Consider an idealised hydrogen atom in the state $|\psi\rangle=\frac{1}{\sqrt{6}}(2|1,0,0\rangle-|2,1,0\rangle+|2,1,1\rangle)$ where $|n,l,m\rangle$ are the normalised eigenstates of the Hamiltonian $\hat{H}$ and of $|\hat{\vec{L}}^{2}|, \hat{L}_z$ where $\hat{L}$ is the angular momentum operator.

Assume now that we measure $\hat{L}_z$ with outcome $m=0$. What state is the system in after the measurement?

solution from my lecture notes: By the form of $|\psi\rangle$, the state after the measurement is $|\phi\rangle=\lambda(2|1,0,0\rangle-|2,1,0\rangle)$ with $\lambda$ the normalisation constant (so $\lambda=1/\sqrt{5}$).

Question: Why are the $|n,l,m\rangle$ in the expansion of $|\phi\rangle$ only ones that appear in the expansion of $|\psi\rangle$ and why are their coefficients preserved? Why not the other $|n,l,m\rangle$ with $n=1,2,\ldots$ and $l=0,1,\ldots,n-1$? I can see that their dot product with $|\psi\rangle$ vanishes and hence adding such a state to $|\phi\rangle$ does not change the probability of measuring this state. How do I continue?

  • 4
    $\begingroup$ The measurement can be mathematically enforced by a projection operator onto the space with eigenvalue $m=0$. Your first two kets are already in this subspace, so they don't change when projected, and since your states are orthogonal, the last ket vanishes upon projection. This projection agrees with the Born rule. $\endgroup$ Jun 28, 2021 at 11:09
  • $\begingroup$ This sounds like an answer. $\endgroup$
    – garyp
    Jun 28, 2021 at 11:21
  • $\begingroup$ @MariusLadegårdMeyer Thanks for the comment/answer but since projection is linear, shouldn't the $1/\sqrt{6}$ be preserved as well? $\endgroup$
    – test123
    Jun 28, 2021 at 11:24
  • 2
    $\begingroup$ @test123 if you want you can keep it, then the normalisation constant will be $\lambda=\sqrt{6/5}$ $\endgroup$
    – Andrea
    Jun 28, 2021 at 11:26

1 Answer 1


Strictly speaking, yes. As mentioned in the comments, a measurement can be modeled by the application of a projection operator onto some subspace of the Hilbert space corresponding to the measurement result. However, projection operators do not preserve normalization, so if $P$ projects onto the $m=0$ subspace, then

$$P|\psi\rangle =\frac{1}{\sqrt{6}}\big(2|1,0,0\rangle-|2,1,0\rangle\big)$$

which clearly is no longer normalized. This isn't a problem, because physical states are only defined up to a multiplicative constant, meaning that $|\phi\rangle$ and $\lambda |\phi\rangle$ represent exactly the same physical state. In that sense, representing the state of your system as a normalized vector is just a useful convention which you don't need to follow.

If you do choose to follow that convention, then you can re-normalize your state after the projective measurement by multiplying by an overall normalization constant, in this case given by $\sqrt{6/5}$. Obviously that changes the coefficients, but not the physical state.


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