# Peskin and Schroeder QFT problem 3.5 (c)

On Peskin and Schroeder's QFT book, page 75, on problem 3.5 (c) (supersymmetry),

The book first ask us to prove following Lagrangian is supersymmetric: \begin{aligned} &\mathcal{L}=\partial_\mu \phi_i^* \partial^\mu \phi_i+\chi_i^{\dagger} i \bar{\sigma} \cdot \partial \chi_i+F_i^* F_i \\ &+\left(F_i \frac{\partial W[\phi]}{\partial \phi_i}+\frac{i}{2} \frac{\partial^2 W[\phi]}{\partial \phi_i \partial \phi_j} \chi_i^T \sigma^2 \chi_j+\text { c.c. }\right), \end{aligned} \tag{A} To prove this, we need to show the variation of this Lagrangian can be arranged to a total divergence, and we need to use the relationship in 3.5 (a) \begin{aligned} \delta \phi &=-i \epsilon^T \sigma^2 \chi \\ \delta \chi &=\epsilon F+\sigma \cdot \partial \phi \sigma^2 \epsilon^* \\ \delta F &=-i \epsilon^{\dagger} \bar{\sigma} \cdot \partial \chi \end{aligned} \tag{B}

I can show that $$\delta\mathcal{L}$$ really can be arranged to a total divergence: we see that this Lagrangian is supersymmetric under (B).

What really troubles me is the book's following argument: "For the simple case $$n=1$$ and $$W=g\phi^3/3$$, write out the field equations for $$\phi$$ and $$\chi$$ (after elimination of $$F$$)"

If I use the above condition, I will get $$\mathcal{L}=\partial_\mu \phi^* \partial^\mu \phi+\chi^{\dagger} i \bar{\sigma}^\mu \partial_\mu \chi+F^* F+\left(g F \phi^2+i g\phi \chi^T \sigma^2 \chi+\text { c.c. }\right) \tag{C}.$$

But now why can we get the E.O.M of $$F$$, and further $$\phi$$ and $$\chi$$? Previously, we knew that we could get the E.O.M by variation of Lagrangian, but that needed that $$\delta F$$, $$\delta \phi$$ and $$\delta \chi$$ be independent.

But now, the situation is different : these quantities are not independent; they are connected via (B).

Actually, in my understanding, (B) already has the function of E.O.M, so we can get a total divergence after derivation, so I am really lost about the book's logic.

• Forget (B), apply EOM, and then apply (B): the two are unrelated. Sep 11, 2022 at 14:52
• @CosmasZachos Does this mean that (B) is just a transformation of the fields, but $\delta \phi$, $\delta \chi$, $\delta F$ still independent? Sep 12, 2022 at 3:30