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I would like to find the helmholtz free energy of a system

$F=-T ln(Tr[e^{\frac{-H}{T}}])$

namely a bcs superconductor (using Annett's notation)

$H=\sum E_k (b_{k\uparrow}^\dagger b_{k\uparrow}+b_{-k\downarrow}^\dagger b_{-k\downarrow})$

I've managed to get my expression for F down to

$F=-T \sum ln(Tr(e^{-\frac{E_k}{T}b_{k\uparrow}^\dagger b_{k\uparrow}})Tr(e^{-\frac{E_k}{T}b_{-k\downarrow}^\dagger b_{-k\downarrow}})) $

ie

$F=-T \sum ln(Tr(e^{\frac{-E_k}{T}n_{k\uparrow}})Tr(e^{\frac{-E_k}{T}n_{-k\downarrow}})) $

But I do not know how to evaluate the trace of operators in the second quantization formalism.

My hunch tells me that $Tr[e^{\frac{-E_k}{T}n_k}]$ is $1+e^{\frac{-E_k}{T}}$, because these are fermions, so the eigenstates are $|0\rangle$ and $|1\rangle$. Is this correct?

If not, how should I proceed from this point? Have I attempted to take a dead end approach?

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  • $\begingroup$ $Tr~ (e^{\frac{H}{T}})$ (first equation) is not equals to $Tr~ (\frac{H}{T})$ (third equation). And there is no $T$ in your second equation. $\endgroup$ – Trimok Nov 23 '13 at 12:15
  • $\begingroup$ You're right, there is no T in the Hamiltonian, that was a mistake. $\endgroup$ – Joel Klassen Nov 23 '13 at 16:10
  • $\begingroup$ and I made a mistake in using the identity $det(e^A)=e^{Tr(A)}$. I'll have to change the phrasing of my question, since I'm still curious about the trace of the number operator. $\endgroup$ – Joel Klassen Nov 23 '13 at 16:21
  • $\begingroup$ You should expand the trace on the Fock state basis. That's the reason why the Fermion number operator $n$ has only two possibilities, as you noticed already. Then you can use that $E_{k}=\sqrt{\left(\left(\hbar k\right)^{2}/2m\right)+\Delta^{2}}$ with $\Delta$ the superconducting gap. You should also simplify since $E_{k}=E_{-k}$. $\endgroup$ – FraSchelle Nov 25 '13 at 14:23

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