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.

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

1 Answer 1


Frankly, I am lost about your logic trail, not the book's. You seem to somehow connect the (super)symmetry variations (B) to the dynamical variations yielding the EOM, and predicate one on the other? Nothing of the sort is even implied in that book.

(A) is invariant under (B), as you confirmed. Its equations of motion need not take cognizance of (B), and hold whether you've "noticed" the symmetry (B) or not.

Your next step asks you to particularize (A) to (C), as I've corrected it. It, too, has the supersymmetry (B), but you are not asked to consider, or even recognize, that, up front. You are asked to find its equations of motion, and observe how "easy" elimination of F is in them, reducing three such to two, slightly messier ones. At no point have you connected to (B); it is a merely "optional" observation, basically at a "right angle" to the EOM. (Later on, you would use these EOM to confirm the on-shell conservation of the supercharge, but this is not apparent at the step you are working on.)


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