[Physics] Probability of measuring eigenvalue of non-normalised eigenstate

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This came up while working on a question about measuring the angular momentum of a particle in a superposition of angular momentum eigenstates:

Given that:
$$\langle\theta,\phi|\psi\rangle \propto \sqrt{2} \cos(\theta) + \sin(\theta)e^{-i\phi} – \sin(\theta)e^{i\phi}$$

What are the possible results and the corresponding probabilities for measurements of $\hat{L}^2$ and $\hat{L}_z$?
$\hat{L}^2$ is simply $2\hbar^2$ as all three terms are eigenstates of $\hat{L}^2$ with eigenvalues $2\hbar^2$

However the three terms are eigenfunctions $\hat{L}_z$ with different eigenvalues, namely $0$, $\hbar$ and $-\hbar$. Now my question is whether I first have to normalise the eigenfunctions and then take the modulus squared of the coefficients to find the probabilites of measuring the corresponding eigenvalue, or whether it is possible to straight away write down: $$p(L_z=0)=\frac{|\sqrt{2}|^2}{|\sqrt{2}|^2+|1|^2+|-1|^2}$$

So basically my question is:

Given a wave function $|\psi\rangle$ and an operator $\hat{A}$, with eigenvalues $\lambda_i$ and non-normalised eigenfunction $|a_i\rangle$, and: $$|\psi\rangle = \sum_i{c_i|a_i\rangle}$$
Is it still true that the probability of obtaining a measurement $\lambda_i$ is given by $p_i=|c_i|^2$?

Best Answer

Yes and no. You can just normalize the results with:

$$ \langle \psi | \psi \rangle$$

but you have computed that incorrectly. Remember, the differential solid angle is:

$$ d\Omega = d(\cos{\theta})d\phi, $$

you have used:

$$ d\Omega = d\theta d\phi.$$

I suggest you verify with a table of $l=1$ spherical harmonics.

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