \qquad \mathcal{M} & = \Big\{\phi^Q, \phi^{\tilde{Q}} ~:~ \phi^Q = \sqrt{1+\frac{\xi}{|\phi^{\tilde Q}|^2}} \phi^{\tilde{Q}}\Big\} = \mathbb{C}~, \\ \xi < 0 ~:~ \qquad \mathcal{M} & = \Big\{\phi^Q, \phi^{\tilde{Q}} ~:~ \tilde{\phi}_Q = \sqrt{1-\frac{\xi}{|\phi^Q|^2}}\phi^Q\Big\} = \mathbb{C}~. \end{split}$ The vacuum expectation value of $$M = \tilde{Q} Q$$ is $$\langle M \rangle = \langle \tilde{Q}\rangle \langle Q\rangle = \langle \phi^{\tilde{Q}}\rangle\langle \phi^Q \rangle$$, hence takes all complex values on $$\mathcal{M}$$.

• For $$m \neq 0$$, $$\xi = 0$$, we have $$\bar F_Q = m \phi^{\tilde{Q}}$$, $$\bar F_{\tilde{Q}} = m \phi^Q$$ and $$D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2)$$. Solving $$\bar F_Q = \bar F_{\tilde{Q}} = 0$$ gives $$\phi^Q = \phi^{\tilde{Q}} = 0$$, for which $$D = 0$$. The $$\mathrm{U}(1)$$ gauge symmetry has a trivial action on $$\phi^Q = \phi^{\tilde{Q}} = 0$$. Therefore, $\mathcal{M} = \Big\{\phi^Q = 0, \phi^{\tilde{Q}} = 0 \Big\} = \cdot ~,$ i.e. a single point. The vacuum expectation value of $$M = \tilde{Q} Q$$ is $$\langle M \rangle = \langle \tilde{Q}\rangle \langle Q\rangle = \langle \phi^{\tilde{Q}}\rangle\langle \phi^Q \rangle = 0$$.

• For $$m \neq 0$$, $$\xi \neq 0$$, we have $$\bar F_Q = m \phi^{\tilde{Q}}$$, $$\bar F_{\tilde{Q}} = m \phi^Q$$ and $$D = -g^2 (|\phi^Q|^2 - |\phi^{\tilde Q}|^2 -\xi)$$. Solving $$\bar F_Q = \bar F_{\tilde{Q}} = 0$$ gives $$\phi^Q = \phi^{\tilde{Q}} = 0$$, for which $$D = g^2\xi \neq 0$$. Therefore, there are no supersymmetric vacua $\mathcal{M} = \emptyset ~,$ and supersymmetry is broken spontaneously.

A non-abelian vector superfield $$V$$ transforms under supersymmetric gauge transformations as $e^{2V} \to e^{i\bar\Delta} e^{2V} e^{-i\Delta} ~,$ where $$\Delta$$ is a chiral superfield valued in the Lie algebra of the gauge group, so that $$e^{i\Delta}$$ is a chiral superfield valued in the gauge group itself. $$\bar\Delta = \Delta^\dagger$$ is the conjugate antichiral superfield.

Check that the gaugino superfield $W_\alpha = -\frac{1}{8} \bar D^2(e^{-2V}D_\alpha e^{2V}) ~,$ transforms as $$W_\alpha \to e^{i\Delta} W_\alpha e^{-i \Delta}$$ under supersymmetric gauge transformations.

Derive the component field expansion of $$W_\alpha$$ in the Wess-Zumino gauge $W_{\text{WZ}}{}_\alpha(y,\theta) = - i \lambda_\alpha(y) + \theta_\alpha D(y) + i (\sigma^{\mu\nu}\theta)_\alpha F_{\mu\nu}(y) + (\theta\theta)(\sigma^\mu D_\mu \bar \lambda(y))_\alpha ~,$ where $F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y) - i [A_\mu(y),A_\nu(y)] ~, \qquad D_\mu \bar\lambda(y) = \hat\partial_\mu \bar\lambda(y) - i [A_\mu(y),\bar\lambda(y)] ~,$ are the non-abelian field strength and the gauge covariant derivative in the adjoint representation of $$\bar\lambda$$ respectively. Recall that $$\hat\partial_\mu$$ just denotes the partial derivative with respect to $$y^\mu$$.

Under supersymmetric gauge transformations $\begin{split} W_\alpha & \to -\frac{1}{8}\bar D^2(e^{i\Delta} e^{-2V} e^{-i\bar\Delta} D_\alpha (e^{i\bar\Delta} e^{2V} e^{-i \Delta})) = -\frac{1}{8} e^{i\Delta} \bar D^2 (e^{-2V} D_\alpha(e^{2V} e^{-i\Delta})) \\ & \qquad = -\frac{1}{8} e^{i\Delta} \bar D^2(e^{-2V}(D_\alpha e^{2V})e^{-i\Delta}) = -\frac{1}{8} e^{i\Delta} \bar D^2(e^{-2V}(D_\alpha e^{2V}))e^{-i\Delta} \\ & \qquad = e^{i\Delta} W_\alpha e^{-i\Delta} ~, \end{split}$ where we have used that $$\bar D_{\dot\alpha} e^{\pm i\Delta} = D_\alpha e^{\pm i\bar\Delta} = 0$$ and $\bar D^2 D_\alpha e^{-i\Delta} = -\bar D^{\dot\alpha} \bar D_{\dot\alpha} D_\alpha e^{-i\Delta} = -\bar D^{\dot\alpha}\{\bar D_{\dot\alpha}, D_\alpha\} e^{-i\Delta} = -2i (\sigma^\mu)_{\alpha\dot\alpha} \bar D^{\dot\alpha} \partial_\mu e^{-i\Delta} = 0 ~,$ since $$\Delta$$ and $$\bar\Delta$$ are chiral and antichiral superfields respectively. Therefore, we see that $$W_\alpha$$ is gauge covariant.

