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I have two annihilation operator $\hat{a}_i$ and $\hat{a}_s$.

I want to calculate the expected value of $\langle \hat{a}_i \hat{a}_s \rangle$. How should proceed with this. The other information I have is that I know $\langle \hat{a}_i^\dagger \hat{a}_i \rangle = N_i$, i.e. mean photon number of the first annihilation operator is $N_i$ and similarly for the second annihilation operator, it is $N_s$. What I am looking for is $\langle \hat{a}_i \hat{a}_s \rangle$ in terms of $N_s$ and $N_i$.

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    $\begingroup$ What does your notation $\langle \dots \rangle$ mean? $\endgroup$
    – Prahar
    Commented Apr 16, 2021 at 8:55
  • $\begingroup$ Expectation @PraharMitra $\endgroup$ Commented Apr 18, 2021 at 7:29
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    $\begingroup$ Expectation value in which state? $\endgroup$
    – Prahar
    Commented Apr 18, 2021 at 7:39

1 Answer 1

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The result depends on the kind of averaging used. In many problems the averaging is done over the states which are also the eigenstates of the number operators, $N_i=a_i^\dagger a_i$, which means that a state with two particles annihilated will necessarily have different quantum numbers from the original states, in other words - the two states are orthogonal and the result is zero.

This is manifestly not the case in superconducting ro superfluid states, which have definite phase, but uncertain number of particles. However, the average $\langle a_i^\dagger a_i\rangle$ is still a meaningful quantity. In other words, the OP tehcnically does not contain enough information for a clear answer.

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