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  Calculation of free energy for Bloch electrons

+ 2 like - 0 dislike

The question has been asked a couple of days ago at https://physics.stackexchange.com/questions/455122/calculation-of-free-energy-for-bloch-electrons

So please excuse me if you have seen it there. Below is the question.

The question is based on Eq. 6-9 in this paper https://arxiv.org/abs/0704.3824

Basically the free energy is defined as 

K({\bf r}) =\sum_{n{\bf k}} f_{n{\bf k}}\text{Re}\left\{\psi_{n{\bf k}}^*({\bf r})\hat{K}\psi_{n{\bf k}}({\bf r})\right\}

where $\hat{K}=\hat{H}-\mu\hat{N}$ is the free energy operator. I will omit the summation and Fermi-Dirac distribution below. 

Immediately the question is why we need the real part at all because that looks real by definition. So I went on to calculate the part with minimal coupling as given by Eq. 6-7 in the paper

\langle\psi_{n{\bf k}}({\bf r})|\frac{e}{2q}\left(\hat{v}_x\sin qy+\sin qy \hat{v}_x\right)|\psi_{n{\bf k}}({\bf r})\rangle \\
=\frac{e}{4iq}\langle u_{n{\bf k}}({\bf r})|e^{-i{\bf k}\cdot {\bf r}}\left(\hat{v}_x(e^{iqy}-e^{-iqy})+(e^{iqy}-e^{-iqy})\hat{v}_x\right)e^{i{\bf k}\cdot {\bf r}}|u_{n{\bf k}}({\bf r})\rangle \\
=\frac{e}{4iq}\langle u_{n{\bf k}}({\bf r})|e^{iqy}(v_x({\bf k}+{\bf q})+v_x({\bf k}))-e^{-iqy}(v_x({\bf k}-{\bf q})+v_x({\bf k}))|u_{n{\bf k}}({\bf r})\rangle 

Here $\hat{v}_x$ is the velocity operator ( in the differential sense) with $e^{i{\bf k}\cdot {\bf r}}\hat{v}_xe^{-i{\bf k}\cdot {\bf r}}=v_x({\bf k})$ being the velocity operator ( in the matrix sense ) in the Bloch bands. We understand that ${\bf k}\pm {\bf q}$ in $v_x({\bf k}\pm {\bf q})$ means $(k_x ,k_y\pm q_y,k_z)$. The expression will be real iff $v_x({\bf k}+ {\bf q}) =v_x({\bf k}- {\bf q})$ which is in general not true. So my question is what is wrong with the above calculation? 

Another related question would be the second line in Eq. 8. Can we make the following simplification?

f_{n{\bf k}}\text{Re}\left\{\langle\delta\psi_{n{\bf k}}(r)|\hat{K}_0|\psi_{n{\bf k}}(r)\rangle+\langle\psi_{n{\bf k}}(r)|\hat{K}_0|\delta\psi_{n{\bf k}}(r)\rangle\right\} =2 f_{n{\bf k}}\text{Re}\langle\psi_{n{\bf k}}(r)|\hat{K}_0|\delta\psi_{n{\bf k}}(r)\rangle

where I just take the hermitian conjugation for the first term. Is it legitimate? 

asked Jan 20, 2019 in Theoretical Physics by fagd (20 points) [ revision history ]
recategorized Jun 11, 2019 by Arnold Neumaier

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