To derive the component field expansion of $$W_\alpha$$ in the Wess-Zumino gauge, we start by expanding out the exponentials $\begin{split} W_{\text{WZ}}{}_\alpha & = - \frac{1}{8} \bar D^2 ((1-2 V_{\text{WZ}} + 2 V_{\text{WZ}}^2)D_\alpha(1+2 V_{\text{WZ}} + 2 V_{\text{WZ}}^2)) ~, \\ & =- \frac{1}{4} \bar D^2(D_\alpha V_{\text{WZ}} -2 V_{\text{WZ}} D_\alpha V_{\text{WZ}} + D_\alpha (V_{\text{WZ}}^2)) \\ & =- \frac{1}{4} \bar D^2(D_\alpha V_{\text{WZ}} +[ D_\alpha V_{\text{WZ}},V_{\text{WZ}}]) \end{split}$ where we have used that $$V_{\text{WZ}}^3 = (D_\alpha V_{\text{WZ}}) V_{\text{WZ}}^2 = V_{\text{WZ}} (D_\alpha V_{\text{WZ}}) V_{\text{WZ}} = V_{\text{WZ}}^2 (D_\alpha V_{\text{WZ}}) = 0$$. In the Wess-Zumino gauge $V_{\text{WZ}}(x,\theta,\bar\theta) = (\theta\sigma^\mu\bar\theta) A_\mu(x) + i (\theta\theta)(\bar\theta\bar\lambda(x)) - i (\bar\theta\bar\theta)(\theta\lambda(x)) + \frac{1}{2}(\theta\theta)(\bar\theta\bar\theta) D(x) ~.$ In terms of the coordinates $$(y,\theta,\bar\theta)$$ we have $$D_\alpha = \hat\partial_\alpha + 2i(\sigma^\mu\bar\theta)_\alpha \hat\partial_\mu$$, $$\bar D_{\dot\alpha} = \hat{\bar\partial}_{\dot\alpha}$$ and $\begin{split} V_{\text{WZ}}(y,\theta,\bar\theta) & = (\theta\sigma^\mu\bar\theta) \big(A_\mu(y)-i (\theta\sigma^\nu\bar\theta)\hat\partial_\nu A_\mu\big) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) D(y) \\ & = (\theta\sigma^\mu\bar\theta) A_\mu(y) + i (\theta\theta) (\bar\theta\bar\lambda(y)) - i (\bar\theta\bar\theta)(\theta\lambda(y)) + \frac{1}{2} (\theta\theta) (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) ~. \end{split}$ Therefore, $\begin{split} D_\alpha V_{\text{WZ}}(y,\theta,\bar\theta) & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + 2i (\sigma^\nu\bar\theta)_\alpha(\theta\sigma^\mu\bar\theta) \hat\partial_\nu A_\mu(y) + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) - 2 (\sigma^\mu\bar\theta)_\alpha (\theta\theta) (\bar\theta\hat\partial_\mu \bar\lambda(y)) \\ & \quad - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) \big(D(y)-i \hat\partial_\mu A^\mu(y)\big) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^\nu\bar\sigma^\mu\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + \frac{i}{2} ((2\eta^{\mu\nu}\mathbf{1}- 4\sigma^{\mu\nu})\theta)_{\alpha} (\bar\theta\bar\theta) \hat\partial_\nu A_\mu(y) -i \theta_\alpha (\bar\theta\bar\theta) \hat\partial_\mu A^\mu(y) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) \\ & = (\sigma^\mu\bar\theta)_\alpha A_\mu(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} (\bar\theta\bar\theta) (\hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)) \\ & \quad + 2i \theta_\alpha (\bar\theta\bar\lambda(y)) + (\theta\theta)(\bar\theta\bar\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} - i (\bar\theta\bar\theta)\lambda_\alpha(y) + \theta_\alpha (\bar\theta\bar\theta) D(y) ~, \end{split}$ where we recall that \begin{aligned} \sigma^\mu\bar\sigma^\nu - \sigma^\nu\bar\sigma^\mu & = 4\sigma^{\mu\nu} ~, \qquad & \bar\sigma^\mu\sigma^\nu - \bar\sigma^\nu\sigma^\mu & = 4\bar\sigma^{\mu\nu} ~, \\ \sigma^\mu\bar\sigma^\nu + \sigma^\nu\bar\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~, \qquad & \bar\sigma^\mu\sigma^\nu + \bar\sigma^\nu\sigma^\mu & = 2\eta^{\mu\nu}\mathbf{1}~. \end{aligned} We also have $\begin{split} [D_\alpha V_{\text{WZ}},V_{\text{WZ}}] (y,\theta,\bar\theta) & = [(\sigma^\mu\bar\theta)_\alpha A_\mu(y)+ 2i \theta_\alpha (\bar\theta\bar\lambda(y)),(\theta\sigma^\nu\bar\theta) A_\nu(y) + i (\theta\theta)(\bar\theta\bar\lambda(y))] \\ & = (\sigma^\mu\bar\theta)_\alpha(\theta\sigma^\nu\bar\theta) [A_\mu(y),A_\nu(y)] \\ & \quad + i (\sigma^\mu\bar\theta)_\alpha (\theta\theta) [A_\mu(y),(\bar\theta\bar\lambda(y))] + 2i (\theta\sigma^\nu\bar\theta)\theta_\alpha[ (\bar\theta\bar\lambda(y)), A_\nu(y) ] \\ & = \frac{1}{2}(\sigma^\mu\bar\sigma^\nu\theta)_\alpha (\bar\theta\bar\theta) [A_\mu(y),A_\nu(y)] \\ & \quad + i (\sigma^\mu\bar\theta)_\alpha (\theta\theta) [A_\mu(y),(\bar\theta\bar\lambda(y))] - i (\sigma^\nu\bar\theta)_\alpha(\theta\theta)[ (\bar\theta\bar\lambda(y)), A_\nu(y) ] \\ & = (\sigma^{\mu\nu}\theta)_\alpha (\bar\theta\bar\theta) [A_\mu(y),A_\nu(y)] - i (\theta\theta)(\bar\theta\bar\theta)[A_\mu(y), (\sigma^\mu\bar\lambda(y))_\alpha] ~. \end{split}$ Now using that $$\hat{\bar\partial}^2(\bar\theta\bar\theta) = -4$$ we find $\begin{split} W_\alpha(y,\theta,\bar\theta) & = -\frac{1}{4} \bar D^2 (D_\alpha V_{\text{WZ}}+[D_\alpha V_{\text{WZ}}, V_{\text{WZ}}])(y,\theta,\bar\theta) \\ & = - i \lambda_\alpha(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} (\hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)) + \theta_\alpha D(y) + (\theta\theta) (\sigma^\mu\hat\partial_\mu\bar\lambda(y))_{\alpha} ~. \\ & \quad + (\sigma^{\mu\nu}\theta)_\alpha [A_\mu(y),A_\nu(y)] - i (\theta\theta)[A_\mu(y), (\sigma^\mu\bar\lambda(y))_\alpha] \\ & = - i \lambda_\alpha(y) + i (\sigma^{\mu\nu}\theta)_{\alpha} F_{\mu\nu}(y) + \theta_\alpha D(y) + (\theta\theta) (\sigma^\mu D_\mu\bar\lambda(y))_{\alpha} ~, \end{split}$ where $$F_{\mu\nu}(y) = \hat\partial_\mu A_\nu(y) - \hat\partial_\nu A_\mu(y)- i [A_\mu(y),A_\nu(y)]$$ is the non-abelian field strength and $$D_\mu\bar\lambda(y) = \hat\partial_\mu \bar\lambda(y) - i [A_\mu(y),\bar\lambda(y)]$$ is the gauge covariant derivative in the adjoint representation of $$\bar\lambda$$.

