# Why in the BCS ground state the probability amplitudes are taken real?

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.

• This is clearly explained here scholarpedia.org/article/Bardeen-Cooper-Schrieffer_theory Jan 31, 2019 at 13:04
• 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: $u=\sqrt{1-h}; v=\sqrt{h}\exp(i\phi)$ Jan 31, 2019 at 13:10
• 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 Jan 31, 2019 at 13:31
• 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? Jan 31, 2019 at 13:46
• 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. Jan 31, 2019 at 14:33

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