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\section{Functional Integrals for Fermions}

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This approach can be extended to Fermions in two formally similar but
conceptually inequivalent ways :--

In the first we define coherent states of Fermions by allowing to take
linear combination of states with coefficients valued in a Grassmann
algebra, i.e., the algebra generated by $2N$ anticommuting objects $\left\{
\xi _{i}\xi _{i}^{*}; i=1,\cdot \cdot \cdot N\right\} .$ Any member of this
algebra is called a Grassmann number. $\int d\xi _{i}\equiv $ $\frac{%
\partial }{\partial \xi _{i}}$ anticommutes with $\xi _{j}$, $j\neq $ $i$

We define a fermion coherent state as 
\[
\left| \xi \right\rangle \equiv \exp \left( -\sum_{\alpha }\xi _{\alpha
}a_{\alpha }^{\dagger }\left| 0\right\rangle \right) =\prod_{\alpha }\left(
1-\xi _{\alpha }a_{\alpha }^{\dagger }\right) \left| 0\right\rangle 
\]
We require that $\left\{ \xi ,a\right\} =0$ and $\left( \xi a\right)
^{\dagger }=a^{\dagger }\xi ^{*}$

Then it is clear that 
\[
a_{\alpha }\left| \xi \right\rangle =\xi _{\alpha }\left| \xi \right\rangle 
\]
\[
\left\langle \xi \right| a_{\alpha }^{\dagger }=\left\langle \xi \right| \xi
_{\alpha }^{*}
\]
To understand the properties of the Fermionic coherent states consider first
a single Fermion degree of freedom $\{a,a^{\dagger }\}$. Define
\[
\left| \xi \right\rangle \equiv e^{-\xi a^{\dagger }}\left| 0\right\rangle 
\]
$\left\{ \xi ,a\right\} =0$ and $\left( \xi a\right) ^{\dagger }=a^{\dagger
}\xi ^{*}$. The overcompleteness relations are easily worked out
\begin{eqnarray*}
\int d\xi ^{*}d\xi \left| \xi \right\rangle \left\langle \xi \right|  &=&%
\frac{\partial }{\partial \xi ^{*}}\frac{\partial }{\partial \xi }(\left|
0\right\rangle -\xi a^{\dagger }\left| 0\right\rangle )(\left\langle
0\right| -\left\langle 0\right| a\xi ^{*})(1-\xi ^{*}\xi ) \\
&=&\left| 0\right\rangle \left\langle 0\right| +a^{\dagger }\left|
0\right\rangle \left\langle 0\right| a=I
\end{eqnarray*}
Notice that $\xi $ anticommutes with vectors in the Hilbert space containing
an odd number of Fermions in which case
\[
\xi \left| \psi \right\rangle =-\left| \psi \right\rangle \xi 
\]
\begin{eqnarray*}
\left\langle \eta |\xi \right\rangle  &=&(\left\langle 0\right|
-\left\langle 0\right| a\eta ^{*})(\left| 0\right\rangle -\xi a^{\dagger
}\left| 0\right\rangle ) \\
&=&1+\eta ^{*}\xi =e^{\eta ^{*}\xi }
\end{eqnarray*}
Finally, notice that if $\left| \psi _{i}\right\rangle $ are vectors in the
Fock space with a definite number of Fermions then $\left\langle \xi |\psi
_{i}\right\rangle $ is a grassmann variable which contains an odd number of
$\eta_\alpha $ if $\left| \psi _{i}\right\rangle $ has an odd number 
of Fermions and
contains an even number of $\eta_\alpha $ if $\left| \psi _{i}\right\rangle $ has an
even number of Fermions. Hence:
\begin{eqnarray*}
\left\langle \xi |\psi _{i}\right\rangle \left\langle \psi _{j}|\xi
\right\rangle  &=&\left\langle \psi _{j}|\xi \right\rangle \left\langle \xi
|\psi _{i}\right\rangle (-1)^{F} \\
&=&\left\langle \psi _{j}|\xi \right\rangle \left\langle \psi _{i}|-\xi
\right\rangle =\left\langle \psi _{j}|-\xi \right\rangle \left\langle \psi
_{i}|\xi \right\rangle 
\end{eqnarray*}
This gives an expression for the trace of an operator $A$ which acts on the
Fock space and preserves the number of particles.
\begin{eqnarray*}
\rm{Tr }A &=&\sum_{n}\left\langle n|A|n\right\rangle =\sum_{n}\int
