3
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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
    Aug 7, 2020 at 13:15
  • $\begingroup$ Extended my answer, let me know if anything is unclear. $\endgroup$
    – Kosm
    Aug 7, 2020 at 19:09

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ updated again, clarified the spinor notation $\endgroup$
    – Kosm
    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
    Aug 7, 2020 at 22:29
  • $\begingroup$ @Stephan transposition is also useful to include, because we construct scalar Lagrangians after all. $\endgroup$
    – Kosm
    Aug 7, 2020 at 22:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.