I have some questions about the BCS groundstate. This is given by

$$\prod_k (u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger) |0\rangle$$

I've heard many times that $|v_k|^2$ is the probability that the pair $(k\uparrow, -k\downarrow$) is occupied and $|u_k|^2$ similarly for unoccupied. However, that does not seem obvious to me. If there was only one momentum state, this interpretation is clear; however, it seems that the product complicates this. Here is a calculation that explains what I mean. The probability that I have a cooper pair with one of the electrons having momentum $l$ is the mod squared of

$$\langle 0|c_{l\uparrow} c_{-l\downarrow} \prod_k (u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger) |0\rangle$$

$$=\langle 0|[\prod_{k<l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] c_{l\uparrow} c_{-l\downarrow} (u_l + v_l c_{l\uparrow}^\dagger c_{-l\downarrow}^\dagger) [\prod_{k>l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] |0\rangle$$

$$=\langle 0|[\prod_{k<l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] (u_l c_{l\uparrow} c_{-l\downarrow} + v_l(1-n_{l\uparrow} - n_{-l\downarrow} + n_{-l\downarrow}n_{l\uparrow})) [\prod_{k>l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] |0\rangle$$

$$=(\prod_{k\neq l} u_k) v_l$$

But we just want the answer to be $v_l$. I realize that the state with a pair occupied is not only the state where there is only one pair, but I am not sure how to cleanly implement that.

I seems like when most people interpret $u_k$ and$v_k$, they make it sounds quite obvious, so there is probably a simple explanation. But if it would help if I was able to show this explicitly.


1 Answer 1


A good analogy could be thinking of a bosonic field with operators $b_k$. A superposition of states with mode $k$ empty and occupied is $$(u_k + v_kb_k^\dagger)|0\rangle = u_k|0\rangle + v_kb_k^\dagger|0\rangle$$ If we can have such a superposition for any state, we write $$|\Psi\rangle = \prod_k(u_k + v_kb_k^\dagger)|0\rangle$$ If we want to calculate the probability that there is a pair in state $q$, we follow the usual quantum mechanical prescription, taking the average of the particle number operator $n_q = b_q^\dagger b_q$: $$\langle \Psi|n_q|\Psi\rangle = \langle \Psi|b_q^\dagger b_q|\Psi\rangle = \langle 0|(u_q^*+b_q v_q^*)b_q^\dagger b_q(u_q+v_qb_q^\dagger)|0\rangle\times\\\prod_{k\neq q} \langle 0|(u_k^*+b_k v_k^*)(u_k+v_kb_k^\dagger)|0\rangle = |v_q|^2\prod_{k\neq q}(|u_k|^2 + |v_k|^2)$$ Note that here we will have a sum of averages over all the multiparticle states that have a particle in state $q$. This is different from projecting on a state $b_q^\dagger|0\rangle$, where we have only one particle, as is done in the OP. The difference is most clearly seen, if we use the resolution of identity in the Fock space, and write $$ \langle \Psi|b_q^\dagger b_q|\Psi\rangle = \sum_\phi\langle \Psi|b_q^\dagger|\phi\rangle\langle\phi b_q|\Psi\rangle =\sum_\phi|\langle\phi|b_q|\Psi\rangle|^2$$ The equation given in the OP is equivalent of taking only one term of the sum with $|\phi\rangle = b_q^\dagger|0\rangle$.

  • $\begingroup$ I just went through the calculation of $\langle \Psi | c^\dagger c |\Psi \rangle$ and got the correct answer of $|v_k|^2$, which makes sense! One last question would be, is it evident by looking at this wavefunction we will get this result? Or do you have to go through the ugly computation to see it? $\endgroup$
    – Tabin
    Jul 9, 2021 at 14:31
  • $\begingroup$ @Tabin I suppose one needs to have experience with this wave function, to see it right away. $\endgroup$ Jul 9, 2021 at 14:33
  • $\begingroup$ I see, thank you for the good answer. $\endgroup$
    – Tabin
    Jul 9, 2021 at 14:34

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