1
$\begingroup$

Suppose we want to study many-fermions quantum mechanics on a lattice, so we start with a finite-dimensional Hilbert space $\mathcal{H}$ and go to its Fock space $\mathcal{F}_{\text{fer}}(\mathcal{H})$. If $H$ is the Hamiltonian of the one-particle system in $\mathcal{H}$, let $d\Gamma(H)$ be the Hamiltonian on $\mathcal{F}_{\text{fer}}(\mathcal{H})$. If $A$ is an operator on $\mathcal{F}_{\text{fer}}(\mathcal{H})$, then the basic hypothesis of the theory is that the average value of $A$ is given by: $$\langle A \rangle_{\beta,\mu} := \frac{1}{Z}_{\beta,\mu}\text{Tr} \bigg{(}e^{-\beta (d\Gamma(H)-\mu N)}A\bigg{)} \tag{1}\label{1}$$ where $\beta > 0$ is the inverse temperature, $\mu$ is the chemical potential, $N$ the number operator and $Z$ the partition function.

The above is saying that the equilibrium state is described by the operator $e^{-\beta(d\Gamma(H)-\mu N)}$.

Now, set $K:= d\Gamma(H)-\mu N$. On page 20 of this book, the following formula is displayed: $$(-1)^{p}\langle \Omega, \mathbb{T}\prod_{l=1}^{p}\bar{\varphi}(x_{l},\sigma_{l})\Omega \rangle = \frac{\int \prod_{l=1}^{p}\bar{\psi}_{x_{l},\sigma_{l}}e^{\mathcal{A}(\bar{\psi},\psi)}\prod_{x,\sigma}d\bar{\psi}_{x,\sigma}d\psi_{x,\sigma}}{\int e^{\mathcal{A}(\bar{\psi},\psi)}d\bar{\psi}_{x,\sigma}d\psi_{x,\sigma}} \tag{2}\label{2}$$ where $\mathbb{T}$ denotes the time-ordering operator and $\bar{\varphi}$ on the left hand side can be either a creation operator $\varphi^{\dagger}(x_{l},\sigma_{l})$ or annihilation operator $\varphi(x_{l},\sigma_{l})$, while the $\bar{\psi}$ stands for the associated Grassmann variable. From the text, $\Omega$ is a eigenstate of $K$, but it is not clear if it is the ground state of $H$ as well. Thus, the right hand side of (\ref{2}) is just a path integral representation.

I want to reconcile formulas (\ref{1}) and (\ref{2}). I recognize the right hand side of (\ref{2}) as the mean value of the product $\prod_{l=1}^{p}\bar{\varphi}(x_{l},\sigma_{l})$ written in terms of a path integral. But then, the right hand side of (\ref{2}) is: $$\langle \prod_{l=1}^{p}\bar{\varphi(x_{l},\sigma_{l})}\rangle_{\beta,\mu} \tag{3}\label{3}$$ How can (\ref{3}) be also the average with respect to the eigenstate $\Omega$? The expression on (\ref{3}) carries a sum over all basis states, how can it be that it is written just with respect to $\Omega$ as well? What is wrong with my reasoning?

$\endgroup$

1 Answer 1

1
$\begingroup$

You can reconcile the formulas by specifying the boundary conditions of your path integral. For (2), the time boundaries (at $0,\beta$) are assumed to be in state $|\Omega\rangle$ but to retrieve (1), you only assume periodic boundary solutions. Things usually get clearer once you've actually computed a concrete case (which is doable when your hamiltonian is quadratic since you only have gaussian integrals to calculate).

Check out QFT in a Nutshell by A. Zee for a good introduction (chapter V.2) on the matter.

Hope this helps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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