Evaluation of thermal average with path integral

Question

I want to evaluate the thermal average

$$<\hat{\phi}(0)\hat{\phi}(0)\hat{\phi}(x)\hat{\phi}(x)>$$

with the path integral. $\phi$ is a real scalar field.

In general:

$$<\hat{\phi}(0)\hat{\phi}(0)\hat{\phi}(x)\hat{\phi}(x)>=\frac{1}{Z}{\rm Tr}[e^{-\beta\hat{H}}\hat{\phi}(0)\hat{\phi}(0)\hat{\phi}(x)\hat{\phi}(x)]$$

where Tr means trace, $Z$ is the partition function and $\hat{H}$ is the Hamilton operator. I know how to evaluate this with eigenstates of the Hamiltonian $\hat{H}|n>=E_n|n>$:

$$<\hat{\phi}(0)\hat{\phi}(0)\hat{\phi}(x)\hat{\phi}(x)>=\frac{1}{Z}\sum_n <n|e^{-\beta\hat{H}}\hat{\phi}(0)\hat{\phi}(0)\hat{\phi}(x)\hat{\phi}(x)|n>$$

However I want to use the path integral approach to evaluate the trace. Therefore

$$<\hat{\phi}(0)\hat{\phi}(0)\hat{\phi}(x)\hat{\phi}(x)>=\frac{1}{Z}\int d\phi <\phi|e^{-\beta\hat{H}}\hat{\phi}(0)\hat{\phi}(0)\hat{\phi}(x)\hat{\phi}(x)|\phi> \tag{*}$$

where $|\phi>$ is an eigenstate of the Schrödinger-picture field operator

$$\hat{\phi}(\vec{x},0)|\phi>=\phi(\vec{x})|\phi>.~~ (**)$$

Question: If $\hat{\phi}(\vec{x},0)$ is in the Schrödinger picture should the states not depend on time? I thought Schrödinger picture means that the operators are time independent but the states carry the time dependence? However the book that I am reading Gale, Kapusta (page 12) does not label the states with time?

The operators that appear in $(*)$ carry a time dependence $\hat{\phi}(x)=\hat{\phi}(\vec{x},t)$. So I interpret them as being in the Heisenberg picture.

Next I take $(**)$ and insert twice $e^{-i\hat{H}t}e^{i\hat{H}t}$:

$$e^{-i\hat{H}t}e^{i\hat{H}t}\hat{\phi}(\vec{x},0)e^{-i\hat{H}t}e^{i\hat{H}t}|\phi>=\phi(\vec{x})|\phi>$$.

I define $e^{i\hat{H}t}|\phi>=:|\phi,t>$ and $e^{i\hat{H}t}\hat{\phi}(\vec{x},0)e^{-i\hat{H}t}=:\hat{\phi}(\vec{x},t)$ to end up with

$$\hat{\phi}(\vec{x},t)|\phi,t>=\phi(\vec{x})|\phi,t>~~ (***)$$.

Question: is this correct? For me it looks wrong that on the RHS there is still $\phi(\vec{x})$ and not $\phi(\vec{x},t)$?

Next to evaluate $(*)$ I need completeness relations. I guess the following should be correct:

$$\int d\phi(t) |\phi,t><\phi,t|$$

I insert this into $(*)$:

$$\frac{1}{Z}\int d\phi \int d\phi(t) \int d\phi_0 <\phi|e^{-\beta\hat{H}}\hat{\phi}(0)\hat{\phi}(0)|\phi,0><\phi,0|\hat{\phi}(x)\hat{\phi}(x)|\phi,t><\phi,t|\phi>$$

Now contrary to what I found in $(***)$ I use:

$$\hat{\phi}(\vec{x},t)|\phi,t>=\phi(\vec{x},t)|\phi,t>$$

to get:

$$(*)=\frac{1}{Z}\int d\phi \int d\phi(t) \int d\phi_0 <\phi|e^{-\beta\hat{H}}|\phi,0>\phi(0)\phi(0)\phi(x)\phi(x)<\phi,0|\phi,t><\phi,t|\phi>$$

after using the orthogonality of the states I get:

$$(*)=\frac{1}{Z}\int d\phi <\phi|e^{-\beta\hat{H}}|\phi>\phi(0)\phi(0)\phi(\vec{x},t)\phi(\vec{x},t)$$

Question this is the result which I think has to come out. Could you confirm that this is correct? From this point I know how to proceed, i.e. $<\phi|e^{-\beta\hat{H}}|\phi>$ can be represented as path integral after switching to imaginary time $<\phi|e^{-\beta\hat{H}}|\phi>=\int D\phi(t) ...$

Answer 1

When you want to find the correlation function of operators at different time arguments, it doesn't make much sense to use Schrodinger picture anymore. So you should work in the Heisenberg picture.

Also, when you are working with mixed states (given by density matrix) in the Schrodinger picture, the time evolution of states are captured by the time evolution of the density matrix: $$ \rho(t)=e^{-iHt}\rho_0 e^{+iHt}$$

But again, you cannot assign time arguments to operators anymore when working in Schrodinger picture.

The most straightforward way to calculate that correlation function is to use Heisenberg picture and Keldysh formalism (if there are interactions).