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I'm studying the hydrogen atom. My professor says that in a spherically symmetric problem, the choice of the $z$-axis is arbitrary and so, given a fixed value of the orbital angular momentum quantum number $l$, all the states with same $l$ and different magnetic quantum number $m$ have to be equiprobable and this should also be the reason why the charge density of a full subshell is spherically symmetric.

Why is this the case? I don't understand why equiprobability is assumed here. If, for example, i have an electron in the hydrogen atom with a fixed value of the modulus of the orbital angular momentum $L$, then the orbital angular momentum quantum number $l$ is fixed, but the $z$-component of the orbital angular momentum can have $2l+1$ different values with probabilities depending on the coefficient multiplying the spherical harmonics in the eigenfunction expansion of the state function $\psi$. As far as i know, the only requisite is that the sum of the modulus squared of the coefficients of the eigenfunction expansion has to be equal to 1. How do I get this equiprobability condition?

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When we have degenerate or quasi-degenerate states, even a very small perturbation of a system prepared in a well-defined quantum state is enough to allow transitions to the other degenerate states. It is then reasonable to assume that a system, weakly interacting with a thermal environment, could be found in each of its degenerate eigenstates with equal probability.

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  • $\begingroup$ Why does this happen? In this chapter I'm treating the hydrogen atom by considering only the coulomb potential and no other perturbations $\endgroup$
    – Tom Avery
    Commented Jul 5, 2023 at 11:24
  • $\begingroup$ The real world is somewhat more complex than the idealized situations described in physics textbooks. The statement about the spherical symmetry of the electronic density is to explain what people would find in a real measurement of a non-prepared status. Of course, if one prepares an eigenstate of $l_z$, things will be different. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jul 5, 2023 at 11:29
  • $\begingroup$ I don't understand why transitions are involved here. If I measure the modulus of the orbital angular momentum and I get a certain value, why is there equiprobability that the z-component is in any of its possible values? $\endgroup$
    – Tom Avery
    Commented Jul 5, 2023 at 11:47
  • $\begingroup$ @TomAvery If the system has not been prepared in a specific state (for instance, with $L_z=1$, it is not possible to use the wavefunction description, and one needs to use the density matrix. Within the density matrix formalism, some weak assumption of coupling between the quantum system and the external world has to be introduced. If the system has time to equilibrate, the equiprobability of the degenerate states with the same energy is probably the simplest, and it is in agreement with the facts. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Jul 5, 2023 at 12:16
  • $\begingroup$ I see. In my course we only treated the wavefunction description, so I thought there was an explanation involving that description.Is there a simpler way or density matrices are strictly required? $\endgroup$
    – Tom Avery
    Commented Jul 5, 2023 at 12:55

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