If a hydrogen atom is in the state $\Psi = \frac{1}{\sqrt{3}}(\psi_{100} + \psi_{211}+\psi_{21-1})$ and the square of the angular momenta was measured to be $2\hbar^2$. After the measurement will the state be $\Psi' = \frac{1}{\sqrt{2}}[\psi_{211}+\psi_{21-1}]$ or $\frac{1}{\sqrt{2}}[Y^{1}_{1}+Y^{1}_{-1}]$ with $Y_m^l$ a spherical harmonic function?


1 Answer 1


After the measurement the state of your Hydrogen atom will be $\frac{1}{\sqrt{2}}[\psi_{211}+\psi_{21-1}]$. With the measurement you are projecting onto the states with angular momentum quantum number $l=1$ as the eigenvalues of $L^2$ are $l(l+1)\hbar$.

The state of your Hydrogen atom remains of course in a "state of the Hydrogen atom". Which contains the Spherical Harmonics, but also information about the radial component. Therefore it is not correct to write the remaining state solely in Spherical Harmonics.

You can make that explicit by writing $\psi_{n,l,m}(r,\theta,\phi)$ and then you see that you would loose the $r$ if you take your proposition with solely Spherical Harmonics.

  • $\begingroup$ Yes, I thought so, but I saw different answer in some test. Thanks! $\endgroup$ Commented Nov 15, 2022 at 14:45
  • $\begingroup$ @NadavAmitai If this answered your question, consider to accept it. $\endgroup$ Commented Nov 15, 2022 at 14:47
  • $\begingroup$ @NadavAmitai Think of the wave function as a description of the state of your atom. It does not make any sense to talk about the state of an atom without the information about the radial distribution. $\endgroup$
    – jan0155
    Commented Nov 15, 2022 at 14:50

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