Do derivatives anticommute with Grassmann variables and complex numbers in a many-body path integral? I'm trying to learn how to do a many-body path integral for both fermions and bosons, and I'm stuck. I'm following Altland and Simons - Condensed Matter Field Theory, chapter 4. On page 167, equation 4.27 is
\begin{equation}
Z = \int \prod_{n=1}^N d(\bar{\psi}^n,\psi^n) e^{-\delta \sum_{n=0}^{N-1}[\delta^{-1}(\bar{\psi}^n - \bar{\psi}^{n+1}).\psi^n + H(\bar{\psi}^{n+1},\psi^n)]}
\end{equation}
(I've set $\mu=0$ from the equation in the book). The limit $N \rightarrow \infty$ is then taken which involves various things, but the part I don't understand is this:
\begin{equation}
\lim_{N \rightarrow \infty} \delta^{-1}(\bar{\psi}^n - \bar{\psi}^{n+1})) \rightarrow -\partial_\tau \bar{\psi}
\end{equation}
which is fine, but the next is 
\begin{equation}
Z = \int D(\bar{\psi},\psi) e^{-S[\bar{\psi},\psi]}, \hspace{4mm} S[\bar{\psi},\psi] = \int_0^\beta [\bar{\psi} \partial_\tau \psi + H(\bar{\psi},\psi)]
\end{equation}
My question is how do you get from $-\partial_\tau \bar{\psi}$, which is $-\partial_\tau \bar{\psi} \psi$ in $Z$, to $+\bar{\psi} \partial_\tau \psi$? If this was only for fermions, I would guess that the Grassmann variable $\psi$ and the derivative $\partial_\tau$ anticommute which is where the minus sign comes from. But the book says it's valid for bosons as well as fermions; for bosons, the $\psi$ is a complex number, and so I wouldn't expect the minus sign.
Any help would be much appreciated!
 A: Jane, $\partial_\tau$ is clearly a derivative with respect to a bosonic time $\tau$, so it commutes with everything else (except for functions of $\tau$ itself, with which it has a nonzero commutator), rather than anticommutes. Only if both objects have a fermionic character (if both of them are Grassmann-odd), they anticommute with one another (or they have an anticommutator that can be evaluated).
There is no sign error in the formulae, however. You asked a good question: how do you get from 
$$-\partial_\tau \bar\psi \psi$$
to 
$$+\bar\psi\partial_\tau \psi$$
One needs a bit of patience to answer this question. Note that in the two expressions, a different variable is differentiated. In the first one, it's $\bar\psi$ that is differentiated; in the second one, it's $\psi$.
You can't just move derivatives around. Even for bosonic functions, $uv'$ is something else than $u'v$, isn't it?
So the two expressions are not "obviously equal", not even up to a sign, and to convert one to the other, you must carefully integrate by parts. Note that
$$\partial_\tau (\bar\psi \psi) = \partial_\tau\bar\psi \psi + \bar\psi\partial_\tau\psi. $$
This "Leibniz rule" proceeded just like for the derivative of products of bosonic factors because I had to bubble $\partial_\tau$ through the $\psi$'s, and $\partial_\tau$ is a bosonic object. If I were writing down a Leibniz rule for a Grassmannian derivative, I would have to change the sign everytime the derivative would bubble through a Grassmann-odd factor.
But here we deal with bosonic $\tau$-derivatives so the Leibniz rule is just like it has always been. So it implies that up to a total derivative - namely the left-hand side  $\partial_\tau (\bar\psi \psi)$ that integrates to zero over the periodic Euclidean time - the two terms on the right hand side are opposite to one another. That's where the minus sign came from.
A: The answer to your question don`t refer $\psi$ being a Grassman variable. Let's do this:
Start with:
\begin{equation}
\lim_{N \to \infty} \sum_n (\bar{\psi}^n - \bar{\psi}^{n+1}) \psi^n
=\lim_{N \to \infty} \sum_n (\bar{\psi}^n\psi^n - \bar{\psi}^{n+1}\psi^n)
\end{equation}
Before taking the limit $N\to \infty$ shift the dummy variable $n$ in the second term to $n-1$ (I dont remember but maybe its necessary to use $\psi(\beta) = -\psi(0)$ at some point), we get:
\begin{equation}
\lim_{N \to \infty} \sum_n (\bar{\psi}^n\psi^n - \bar{\psi}^{n}\psi^{n-1}) = 
\lim_{N \to \infty} \delta \sum_n \bar{\psi}^n \dfrac{(\psi^n - \psi^{n-1})}{\delta}
\end{equation}
Therefore when you apply the limit $N\to\infty$ you get:
\begin{equation}
\int d\tau \ \bar{\psi}(\tau) \partial_\tau \psi(\tau),
\end{equation}
solving the problem.
However, if you want an elegant way to transform $\partial_\tau\bar{\psi}\psi$ to $-\bar{\psi}\partial_\tau\psi$ you define the derivative of a Grassman product by a normal variable (by normal I mean anything that is not Grassman) as: 
$$ \partial_\tau (\eta \eta\prime) = \partial_\tau\eta \eta\prime + \eta\partial_\tau\eta\prime
\longrightarrow
\partial_\tau\eta \eta\prime = \partial_\tau (\eta \eta\prime) - \eta\partial_\tau\eta\prime
$$
therefore:
\begin{equation}
\int d\tau \  \partial_\tau\bar{\psi}(\tau) \psi(\tau) 
= \left[\bar{\psi}(\tau) \psi(\tau)\right]^\beta_0
-\int d\tau \ \bar{\psi}(\tau) \partial_\tau \psi(\tau)
=-\int d\tau \ \bar{\psi}(\tau) \partial_\tau \psi(\tau),
\end{equation}
also solving the problem.