# 7 Spontaneous supersymmetry breaking

The O’Raifeartaigh model is a Wess-Zumino model of three chiral superfields with superpotential $W(X,Y,Z) = Y(a^2 - X^2) + b XZ ~,$ where $$a$$ and $$b$$ are non-zero complex parameters. Show that supersymmetry is spontaneously broken. Assuming $$2|a|^2<|b|^2$$, find the global minimum (or minima) of the scalar potential and determine the vacuum expectation values of the scalar potential.

For the superpotential $W(X,Y,Z) = Y(a^2 - X^2) + b XZ ~,$ we have $\begin{split} \bar F_X = \frac{\partial W}{\partial X}\Big|_{X = \phi^X,Y=\phi^Y,Z=\phi^Z} & = - 2 \phi^X \phi^Y +b \phi^Z ~, \\ \bar F_Y = \frac{\partial W}{\partial Y}\Big|_{X = \phi^X,Y=\phi^Y,Z=\phi^Z} & = a^2 - (\phi^X)^2 ~, \\ \bar F_Z = \frac{\partial W}{\partial Z}\Big|_{X = \phi^X,Y=\phi^Y,Z=\phi^Z} & = b \phi^X ~. \end{split}$ $$\bar F_Z = 0$$ implies that $$\phi^X = 0$$. However, this implies that $$\bar F_Y = a^2 \neq 0$$. Therefore, supersymmetry is spontaneously broken.

