Statistical and spectral function in thermal state

[physicus]

Homework Statement

The statistical and spectral functions for bosonic operators [itex]\phi_a[/itex] are:
[itex] G_{ab}(t,\vec{x})=\frac{1}{2}\langle \{\phi_a(t,\vec{x}),\phi_b(0,\vec{0})\}\rangle [/itex],
[itex] \rho_{ab}(t,\vec{x})=\langle [\phi_a(t,\vec{x}),\phi_b(0,\vec{0})]\rangle [/itex].
The expectation values are in static thermal equilibrium in the grand canonical ensemble with Hamiltonian [itex]H'=H-\mu Q[/itex].

Show that
[itex] G_{ab}(\omega,\vec{k})=\frac{1}{2}\frac{1+e^{-\beta \omega}}{1-e^{-\beta\omega}}\rho_{ab}(\omega,\vec{k})[/itex]
using the definitions of [itex]G_{ab}(\omega,\vec{k})[/itex] and [itex]\rho_{ab}(\omega,\vec{k})[/itex] in the basis of H' and inserting a complete set of states.

Homework Equations

The Attempt at a Solution

[itex]G_{ab}(\omega,\vec{k})[/itex] and [itex]\rho_{ab}(\omega,\vec{k})[/itex] re defined as Fourier transforms of the above expressions. The expectation values are given as
[itex] \langle A \rangle = \frac{tr\, e^{\beta H'}A}{tr\, e^{\beta H'}}[/itex]
so
[itex] \rho_{ab}(\omega,\vec{k})=\int d^dx dt\, \langle [\phi_a(t,\vec{x}),\phi_b(0,\vec{0})] \rangle e^{-i\vec{k}\vec{x}+i\omega t}[/itex]
[itex] = \frac{1}{tr\, e^{\beta H'}}\int d^dx dt \sum_\alpha \langle\alpha|e^{\beta H'}[\phi_a(t,\vec{x}),\phi_b(0,\vec{0})]|\alpha\rangle e^{-i\vec{k}\vec{x}+i\omega t}[/itex]
[itex] = \frac{1}{tr\, e^{\beta H'}}\int d^dx dt \sum_{\alpha,\gamma} \langle\alpha|e^{\beta \omega_\gamma}|\gamma\rangle\langle\gamma|[\phi_a(t,\vec{x}),\phi_b(0,\vec{0})]|\alpha\rangle e^{-i\vec{k}\vec{x}+i\omega t}[/itex]
[itex] = \frac{1}{tr\, e^{\beta H'}}\int d^dx dt \sum_\alpha e^{\beta \omega_\alpha}\langle\alpha|[\phi_a(t,\vec{x}),\phi_b(0,\vec{0})]|\alpha\rangle e^{-i\vec{k}\vec{x}+i\omega t}[/itex]

However, I don't know how the operators [itex]\phi_a(t,\vec{x})[/itex] act on the eigenstates of H' which I called [itex]|\alpha\rangle[/itex] here. So I don't really know how to proceed.

 

[mark726]

Dear fellow forum member,

Thank you for your response to my post. I am glad to see that you have made some progress in your attempt to solve the problem. However, I would like to offer some suggestions and clarifications to help you further.

Firstly, in your attempt, you have used the operators \phi_a(t,\vec{x}) in the basis of H', which is a good approach. However, you have not fully utilized the fact that the expectation values are in thermal equilibrium in the grand canonical ensemble with Hamiltonian H'. This means that we can write the expectation values in terms of the thermal average over all possible states, which can be expressed as a trace over the states. In other words, we can write:

\langle A \rangle = \frac{1}{Z}\mathrm{Tr}\left(e^{-\beta H'}A\right)

where Z = \mathrm{Tr}\left(e^{-\beta H'}\right) is the partition function. This is a useful expression, as it allows us to express the expectation values in terms of the thermal average of the operator A. This is particularly helpful when dealing with operators that do not commute, as in the case of \phi_a(t,\vec{x}) and \phi_b(0,\vec{0}). Therefore, I would suggest that you re-express your expression for \rho_{ab}(\omega,\vec{k}) using this approach.

Secondly, in your attempt, you have inserted a complete set of states, which is a good step. However, you have not fully utilized the fact that the operators \phi_a(t,\vec{x}) and \phi_b(0,\vec{0}) are bosonic operators. This means that they can be expressed in terms of creation and annihilation operators, which can be used to create and destroy particles respectively. In other words, we can write:

\phi_a(t,\vec{x}) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\vec{p}}}}\left(a_{\vec{p}}e^{-ip\cdot x}+a_{\vec{p}}^\dagger e^{ip\cdot x}\right)

where a_{\vec{p}} and a_{\vec{p}}^\dagger are the annihilation and creation operators respectively, and \omega_{\vec{p}}