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Suppose I have 2 electrons under an harmonic potential: $$H=\frac{1}{2m}(p_1^2+p_2^2)+\frac{1}{2}m\omega ^2 (x_1^2+x_2^2)$$ Now let's think about the eigenvalue and the eigenfunctions of energy of the first excited level (the one above the ground state): well since the Hamiltonian is separable the energy must be: $$E_{n_1,n_2}=E_{n_1}+E_{n_2}=\hbar \omega (n_1+n_2+1)$$ So for the first excited level we get: $$E_{1,0}=E_{0,1}=2\hbar \omega$$ No problem here. But now, again for the separability of the Hamiltonian, the eigenfunctions must be: $$\psi(x_1,x_2)=u(x_1)u(x_2)$$ where with $u_n(x)$ we denote the the nth eigenfunction of the harmonic oscillator. So in our case we get: $$\psi(x_1,x_2)=u_1(x_1)u_0(x_2) \ \ \ \ \ or \ \ \ \ \ \psi(x_1,x_2)=u_0(x_1)u_1(x_2) \tag{1}$$ but now comes the problem: I have seen lectures in which the eigenfunctions are then written as combinations of this two solutions: $$\psi_{sym}(x_1,x_2)=\frac{1}{\sqrt{2}}[u_1(x_1)u_0(x_2)+u_0(x_1)u_1(x_2)] \tag{2}$$ $$\psi_{asym}(x_1,x_2)=\frac{1}{\sqrt{2}}[u_1(x_1)u_0(x_2)-u_0(x_1)u_1(x_2)] \tag{3}$$ it is claimed that this two are the two true solutions for the eigenfunctions of energy and not (1). This is of course very convenient since we have a symmetric and an antisymmetric solution, so it is now easy to introduce the spin and assign the singlet to the first and the triplet to the second. (Since we are dealing with fermions that must have antisymmetric wavefunctions). So I get how to solve the exercise.

My problem is that I don't understand why the solution are (2) and (3) instead of (1). Why are we doing linear combinations of the solutions? Why specifically in this way? (I get that $1/\sqrt{2}$ is a normalization constant) What postulate/theorem is telling us to do this instead of accepting (1) as the solutions? And also if we are doing linear combinations why we get only two liner combination, the symmetric and the antisymmetric one, instead of a bunch of those? Like for example: $$\psi_{sym}(x_1,x_2)=\frac{1}{\sqrt{2}}[-u_1(x_1)u_0(x_2)-u_0(x_1)u_1(x_2)]$$


In response to the current answers: One of the problems I have with this is that I do not understand the need to combine the two solution present in (1) to get the true solution. More details on what does not convince me are present in the comments I left under the answers of noah and Ozz.

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  • $\begingroup$ Well, you basically answered your question on your own (if I understood it correctly): The many-body wave function of fermions must be anti-symmetrized... $\endgroup$ Feb 5, 2021 at 11:02

3 Answers 3

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The postulate forcing this form is the indistinguishability of identical particles. It states that identical particles (say, two electrons) do not carry a "label" identifying them as "electron 1" or "electron 2", and one can truly not tell them apart.

We can introduce an exchange operator $P_{1,2}$, that just flips the coordinates of the two particles ($x_1 \leftrightarrow x_2$). Exchanging them twice clearly gives the same situation again (let's ignore anyons in this context), so the eigenvalues of $P_{1,2}$ are $\pm 1$. Turns out that the particles having the $+$ are bosons, fermions have the negative eigenvalue.

So after all this, we know the solution must only change sign (but everything else stays the same) upon exchange of the coordinates, which means $\psi(x_1, x_2) \overset{!}{=} -\psi(x_2, x_1)$. This is not the case in (1). However, it is true in (2) and (3) if we also included the spin part (the symmetric solution will have an antisymmetric spin part).

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  • $\begingroup$ And the fact that the general solution is the linear combination of particular solutions follows from the linearity of the Schrodinger equation? I would like to understand why are we authorize to get (2) and (3) from one. Because another possibility is to state that there are no solution at all since (1) is not in agreement with the postulate of indistinguishability. You have explained to me why (1) is not an acceptable solution. But I also would like to understand precisely how we derive (2) and (3). $\endgroup$
    – Noumeno
    Feb 5, 2021 at 11:28
  • $\begingroup$ And also: Solution (1) is symmetric right? Why can't we just attach the antisymmetric singlet to it, making it antisymmetric, and call it a day? $\endgroup$
    – Noumeno
    Feb 5, 2021 at 11:34
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    $\begingroup$ (1) is not symmetric, $u_0(x_1)u_1(x_2) \neq u_0(x_2)u_1(x_1)$ because $u_0 \neq u_1$. $\endgroup$
    – noah
    Feb 5, 2021 at 11:38
  • $\begingroup$ Try writing down another normalized symmetric solution except (2). $\endgroup$
    – noah
    Feb 5, 2021 at 11:40
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I cannot follow your reasoning for assuming the two solutions (1). These are special solutions, but any linear combination of those is also a solution (in fact, a more general one). The last solution you give is the same as the original $\psi_{sym}$, up to a sign. This sign is irrelevant since we are only interested in quantities like $|\psi_{sym}|^2$. The two solutions (2) and (3) are precisely such, so that they describe (as you say) fermions.

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  • $\begingroup$ I would like a lengthier explanation. For example as you say my solution (1) are in fact solution of the problem, (1) are eigenfunctions of energy, but they are then discarded! In fact solution of the form (2) and (3) are not equivalent to (1). Of course if we would have done a proper linear combination $au_1(x_1)u_0(x_2)+bu_0(x_1)u_1(x_2)$ then (1) would be included with $a$ or $b$ equal to zero! But with (2) and (3) seem to me that we are discarding solutions! And also who is telling us that a liner combination is a more general solution? is it the linearity of the schrodinger equation? $\endgroup$
    – Noumeno
    Feb 5, 2021 at 11:17
  • $\begingroup$ @Noumeno : As I commented, the wave function of fermions must be anti-symmetrized, which in particular implies for your example that a solution of the Schrödinger equation $\Psi(x_1,x_2)$ must fulfill $\Psi(x,x) = 0$. Do your solutions in $(1)$ obey this property? $\endgroup$ Feb 5, 2021 at 11:19
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It's just a fact that the physical states of a system of $N$ indistinguishable particles are eigenstates of the exchange operator $\chi |1\rangle |2\rangle = |2\rangle |1\rangle$.

With this in mind, if you want eigenstates of some Hamiltonian $H$ that are also physical states, then you are faced with the task of simultaneously diagonalising $\chi,H$. Of course the first two states you said are eigenstates of $H$, but they aren't of $\chi$. If they were distinguishable, we wouldn't require them to be eigenstates of $\chi$ and you could use whatever basis you want in that degenerate 2-dimensional subspace, but here they are indistinguishable and the subspace breaks up into two parts with eigenvalues $\pm 1$ for $\chi$.

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