The scalar potential is given by $\begin{split} V(\phi^X,\phi^Y,\phi^Z,\bar\phi_X,\bar\phi_Y,\bar\phi_Z) & = \bar F_X F^X + \bar F_Y F^Y + \bar F_Z F^Z ~. \end{split}$ The stationary points of the potential solve $\begin{split} \frac{\partial V}{\partial \phi^X} & = -2 \phi^Y(-2 \bar\phi_X\bar\phi_Y + \bar b \bar\phi_Z) - 2 \phi^X(\bar a^2 - (\bar\phi_X)^2) + b \bar b \bar\phi_X = 0 ~, \\ \frac{\partial V}{\partial \phi^Y} & = -2 \phi^X(-2 \bar\phi_X\bar\phi_Y + \bar b \bar\phi_Z) = 0~, \\ \frac{\partial V}{\partial \phi^Z} & = b(-2 \bar\phi_X\bar\phi_Y + \bar b \bar\phi_Z) = 0~. \end{split}$ From the third equation we find that $\phi^Z = \frac{2\phi^X \phi^Y}{b} ~.$ The second equation is then solved, while the first equation becomes $2 \bar a^2 \phi^X = (2 |\phi^X|^2 + |b|^2) \bar\phi_X ~.$ This implies $2 |a|^2 |\phi^X| = (2 |\phi^X|^2 + |b|^2 ) |\phi^X| \qquad \Rightarrow \qquad (2 |\phi^X|^2 + |b|^2 -2 |a|^2) |\phi^X| = 0~.$ Given $$|\phi^X|^2 > 0$$ and we assume that $$2|a|^2 < |b|^2$$, the only solution to this equation is $$|\phi^X| =0$$, which implies that $$\phi^X = 0$$. It follows that the stationary points of the potential are $\langle\phi^X\rangle = \langle\phi^Z\rangle = 0 ~, \qquad \langle\phi^Y\rangle ~\text{arbitrary} ~.$ Therefore, $$\langle\bar F_X\rangle = \langle\bar F_Z\rangle = 0$$ and $$\langle\bar F_Y\rangle = a^2$$, hence the vacuum expectation value of the scalar potential is $\langle V \rangle = |a|^4 ~.$ Since the scalar potential is bounded below and $$\langle V\rangle$$ takes the same value at all the stationary points, it follows that the stationary points are minima, i.e. stable vacua. Note that $$\phi^Y$$ is a flat direction of the potential.

The Polonyi model is a Wess-Zumino model of a single chiral superfield $$X$$ with linear superpotential $W(X) = f X ~,$ where $$f$$ is a non-zero complex parameter. The Polonyi model can be modified by adding a small cubic term to the superpotential $W_\epsilon (X) = f X + \epsilon \frac{\lambda}{3} X^3 ~, \qquad 0 < \epsilon \ll 1 ~,$ where $$\lambda$$ is a non-zero complex parameter. Show that supersymmetry is broken spontaneously in the Polonyi model and compute the mass spectrum of bosons and fermions. Show that supersymmetry is restored in the modified Polonyi model and that $$\langle \phi^X \rangle = 0$$ is an unstable vacuum, i.e. a stationary point of the potential that is not a minimum. What happens to the supersymmetric vacua and to the scalar potential of the modified Polonyi model in the limit $$\epsilon \to 0$$?

For the superpotential of the Polonyi model $W(X) = f X ~,$ we have $\bar F_X = \frac{\partial W}{\partial X}\Big|_{X = \phi^X} = f ~.$ Since $$\bar F_X \neq 0$$ supersymmetry is spontaneously broken. The action of the Wess-Zumino model in terms of component fields is given by $\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi_X \partial^\mu\phi^X - i\bar\psi_X \bar\sigma^\mu \partial_\mu \psi^X \\ & \hspace{50pt} - W'(\phi^X) \bar W'(\bar\phi_X) -\frac{1}{2}W''(\phi^X)\psi^X\psi^X -\frac{1}{2}\bar W''(\bar\phi_X)\bar\psi_X\bar\psi_X \big) ~. \end{split}$ Using that $$W'(\phi^X) = f$$ and $$W''(\phi^X) = 0$$ the action becomes $\begin{split} \mathcal{S}& = \int d^4x \, \big(\partial_\mu\bar\phi_X \partial^\mu\phi^X - i\bar\psi_X \bar\sigma^\mu \partial_\mu \psi^X - f\bar f \big) ~. \end{split}$ Therefore, we can straightforwardly read off that the complex scalar and Majorana fermion are both massless.

