# Problem in deriving equations of motion for a free chiral superfield

### Setting of the problem

We are in $$\mathcal{N}=1$$ SUSY and we are using superspace formalism in the set of coordinates $$(x^\mu,\theta_\alpha,\bar\theta_{\dot{\alpha}})$$ where $$\theta_\alpha$$ and $$\bar\theta_{\dot{\alpha}}$$, with $$\alpha,\dot{\alpha}=1,2$$, are Grassmann numbers. $$x^\mu$$, with $$\mu=0,1,2,3$$ are the usual Minkowski spacetime coordinates. I'm following Bertolini's notes on supersymmetry, and I use the same conventions. At the end I list the ones I use for the problem plus some definitions and results (labeled from (A) to (G)).

### The problem

Let's consider the lagrangian $$$$\tag{1} \mathcal{L}=\int\mathrm{d}^2\theta\mathrm{d}^2\bar\theta\,\bar\Phi\Phi \ ,$$$$ where $$\Phi$$ a chiral superfield, wich expansion reads $$\begin{multline} \tag{2}\Phi(x,\theta,\bar\theta)=\phi(x)+\sqrt{2}\theta\psi(x)+i\theta\sigma^\mu\bar\theta\partial_\mu\phi(x)-\theta\theta F(x) \\-\frac{i}{\sqrt2}\theta\theta\,\partial_\mu\psi(x)\sigma^\mu\bar\theta-\frac{1}{4}\theta\theta \,\bar{\theta}\bar{\theta}\square\phi(x) \ , \end{multline}$$ where $$\phi$$ and $$F$$ are scalar fields and $$\psi_\alpha$$ is a Weyl spinor. I am interested in working out the equations of motion for $$\Phi$$; two ways are possible:

1. Using the expansion (2) in (1), after some work one can find that $$\mathcal{L}=\partial_\mu\bar{\phi}\partial^\mu\phi+\frac{i}{2}\Big(\partial_\mu\psi\,\sigma^\mu\bar\psi-\psi\sigma^\mu\partial_\mu\bar\psi\Big)+\bar{F}F$$ up to total derivatives. From this I find the following equations of motion for $$\phi$$, $$\psi$$ and $$F$$, $$\square\phi=0\quad,\quad \partial_\mu\psi^\alpha\sigma^\mu_{\alpha\dot{\beta}}=0\quad,\quad F=0 \ . \tag{eom}$$

2. The lagrangian (1) is a constrained one, because of the chiral constraint $$\bar{D}_\alpha\Phi=0$$, so we can't derive directly from it the equations of motions by means of a variational principle. First, we have to rewrite (1) as $$\mathcal{L}=-\frac{1}{4}\int \mathrm{d}^2\bar\theta \bar\Phi D^2\Phi +\text{total derivative} \ ,$$ where $$D^2=D^\alpha D_\alpha$$ and we have used (A), (C) and the fact that $$D\bar\Phi=0$$. Now, varying with respect to $$\bar\Phi$$, we get the equations of motion for $$\Phi$$ $$$$D^2\Phi=0 \ .\tag{eom'}$$$$ Expanding $$\Phi$$ by means of (2) and $$D^2$$ by means of (A), one should recover (eom) from the latter one.

### My problem

Here comes my problem: I am unable to recover (eom) from (eom') as I should. I will now carefully expose my work.

