I saw the following argument for calculating the energy levels of a helium atom.

First, ignore the Coulomb interaction term between two electrons. For this simplified model, we have the same solutions as in the hydrogen atom: so let's say that electron 1 is in $\psi_1=\psi_{n,l,m}$ and electron 2 is in $\psi_2 = \psi_{n', l', m'}$.

Now since these electrons are indistinguishable, we must antisymmetrize or symmetrize the spatial wavefunction. (These correspond to orthohelium/parahelium respectively.)

$$ \psi = \frac 1 {\sqrt 2}(\psi_1(r_1)\psi_2(r_2)\pm \psi_2(r_1)\psi_1(r_2))$$

Then according to the first-order perturbation theory, we calculate

$$\left\langle \psi \left| \frac{ e^2 }{4 \pi \epsilon_0 |r_1 - r_2| } \right| \psi \right\rangle.$$

However, since the energy levels are degenerate, we must find some 'good quantum numbers' for the above formula to be correct. I cannot see what are these 'good quantum numbers' for the helium atom. This is because the perturbation $H' = \frac{ e^2 }{4 \pi \epsilon_0 |r_1 - r_2| }$ does not commute with the usual operators for $n,l, m$.

Could somebody help me, please?

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    $\begingroup$ Perturbation theory does not require the perturbation to commute with the unperturbed hamiltonian. Why do you think so? $\endgroup$ – Emilio Pisanty Jan 11 at 8:15
  • $\begingroup$ As far as I know: In the degenerate case, not all eigenfunctions of the Hamiltonian can be used as a starting point. That is, even if $H^{(0)} \psi^{(0)} = E^{(0)} \psi^{(0)}$, there might not exist a curve $\psi(\lambda) = \psi ^{(0)} + \lambda \psi^{(1)} + \lambda ^2 \psi^{(2)} + \cdots$ which is a eigenfunction of $ H^{(0)} + \lambda H'$ for each $\lambda \in [0,1]$. $\endgroup$ – Moca Aoba Jan 11 at 12:15
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    $\begingroup$ The terminology 'good quantum number' is standard. You should also look up the term 'complete set of commuting observables' (CSCO), which is the correct formal description for your $A_i$. $\endgroup$ – Emilio Pisanty Jan 11 at 14:50
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    $\begingroup$ The condition you've stated is excessively restrictive, by far. If there is a CSCO which commutes with both the unperturbed hamiltonian $H^{(0)}$ and the perturbation $H'$ (i.e. $[A_i,H']=0$) then there is no need for perturbation theory $-$ $H'$ is already diagonal in the CSCO eigenbasis. (To prove this, take any two basis states $|\phi⟩$, $|\psi⟩$, and $A_i$ with distinct eigenvalues for both, from which $⟨\phi|[H',A_i]|\psi⟩=0$ implies $⟨\phi|H'|\psi⟩=0$, i.e. all off-diagonal elements of $H'$ are zero.) Where did you find the claim that this condition is necessary? It is not in Griffiths. $\endgroup$ – Emilio Pisanty Jan 11 at 15:00
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    $\begingroup$ ... and neither is the claim that the curve $\psi(\lambda)$ might not exist. What specific source are you taking that from? (It would also help to be very specific regarding whether this is for degenerate or non-degenerate perturbation theory.) $\endgroup$ – Emilio Pisanty Jan 11 at 15:01

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