# Prove that this operator is unitary

$$\hat{O}\equiv(1/\sqrt{2\pi})\int e^{-iNz}dz$$

$$\hat{O}^\dagger\equiv(1/\sqrt{2\pi})\int e^{iN'x}dx$$

We have the operator $\hat{O}$ and its Hermitian adjoint $\hat{O}^\dagger$, 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 $\hat{O}^\dagger\hat{O}=\hat{I}$, I got:

$\hat{O}^\dagger\hat{O}=(1/2\pi)\int\int e^{i(N'-N)x}dxdx$

I did the double integral and the answer does not turn out nice. I know the periodicity of the function is $2\pi$, 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 $\hat{O}^{-1}=\hat{O}^*$. That did not turn out well either.

How should I go about proving that $\hat{O}$ is unitary?

• what is $N$? is it a number? if it is then your operator is the multiplication by a number (not very interesting) and it is unitari only if it is a complex number of modulus one. but I suspect the definition you gave is not so precise...and anyways, as pointed out, the solution below does not seem to be correct – yuggib Mar 21 '15 at 7:58

$$\delta (x-\alpha )={\frac {1}{2\pi }}\int _{{-\infty }}^{\infty }e^{{ip(x-\alpha )}}\ dp$$
$$\hat{O}^\dagger\hat{O}=(1/2\pi)\int\int e^{i(N'-N)x}dxdx = \int \delta(N-N') dx = 1$$ for N = N'.
• $dx dx$? I think you made a typo... – Danu Mar 21 '15 at 2:27
• This is wrong. This is not the correct answer and should not have been marked as such. You have written $\int \delta(N-N')dx=1$. That makes no sense. – hft Mar 21 '15 at 6:45