Bosonic operators and fourier transformation.

[barnflakes]

If $$a_m = \frac{1}{\sqrt{N}} \sum_k e^{-ikm}a_k$$
where $$a_k$$ is a bosonic operator fulfilling $$[a_k, a_{k'}^{\dagger}] = \delta_{kk'}$$

then is the product $$a_m a_{m+1} = \frac{1}{N} \sum_k e^{-ikm}e^{-ik(m+1)}a_k a_{k+1}$$

? Because that's what I'm doing but it doesn't lead me anywhere near to the correct answer in my textbook.

 

[mathman]

NO! You need to first express it as a double sum. Then this can be collapsed using the δ function.

Also (I believe) the exponent should be -ikm+ik(m+1) - assuming you are using the bracket function. a_m+1 should be a complex conjugate.

 

[barnflakes]

OK I fixed that, thanks.

One thing is that my notes often say $$(a_k)^{\dagger}a_k = (a_{-k})^{\dagger}a_{-k}$$ but I did some calculations with the definition of a_k and didn't find this to be true. Any idea what is going on?

 

[mathman]

I am not familiar with the physics involved, so I don't know the definition for a_k.