Let
$$
H_{T}=\dot{x}^{I}\frac{\partial}{\partial \psi^{I}}T(x,\psi)
$$
and consider the transformation:
$$
x^{I}\mapsto x^{I}+i\epsilon\psi^{I}
\\
\psi^{I}\mapsto\psi^{I}-2\epsilon\dot{x}^{I}
$$
where $T$, $\epsilon$ and $\psi^{I}$ are fermionic and the $x^{I}$ are bosonic (note that $T$ need not commute with its derivatives, but will commute with the $x^{I}$'s and anticommute with the $\psi^{I}$'s).
I am having a really frustrating, but probably very simple issue. In short, I have calculated $\delta H_{T}$ for a specific $T$ in two different ways, and cannot reconcile the results.
To begin with, I derive a general form of $\delta H_{T}$. It is not hard to show that under the above transformation:
$$
\delta H_{T}
=
i\epsilon\left(
\dot{\psi}^{I}\frac{\partial T}{\partial \psi^{I}}
–
\dot{x}^{I}\frac{\partial T}{\partial x^{I}}
+
\psi^{J}\dot{x}^{I}\frac{\partial^{2}T}{\partial\psi^{I}\partial x^{J}}
\right).\tag{1}
$$
So far so good. I now want to take $T(x,\psi)=S(x)-\psi^{I}A_{I}(x)$, where $S$ is fermionic and $A_{I}$ is bosonic. Note that as with $T$ itself, $S$ and $A_{I}$ need not commute with each other or with their derivatives.
It is easy to show directly that with this choice,
$
H_{T}
=
-\dot{x}^{I}A_{I}
$, and so $$\delta H_{T}=-i\epsilon(\dot{\psi}^{I} A_{I}+\dot{x}^{I}\psi^{J}\partial_{J} A_{I}).$$ My issue is with verifying this calculation with the general result $(1)$. Plugging $T$ into $(1)$, we find:
$$
\delta H_{T}
=
-i\epsilon
\left(
\dot{\psi}^{I}A_{I}
+
\dot{x}^{I}\psi^{J}\partial_{J} A_{I}
\right)
-i\epsilon\dot{x}^{I}(\partial_{I} S-\psi^{J}\partial_{I} A_{J})
$$
Which clearly disagrees with my previous result. In fact, one result is $S$-dependent while the other is not! The only thing I can think of is that I have somehow made a mistake deriving $(1)$, but I have retried it many times with no luck. This is really frustrating and probably just a silly oversight, but I have spent many hours staring at this and would really appreciate some help.
EDIT:
I will now add some additional workings in response to Qmechanics's comment.
The second term in $(1)$ arose in my calculations as follows. Firstly, expanding $T$ in a truncating Taylor series shows that:
$$
\delta T = i\epsilon\psi^{I}\frac{\partial T}{\partial x^{I}}-2\epsilon\dot{x}^{I}\frac{\partial T}{\partial\psi^{I}}
$$
and so:
$$
\delta\left(\frac{\partial}{\partial\psi^{I}}T(x,\psi)\right)
=
-i\epsilon\frac{\partial T}{\partial x^{I}}
+
i\epsilon\psi^{J}\frac{\partial^{2}T}{\partial\psi^{I}\partial x^{J}}
+
2\epsilon\dot{x}^{J}\frac{\partial^{2}T}{\partial\psi^{I}\partial\psi^{J}}
$$
The term in question then arises from the $\dot{x}^{I}\delta\left(\frac{\partial}{\partial\psi^{I}}T(x,\psi)\right)$ term of $\delta H_{T}$.
One possibility that I am considering is that when calculating $\delta\left(\frac{\partial}{\partial\psi^{I}}T(x,\psi)\right)$, I need to somehow vary the differentiation variable at the same time, however this does not seem to make much sense.
Best Answer
I think your variation does not commute with the derivative $\frac{\partial}{\partial \psi}$. You've shown what happens if you take $H_T = S - \psi A$, do the derivative with respect to $\psi$, and then perform the variation. However if you do things in the opposite order (first vary and then differentiate) I think you get a term proportional to $\frac{\partial S}{\partial x}$. When you apply the variation to $S$, you'll get a term proportional to $i \epsilon \psi \frac{\partial S}{\partial x}$, which then survives when you hit it with $\frac{\partial}{\partial \psi}$. Therefore, I think the issue is simply that you have to be careful about the ordering of the variation and derivative when you calculate the general expression.