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I am asked to show that an new defined operator:

$$U_{\beta} = \exp(\displaystyle\frac{i\beta L_z}{\hbar})$$

is unitary, where $$L_z = -i\hbar\,\,(x\displaystyle\frac{\partial}{\partial y} - y \frac{\partial}{\partial x }).$$

I tried the following:

$$ U_{\beta}^{\dagger} U_{\beta} = \exp \left( \frac{i\beta(-L_z^{\dagger}+L_z)}{\hbar} \right)$$

So I couldn't make the inside of exponential zero.

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Try explicitly calcultaing $L_z^{\dagger}$ from the differential representation you correctly gave for $L_z$.

Remember that the derivative is by definition an antihermitian operator (ie $\partial_x^{\dagger}=-\partial_x$)

Which relation you find between $L_z$ and $L_z^{\dagger}$?

How is this useful to solve your problem?

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  • $\begingroup$ Yes. I think I forgot to utilize $\partial_x^{\dagger} = -\partial_x$. When I did like that, I found $-L_z^{\dagger} + L_z = 0$. $\endgroup$ – Engin Eren Oct 26 '13 at 18:54
  • $\begingroup$ Yes, it is right. Furthermore it is very important from a physical point of view what @Danu was suggesting to you. $L_z$ is the third component of the angular momentum vector-operator in quantum mechanics. Since it is an observable (you can measure it) it HAS to be self-adjoint. It is one of the axioms of QM that all observables are self-adjoint. $\endgroup$ – Federico Carta Oct 26 '13 at 18:57
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You are talking about the operator $L_z$ here. As you probably know, $L_z$ is an observable (why?). What is the defining characteristic of observables in quantum mechanics?

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  • $\begingroup$ Right! I read my QM book (formalism chapter). I saw that "Observables are represented by hermitian operators". That is the same comment that @Federico make. I really appreciate for your help. $\endgroup$ – Engin Eren Oct 26 '13 at 19:13

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