ˆO≡(1/√2π)∫e−iNzdz
ˆO†≡(1/√2π)∫eiN′xdx
We have the operator ˆO and its Hermitian adjoint ˆO†, in the one dimensional space where x is position. I am trying to prove that this is a unitary operator. I'm told that N′ does not necessarily equal N. So when I tried the old ˆO†ˆO=ˆI, I got:
ˆO†ˆO=(1/2π)∫∫ei(N′−N)xdxdx
I did the double integral and the answer does not turn out nice. I know the periodicity of the function is 2π, but I'm not sure how that helps cancel the denominator. Also confused on what I'm supposed to do with the N terms.
Also tried using ˆO−1=ˆO∗. That did not turn out well either.
How should I go about proving that ˆO is unitary?
Answer
This is a well known definition of a delta function:
δ(x−α)=12π∫∞−∞eip(x−α) dp
therefore:
ˆO†ˆO=(1/2π)∫∫ei(N′−N)xdxdx=∫δ(N−N′)dx=1
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