I have a question regarding the diagonalization of the BCS-Hamiltonian using the Bogoliubov-DeGennes-transformation. I hope someone can help me, so I start with the following Hamiltonian, it is related to the kinetic term in the equation and follows Tinkham's book.

$$ \sum_{\sigma k} \xi_k c_{\sigma k}^\dagger c_{\sigma k} - \dots $$

we go on to define the quasiparticles

$$ c_{k\uparrow} = u^*_{k}\gamma_{k0} + v_{k}\gamma^\dagger_{k1}, $$


$$ c^\dagger_{-k\downarrow} = -v^*_{k}\gamma_{k0} + u_{k}\gamma^\dagger_{k1} $$

where $\gamma_{k0}$ participates in destroying an electron with $k\uparrow$ or creating one in $-k\downarrow$.

Now, if I do the spin summation I will get one term for up and one for down. The down-term is

$$ \sum_k \xi_k c^\dagger_{k\downarrow} c_{k\downarrow} $$

In the subscript there is no minus-sign, and so it does not "fit" with any of the defined quasiparticles. I could of course use the latter one, and let $k \rightarrow -k$ but then I should be left with terms proportional to $v_{-k}$ and so on, however in the final solution there are no such terms. What am I missing?

The final expression (the kinetic part) is

$$ \sum_k \xi_k [(|u_k|^2 - |v_k|^2)(\gamma^\dagger_{k0}\gamma_{k0} + \gamma_{k1}^\dagger \gamma_{k1}) + 2|v_k|^2 + 2u_k^*v_k^* \gamma_{k1}\gamma_{k0} + 2u_k v_k \gamma_{k0}^\dagger \gamma_{k1}^\dagger]. $$

  • 1
    $\begingroup$ Could you write down what the final expression in terms of quasi-particle operators is supposed to be? $\endgroup$
    – Nephente
    Commented Apr 10, 2015 at 15:31
  • $\begingroup$ Yes, I added it now, and I also changed the notation so that it is exactly the same as Tinkham. But still I dont know how to handle the spin-down, should I flip the spin in the first particle or invert the momentum in the latter? $\endgroup$
    – camzor00
    Commented Apr 10, 2015 at 15:42
  • $\begingroup$ You're on the right track, but notice that the RHS of the transformations contain no spin indices. So flipping the spin in the first expression doesn't mean anything. Inverting momentum is the only trick you've got. $\endgroup$
    – wsc
    Commented Apr 10, 2015 at 16:17
  • $\begingroup$ You'll probably find it helpful to assume that the amplitudes are even in $k$, e.g. $v_k = v_{-k}$. $\endgroup$ Commented Apr 10, 2015 at 16:24
  • 1
    $\begingroup$ It's probably easiest to assume $\xi_k=\xi_{-k}$. This plus letting $k$ go to $-k$ in the sum over spin-down. $\endgroup$
    – Nephente
    Commented Apr 10, 2015 at 16:36

2 Answers 2


I think you are missing the meaning of a Bogoliubov-DeGennes transformation. This transformation is useful to write a Hamiltonian in its diagonal pertubations coordinates. The pertubations must have some conserved properties, for example:

Supose that $c_{\textbf{k} \uparrow}^\dagger$ creates a pertubation in some state (in your case it creates a electron with energy $\epsilon_\textbf{k}$ in a vaccum state, but pay attention to the general discussion). The Bogoliubov-DeGennes idea is to write that pertubation as a linear combination of other two: $$c_{\textbf{k} \uparrow}^\dagger=\alpha_\textbf{k} \ d_{\textbf{k} \uparrow}^\dagger+\beta_\textbf{k} \ d_{-\textbf{k} \downarrow}$$

Look what the "other two" have in common: the $c_{\textbf{k} \uparrow}^\dagger$ pertubation creates a linear combination of a $d$ particle with spin $\uparrow$ and momentum $\textbf{k}$ and a $d$ particle with spin $\downarrow$ and momentum $-\textbf{k}$, so the net spin flux remains the same (the spin $\downarrow$ goes in $-\textbf{k}$ direction, equivalent to the spin $\uparrow$ going in $\textbf{k}$ direction). Assuming the $d$ particles have the same commutation properties than the $c$, you find that this transformation must be unittary. In a general form: $$ \quad \begin{pmatrix} c_{\textbf{k} \uparrow}^\dagger \\ c_{-\textbf{k} \downarrow} \end{pmatrix} = \begin{pmatrix} \alpha_\textbf{k} & \beta_\textbf{k} \\ \gamma_\textbf{k} & \upsilon_\textbf{k} \end{pmatrix} \begin{pmatrix} d_{\textbf{k} \uparrow}^\dagger \\ d_{-\textbf{k} \downarrow} \end{pmatrix} = P \begin{pmatrix} d_{\textbf{k} \uparrow}^\dagger \\ d_{-\textbf{k} \downarrow} \end{pmatrix}, $$

where $P$ matrix must be unitary. The constants must also satisfy other condition you fix on then (usually, equations that diagonalize the Hamiltonian).


Consider $c^+_{\boldsymbol{k}\uparrow}=(c_{\boldsymbol{k}\uparrow})^+=(u^*_\boldsymbol{k}γ_\boldsymbol{k0}+v_\boldsymbol{k}γ^+_\boldsymbol{k1})^+=u_{\boldsymbol{k}}γ^{+}_\boldsymbol{k0}+v^{*}_\boldsymbol{k}γ_\boldsymbol{k1}$, it yields the right result.


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