# Why is this function an eigenfunction of $\hat{L}_{z}$?

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

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.

• If both sides $\uparrow$ are not same, it clearly means the given is not an eigenfunction. You don't need any expansions and such. – Sunyam Jan 9 '19 at 16:46
• This function is not an eigenfunction of $\hat{L}_z$. You have probably missed $i=\sqrt{-1}$ in your formulas. – Gec Jan 9 '19 at 16:47
• @Gec I thought about that $i$ complex factor so that either the notes are incorrect or the function is not well expressed. – S. Rubio Jan 9 '19 at 17:07