I want to show that the free Wess-Zumino Lagrangian is invariant under a SUSY transformation, e.g. following this reference (section 3.1).

However, I have a hard time understanding the daggers and stars on the fields. In particular, with the fermionic fields. The fermion Lagrangian looks like this: $$ \mathcal L_\text{fermion}=\text{i} \psi^\dagger \bar\sigma^\mu \partial_\mu \psi. \tag{3.1.2} $$ In index notation, this should be $\text{i} \bar \psi_{\dot a} (\bar\sigma^\mu)^{\dot aa} \partial_\mu \psi_a$. If we start with $$ \delta\psi_a = -\text{i} (\sigma^\mu \epsilon^\dagger)_a \partial_\mu\phi+\epsilon_aF = -\text{i} (\sigma^\mu)_{a\dot a} \bar\epsilon^{\dot a} \partial_\mu\phi+\epsilon_aF \tag{3.1.15}, $$ then my guess for the conjugate transformation $\delta\bar\psi_{\dot a}$ would be: $$ \begin{align}\delta\bar\psi_{\dot a} &= \text{i} \big((\sigma^\mu)_{a\dot a} \bar\epsilon^{\dot a}\big )^* \partial_\mu\phi^* +\bar\epsilon_{\dot a}F^* \\&= \text{i} (\sigma^\mu)_{\dot aa} \epsilon^{a} \partial_\mu\phi^* +\bar\epsilon_{\dot a}F^* \\&= \text{i} \epsilon^{a}(\sigma^\mu)^T_{a\dot a} \partial_\mu\phi^* +\bar\epsilon_{\dot a}F^* \\& = \text{i} (\epsilon \sigma^{\mu T})_{\dot a}\partial_\mu\phi^* +\bar\epsilon_{\dot a}F^* \end{align}$$ where I used the fact that the Pauli matrices are hermitian (therefore, complex conjugation becomes a transpose). However, it should actually be $$ \delta\bar\psi_{\dot a} = \text{i} (\epsilon \sigma^{\mu})_{\dot a}\partial_\mu\phi^* +\bar\epsilon_{\dot a}F^* \tag{3.1.15}$$ i.e. without the transpose on the $\sigma^\mu$ matrix.

Where's my mistake? I feel like I don't really understand the spinor index notation.

For what it's worth, I'm using these assignments in order to use index notation, $$ \begin{align} \psi &\sim \psi_a \\ \bar\psi = \psi^* &\sim \bar\psi_{\dot a} \\ \psi^T &\sim \psi^a \\ \bar\psi^T=\psi^\dagger &\sim \psi^{\dot a} \end{align} $$ as well as contracting indices like ${}^a{}_a$ and ${}_{\dot a}{}^{\dot a}$.

I have already considered these questions [1, 2, 3, 4], but did not find a solution to my problem.

  • $\begingroup$ Having seen your component notation for Weyl fermions, I think it's not right too. I will update my answer a bit later. $\endgroup$
    – Kosm
    Commented Aug 7, 2020 at 13:15
  • $\begingroup$ Extended my answer, let me know if anything is unclear. $\endgroup$
    – Kosm
    Commented Aug 7, 2020 at 19:09

1 Answer 1


First, for fermion component notation in the Martin's textbook. Forget your notations for a while, and start from the beginning. For Weyl spinors, let me replace the dagger (h.c.) with the bar to avoid clutter (which is quite a common practice). This bar (or dagger) always accompanies the dotted indices, upper or lower, while undotted indices are always unbarred. Lower undotted index represents a left-handed column spinor, while upper undotted index represents a left-handed row spinor. Conversely, lower dotted index represents a right-handed row spinor, while upper dotted index - right-handed column spinor. Indices (as you have probably read) are raised and lowered by antisymmetric tensors ($\varepsilon_{ab}$ or $\varepsilon_{\dot{a}\dot{b}}$). To summarize: $$ \psi_a= \begin{pmatrix} \psi_{1} \\ \psi_{2} \end{pmatrix}~,~~~ \psi^a=(\psi_2,~-\psi_1)~, $$ and for the right-handed spinor $$ \bar{\chi}_\dot{a}=(\bar{\chi}_1,~\bar{\chi}_2),~~~ \bar{\chi}^\dot{a}= \begin{pmatrix} \bar{\chi}_{2} \\ -\bar{\chi}_{1} \end{pmatrix},~~~ $$ where I used $\varepsilon^{12}=\varepsilon_{21}=1$ (same for dotted and undotted indices) and minus one for switched indices. According to the textbook, we also have $(\psi_a)^\dagger=\bar{\psi}_\dot{a}$, where the bar is the same as the dagger in my notation, as I mentioned. Then, from the above definition of $\psi$, $$ \bar{\psi}_\dot{a}=(\psi_1^*,~\psi_2^*),~~~ \bar{\psi}^\dot{a}= \begin{pmatrix} \psi_2^*\\ -\psi_1^* \end{pmatrix}, $$ where $\dagger=*$ for each particular component.

As for the Pauli matrices, there is the following "bar" notation, where the bar accompanies the matrix components with upper indices: $$ \bar{\sigma}^{\dot{a}a}=\varepsilon^{\dot{a}\dot{b}}\varepsilon^{ab}\sigma_{b\dot{b}} $$ suppressing the spacetime index. The matrix components with lower indices are always unbarred.

Finally to the question itself, the quantity $(\sigma^{\mu}_{a\dot{a}}\bar{\epsilon}^\dot{a})$ is a spinor (component), so we are interested in the Hermitian conjugate ($\dagger$, or bar in my notation) instead of * (bar in your notation). So the quantity under question must be treated as $$ (\sigma^{\mu}_{a\dot{a}}\bar{\epsilon}^\dot{a})^\dagger=(\sigma^\mu\bar{\epsilon})_{a}^\dagger=(\epsilon\sigma^\mu)_\dot{a}=\epsilon^a\sigma^{\mu}_{a\dot{a}}. $$ The reason there is no bar on the $\sigma$ is that it has lower spinor indices, so by the convention, it is "unbarred".

In addition: in your derivation of $\delta\bar{\psi}_\dot{a}$ there should be Hermitian conjugation, i.e. in matrix notation $$ \delta\bar{\psi}=i(\sigma^\mu\bar{\epsilon})^\dagger\partial_\mu\phi^*+\bar{\epsilon}F^*= i(\epsilon{\sigma^\mu}^\dagger)\partial_\mu\phi^*+\bar{\epsilon}F^*~,\tag{1} $$ and because Pauli matrices are Hermitian $\sigma=\sigma^\dagger$, you have the expression (3.1.15). And by the way, the bar notation for Pauli matrices I wrote above gives the components of the transpose (or complex conjugate) Pauli matrix, but in equation (1) there is Hermitian conjugation, therefore no barred Pauli matrices in the final result. I think this is the main point.

  • $\begingroup$ updated again, clarified the spinor notation $\endgroup$
    – Kosm
    Commented Aug 7, 2020 at 14:46
  • $\begingroup$ Thanks for the detailed answer, really appreciate it! I noticed you use a dagger to switch between left- and right-handed, why is that? The (1/2,0) and (0,1/2) reps of the Lorentz group are related by a complex conjugation, so wouldn‘t a star be enough instead of the dagger? $\endgroup$
    – ersbygre1
    Commented Aug 7, 2020 at 22:29
  • $\begingroup$ @Stephan transposition is also useful to include, because we construct scalar Lagrangians after all. $\endgroup$
    – Kosm
    Commented Aug 7, 2020 at 22:39

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