# How to correctly apply the $L^2$-operator to a wave function?

If I have a wave function that says $$\psi = \alpha Y_1^1 + \beta Y_1^0 + \gamma Y_1^{-1},$$ then it is clearly that this wave function is an eigenfunction of $$\hat{L}^2$$ with its $$l$$-value being $$1$$. However, when I do $$\hat{L}^2\psi$$, I get something like $$\hat{L}^2\psi = \hbar^2\left[1(1 + 1) + 1(1+1) + 1(1+1) \right] = 6\hbar^2\psi$$ (note: here I am just apply $$\hat{L}^2$$ to the three spherical harmonics individually using $$\hat{L}^2Y_l^m = \hbar^2l(l+1)Y_l^m$$), which implies that the $$l$$-value is $$2$$. So am I applying the operator incorrectly?

• Your second equation should read $\hat{L}^2\psi = \hbar^2\left[1(1 + 1)\alpha Y_1^1 + 1(1+1) \beta Y_1^0 + 1(1+1) \gamma Y_1^{-1} \right] = 2\hbar^2(\alpha Y_1^1 + \beta Y_1^0 + \gamma Y_1^{-1})$. – gj255 Jan 25 at 9:57
• Okay. This answers my question. – Kane Billiot Jan 25 at 12:18

I realized what I did wrong. When I apply the operator to the wave function, I should have gotten $$\hat{L}^2\psi = \hbar^2 \left[1(1+1)\alpha Y_1^1 + 1(1+1)\beta Y_1^0 + 1(1+1)\gamma Y_1^{-1} \right]$$ like @gj255 pointed out in the comment. This means I would get $$2\hbar^2$$ as my eigenvalue, which is what we expected.