When diagonalising the mean-field Hamiltonian in BCS-theory, there is some freedom as to how one defines the Bogoliubov-operators. One convenient choice is to let \begin{align*} \gamma_{\mathbf{k}0}^\dagger &= u_{\mathbf{k}}^* c_{\mathbf{k}\uparrow}^\dagger - v_{\mathbf{k}}c_{-\mathbf{k}\downarrow} \\ \gamma_{-\mathbf{k}1}^\dagger &= u_{\mathbf{k}}^* c_{-\mathbf{k}\downarrow}^\dagger + v_{\mathbf{k}}c_{\mathbf{k}\uparrow}, \end{align*}

since these operators creates spin $\pm1/2$ excitations on top of the BCS ground state, and their adjoints annihilate it. When diagonalising the Hamiltonian one arrives at the conclusion that the excitation spectrum of these two particles is the same, $$ H = E_0 + \sum_{\mathbf{k}} \sum_{\alpha = 0,1} E_{\mathbf{k}} \gamma_{\mathbf{k}\alpha}^\dagger \gamma_{\mathbf{k}\alpha}. $$ Another choice is simply to define operators $\eta_{\mathbf{k}0}^\dagger = \gamma_{\mathbf{k}0}^\dagger$ and $\eta_{\mathbf{k}1}^\dagger = \gamma_{\mathbf{k}1}$, which evidently means that their excitation spectra will have opposite signs.

Now, computing the free energy of a fermionic system one uses the grand canonical ensemble $$ Q = e^{-\beta E_0}\prod_{\mathbf{k},\alpha} \ln(1 + \mathrm{e}^{-\beta \epsilon_\mathbf{k,\alpha}}), $$ where $\epsilon_\mathbf{k,\alpha}$ is the excitation energy of the fermions of type $\alpha$.

My question is essentially: which of the excitation spectra should one use in this case, and why? When using those of the $\eta$-operators, I get a free energy which upon minimisation yields the correct self-consistency equation for the gap, since the derivative of $F$ gives (amongst other terms) the terms $$ \frac{\partial E_\mathbf{k}}{\partial \Delta} \left( \frac{(-1)\mathrm{e}^{-\beta E_\mathbf{k}}}{\mathrm{e}^{-\beta E_\mathbf{k}} + 1} + \frac{\mathrm{e}^{\beta E_\mathbf{k}}}{\mathrm{e}^{\beta E_\mathbf{k}} + 1} \right) $$

which add up to the $\tanh$ in the self-consistency equation with the correct prefactor. Using directly the excitation spectra of the $\gamma$-operators does not give the same function. This seems to suggest that we cannot naively treat the excitations from the $\gamma$-operators as fermionic excitations. I cannot seem to figure out why this has to be so.

  • 1
    $\begingroup$ Using $\eta$ or $\gamma$ should lead to the same answer, since it's just change of variables. In fact, when going from $\gamma$ to $\eta$, you need to move a piece of $E_k$ to $E_0$. As to why you didn't get the correct gap equation using $\gamma$, most likely you haven't considered the $\Delta$ dependence of $E_0$. $\endgroup$
    – Meng Cheng
    Jul 10, 2022 at 14:10

1 Answer 1


As mentioned in the comments, moving from the $\gamma$-representation to the $\eta$-representation adds a term $\sum_{\mathbf{k}} E_{\mathbf{k}}$ to $E_0$, which effectively removes the $\Delta$-dependence of $E_0$ apart from the term $V^{-1}\Delta^2$ in the latter - thereby explaining why it gives the correct gap equation - while in the former one gets an additional term to the equation for $\partial F/\partial \Delta$ : $$ \color{red}{\frac{\partial E_{\mathbf{k}}}{\partial \Delta}} -\frac{\partial E_\mathbf{k}}{\partial \Delta} \left( \frac{\mathrm{e}^{-\beta E_\mathbf{k}}}{\mathrm{e}^{-\beta E_\mathbf{k}} + 1} + \frac{\mathrm{e}^{-\beta E_\mathbf{k}}}{\mathrm{e}^{-\beta E_\mathbf{k}} + 1} \right) = \frac{\partial E_{\mathbf{k}}}{\partial \Delta} \tanh\left(\frac{\beta E_{\mathbf{k}}}{2}\right), $$ thereby giving rise to the correct $\tanh$ dependence.


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