4
$\begingroup$

I am following the discussion of fermionic path integrals and Grassmann variables in QFT for the Gifted Amateur (ch. 28). It defines a coherent state for fermions $\rvert \eta \rangle$ as \begin{align} \rvert \eta \rangle &= e^{-\eta \hat{c}^\dagger} \rvert 0 \rangle \\ &= \rvert 0 \rangle - \eta \rvert 1 \rangle \end{align} where $\hat{c}^{\dagger}$ is the fermion creation operator and $\eta$ is a Grassmann number.
I do see how this is a coherent state, since $$\hat{c}\rvert \eta \rangle = \eta \rvert \eta \rangle. $$

Then in the next section, 28.3 The Path Integral for Fermions, the book writes:

We start by evaluating the Gaussian integral for coherent states $$ \int d\eta d\bar{\eta}e^{\bar{\eta}a\eta} = \int d\eta d\bar{\eta}\left( 1+ \bar{\eta}a\eta \right) =\int d\eta a\eta = a $$ This carries over to the case of Grassmann-valued (N-component) vectors $\boldsymbol{\eta} = \left(\eta_1,\eta_2,...,\eta_N \right)$ and $\bar{\boldsymbol{\eta}} = \left( \bar{\eta}_1,\bar{\eta}_2,...,\bar{\eta}_N \right)$ and we find that for a matrix $\boldsymbol{A}$: $$\int d^N\eta d^N\bar{\eta}e^{ \bar{\boldsymbol{\eta}}\boldsymbol{A}\boldsymbol{\eta}} = det\boldsymbol{A},$$ where $ d^N\eta d^N\bar{\eta} = d\eta_1 d\bar{\eta}_1 \cdot\cdot\cdot d\eta_N d\bar{\eta}_N $

I'm not sure how the above only applies to coherent states, I thought that it would work for any set of Grassmann variables.

The book then does a calculation for a Gaussian integral and finds that the generating functional for fermions is $$ Z\left[ \bar{\eta},\eta \right] = \int \mathcal{D}\psi \mathcal{D}\bar{\psi} e^{ i \int d^4x \left[ \mathcal{L}\left( \bar{\psi},\psi \right) + \bar{\eta}(x)\psi(x)+\eta(x)\bar{\psi}(x) \right] } $$

What I would like to know is, how much of this depends on the fact that we are (supposedly) using coherent states? I suspect it involves the fact that the identity can be represented by $$ 1 = \int d\eta d\bar{\eta} e^{\bar{\eta}\eta} \rvert \bar{\eta} \rangle \langle \eta \lvert \quad = \quad \rvert 0 \rangle \langle 0 \lvert + \rvert 1 \rangle \langle 1 \lvert $$ (this is one of the problems at the end of the chapter) and perhaps this is involved in the derivation of the generating functional, and the book is just not describing that step.

$\endgroup$
2
$\begingroup$

The integral formula that you write $\int d^N\eta d^N\bar{\eta} e^{\bar{\eta}A\eta}=\det A$ is indeed true for any Grassman numbers, and does not involve coherent states. The Grassman resolution of the identity, on the other hand, is given by \begin{equation} I=\int\prod_{\alpha} d \xi_{\alpha}^* d \xi_{\alpha}\, e^{\sum_{\alpha}\xi_\alpha \xi_\alpha^\star}|\xi><\xi|, \end{equation} where $|\xi>$ is a coherent state. If you insert this resolution of the identity at each time slice, then you should recover the coherent state path integral, similarly to the case for a bosonic path integral. This requires a Hamiltonian formulation, which is then converted to a Lagrangian when the momenta are integrated over, which should be described in most QFT books.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.