- The variation $\delta F$ for any field (or degree of freedom) $F$, given an infinitesimal transformation, is
*always* calculated as the commutator
$$ \delta F = [ \bar\epsilon Q, F ] $$
where $\bar \epsilon$ is a parameter ("angle" or "shift" or some generalization) of the transformation and $Q$ is the generator. (Those may be replaced by other letters.)

This is the usual Lie-algebra-based way how operators transform. The finite (but very close to identity) transformation may be said to be
$$ U = \exp(\bar\epsilon Q) = 1 + \bar\epsilon Q + o(\epsilon) $$
and the difference of the conjugated $F$ from the original one is the variation
$$\delta F = U F U^{-1} - F $$
So these totally general rules that are already taught in undergraduate quantum mechanics etc. are just applied to the generator $Q$, the infinitesimal "superangle" $\bar\epsilon$, and operators like $\theta^A$, $\sigma$, and $Y$...

Note that the product $\bar\epsilon$ is "bosonic", so its commutators, and not anticommutators, enter the formulae. However, they may be decomposed to anticommutators.

This explains the first "equation" in (4.21) and (4.22). The following ones are the actual calculations, using (4.20). The second term in $Q$ according to (4.20), one which contrains $\partial_\alpha$, doesn't contribute anything to (4.21) because $\theta^A$ and $\sigma^\alpha$ are independent coordinates of the superspace (super world sheet), so the partial derivative of one with respect to the other vanishes.

Analogously, the first term vanishes and only the second term contributes in (4.22).

- In (4.23), the expression $QY^\mu$ simply means the same as $[Q,Y^\mu]$: it is the differential operators in $Q$, with all the right coefficients, acting on $Y^\mu$. It is similar to differentiating functions of positions in ordinary quantum mechanics. Imagine that you have a function $V(x)$ of the operator $x$. Then you may write $V'(x)$, a different, differentiated function of the same operator $x$, as $i/\hbar$ times $[p,V(x)]$. The commutator of $p$ (the $x$-derivative) with the operator does the differentiating of the functions. On state vectors, derivatives act simply from the left, but the analogous action on the operators has to be written as commutators.

The other term $[\bar\epsilon,Y^\mu]=$ doesn't contribute, it is zero, because $\bar\epsilon$ is a (Grassmannian but still) $c$-number. So this analogous is zero much like the commutator $[5,x]$ in quantum mechanics.

- The derivative of $\bar\theta^A \theta$ with respect to $\bar\theta^A$ is $+\theta$, like division, and one may get a factor of $2$ there is a sum over $A$. It is, up to possible signs, the same claim as that the $x$-derivative of $xy$ is $y$. You must have missed some prefactors $\theta$ in some terms when you decided about the incorrect result of the derivative.

The name of the male co-author is John Schwarz, not Schwartz.

This post imported from StackExchange Physics at 2015-05-02 20:48 (UTC), posted by SE-user Luboš Motl