Projecting onto an eigenstate when we make a measurement

[AxiomOfChoice]

Suppose we prepare a system in some properly normalized superposition of the spherical harmonics: A|11> + B|10> + C|1-1>. One of the fundamental results of quantum mechanics is that, if we measure L_z, we will collapse the state of the system onto an eigenstate of the eigenvalue we measure. My question is this: Suppose we measure L_z and we get 0. Which eigenstate of L_z do we collapse to? There are infinitely many spherical harmonics for which m = 0!

Thanks.

 

[jamesmaxwell]

If the original wave function is A|11> + B|10> + C|1-1> and it is normalizated, eigenfunction you can get when measurement of Lz is zero is l1 0>

I hope my reply help you

 

[StatusX]

The general rule is that when your measurement returns one of the eigenvalues, the state gets projected onto the corresponding eigenspace. That is, if there are n (orthonormal) eigenstates $|\lambda_i>$ corresponding to the eigenvalue $\lambda$, then if we define:

$$P_\lambda = |\lambda_1><\lambda_1| + |\lambda_2><\lambda_2| + ... + |\lambda_n><\lambda_n|$$

Then if a measurement of $|\psi>$ returns $\lambda$, then the state collapses to $P_\lambda |\psi>$.

If you have this degeneracy, it means there are other commuting observables you can still measure to distinguish between the different $|\lambda_i>$. In your case, one such observable would be the total angular momentum, and your state is already in an eigenstate of this, so there's only one $|\lambda_i>$ possible, |1 0>.