1. Expansion of $$D^2$$. Using (A), \begin{align} D^2=D^\alpha D_{\alpha}&=\Big(-\partial^\alpha-i\bar\theta_{\dot{\lambda}}(\bar{\sigma}^\mu)^{\dot{\lambda}\alpha}\partial_\mu\Big)\Big(\partial_\alpha+i\sigma^\nu_{\alpha\dot{\beta}}\bar\theta^{\dot{\beta}}\partial_\nu\Big)\\ &=-\partial\partial-i\partial \sigma^\mu \bar\theta \, \partial_\mu-i\bar\theta\bar\sigma^\mu\partial \, \partial_\mu+\bar\theta\bar\sigma^\mu\sigma^\nu \bar\theta\, \partial_\mu\partial_\nu \ , \end{align} where I have defined $$\partial \sigma^\mu \bar\theta\equiv\partial^\alpha\sigma_{\alpha\dot{\beta}}^\mu\bar\theta^{\dot{\beta}}$$ and similarly for $$\bar\theta\bar\sigma^\mu\partial$$. Now I show that $$\partial \sigma^\mu \bar\theta=\bar\theta\bar\sigma^\mu\partial$$, in fact, \begin{align} \partial \sigma^\mu \bar\theta&=\partial^\alpha\sigma_{\alpha\dot{\beta}}^\mu\bar\theta^{\dot{\beta}}\\ &=-\partial_\lambda \epsilon^{\dot{\beta}\dot{\rho}}\epsilon^{\alpha\lambda}\sigma_{\alpha\dot{\beta}}^\mu\bar\theta_{\dot{\rho}}\\ &=-\partial_\lambda \epsilon^{\dot{\rho}\dot{\beta}}\epsilon^{\lambda\alpha}\sigma_{\alpha\dot{\beta}}^\mu\bar\theta_{\dot{\rho}}\\ &=-\partial_\lambda (\bar\sigma^\mu)^{\dot{\rho}\lambda}\bar\theta_{\dot{\rho}}\\ &=\bar\theta_{\dot{\rho}} (\bar\sigma^\mu)^{\dot{\rho}\lambda}\partial_\lambda \\ &= \bar\theta\bar\sigma^\mu\partial \ . \end{align} Using this result I can rewrite $$D^2=-\partial\partial-2i\bar\theta\bar\sigma^\mu\partial \, \partial_\mu+\bar\theta\bar\sigma^\mu\sigma^\nu \bar\theta\, \partial_\mu\partial_\nu \ .$$ Using (D) and the antisymmetry of $$\bar\sigma^{\mu\nu}$$, we get $$\tag{3} D^2=-\partial\partial-2i\bar\theta\bar\sigma^\mu\partial \, \partial_\mu+\bar\theta\bar\theta\, \square \ ,$$ where $$\square=\partial^\mu\partial_\mu$$ is the usual D'Alembertian.