For the superpotential of the modified Polonyi model $W_\epsilon (X) = f X + \epsilon \frac{\lambda}{3} X^3 ~, \qquad 0 < \epsilon \ll 1 ~,$ we have $\bar F_X = \frac{\partial W}{\partial X}\Big|_{X = \phi^X} = f + \epsilon \lambda (\phi^X)^2 ~.$ Therefore, we can solve $$\bar F_X = 0$$ and we have the supersymmetric vacuum $\langle \phi^X \rangle = \sqrt{-\frac{f}{\epsilon\lambda}} ~.$ The scalar potential is given by $V(\phi^X,\bar\phi_X) = \bar F_X F^X = (f + \epsilon \lambda (\phi^X)^2 )(\bar f + \epsilon \bar \lambda (\bar \phi_X)^2 ) ~.$ The stationary points of the potential solve $\frac{\partial V}{\partial \phi^X} = 2\epsilon \lambda \phi^X(\bar f + \epsilon \bar \lambda (\bar \phi_X)^2 ) = 0~.$ In addition to the supersymmetric vacuum, we have the additional stationary point $\langle \phi^X \rangle = 0 ~.$ Setting $$\phi^X = \frac{1}{\sqrt{2}} (\varphi_r^X + i \varphi_i^X)$$, the Hessian matrix at this stationary point is $\begin{pmatrix} \frac{\partial^2 V}{\partial \varphi_r^X\partial \varphi_r^X } & \frac{\partial^2 V}{\partial \varphi_r^X\partial \varphi_i^X } \\ \frac{\partial^2 V}{\partial \varphi_i^X\partial \varphi_r^X } & \frac{\partial^2 V}{\partial \varphi_i^X\partial \varphi_i^X } \end{pmatrix} \Bigg|_{ \varphi_r^X = \varphi_i^X = 0} = \begin{pmatrix} \epsilon(\bar f\lambda+f\bar\lambda) & i\epsilon(\bar f\lambda-f\bar\lambda) \\ i\epsilon(\bar f\lambda-f\bar\lambda) & -\epsilon(\bar f\lambda+f\bar\lambda) \end{pmatrix} ~,$ which has eigenvalues $$\pm 2 \epsilon |f| |\lambda|$$. Since one of these eigenvalues is negative it follows that the vacuum $$\langle \phi^X \rangle = 0$$ is unstable, i.e. it is not a minimum of the potential. As $$\epsilon \to 0$$, the supersymmetric vacuum $$\langle \phi^X \rangle = \sqrt{-\frac{f}{\epsilon\lambda}} \to \infty\sqrt{-\frac{f}{\lambda}}$$, while the scalar potential becomes a constant $V(\phi^X,\bar\phi_X) = (f + \epsilon \lambda (\phi^X)^2 )(\bar f + \epsilon \bar \lambda (\bar \phi_X)^2 ) \to f \bar f ~.$

Consider massive SQED with matter fields $$Q$$ of charge 1 and $$\tilde{Q}$$ of charge $$-1$$, and a non-zero FI term. Assume that the complex mass $$m \neq 0$$ and that the FI parameter $$\xi > 0$$. Show that $$F^Q$$, $$F^{\tilde{Q}}$$ and $$D$$ cannot be set to zero at the same time and explain why this implies that supersymmetry is broken spontaneously.

Let $$X = \frac{g^2\xi}{|m|^2}$$. Find the global minimum of the scalar potential and determine the vacuum expectation values $$\langle Q\rangle$$ and $$\langle\tilde{Q}\rangle$$, up to gauge freedom, as functions of $$m$$, $$g$$ and $$X$$. For $$X \leq 1$$ and $$X > 1$$ determine whether D-term or F-term spontaneous supersymmetry breaking occurs, or both.

Calculate the mass spectra of bosons and fermions and identify the Goldstino, i.e. the massless Goldstone fermion associated to spontaneously broken supersymmetry, in the two regimes $$X \leq 1$$ and $$X > 1$$.

The action of massive QED with matter fields $$Q$$ of charge 1 and $$\tilde{Q}$$ of charge $$-1$$, and a non-zero FI term is $\begin{split} \mathcal{S}& = \mathop{\mathrm{Im}}\Big(\int d^4 x d^2\theta \, \frac{\tau g^2}{8\pi} W^\alpha W_\alpha \Big) + \int d^4x d^2\theta d^2\bar\theta \, \big(\bar Q e^{2gV} Q + \bar{\tilde{Q}} e^{-2gV} \tilde{Q} - 2g \xi V\big) \\ & \quad + \int d^4 x d^2\theta \, m \tilde{Q} Q + \int d^4x d^2\bar\theta \, \bar m \bar{\tilde{Q}} \bar{Q} ~, \end{split}$ where $$m \neq 0$$ is the complex mass parameter, $$\tau = \frac{4\pi i}{g^2} + \frac{\theta}{2\pi}$$ is the complexified gauge coupling and $$\xi > 0$$ is the real Fayet-Iliopoulos (FI) parameter.

The component field expansion of the chiral superfields $$Q$$ and $$\tilde{Q}$$ is $Q = \phi^Q + \sqrt{2} \theta\psi^Q - \theta\theta F^Q ~, \qquad \tilde Q = \phi^{\tilde Q} + \sqrt{2} \theta\psi^{\tilde Q} - \theta\theta F^{\tilde Q} ~,$ The scalar potential is then given by $V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = \bar F_Q F^Q + \bar F_{\tilde{Q}} F^{\tilde{Q}} + \frac{1}{2} D^2 ~,$ where \begin{aligned} F^Q & = \partial_{\bar {\phi}_Q}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{\tilde{Q}} ~, \qquad & \bar F_Q & = \partial_{\phi^Q}(m \phi^{\tilde Q}\phi^{Q}) = m \phi^{\tilde{Q}} ~, \\ F^{\tilde{Q}} & = \partial_{\bar{\phi}_{\tilde{Q}}}(\bar m \bar\phi_{\tilde{Q}} \bar\phi_Q) = \bar m \bar\phi_{Q} ~, \qquad & \bar F_{\tilde{Q}} & = \partial_{\phi^{\tilde{Q}}}(m \phi^{\tilde Q} \phi^{Q}) = m \phi^Q ~, \\ D & = -g (\bar \phi_Q \phi_Q - \bar\phi_{\tilde{Q}} \phi_{\tilde Q} - \xi) ~. && \end{aligned} Therefore, $V(\phi^Q, \phi^{\tilde{Q}}, \bar\phi_Q, \bar\phi_{\tilde{Q}}) = |m|^2(|\phi^Q|^2 + |\phi^{\tilde Q}|^2) + \frac{g^2}{2}(|\phi^Q|^2 - |\phi^{\tilde Q}|^2 - \xi)^2 ~.$

