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In section 3.7 of his book Introduction to Superconductivity (2nd Ed.), Tinkham states that

[...] we note that S has the eigenvalue $e^{i\varphi}$ in a BCS state in which the the phase of $\Delta$ [...] is $\varphi$.

where S is defined simply as an operator

[...] which annihilates a Cooper pair [...].

I'm trying to prove the above statement. Since Tinkham does not give a explicit definition of this S operator, nor do the references which he cites (Josephson and Bardeen), I am supposing that he meant something in the likings of \begin{align} \hat{S} = \sum_{k} \Phi_k^* \hat{c}_{k \uparrow} \hat{c}_{-k\downarrow}. \end{align}

With this definition, the BCS groundstate can be written, up to normalization, as \begin{align} \vert BCS \rangle &\propto \exp(\hat{S}^{\dagger}) \vert 0 \rangle, \end{align} where $\vert 0 \rangle$ is the vacuum. By applying the Cooper pair destruction operator and after some calculation, however, I find that \begin{align} \hat{S}\vert BCS \rangle &= \exp(\hat{S}^{\dagger}) \sum_k \vert \Phi_k \vert^2 \left( 1 - \Phi_k \hat{c}^{\dagger}_{k \uparrow} \hat{c}^{\dagger}_{-k \downarrow} \right)\vert 0 \rangle. \end{align}

The first term inside the parenthesis gives origin to a term parallel to the BCS state, which is what we wanted. The second term, however, does not seem to do so. This means that the overall state is not parallel to the BCS state and thus the BCS state is not an eigenstate of $\hat{S}$. I will be happy to share any part of the calculations that seem necessary. In particular, the second term that I found seems to stem from the fact that the Cooper pairs are not legitimate bosons, that is they don't obey the usual bosonic commutation relation \begin{align} [\hat{b}, \hat{b}^{\dagger}] &= 1, \end{align} but a modified version \begin{align} [\hat{S}, \hat{S}^{\dagger}] &= \sum_k \vert \Phi_k \vert^2 \left( 1 - \hat{n}_{k \uparrow} - \hat{n}_{k \downarrow} \right). \end{align}

I have checked multiple times my calculations and was not able to find a mistake, which leaves me to suspect of two options:

  • Either the definition of $\hat{S}$ I used is not the one used by Tinkham;

or

  • There is an approximation that should be made somewhere so that the second term vanishes.

Maybe both, who knows. My question is then:

Do any and which of the above statements apply? Please, provide sources or own derivation.

Lucas Baldo
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  • In Quantum Field Theory for the Gifted Amateur, by Lancester and Blundell, chapter 44, the pair creation operator is defined by $P^\dagger_p \equiv c^\dagger_{p\uparrow}c^\dagger_{-p\downarrow}$, such that $|\mathrm{BCS}\rangle = \prod\limits_p C_p,e^{\alpha_p P^\dagger_p} |0\rangle $. – Tobias Fünke Oct 12 '21 at 20:05
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    @Jakob yes, but you can put the product into the exponent as a sum and then you get the same expressions, no? – Lucas Baldo Oct 12 '21 at 20:37
  • Have you tried to compute the expectation value of $S$ ($P_p$ respectively) in the BCS state? Does it coincide with the literature? – Tobias Fünke Oct 13 '21 at 12:12
  • @Jakob where can I find this value in literature? Does Blundell calculate it? I don't think Tinkham does it. – Lucas Baldo Oct 13 '21 at 13:18
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    I think it is something like $u_p^* , v_p$ if you write $|\mathrm{BCS}\rangle = \prod_p \left(u_p + v_p, P_p^\dagger \right) |0\rangle$. Check: these lecture notes right below equation 95. However, note that they presumably assume that $u_k$ and $v_k$ are real. Actually I think this is a really nice question, since in the literature (as far as I can see) it is never demonstrated that $|\mathrm{BCS}\rangle$ is an eigenstate of the pair annihilation operator, but it is claimed that it is an coherent state... – Tobias Fünke Oct 13 '21 at 13:21
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    See also these lecture notes by S. Simon, equations 11.12 and 11.13 for the anomalous expectation values. Check also section 11.7 for a brief discussion concerning coherent states and the Josephson effect (which I think is also what Tinkham refers to, no?). – Tobias Fünke Oct 13 '21 at 13:55
  • Thanks, I will take a look at those references. It is really weird if there is really no proof of the BCS being an eigenstate in the literature... And yes, Tinkham makes this statement in preparation to the Josephson effect, I believe. – Lucas Baldo Oct 13 '21 at 16:25
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    Indeed. I tried to prove it, but failed. However, I obtained the correct expectation value... So either $|\mathrm{BCS}\rangle$ is really not an eigenstate of the pair annihilation operator and is called coherent state for other similarities with the coherent state of e.g. a superfluid or QHO (and Tinkham refers to the states just before he mentions this and not to $|\mathrm{BCS}\rangle$?)... or perhaps the proof is trivial and I'm doing something wrong. In any case, let me know if you made some progress. – Tobias Fünke Oct 13 '21 at 16:33

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