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In the approximation that the two electrons of the He atom moved independently of each other, we can say that electron 1 is in state $\psi_a(1)$ where $a$ represents the orbital quantum numbers $nlm$ for electron 1 and that electron 2 is in the state $\psi_b(2)$ where $b$ designates the $nlm$ values for electron 2.

Thus the atomic wavefunction can be $$\psi_{atom}=\psi_a(1) \psi_b(2).$$

I read that this expression for $\psi_{atom}$ is incorrect because if we let electron 2 in state $a$ and electron 1 in state $b$ instead, i.e. $$\psi_{atom}=\psi_a(2) \psi_b(1),$$ one see that the probability distribution $|\psi_{atom}|^2$ is not the same in both cases.

I have trouble seeing why $|\psi_{atom}|^2$ is not the same. Is it because that $\psi_a(1) \neq \psi_a(2)$? If so, why are they not equal? Electrons 1 and 2 are in the same physical state $a$ so intuitively I think that they should have the same wavefunctions.

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  • $\begingroup$ I think it is a fine wavefunction as long as you anti-symmetrize it - that is, subtract your second wavefunction from the first and divide the lot by the square root of two. $\endgroup$ – Paul Young Mar 6 at 16:12
  • $\begingroup$ The Hartree wave function $\psi_a(r_1) \psi_b(r_2)$ is not very accurate because it does not account for Pauli correlation between the electrons. $\endgroup$ – my2cts Mar 6 at 19:11
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The reason that you recognize that $|\psi_{\text{atom}}|$ should be the same if we interchange the two electrons among the states is precisely at the heart of the resolution. Well, one should actually realize that upon interchanging the two electrons, not only the norm of the wave-function should remain unchanged but the wave-function itself should remain physically unchanged. The reason is that the two electrons are completely identical particles and their interchange shouldn't change anything physical because there is no physical difference between the two electrons. This means that $\psi_{\text{atom}}$ can only pick up an overall phase factor.

Let's say that $\psi_{\text{atom}}$ is the most general linear combination of $\psi_a(1)\psi_b(2)$ and $\psi_a(2)\psi_b(1)$, i.e., $$\psi_{\text{atom}}=\alpha\psi_a(1)\psi_b(2)+\beta \psi_a(2)\psi_b(1) $$where $\alpha,\beta$ are some general complex coefficients. It can be easily shown (do it yourself!) that upon the interchanging of electrons, the new wave-function will differ from the old one only by an overall phase-factor if and only if $\alpha=\pm\beta$. The indistinguishability of particles forces this restriction upon the composite wave-function. The "plus-case" is the right choice for the Bosonic particles (you can take this as the definition of Bosonic particles) and the "minus-case" is the right choice for the Fermionic particles. Since the electrons are Fermions, one must go with $\alpha=-\beta$. This fixes the overall wave-function up to a physically irrelevant overall phase (provided you choose the obvious normalization for $\alpha,\beta$).

As you can see, this automatically resolves the issue of the norm being invariant under the interchange of particles--but not only that, it keeps the whole wave-function physically intact under such an interchange.

Edit

I noticed that I have used the subscripts $a,b$ to denote all the relevant quantum numbers associated with the two electrons--in particular, these indices include both the spatial and internal quantum numbers. Thus, the result that the Fermionic states should be anti-symmetric applies to this whole wave-function. As explained in this answer, it could very well be the case that the spatial parts and the spin parts of the wave-function are not individually anti-symmetric as long as the full wave-function is anti-symmetric.

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  • $\begingroup$ thanks for the answer. But i am still unsure how $\psi_a(1)$ differs from $\psi_a(2)$. I think that they are the same wavefunction of an electron in state a. Could you perhaps give an example of what exactly can $\psi_a(1)$ and $\psi_a(2)$ be? $\endgroup$ – TaeNyFan Mar 7 at 13:49
  • $\begingroup$ @TaeNyFan $\psi_a(1)$ and $\psi_a(2)$ are completely different because they simply don't live in the same Hilbert space. The first lives in the single-particle Hilbert space of the first particle while the second in the single-particle Hilbert space of the second particle. $\endgroup$ – Feynmans Out for Grumpy Cat Mar 7 at 15:11

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