To determine the moduli space of supersymmetric vacua we solve $$F^Q = F^{\tilde{Q}} = D = 0$$ and quotient out by the $$\mathrm{U}(1)$$ gauge symmetry \begin{aligned} Q & \to e^{i\alpha(x)} Q ~, \qquad & \bar Q & \to e^{-i\alpha(x)} \bar Q ~, \\ \tilde{Q} & \to e^{-i\alpha(x)} \tilde{Q} ~, \qquad & \bar{\tilde{Q}} & \to e^{i\alpha(x)} \bar{\tilde{Q}} ~. \end{aligned} For $$m \neq 0$$ and $$\xi > 0$$, we have $$F^Q = \bar m \bar \phi_{\tilde{Q}}$$, $$F^{\tilde{Q}} = \bar m \bar \phi_Q$$ and $$D = -g (|\phi^Q|^2 - |\phi^{\tilde Q}|^2 -\xi)$$. Solving $$F^Q = F^{\tilde{Q}} = 0$$ gives $$\phi^Q = \phi^{\tilde{Q}} = 0$$, for which $$D = g\xi \neq 0$$. Therefore, there are no supersymmetric vacua and supersymmetry is broken spontaneously. This follows since the vacuum energy is non-zero, while for supersymmetric vacua it should vanish. Furthermore, the supersymmetry transformations of the vacuum expectation values of the fermions are $\delta_{\varepsilon,\bar\varepsilon} \langle \psi^Q \rangle \sim \varepsilon \langle F^Q \rangle ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \langle \psi^{\tilde Q} \rangle \sim \varepsilon \langle F^{\tilde Q} \rangle ~, \qquad \delta_{\varepsilon,\bar\varepsilon} \langle \lambda \rangle \sim \varepsilon \langle D \rangle ~,$ hence we see explicitly that the vacuum is supersymmetric if and only if $$\langle F^Q \rangle = \langle F^{\tilde Q} \rangle = \langle D \rangle = 0$$.

The stationary points of the potential solve $\begin{split} \frac{\partial V}{\partial \phi^Q} & = \big(|m|^2 + g^2 (|\phi^Q|^2 - |\phi^{\tilde{Q}}|^2 - \xi)\big)\bar\phi_Q = 0 ~, \\ \frac{\partial V}{\partial \phi^{\tilde Q}} & = \big(|m|^2 - g^2 (|\phi^Q|^2 - |\phi^{\tilde{Q}}|^2 - \xi)\big)\bar\phi_{\tilde Q} = 0 ~. \end{split}$ Given that $$m\neq 0$$, let us define $U = \frac{g \phi^Q}{m} ~, \qquad \tilde U = \frac{g\phi^{\tilde Q}}{m} ~, \qquad X = \frac{g^2 \xi}{|m|^2} > 0 ~,$ so that these equations simplify to $(1 + |U|^2 - |\tilde{U}|^2 - X)\bar U = 0 ~, \qquad (1 - |U|^2 + |\tilde{U}|^2 + X) \bar{\tilde{U}} = 0 ~.$ Depending on the value of $$X$$, there are either one or two solutions to these equations. These are solution (i), $$U = \tilde U = 0$$, which leads to the vacuum expectation values $\begin{gathered} \langle \phi^Q \rangle = \langle \phi^{\tilde Q} \rangle = 0 ~, \qquad \langle F^Q \rangle = \langle F^{\tilde Q} \rangle = 0 ~, \qquad \langle Q \rangle = \langle \tilde{Q} \rangle = 0 ~, \\ \langle D \rangle = \frac{|m|^2}{g} X ~, \qquad \langle V\rangle = \frac{|m|^4}{2g^2} X^2 ~, \end{gathered}$ and, for $$X\geq 1$$, solution (ii), $$|U| = \sqrt{X-1}$$, $$\tilde U = 0$$, which leads to the vacuum expectation values $\begin{gathered} \langle \phi^Q \rangle = \frac{m e^{i\alpha(x)}}{g} \sqrt{X-1} ~, \qquad \langle \phi^{\tilde Q} \rangle = 0 ~, \qquad \langle F^Q \rangle = 0 ~, \qquad \langle F^{\tilde Q} \rangle = \frac{|m|^2 e^{-i\alpha(x)}}{g} \sqrt{X-1} ~, \\ \langle Q \rangle = \frac{m e^{i\alpha(x)}}{g} \sqrt{X-1} ~, \qquad \langle \tilde Q \rangle = -\theta\theta \frac{|m|^2 e^{-i\alpha(x)}}{g} \sqrt{X-1} ~, \\ \langle D \rangle = \frac{|m|^2}{g} ~, \qquad \langle V \rangle = \frac{|m|^4}{2g^2} (2X - 1) ~, \end{gathered}$ where the phase $$\alpha(x)$$ parametrises the $$\mathrm{U}(1)$$ gauge freedom. This is only a solution for $$X\geq 1$$ since $$|U| \in \mathbb{R}$$ and $$\sqrt{X-1}$$ is imaginary for $$X<1$$. For $$X=1$$ solution (ii) coincides with solution (i). In principle there could be two other solutions to these equations $\begin{gathered} U = 0 ~, \qquad |\tilde U| = \sqrt{-X-1} ~, \\ |U|^2 - |\tilde U|^2 = X-1 = X+1 ~. \end{gathered}$ The first of these is not a solution since $$|\tilde U| \in \mathbb{R}$$ and $$X > 0$$ implies that $$\sqrt{-X-1}$$ is imaginary, while the second immediately leads to a contradiction. When computing the vacuum expectation values of $$Q$$ and $$\tilde Q$$ above, we have used that the vacuum expectation values of fermions vanish by the Poincaré invariance of the vacuum.

