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

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

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?

  • $\begingroup$ 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 $\endgroup$ – yuggib Mar 21 '15 at 7:58

This is a well known definition of a delta function:

$$ \delta (x-\alpha )={\frac {1}{2\pi }}\int _{{-\infty }}^{\infty }e^{{ip(x-\alpha )}}\ dp $$


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

  • 1
    $\begingroup$ $dx dx$? I think you made a typo... $\endgroup$ – Danu Mar 21 '15 at 2:27
  • $\begingroup$ @Danu, hmm not sure about this I have to admit. Don't they operators have to be considered at equal positions? $\endgroup$ – Constandinos Damalas Mar 21 '15 at 2:34
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    $\begingroup$ 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. $\endgroup$ – hft Mar 21 '15 at 6:45
  • $\begingroup$ It did originally appear in the question, but yeah, that's something an answer should probably point out. You can't integrate over the same variable twice. (Not to be confused with two variables that share the same label, but that's not what's going on here.) $\endgroup$ – David Z Mar 21 '15 at 7:04

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