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A coherent state is defined as the eigenstate of the annihilation operator $\hat{a}$. It can be obtained from the vacuum of the number operator by acting with displacement operator: $$|z\rangle=\hat{D}(z)|0\rangle=e^{-|z|^2/2}\exp[z\hat{a}^\dagger]\exp[-z^*\hat{a}|0\rangle\\ ~~~~~~~ =e^{-|z|^2/2}\exp[z\hat{a}^\dagger]|0\rangle$$ where in writing the last step we used $\hat{a}|0\rangle=0.$

Now the BCS ground state of a superconductor, written as (See Superconductivity and SUperfluidity by Annett or QFT by Blundell) $$|\Psi_{\rm BCS}\rangle=\prod_{\vec p}\frac{\exp\left[\alpha_{\vec p}\hat{P}^\dagger_{\vec p}\right]}{\sqrt{1+|\alpha_{\vec p}|^2}}|0\rangle$$ where $\hat{P}^\dagger_{\vec p}=\hat{c}_{\vec p}^\dagger \hat{c}^\dagger_{-\vec p}$ is the Cooper pair creation operator, is also called the BCS coherent state. But it does not look like the usual coherent state that I have written above. Then, why is this called a coherent state?

Solidification
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    I have the same doubt. I have asked a somehow related question here https://physics.stackexchange.com/questions/614108/ground-state-of-composite-and-non-interacting-bosons-on-a-lattice even though my focus was on the interpretation of strongly interacting electronic pairs as bosons in a superfluid state.

    If anyone can answer this question, then please check out also mine! thanks :)

     – Matteo Feb 20 '21 at 13:45
  • It's constructed by exponentiating creation operators. Have you tried acting on it with an annihilation operator? Can you expand a bit on why you think it doesn't look like a field theory generalization of a coherent state, where there are now creation and annihilation operators for each momentum? – J. Murray Feb 20 '21 at 16:06
  • @J.Murray Which annihilation operator? There are many pair annihilation operator $P_{\vec p}$, one for each ${\vec p}$. Things are also complicated because $[P_{\vec p},P^\dagger_{\vec q}]\neq \delta(\vec p -\vec q)$. I am not aware of what a field theory generalization of coherent state is. – Solidification Feb 20 '21 at 18:31
  • I think my answer to a similar question may cover this: https://physics.stackexchange.com/questions/546836/how-to-understand-the-condensation-of-cooper-pair-in-bcs-theory/547280#547280

    In particular, note how the BCS state is shown (in a grand canonical picture) to have a non-vanishing expectation value of an annihilation operator, as you'd want for a coherent state.

     – Rococo Feb 21 '21 at 17:41
  • Related but no answer so far. – Tobias Fünke Oct 13 '21 at 17:18
  • Perhaps it is meant that it is an eigenstate of the fermionic annihilation operator $c_p$? Have you tried that? – Tobias Fünke Dec 02 '23 at 21:20

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