For $$X<1$$ the only solution is solution (i), which is the global minimum. For $$X>1$$ we have that $(X-1)^2 > 0 \qquad \Rightarrow \qquad X^2 > 2X-1 ~,$ hence the global minimum is given by solution (ii), while solution (i) is a local maximum. For $$X=1$$ the global minimum is given by both solutions, which coincide. It follows that for $$X\leq 1$$ we have D-term spontaneous supersymmetry breaking since $$\langle F^Q \rangle = \langle F^{\tilde Q} \rangle = 0$$ and $$\langle D \rangle \neq 0$$, while for $$X > 1$$ we have both F-term and D-term spontaneous supersymmetry breaking.

Using the results of exercise 6.2 to compute the component field expansion of the action, expanding around a general vacuum, and extracting those terms quadratic in the fields, we find $\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} \big) \\ & \quad + \int d^4x \, \big( (\partial_\mu \bar\phi_Q + i g \langle \bar\phi_Q\rangle A_\mu ) (\partial^\mu \phi^Q - i g \langle \phi^Q\rangle A^\mu) + (\partial_\mu \bar\phi_{\tilde Q} - i g A_\mu \langle \bar\phi_{\tilde Q}\rangle) (\partial^\mu \phi^{\tilde Q} + i g A^\mu \langle \phi^{\tilde Q}\rangle) \big) \\ & \quad - \int d^4 x \, \frac{g^2}{2} \begin{pmatrix} \phi^Q & \phi^{\tilde Q} & \bar\phi_Q & \bar\phi_{\tilde Q} \end{pmatrix} M_{\text{scalar}} \begin{pmatrix} \phi^Q & \phi^{\tilde Q} & \bar\phi_Q & \bar\phi_{\tilde Q} \end{pmatrix}^\mathrm{t} \\ & \quad + \int d^4 x\, \big( - i \bar\lambda \bar\sigma^\mu \partial_\mu\lambda -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\tilde Q} \bar\sigma^\mu \partial_\mu \psi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( + i g \sqrt{2} \langle \bar\phi_Q\rangle \lambda \psi^Q - i g \sqrt{2} \langle \bar\phi_{\tilde Q}\rangle \lambda \psi^{\tilde Q} - i g \sqrt{2} \langle \phi^Q\rangle \bar \lambda \bar \psi_Q + i g \sqrt{2} \langle \phi^{\tilde Q}\rangle \bar \lambda \bar \psi_{\tilde Q} \\ & \hspace{300pt} - m \psi^Q \psi^{\tilde Q} - \bar m \bar\psi_Q \bar\psi_{\tilde Q} \big) ~, \end{split}$ where $M_{\text{scalar}} = \small \begin{pmatrix} \langle \bar\phi_Q\rangle^2 & - \langle \bar\phi_Q\rangle\langle \bar\phi_{\tilde Q} \rangle & |\langle \phi^Q\rangle|^2 +\frac{|m|^2 - g\langle D\rangle}{g^2} & - \langle \bar\phi_Q\rangle\langle \phi^{\tilde Q} \rangle \\ - \langle \bar\phi_Q\rangle\langle \bar\phi_{\tilde Q} \rangle & \langle \bar\phi_{\tilde Q}\rangle^2 & - \langle \bar\phi_{\tilde Q}\rangle\langle \phi^{Q} \rangle & |\langle \phi^{\tilde Q}\rangle|^2 +\frac{|m|^2 + g\langle D\rangle}{g^2} \\ |\langle \phi^Q\rangle|^2 +\frac{|m|^2 - g\langle D\rangle}{g^2} & - \langle \bar\phi_{\tilde Q}\rangle\langle \phi^{Q} \rangle & \langle \phi^Q\rangle^2 & - \langle \phi^Q\rangle \langle \phi^{\tilde Q}\rangle \\ - \langle \bar\phi_Q\rangle\langle \phi^{\tilde Q} \rangle & |\langle \phi^{\tilde Q}\rangle|^2 +\frac{|m|^2 + g\langle D\rangle}{g^2} & - \langle \phi^Q\rangle \langle \phi^{\tilde Q}\rangle & \langle \phi^{\tilde Q}\rangle^2 \end{pmatrix} \normalsize ~.$

