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?

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    $\begingroup$ Your second equation should read $\hat{L}^2\psi = \hbar^2\left[1(1 + 1)\alpha Y_1^1 + 1(1+1) \beta Y_1^0 + 1(1+1) \gamma Y_1^{-1} \right] = 2\hbar^2(\alpha Y_1^1 + \beta Y_1^0 + \gamma Y_1^{-1})$. $\endgroup$ – gj255 Jan 25 '19 at 9:57
  • $\begingroup$ Okay. This answers my question. $\endgroup$ – Kane Billiot Jan 25 '19 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.

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