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I was doing a calculation in Giamarchi's Introduction to Many Body Physics, chapter 3, on BCS theory and second quantization, and ran into some confusion with the BCS Hamiltonian. The pdf is here for your reference: http://dpmc.unige.ch/gr_giamarchi/Solides/Files/many-body.pdf

The main confusion comes with eqn. 3.154. Here, the BCS Hamiltonian is given by

$$ H=\sum_k \left( A(k)(\beta_k^\dagger \beta_k-\alpha_k^\dagger \alpha_k)+\Delta(\alpha_k^\dagger \beta_k+\beta_k^\dagger \alpha_k)\right)+ \sum_k\xi(k) $$

Where $\xi(k)$ and $A(k)$ are functions of $k$ and $\alpha_k$ and $\beta_k$ are fermionic operators. Now, I know that the tight-binding Hamiltonian with a periodic potential is given by eqn. 3.128: $$ H=\sum_{k}\bigg( A(k)(\beta_k^\dagger \beta_k-\alpha_k^\dagger \alpha_k)+V(\alpha_k^\dagger \beta_k+\beta_k^\dagger \alpha_k)\bigg) $$ The solution to this is easy to solve with the Bogoliubov transformation, and is given by

$$ E(k)=\sqrt{\xi(k)^2+V^2} $$ I was able to derive this without any problems. However, my question is this: would the solution of the BCS Hamiltonian then be

$$ E_{BCS}(k)=\sqrt{\xi(k)^2+V^2}+\xi(k) $$

Or would it be identical to the tight-binding Hamiltonian? Would the eigenvectors as given by the Bogoliubov transformation change as well?

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It depends if you are interested in the spectrum of the Hamiltonian (the eigenvalues of H) or in the increase of energy when you add one quasi-particle. Usually, one is interested in the latter. –  Adam May 22 at 16:56
    
@Adam So the spectrum is $\sqrt{\xi(k)^2+V^2}$, while the increased energy after adding a quasi-particle is $\sqrt{\xi(k)^2+V^2}+\xi(k)$? –  Bronzeclocksofbenin May 22 at 17:31
    
Rewrite the Hamiltonian after the Bogoliubov transformation (and be careful with the commutation of the ladder operators), and everything should be clear. –  Adam May 22 at 17:48
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I do not see how you can possibly arrive at $\sqrt{\xi^2+V^2}+\xi$. Just remind that the last term $\sum_k\xi_k$ is the vacuum energy, which does not enter either the spectrum or the quasiparticle excitation energy. In any case the spectrum should be $\sqrt{\xi^2+V^2}$ without the additional $\xi$. –  Everett You May 23 at 3:17

1 Answer 1

up vote 2 down vote accepted

As is stated in Giamarchi's document, "This is, up to a simple constant, exactly the Hamiltonian we already examined, and it thus can be solved by exactly the same transformations."

Indeed, $\sum_k\xi(k)$ is a constant: it does not involve any of the operators $\alpha_k$ or $\beta_k$, and may be written $E_0 = \sum_k\xi(k)$

Therefore, the resolution of the BCS Hamiltonian would lead to $$ E_{BCS}(k)=\sqrt{\xi(k)^2+V^2}+ \sum_{k'}\xi(k')=\sqrt{\xi(k)^2+V^2}+ E_o $$

which is just the same energies as TB case with potential above, shifted by a fixed amount, and would lead to the same eigenvectors.

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