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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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1 Answers1

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It is better to understand "coherent" from the Cooper pair operator $$\Lambda^\dagger = \sum_k \alpha_k c^\dagger_{k\downarrow}c^\dagger_{-k\uparrow}$$ which, in real space, creates a pair of electron above Fermi surface $$\Lambda^\dagger = \int d^3x d^3x' \alpha(x - x') \psi_{\uparrow}^\dagger(x) \psi_{\downarrow}^\dagger(x') $$ It is then obvious that the BCS ground state is a coherent state with $|BCS\rangle = e^{\Lambda^\dagger}|0\rangle$ (normalizing factors ignored), as $$|BCS\rangle = \prod_k \exp\left(\alpha_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger \right) |0\rangle = \prod_k \left(1 + \alpha_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger \right) |0\rangle $$ The second equal sign follow from the fact $(c^\dagger_{k\downarrow}c^\dagger_{-k\uparrow})^n = 0 \; (n \geq 2)$ because you cannot create two pairs with same $k$. The last formula has the same form with the usual description using Bogoliubov operators.

This answer is based entirely on this very nice note. I hope it helps.

ninepoints
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