# Bose-Fermi mixture and Grassmann integrals

The problem

Suppose we have a mixture of $N$ fermionic and $M$ bosonic degrees of freedom. The fermionic degrees of freedom are described by $N$ Grassmann variables $ψ_i$, $i = 1 . . . N$ and the bosonic degrees are described by $M$ complex variables $φ_i$, $i = 1 . . . M$. We combine these into a single vector

$$X=\begin{pmatrix}\Psi\\ \Phi\end{pmatrix}\quad\text{with}\quad\Psi=\begin{pmatrix}\psi_1\\ \vdots \\ \psi_N\end{pmatrix}\quad\text{and}\quad\Phi=\begin{pmatrix}\phi_1\\ \vdots \\ \phi_M\end{pmatrix}$$

We also define a matrix $G$ of both complex numbers and Grassmann numbers as

$$G=\begin{pmatrix} A & B\\ C&D \end{pmatrix}$$

where $A$ and $D$ are resp. $N\times N$ and $M\times M$ matrices which consist of normal complex numbers. $B$ and $C$ are resp. $N\times M$ and $M\times N$ matrices of Grassmann numbers.

Now we define 'action' $S(X^\dagger,X)=X^\dagger G X$ and $$Z[G]=\int\mathcal{D}X^\dagger\mathcal{D}X \exp[S(X^\dagger,X)]\quad\text{with}\quad\mathcal{D}X^\dagger\mathcal{D}X=\prod_{i=1}^N d\psi_i^*d\psi_i\prod_{j=1}^M\frac{d\phi_j^*d\phi_j}{2\pi i}$$ which I want to compute.

Attempt at solution

For simplicity I first want to compute $Z[G]$ in the case $N=M=1$, i.e. there is only one fermionic degree and one bosonic degree of freedom. I write

$$G=\begin{pmatrix} \alpha & \beta\\ \gamma & \delta \end{pmatrix}\quad\text{and}\quad X=\begin{pmatrix}\psi\\ \phi\end{pmatrix}$$

The action now reduces to

$$S(X^\dagger,X)=\alpha\psi^*\psi+\delta\phi^*\phi+\psi^*\beta\phi+\phi^*\gamma\psi$$

and the partition function $Z[G]$ becomes

\begin{align} Z[G]&=\int d\psi^*d\psi\int\frac{d\phi^*d\phi}{2\pi i}e^{-\alpha\psi^*\psi}e^{-\delta\phi^*\phi}e^{-\psi^*\beta\phi}e^{-\phi^*\gamma\psi}\\ &=\int d\psi^*d\psi\int\frac{d\phi^*d\phi}{2\pi i}(1-\alpha\psi^*\psi)(1-\psi^*\beta\phi)(1-\phi^*\gamma\psi)e^{-\delta\phi^*\phi}\\ &=\int d\psi^*d\psi\int\frac{d\phi^*d\phi}{2\pi i}(\beta\gamma\psi^*\psi\phi^*\phi-\alpha\phi^*\phi)e^{-\delta\phi^*\phi} \end{align}

where I used that $\psi^2=(\psi^*)^2=0$, the anti-commuting properties of $\beta,\gamma,\psi,\psi^*$ and that integrating over a single $\psi$ or $\psi^*$ yields zero.

If this is all correct, the result would be $$Z[G]=\frac{\alpha\delta-\beta\gamma}{\delta^2} \tag{*}$$

However, in the general case, one is supposed to find $Z[G]=1/\det[G]$. So my guess is that my result is not correct. Any help?

• What general case are you referring to? As far as I know what you say is true for ordinary variables. Gaussian integral over Grassmann variables yield $\sqrt{\det{G}}$ but here you have a mixture of the two, so you have to be more careful. – DelCrosB Oct 19 '16 at 11:41
• The general case I was referring to is the case for general $M$ and $N$ instead of $M=N=1$. – Kevin Peters Oct 19 '16 at 12:03
• I see. But then it only holds for ordinary variables. In any case, I think that Qmechanic's answer solves your issue. What he wrote is the correct generalization when you have a mixture of ordinary and Grassmann variables. – DelCrosB Oct 19 '16 at 12:12

## 1 Answer

Yes, OP's calculation (*) is correct. The partition function/Gaussian integral of the complex supervariables is supposed to be the inverse superdeterminant $$Z(G)~=\frac{1}{\rm sdet(G)},$$ which it is.