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  How to derive electron number equation of Bogoliubov Hamiltonian using thermodynamic relations.

+ 5 like - 0 dislike

My question arise from this article: Edge superconducting correlation in the attractive-U Kane-Mele-Hubbard model. I will describe my question in detail so that you might not need to look into that article. What I want to ask for help is an easy way to derive the electron number equation, may be using some thermodynamic relations just as how I derive the gap equation.

The Hamiltonian considered is the mono-layer graphene with intrinsic SO interaction, plus the negative-U Hubbard term. After the mean field approximation with S-wave superconducting order parameter, we obtain the mean-field Hamiltonian:

$$ H=\sum_k\phi_k^\dagger H_k\phi_k+E_0 $$

where $\phi_k$ is the Nambu spinor $\phi_k^\dagger=(a_{k\uparrow}^\dagger, b_{k\uparrow}^\dagger, a_{-k\downarrow}, b_{-k\downarrow})$, $E_0=2N\Delta^2/U$, $N$ is the number of unit cell, and

$$ H_k=\begin{pmatrix} \lambda_k-\mu & -t\gamma_k & -\Delta & 0 \\ -t\lambda_k^* & -\lambda_k-\mu & 0 & -\Delta \\ -\Delta^* & 0 & -\lambda_k+\mu & t\gamma_k \\ 0 & -\Delta^* & t\gamma_k^* & \lambda_k+\mu \end{pmatrix} $$

The $\mu$ is the chemical potential in the original Hubbard term, $-t\gamma_k$ is the sum of the graphene hopping integral of nearest neighbors, $\gamma_k$ is from SO term. We can just ignore their physical meaning and regard them as some parameters, they are not very important to my question.

Diagonalizing $H_k$, we have four eigenvalues, $\omega_{ks\alpha}=\alpha\omega_{ks}=\alpha\sqrt{(\epsilon_k+s \mu)^2+\Delta^2}$ with $\epsilon_k=\sqrt{\lambda_k^2+t^2|\gamma_k|^2}$, where $s,\alpha$ are $\pm 1$.

Now we have gap equation and electron number equation: $$ \frac{1}{U}=\frac{1}{4N}\sum_{ks}\frac{\tanh{(\beta\omega_{ks}/2)}}{\omega_{ks}} $$ $$ n_e-1=-\frac{1}{N}\sum_{ks}\frac{s\epsilon_k -\mu}{\omega_{ks}}\tanh(\beta \omega_{ks}/2) $$ where $n_e$ is the average electron number on one sublattice.

The following is how I derive the gap equation, the free energy is: $$ F=-\frac{1}{\beta}\sum_{ks\alpha}\ln{(1+\mathrm{e}^{-\beta\omega_{ks\alpha}})}+\frac{2N\Delta^2}{U} $$ the free energy is minimized when $\Delta$ choose to have its true value, i.e. using $\partial F/\partial \Delta=0$ we can derive the gap equation showing above.

How can I derive the electron number equation? I know in principle I can derive it by representing the original electron operators instead of the diagonalized Bogoliubov quasi-particle operators, but this is much too complicated even one trying to derive them using Mathematica.

So just as I said in the beginning of this question:I need your help to get an easy way to derive the electron number equation, may be using some thermodynamic relations just as how I derive the gap equation

This post imported from StackExchange Physics at 2014-08-22 05:03 (UCT), posted by SE-user luming
asked May 25, 2014 in Theoretical Physics by BaBQ (95 points) [ no revision ]
I suggest trying $\partial F/\partial \mu=0$.

This post imported from StackExchange Physics at 2014-08-22 05:03 (UCT), posted by SE-user leongz

1 Answer

+ 2 like - 0 dislike

Thermodynamic relation $N=-\frac{\partial J}{\partial \mu}$ exactly gives you the particle number equation, wherein $J$ is the macroscopic thermodynamic potential, i.e., the quantity $F$ in your question.

By thermodynamics, $dJ=-SdT+Ydy-Nd\mu$ tells you why the partial derivative equation is valid. And in statistical mechanics, macroscopic ensemble formalism defines $J$ as $e^{-\beta J}\equiv\mathrm{Tr}[e^{-\beta (\hat{H}-\mu\hat{N})}]$. Of course, you can derive the partial derivative equation within this formalism as well.

This post imported from StackExchange Physics at 2014-08-22 05:03 (UCT), posted by SE-user huotuichang
answered Jun 18, 2014 by sfman (270 points) [ no revision ]

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