\prod_{\alpha }d\xi _{\alpha }^{*}d\xi _{\alpha }e^{-\sum_{\alpha }\xi
_{\alpha }^{*}\xi _{\alpha }}\left\langle n|\xi \right\rangle \left\langle
\xi |A|n\right\rangle  \\
&=&\int \prod_{\alpha }d\xi _{\alpha }^{*}d\xi _{\alpha }e^{-\sum_{\alpha
}\xi _{\alpha }^{*}\xi _{\alpha }}\sum_{n}\left\langle -\xi
|A|n\right\rangle \left\langle n|\xi \right\rangle  \\
&=&\int \prod_{\alpha }d\xi _{\alpha }^{*}d\xi _{\alpha }e^{-\sum_{\alpha
}\xi _{\alpha }^{*}\xi _{\alpha }}\left\langle -\xi |A|\xi \right\rangle 
\end{eqnarray*}
Equipped with these techniques we can write functional integral expression
for Fermi systems in complete analogy with the Bosonic theory.
\[
U(\tau ,\tau ^{\prime })=\prod_{i=0}^{N-1}U(\tau _{i+1},\tau _{i})
\]
$\tau _{0}=\tau ^{^{\prime }},\tau _{N}=\tau $ . We express
\[
U(\tau _{i+1,}\tau _{i})=\exp -H\left( \frac{\tau _{i+1}+\tau _{i}}{2}%
\right) \Delta \tau 
\]
and insert at each point a resolution of the identity to find
\[
\left\langle \eta _{N}\left| U(\tau ,\tau ^{\prime })\right| \eta
_{0}\right\rangle =\prod_{i=1}^{N-1}\int d\eta _{i}^{*}d\eta _{i}e^{-\eta
_{i}^{*}\eta _{i}}\left\langle \eta _{i+1}\left| U(\tau _{i+1,}\tau
_{i})\right| \eta _{i}\right\rangle \left\langle \eta _{1}\left| U(\tau
_{1,}\tau _{0})\right| \eta _{0}\right\rangle 
\]
Using the expression for the trace of an operator derived earlier one finds
\[
\rm{Tr }U(\tau ,\tau ^{\prime })=\int d\eta _{0}^{*}d\eta _{0}e^{-\eta
_{0}^{*}\eta _{0}}\left\langle \eta _{0}\left| U(\tau _{,}\tau ^{\prime
})\right| \eta _{0}\right\rangle 
\]
If $U$ $\simeq $ $e^{-H\left( \frac{\tau _{i+1}+\tau _{i}}{2}\right) \Delta
\tau }$and $H_{c}(\eta ^{*},\eta )$ with $\eta ^{*}$ and $\eta $ and
replaced by $a^{\dagger }$ and $a$ give the normal ordered form of $H$ we
find 
\begin{eqnarray*}
\rm{Tr }U(\tau ,\tau ^{\prime }) &=&\prod_{i=0}^{N-1}\int d\eta
_{i}^{*}d\eta _{i}e^{-\sum_{i=0}^{N-1}\eta _{i}^{*}\eta _{i}}\left\langle
\eta _{i+1}|\eta _{i}\right\rangle e^{-H_{c}\left( \eta _{i+1}^{*},\eta
_{i}\right) \Delta \tau } \\
&=&\prod_{i=0}^{N-1}\int d\eta _{i}^{*}d\eta _{i}e^{-\sum_{i=0}^{N-1}\eta
_{i}^{*}\eta _{i}}e^{\sum_{i=0}^{N-1}\eta _{i}^{*}\eta
_{i}}e^{-H_{c}\sum_{i=0}^{N-1}\left( \eta _{i+1}^{*},\eta _{i}\right) \Delta
\tau }
\end{eqnarray*}
$(\eta _{N}^{*}\equiv -\eta _{0}^{*})$. Notice that if $H_{c}$ contained
some explicit time dependence due to the source or an auxillary field, it
should be evaluated at time $\tau =\frac{\tau _{i+1}+\tau _{i}}{2}$.
Proceeding heuristically we could define
\[
\rm{Tr }e^{-\beta H}=\int D\eta ^{*}D\eta e^{-\left( \eta ^{*}\frac{%
\partial \eta }{\partial \tau }+H_{c}(\eta ^{*},\eta )\right) }
\]
where the operator $\frac{\partial }{\partial \tau }$ is defined on the
space of functions obeying $\eta (\beta )\equiv -\eta (0)$. 

\section{The Standard Gaussian Integral:}

\[
\int \prod_{i}d\eta _{i}^{*}d\eta _{i}e^{-\eta ^{*}A\eta +\eta
^{*}j+j^{*}\eta }=\int \prod_{i}d\eta _{i}^{*}d\eta _{i}e^{-\eta ^{*}A\eta
}e^{j^{*}A^{-1}j}
\]
using the transformation 
\begin{eqnarray*}
\eta  &\rightarrow &\eta +A^{-1}j \\
\eta ^{*} &\rightarrow &\eta ^{*}+j^{*}A^{-1}
\end{eqnarray*}
Thus
\[
\int \prod_{i}d\eta _{i}^{*}d\eta _{i}e^{-\eta ^{*}A\eta +\eta
^{*}j+j^{*}\eta }\simeq (\det A)e^{j^{*}A^{-1}j}
\]
expanding to second order.
\begin{eqnarray*}
\eta _{i}^{*}j_{i}j_{k}^{*}\eta _{k} &=&j_{k}^{*}A_{kl}^{-1}j_{l} \\
\left\langle \eta _{k}|\eta _{l}^{*}\right\rangle  &=&A_{kl}^{-1}
\end{eqnarray*}

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