2
$\begingroup$

In some references (see for example Ballentine ch. 18.5) the ground state of the BCS theory is assumed to be

\begin{equation} |BCS\rangle = \prod_{\bf k} (u_{\bf k}+v_{\bf k}\hat{c}^{\dagger}_{\bf k,\uparrow}\hat{c}^{\dagger}_{-\bf k,\downarrow})|0\rangle. \end{equation}

with the normalization constraint $u_{\bf k}^2 + v_{\bf k}^2 = 1$.

The same can be found in the original paper by Bardeen, Cooper and Schrieffer where you can read

\begin{equation} |BCS\rangle = \prod_{\bf k}( (1-h_{\bf{k}})^{1/2} + h_{\bf{k}}^{1/2} b^{\dagger}_k |0\rangle. \end{equation}

But in other books like the one by Annett those parameters are considered complex.

Why the parameter $u_{\bf k}$ and $v_{\bf k}$ are taken real? Shouldn't they be taken complex amplitudes? Is this state a classical mixture?

My guess is that this choice is justified a posteriori, but I cannot find any detail about it in any reference I have read.

Improve this question
$\endgroup$
6
  • $\begingroup$ This is clearly explained here scholarpedia.org/article/Bardeen-Cooper-Schrieffer_theory $\endgroup$
    – KF Gauss
    Jan 31, 2019 at 13:04
  • $\begingroup$ The article says that "The phase of the ground state may be chosen so that with no loss of generality $u_k$ is real." Obviously, with a change of basis, you can always take one of the two parameters as real, but you still will have the other parameter $v_k$ complex in general. In fact just below is stated: $u=\sqrt{1-h}; v=\sqrt{h}\exp(i\phi)$ $\endgroup$
    – skdys
    Jan 31, 2019 at 13:10
  • $\begingroup$ Yes, the most general case includes a phase, so u,v are treated as components of a complex number with magnitude 1. Is your question on the physical meaning of a nonzero phase is? If so, you may want to edit your question $\endgroup$
    – KF Gauss
    Jan 31, 2019 at 13:31
  • $\begingroup$ My doubt is why I am allowed to ignore the phase, how can I be sure to come to the correct result? So, as I said in my question, why the parameters can be taken real? $\endgroup$
    – skdys
    Jan 31, 2019 at 13:46
  • $\begingroup$ If you read carefully the link you will see they do not ignore the phase, they simply minimize the free energy with respect to it and find it should be taken to be zero for their Hamiltonian. From there, all the usual BCS behaviors are derived. $\endgroup$
    – KF Gauss
    Jan 31, 2019 at 14:33

1 Answer 1

Reset to default
1
$\begingroup$

I have found the answer to my question on page 86 of the following document https://www.physik.tu-dresden.de/~timm/personal/teaching/thsup_w11/Theory_of_Superconductivity.pdf

Here is clear that we have no reason a priori to assume that the two parameters are real. Therefore we should start with complex amplitudes.

Searching for the energy of the states we get to

\begin{equation} \langle BCS|\hat{H} | BCS \rangle = \sum_{\bf{k}} 2 \xi_k |v_{\bf{k}}|^2+\sum_{\bf{k}} V_{\bf{kk'}} v_{\bf{k}}u^*_{\bf{k}}u_{\bf{k'}}v^*_{\bf{k'}} \end{equation}

From this expression is clear that, in order to avoid a complex energy, $v_{\bf{k}}$ and $u_{\bf{k}}$ should have the same phase. Therefore, without loss of generality, we can take both parameters as real numbers.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.