# Spin-wave approximation - bosonic operator question

Can someone explain the attached image for me please?

I do not understand how $$2\delta_{k, k'}a_{k'}^{\dagger}a_{k}$$ becomes $$a_{k}^{\dagger}a_{k} + a_{-k}^{\dagger}a_{-k}$$ to me it should just be $$2a_{k}^{\dagger}a_{k}$$

and also I do not understand how $$e^{-ik}a_{-k}a_{k} + e^{ik}a_{-k}^{\dagger}a_{k}^{\dagger} = \cos(k) a_{-k}a_{k} + \cos(k) a_{-k}^{\dagger}a_{k}^{\dagger}$$

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You have to remember that this occurs under the summation over the BZ - as the sum includes -k for every k, you can take, for example, $\sum_{k \in BZ} a^\dagger_k a_k \to \sum_{k \in BZ} a^\dagger_{-k} a_{-k}$ with impunity.

That is excellent, thanks theZ. However, it's still not clear to me why $$e^{-ik}a_{-k}a_{k} + e^{ik}a_{-k}^{\dagger}a_{k}^{\dagger} = \cos(k) a_{-k}a_{k} + \cos(k) a_{-k}^{\dagger}a_{k}^{\dagger}$$

Can you explain that? To me it implies that $$a_{-k}a_{k} = a_{-k}^{\dagger}a_{k}^{\dagger}$$ and I don't see why that should be true. Thanks again.

As I said, you must understand the equality after summing together k, -k. Look at the creation and annihilation terms separately. For, say, the creation part, call the term to be summed, as initially written, f(k). Call the term to be summed, as the text has rewritten it, g(k). f(k) + f(-k) = g(k) + g(-k) by the definition of cosine. If the operators were fermionic, you would get i sin(k).

Excellent, I finally got it! Thank you theZ. It makes perfect sense now.

Do you also happen to know about the BSC hamiltonian?

For instance, when the BCS Hamiltonian contains the summation $$\sum_{\vec{k} \sigma} c_{\vec{k}\sigma}^{\dagger}c_{\vec{k}\sigma}$$ does this also imply summation over -k and -sigma?