2. Applying $$D^2$$ to $$\Phi$$. Now I apply $$D^2$$ separately at every term in (2). Then, summing the terms with the same field and putting each results equal to zero, we should recover (eom). My problem is primarly with the equation of motion for $$\psi$$. The terms in (2) involving $$\psi$$ are $$\sqrt{2}\theta\psi$$ and $$-\frac{i}{\sqrt2}\theta\theta\,\partial_\mu\psi\sigma^\mu\bar\theta$$. Applying $$D^2$$ to them using (3), we get \begin{align} D^2(\sqrt{2}\theta\psi)&=-i2\sqrt{2}\bar\theta\bar\sigma^\mu\partial\,\partial_\mu(\theta\psi)+\sqrt{2}\bar\theta\bar\theta\,\theta\square\psi\\ &=-i2\sqrt{2}\bar\theta\bar\sigma^\mu\partial_\mu\psi+\sqrt{2}\bar\theta\bar\theta\,\theta\square\psi \\ &=i2\sqrt{2}\partial_\mu\psi\sigma^\mu\bar\theta+\sqrt{2}\bar\theta\bar\theta\,\theta\square\psi \ , \tag{4} \end{align} where in the last equality we used (E), and \begin{align} D^2\Big(-\frac{i}{\sqrt2}\theta\theta\,\partial_\nu\psi\sigma^\nu\bar\theta\Big)&=\frac{i}{\sqrt{2}} \partial\partial(\theta\theta)\partial_\mu\psi\sigma^\mu\bar\theta-\sqrt{2}\bar\theta\bar\sigma^\mu\partial\,\partial_\mu\Big(\theta\theta\partial_\nu\psi\sigma^\nu\bar\theta\Big)\\ &=i2\sqrt{2}\partial_\mu\psi\sigma^\mu\bar\theta+\underbrace{\bigg[-\sqrt{2}\bar\theta_{\dot{\lambda}}(\bar\sigma^\mu)^{\dot{\lambda}\alpha}\,\partial_\alpha(\theta\theta)\,\partial_\mu\partial_\nu\psi\sigma^\nu\bar\theta\bigg] }_{\equiv A}\ , \end{align} where we used (B). Working out the second term on the RHS, noticing that $$\partial_\alpha(\theta\theta)=2\theta_\alpha$$, we get \begin{align} A&=-2\sqrt{2}\bar\theta_{\dot{\lambda}}(\bar\sigma^\mu)^{\dot{\lambda}\alpha}\theta_\alpha\,\partial_\mu\partial_\nu\psi^\rho\sigma^\nu_{\rho\dot{\beta}}\bar\theta^{\dot{\beta}}\\ &=-2\sqrt{2}\bar\theta_{\dot{\lambda}}\bar\theta^{\dot{\beta}}(\bar\sigma^\mu)^{\dot{\lambda}\alpha}\theta_\alpha\,\partial_\mu\partial_\nu\psi^\rho\sigma^\nu_{\rho\dot{\beta}}\\ &=-2\sqrt{2}\epsilon_{\dot{\lambda}\dot{\gamma}}\bar\theta^{\dot{\gamma}}\bar\theta^{\dot{\beta}}(\bar\sigma^\mu)^{\dot{\lambda}\alpha}\theta_\alpha\,\partial_\mu\partial_\nu\psi^\rho\sigma^\nu_{\rho\dot{\beta}} \ , \end{align} using (F),
\begin{align} \phantom{A}&=-\sqrt{2}\bar\theta\bar\theta \underbrace{\epsilon_{\dot{\lambda}\dot{\gamma}}\epsilon^{\dot{\gamma}\dot{\beta}}}_{\delta_{\dot{\lambda}}^{\dot{\beta}}}(\bar\sigma^\mu)^{\dot{\lambda}\alpha}\theta_\alpha\,\partial_\mu\partial_\nu\psi^\rho\sigma^\nu_{\rho\dot{\beta}}\\ &=-\sqrt{2}\bar\theta\bar\theta \underbrace{\epsilon_{\dot{\lambda}\dot{\gamma}}\epsilon^{\dot{\gamma}\dot{\beta}}}_{\delta_{\dot{\lambda}}^{\dot{\beta}}}(\bar\sigma^\mu)^{\dot{\lambda}\alpha}\theta_\alpha\,\partial_\mu\partial_\nu\psi^\rho\sigma^\nu_{\rho\dot{\beta}}\\ &=\sqrt{2}\bar\theta\bar\theta \partial_\mu\partial_\nu\psi^\rho\sigma^\nu_{\rho\dot{\beta}}(\bar\sigma^\mu)^{\dot{\beta}\alpha}\theta_\alpha\\ &=\sqrt{2}\bar\theta\bar\theta \partial_\mu\partial_\nu\psi\sigma^\nu\bar\sigma^\mu\theta \ , \end{align} using (G) and then (D'), \begin{align} A&=\sqrt{2}\bar\theta\bar\theta \,\theta\sigma^\mu\bar\sigma^\nu\partial_\mu\partial_\nu\psi \\ &=\sqrt{2}\bar\theta\bar\theta \,\theta\square\psi \ . \end{align} Putting all togheter, $$\tag{5} D^2\Big(-\frac{i}{\sqrt2}\theta\theta\,\partial_\nu\psi\sigma^\nu\bar\theta\Big)=2\sqrt{2}\partial_\mu\psi\sigma^\mu\bar\theta+\sqrt{2}\bar\theta\bar\theta \,\theta\square\psi \ .$$ The equations of motion for $$\psi$$ are given putting (4)+(5)=0, so we get $$i4\sqrt{2}\partial_\mu\psi\sigma^\mu\bar\theta+2\sqrt{2}\bar\theta\bar\theta\,\theta\square\psi=0 ,$$ that is, \begin{align} \partial_\mu\psi^\alpha\sigma^\mu_{\alpha\dot{\beta}}&=0 \ ,\tag{6a}\\ \square\psi_\alpha&=0 \tag{6b} \ , \end{align} independently. The first one is the right equation we should get, the one in (eom). The second one, however, is something more and I think it shouldn't come from $$D^2\Phi=0$$.

So, I don't know where I am wrong. It could be some mistake I made in the above calculations or something wrong in the results from (A) to (G) that I listed below. However, those results are also reported in Bertolini's notes (maybe not all of them and maybe some of them are written in another form, like for (F) that in the notes appears its conjugate companion $$\theta^\alpha\theta^\beta=-\frac{1}{2}\epsilon^{\alpha\beta}\theta\theta$$, so I did some calculations to derive them, but I am pretty confident that those are correct). I am a little bit desperate because I did the math over and over getting everytime the same result, so I seek for help.

### Conventions used

• Indices. $$\alpha,\beta,\gamma,...=1,2$$ and $$\dot{\alpha},\dot{\beta},\dot{\gamma},...= \dot{1}, \dot{2}$$ for (Weyl) spinor indices. $$\mu,\nu=0,1,2,3$$ are reserved for spacetime indices.

• Antisimmetric epsilon tensor. $$\epsilon_{12}=\epsilon_{\dot{1}\dot{2}}=-1$$ and $$\epsilon^{12}=\epsilon^{\dot{1}\dot{2}}=+1$$.