For solution (i), i.e. $$X\leq 1$$, we have $$\langle \phi^Q \rangle = \langle \phi^{\tilde{Q}} \rangle = 0$$ and the quadratic action simplifies to $\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} + \partial_\mu \bar\phi_Q \partial^\mu \phi^Q + \partial_\mu \bar\phi_{\tilde Q}\partial^\mu \phi^{\tilde Q} \big) \\ & \quad + \int d^4x \, \big( - i \bar\lambda \bar\sigma^\mu \partial_\mu\lambda -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\tilde Q} \bar\sigma^\mu \partial_\mu \psi^{\tilde Q} - m \psi^Q \psi^{\tilde Q} - \bar m \bar\psi_Q \bar\psi_{\tilde Q} \big) ~. \end{split}$ Therefore, all the fields are massless apart from $$\psi^Q$$ and $$\psi^{\tilde Q}$$, which together form a Dirac fermion of mass $$|m|$$. The gaugino $$\lambda$$ is massless, hence it is the goldstino.

For solution (ii), i.e. $$X>1$$, we have $$\langle \phi^{\tilde{Q}} \rangle = 0$$, $$\langle D \rangle = \frac{|m|^2}{g}$$ and we use the $$\mathrm{U}(1)$$ gauge symmetry to fix $$\phi^Q = \bar\phi_Q = \frac{1}{\sqrt{2}}\varphi \in \mathbb{R}$$, hence $$\langle \phi^Q \rangle = \langle \bar\phi_Q\rangle = \frac{|m|}{g} \sqrt{X-1} \in \mathbb{R}$$. The quadratic action then becomes $\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} +\frac{1}{2}\partial_\mu \varphi \partial^\mu \varphi + \partial_\mu \bar\phi_{\tilde Q}\partial^\mu \phi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( |m|^2(X-1) A_\mu A^\mu - |m|^2 (X-1) \varphi^2 - 2|m|^2\bar\phi_{\tilde Q}\phi^{\tilde Q}\big) \\ & \quad + \int d^4x \, \big( - i \bar\lambda \bar\sigma^\mu \partial_\mu\lambda -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\tilde Q} \bar\sigma^\mu \partial_\mu \psi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( - \psi^Q (m \psi^{\tilde{Q}} - i |m| \sqrt{2} \sqrt{X-1} \lambda ) - \bar\psi_Q (\bar m \bar\psi_{\tilde Q} + i |m| \sqrt{2} \sqrt{X-1} \bar \lambda) \big) ~. \end{split}$ To diagonalise the fermionic mass terms in this quadratic action we define $\psi^{\tilde Q} = \frac{\sqrt{\frac{\bar m}{m}} \psi^{\hat Q} +\sqrt{2}\sqrt{X-1} \psi_{\text{G}}}{\sqrt{2X-1}} ~, \qquad \lambda = \frac{i(\sqrt{2}\sqrt{X-1}\psi^{\hat Q} - \sqrt{\frac{m}{\bar m}} \psi_{\text{G}})}{\sqrt{2X-1}}~,$ so that $\begin{split} \mathcal{S}_2 & = \int d^4x\,\big(-\frac{1}{4}F^{\mu\nu} F_{\mu\nu} + \frac{\theta g^2}{32\pi^2}\tilde F^{\mu\nu}F_{\mu\nu} +\frac{1}{2}\partial_\mu \varphi \partial^\mu \varphi + \partial_\mu \bar\phi_{\tilde Q}\partial^\mu \phi^{\tilde Q} \big) \\ & \quad + \int d^4 x\, \big( |m|^2(X-1) A_\mu A^\mu - |m|^2 (X-1) \varphi^2 - 2|m|^2\bar\phi_{\tilde Q}\phi^{\tilde Q}\big) \\ & \quad + \int d^4x \, \big( - i \bar\psi_{\text{G}} \bar\sigma^\mu \partial_\mu\psi_{\text{G}} -i \bar\psi_Q \bar\sigma^\mu \partial_\mu \psi^Q -i \bar\psi_{\hat Q} \bar\sigma^\mu \partial_\mu \psi^{\hat Q} \big) \\ & \quad + \int d^4 x\, \big( - |m|\sqrt{2X-1} \psi^Q \psi^{\hat Q} - |m| \sqrt{2X-1} \bar\psi_Q \bar \psi_{\hat Q} \big) ~. \end{split}$ This action describes a massive vector $$A_\mu$$ of mass $$\sqrt{2}|m|\sqrt{X-1}$$, a massive real scalar of mass $$\sqrt{2}|m|\sqrt{X-1}$$ and a massive complex scalar of mass $$\sqrt{2}|m|$$. Here, the vector $$A_\mu$$ has acquired its mass via the Higgs mechanism. Together $$\psi^Q$$ and $$\psi^{\hat Q}$$ form a Dirac fermion of mass $$|m|\sqrt{2X-1}$$, while $$\psi_{\text{G}}$$ is the massless goldstino.