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$$\Psi(\varphi)=\frac{1}{\sqrt{2\pi}}(\sin\varphi-\cos\varphi)$$

I am not able to see why the above function is an eigenfunction of $\hat{L}_{z}$ and which is its eigenvalue. I've been trying with the definition of the operator and the following eigenvalue equation:

$$-i\hbar\frac{\partial}{\partial\varphi}\Psi(\varphi)=m\hbar\Psi(\varphi)$$

However I don't get the same expression on both sides of the equation. I've also tried expanding $\Psi(\varphi)$ in spherical harmonics but the integrals associated with $l=0,1$ turned out to be zero.

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  • $\begingroup$ If both sides $\uparrow$ are not same, it clearly means the given is not an eigenfunction. You don't need any expansions and such. $\endgroup$ – Sunyam Jan 9 at 16:46
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    $\begingroup$ This function is not an eigenfunction of $\hat{L}_z$. You have probably missed $i=\sqrt{-1}$ in your formulas. $\endgroup$ – Gec Jan 9 at 16:47
  • $\begingroup$ @Gec I thought about that $i$ complex factor so that either the notes are incorrect or the function is not well expressed. $\endgroup$ – S. Rubio Jan 9 at 17:07

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