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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!

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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.

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  • $\begingroup$ Except I'm fairly sure that Grassman numbers form a differential graded algebra and so obey a graded Leibniz rule... $\endgroup$ – genneth Feb 11 '11 at 21:43
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    $\begingroup$ No, @genneth, you're confused. The operator $d$ on the Wikipedia page you linked is a Grassmann-odd operator (that's why it squares to zero, see the rule (i) on that page) but the operator $\partial_\tau$ in Jane's question is an ordinary Grassmann-even derivative. The factor $(-1)^{|a|}$ in front of the second term in (ii) is, more generally, $(-1)^{|a|\cdot |d|}$, which proves that $d$ is Grassmann-odd over there, but not here. $\endgroup$ – Luboš Motl Feb 11 '11 at 21:45
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    $\begingroup$ @lubos: I stand corrected. You are completely right :-) +1 $\endgroup$ – genneth Feb 11 '11 at 21:46
  • $\begingroup$ @Lubos I've got another related question to this, but I'm not sure it's interesting/different enough to ask it as a new question: we have $\partial_\tau (\bar{\psi} \psi)=(\partial_\tau\bar{\psi})\psi+\bar{\psi}\partial_\tau \psi=0$. Can I write $(\partial_\tau \bar{\psi}) \psi=−\psi(\partial_\tau \bar{\psi})$ by moving the $\psi$ passed the other Grassmann number? Then $\psi(\partial_\tau \bar{\psi})=\bar{\psi}\partial_\tau \psi$. $\endgroup$ – Jane May 4 '11 at 20:31
  • $\begingroup$ i don't know what a 'bosonic time' is. Is there also a 'fermionic time'? or you are just saying bosonic as a synonym of scalar-that-conmutes-with-everything? $\endgroup$ – lurscher Oct 11 '11 at 20:15
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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.

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