• Pauli matrices. $$\sigma^\mu=(\mathbf{1},\sigma^i)$$ and $$\bar\sigma^\mu=(\mathbf{1},-\sigma^i)$$ where $$\sigma^i$$ with $$i=1,2,3$$ are the usual Pauli matrices and $$\mathbf{1}$$ is the $$2\times 2$$ identity matrix. The index structure of $$\sigma^\mu$$ is $$\sigma^\mu_{\alpha\dot{\beta}}$$ while $$(\bar\sigma^\mu)^{\dot{\alpha}\beta}\equiv\epsilon^{\dot{\alpha}\dot{\lambda}}\epsilon^{\beta\rho}\sigma^\mu_{\rho\dot{\lambda}} \ .$$

• Spinors. Given two Weyl spinors $$\chi_\alpha$$ and $$\psi$$, we have $$(\chi_\alpha)^\dagger\equiv \bar{\chi}_{\dot{\alpha}}\quad,\quad(\chi^\alpha)^\dagger\equiv \bar{\chi}^{\dot{\alpha}}\quad,\quad \chi^\alpha\equiv\epsilon^{\alpha\beta}\chi_\beta\quad,\quad\chi_\alpha\equiv\epsilon_{\alpha\beta}\chi^\beta\quad,\quad \bar\chi^{\dot{\alpha}}\equiv\epsilon^{\dot{\alpha}\dot{\beta}}\bar\chi_{\dot{\beta}}\quad,\quad\bar\chi_{\dot{\alpha}}\equiv\epsilon_{\dot{\alpha}\dot{\beta}}\bar\chi^{\dot{\beta}}\\ \psi\chi\equiv\psi^\alpha\chi_\alpha\quad,\quad\bar{\psi}\bar{\chi}\equiv\bar{\psi}_{\dot{\alpha}}\bar{\chi}^{\dot{\alpha}} \quad,\quad \psi\sigma^\mu\bar\chi\equiv \psi^\alpha\sigma^\mu_{\alpha\dot{\beta}}\bar\chi^{\dot{\beta}}\quad,\quad \bar\psi\bar\sigma^\mu\chi\equiv \bar\psi_{\dot{\alpha}}(\bar\sigma^\mu)^{\dot{\alpha}\beta}\chi_{\beta}\\ \psi\sigma^\mu\bar\sigma^\nu\chi\equiv\psi^\alpha\sigma^\mu_{\alpha\dot{\beta}}(\bar\sigma^\nu)^{\dot{\beta}\lambda}\chi_\lambda\quad,\quad \bar\psi\bar\sigma^\mu\sigma^\nu\bar\chi\equiv \bar\psi_{\dot{\alpha}}(\bar\sigma^\mu)^{\dot{\alpha}\beta}\sigma^\nu_{\beta\dot{\lambda}}\bar\chi^{\dot{\lambda}} \ .$$

• Grassmann derivatives and integrals. We define derivative for Grassmann numbers, $${\partial}/{\partial\theta^\alpha}$$, $${\partial}/{\partial\bar\theta^{\dot{\alpha}}}$$ such that $$\frac{\partial}{\partial\theta^\alpha}\theta^\beta=\delta^\beta_\alpha\quad,\quad \frac{\partial}{\partial\theta^\alpha}\bar\theta^{\dot{\beta}}=0\quad,\quad\frac{\partial}{\partial\bar\theta^{\dot{\alpha}}}\bar\theta^{\dot{\beta}}=\delta^{\dot{\beta}}_{\dot{\alpha}}\quad,\quad \frac{\partial}{\partial\bar\theta^{\dot{\alpha}}}\theta^\beta=0 \ .$$ They also anticommute among themselves and satisfy $$\{{\partial}/{\partial\theta^\alpha},\theta^\beta\}=\delta^\beta_\alpha$$, $$\{{\partial}/{\partial\bar\theta^{\dot{\alpha}}},\bar\theta^{\dot{\beta}}\}=\delta^{\dot{\beta}}_{\dot{\alpha}}$$ when acting as an operator, while other combinations anticommutes. Grassmann integrals $$\int \mathrm{d}\theta^\alpha$$ and $$\int \mathrm{d}\bar\theta^{\dot{\alpha}}$$ are defined in such a way that $$\int \mathrm{d}\theta^\alpha = \frac{\partial}{\partial\theta^\alpha}\quad\text{and}\quad \int\mathrm{d}\bar\theta^{\dot{\alpha}} =\frac{\partial}{\partial\bar\theta^{\dot{\alpha}}} \ .$$ The following shortcuts are used, $$\partial_{\alpha}\equiv\frac{\partial}{\partial\theta^\alpha} \quad , \quad \bar{\partial}_{\dot{\alpha}}\equiv \frac{\partial}{\partial\bar{\theta}^{\dot{\alpha}}}\quad ,\quad \partial^\alpha\equiv -\epsilon^{\alpha\beta}\partial_{\beta}\quad,\quad\bar{\partial}^{\dot{\alpha}}\equiv -\epsilon^{\dot{\alpha}\dot{\beta}}\partial_{\dot{\beta}}\quad,\quad\partial\partial\equiv\partial^\alpha\partial_\alpha \\ \mathrm{d}^2\theta\equiv \frac{1}{2}\mathrm{d}\theta^1\mathrm{d}\theta^2\quad,\quad \mathrm{d}^2\bar\theta\equiv \frac{1}{2} \mathrm{d}\bar\theta^{\dot{2}}\mathrm{d}\bar{\theta}^{\dot{1}} \ .$$

