I don't quite understand how the exchange of two indistinguishable particles works. In the image I imagine I have two fermions of spin 1/2 in a kind of Hydrogen atom with two discrete energy levels, $E_1$ and $E_2$. The fermions are non-interacting so the total Hamiltonian is given by the sum of the individual Hamiltonians: $H=H_1+H_2$ and the total eigenfunction is given by $\psi_{\alpha}(\vec{r})=\psi_{\alpha_1}(\vec{r_1})\psi_{\alpha_2}(\vec{r_2})$ where $\alpha_1$ and $\alpha_2$ are the set of quantum numbers of particles $1$ and $2$. From what I know from theory, in the case where the eigenfunction is of the type $\psi_{\alpha}(\vec{r})=\psi_{\alpha_1}(\vec{r_1})\psi_{\alpha_2}(\vec{r_2})$ exchanging two particles means exchanging the set of quantum numbers i.e., if $\hat{C}$ is the exchange operator I have: $\hat{C}\psi_{\alpha_1}(\vec{r_1})\psi_{\alpha_2}(\vec{r_2})=\psi_{\alpha_2}(\vec{r_1})\psi_{\alpha_1}(\vec{r_2})$

enter image description here

Now comes the question:

with reference to my image where the two arrows above the fermions indicate the component along the z-axis of the single particle spin, if I swap the two particles, and from the above this means swapping the two quantum numbers, what do I get? Does it make sense to swap these two identical fermions? I understand that mathematically I get a new eigenfunction using the exchange operator $\hat{C}$ but physically don't I get the same situation as before? That is, a fermion in the $E_1$ level with $S_z=1/2$ and a fermion in the $E_2$ level with $S_z=1/2$? In particular I don't understand what happens when I swap quantum numbers, physically speaking is it like taking one fermion and putting it in place of the other? What does $\vec{r_1}$ mean, is it the coordinate relative to particle $1$? But if the particles are indistinguishable what does it mean to give particle 1 the quantum numbers of particle 2?


1 Answer 1


Your figure assigns the label 1 to the inner particle and the label 2 to the outer. You could have changed this by assigning label 2 to the inner particle and label 1 to the outer one.

Both drawings represent solutions with the same energy since $E=E_1+E_2=E_2+E_1$ so a linear combination of the two solutions is also a solution. In other words, the properly symmetrized solution is not just “your drawing” but a “linear combination of your drawing plus one where the labels on $\vec r$ are interchanged”.

Which linear combination depends on the total spin state. If the electrons are in a state with total spin $S=0$, and thus antisymmetric under permutation of particle labels, then the “linear combination of your drawings” should be symmetric. If the electrons are in a state of total $S=1$, then the spatial part must be antisymmetric.

As a result, because the spatial part is a combination of two drawings, it does not make sense to think of particle 1 at the inner radius since it could equally well be that particle 1 is on the outer radius.

  • $\begingroup$ I don't understand how a linear combination of the two solutions is also a solution. If I physically "exchange" the two particles I do not obtain something different from my initial state, so why do I have to count both situations? In my mind is like saying: "in one case I have a particle with energy $E_1$ and spin $S_z=1/2$ and a particle with energy $E_2$ and spin $S_z=1/2$ BUT in the other case I have a particle with energy $E_1$ and spin $S_z=1/2$ and a particle with energy $E_2$ and spin $S_z=1/2$". How do the two identical cases change my eigenfunctions? $\endgroup$
    – Salmon
    Feb 26, 2022 at 20:29
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    $\begingroup$ Just try for - say - the infinite well. You can easily verify that $\psi_a(x_1)\psi_b(x_2)$ has energy $E_a+E_b$ and $\psi_b(x_1)\psi_a(x_2)$ is also an eigenstate with the same energy. By looking at symmetric or antisymmetric combination, you effectively remove the notion that particle 1 is in state 1 and particle 2 in state 2, which does not make sense if both particles are indistinguishable. The idea of indistinguishability is treated in elementary textbooks. Note that a linear combination of solutions with the same total energy is also a solution. $\endgroup$ Feb 26, 2022 at 21:25
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    $\begingroup$ You're misinterpreting "getting a new physical state": in quantum mechanics, such a statement does not make sense for indistinguishable particles unless the state is properly symmetrized. Thus, you do not get a new physical state if you interchange particles 1 and 2 since you state only picks up a $\pm 1$ and two states differing by an overall phase are equivalent. Additionally, we are not dealing with anything statistical here, much less quantum statistics. see also physics.stackexchange.com/q/614778/36194 $\endgroup$ Feb 27, 2022 at 19:21
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    $\begingroup$ the physical state is the linear combination. $\endgroup$ Feb 28, 2022 at 0:36
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    $\begingroup$ The physical states could be (up to normalization) $\vert\Psi_1\rangle=(\phi_1(r_1)\phi_2(r_2)+\phi_2(r_1)\phi_1(r_2))\left(\vert +\rangle_1\vert -\rangle_2 -\vert -\rangle_1\vert +\rangle_2\right)$ , or $\vert\Psi_2\rangle=(\phi(r_1)\phi_(r_2)-\phi_2(r_1)\phi_1(r_2))\left(\vert +\rangle_1\vert -\rangle_2 +\vert -\rangle_1\vert +\rangle_2\right)$. Both are antisymmetric under interchange of particles $1$ and $2$ but $\vert\Psi_1\rangle$ has total spin $S=0$ whereas $\vert \Psi_2\rangle$ has $S=1$. Both states have the same energy $E_1+E_2$ (assuming no spin dependence in the Hamiltonian). $\endgroup$ Feb 28, 2022 at 16:41

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