- #1
physicus
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- 3
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.