• Superspace covariant derivatives We define $$$$\tag{A} D_\alpha=\partial_\alpha+i\sigma^\mu_{\alpha\dot{\beta}}\bar\theta^{\dot{\beta}}\partial_\mu \quad \text{and}\quad D^\alpha=\epsilon^{\alpha\beta} D_\beta=-\partial^\alpha-i\bar\theta_{\dot{\lambda}}(\bar{\sigma}^\mu)^{\dot{\lambda}\alpha}\partial_\mu$$$$ as the covariant derivatives for the superspace.

• Some results. $$\tag{B} \partial\partial(\theta\theta)=4 \ ,$$ $$\tag{C} \int \mathrm{d}^2\theta = \frac{1}{4} \partial\partial \ ,$$ $$\tag{D} \bar\sigma^\mu\sigma^\nu= \eta^{\mu\nu}\mathbf{1}+2\bar\sigma^{\mu\nu} \ ,$$ $$\tag{D'} \sigma^\mu\bar\sigma^\nu= \eta^{\mu\nu}\mathbf{1}+2\sigma^{\mu\nu} \ ,$$ where $$\eta^{\mu\nu}=\mathrm{diag}(+1,-1,-1,-1)$$ is the Minkowski metric, $$\bar\sigma^{\mu\nu}=\frac{1}{4}(\bar\sigma^\mu\sigma^\nu-\bar\sigma^\nu\sigma^\mu)$$ and $$\sigma^{\mu\nu}=\frac{1}{4}(\sigma^\mu\bar\sigma^\nu-\sigma^\nu\bar\sigma^\mu)$$ . $$\tag{E} \bar\theta\bar\sigma^\mu\psi=-\psi\sigma^\mu\bar\theta \ ,$$ $$\tag{F} \bar\theta^{\dot{\gamma}}\bar\theta^{\dot{\beta}}=\frac{1}{2}\bar\theta\bar\theta \epsilon^{\dot{\gamma}\dot{\beta}} \ ,$$ $$\tag{G} \psi\sigma^\nu\bar\sigma^\mu\theta=\theta\sigma^\mu\bar\sigma^\nu\psi \ .$$

• Why is this second equation, the one with the d'Alembert operator, problematic? This equation just follows from the first, so it anyway doesn't add any information. Mar 11 at 21:44
• @Lunaron Yes you are right! I didn't see that, thanks a lot. Mar 12 at 17:03

Thanks to @Lunaron, that pointed out that equation (6b) follows from (6a), in fact, starting from (6a), we multiply it with $$(\bar{\sigma}^\nu)^{\dot{\beta}\gamma}\partial_\nu$$ and sum over $$\dot{\beta}$$, so we get \begin{align} &\sigma^\mu_{\alpha\dot{\beta}}(\bar\sigma^\nu)^{\dot{\beta}\gamma}\partial_\mu\partial_\nu\psi^\alpha=0\\ \iff\quad&(\sigma^\mu\bar\sigma^\nu)^\gamma_\alpha \partial_\mu\partial_\nu\psi^\alpha=0 \ , \end{align} using now (D'), $$\eta^{\mu\nu}\delta^{\gamma}_\alpha \partial_\mu\partial_\nu\psi^\alpha=0 \ ,$$ i.e. $$\square \psi^\gamma=0 \ ,$$ that is exactly (6b).