In quantum 101, we all learned that identical particles behave strangely, even in the absence of interactions: no two fermions can be in the same state, but bosons love to be in the same state.

But what if you have some fermions, all with the same spin. However, they have other difference(s) that make them distinguishable (i.e. mirror image molecules), but don't affect the overall shape of the wave-function very much.

You put these molecules in a potential well in various states. You have a loose potential well so that the wavelengths of the different energy states are much larger than the size of the molecules; they are effectively point particles. There are too many fermions to all go in the ground state if they were identical, but they all can fit if they "know" that they are different.

My reasoning is that they would initially refuse to drop into the ground state, like fermions, but eventually the tiny differences in mass/binding energy would distinguish them and allow them to settle in. Even if for mirror-image molecules with the same mass, asymmetrical multi-polar moments would introduce interaction energy terms that are different. Thus they would all end up in the ground state, but much more slowly than a single molecule would.

Is this what would actually happen? Is there a way to estimate the "identicalness" (i.e. a difference of 0.001 eV in energy is slower to reach ground by a factor of x compared to a single molecule)?

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    $\begingroup$ Glaring contradiction: "same spin that differ by a tiny amount." It cannot be the same and different. $\endgroup$
    – Kyle Kanos
    Sep 26, 2013 at 2:54
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    $\begingroup$ " no two fermions can be in the same state". This is false. This concerns only identical fermions. $\endgroup$
    – Trimok
    Sep 26, 2013 at 5:46
  • $\begingroup$ @kyle: mirror-image molecules will have the same spin (if they are in the ground state). $\endgroup$ Sep 26, 2013 at 15:37
  • $\begingroup$ @Trimok: the word identical in the same sentence has been italicized so people don't miss it. $\endgroup$ Sep 26, 2013 at 15:40
  • $\begingroup$ @KevinKostlan: If two mirror image molecules have the same spin, then by definition their spins are not different. This is your contradiction: you want the spins to be both the same and different, but this is impossible. $\endgroup$
    – Kyle Kanos
    Sep 26, 2013 at 15:46

1 Answer 1


Short answer: "almost identical" particles are either truly identical, or not identical all for the purposes of quantum statistics. Quantum statistic don't care if your particle has all the same properties as an electron except for a mass $1.000000000000001 m_e$: that counts as a distinguishable particle from any electron. You might complain that this gives us a way to experimentally distinguish masses etc. to arbitrary precision and you would be right: it does allow that. The exception is if nature conspires to make life maximally difficult for us by always entangling particles and their dopplegangers in a way that almost forbids transitions to the ground state. There is no reason to believe this and nothing like it has been seen. The generic situation, even for very similar particles, involves rapid transitions with no Pauli blocking.

Long answer:

Quantum statistics are a particular kind of entanglement and "superselection" rules which say this: the only states which are a part of the Hilbert space of a system are those which are completely (anti)symmetric under the exhange of two identical particles. For this to have any force the particles must be truly identical, not merely "almost" identical, whatever that means.

This is actually an automatic consequence of the formalism of quantum field theory, which is more fundamental and where the required exchange (anti)symmetry is completely manifest and obvious. But in order to address the possibility that nature could be some other way I'll work in a formalism where the right answer is not manifest and obvious: the usual many body quantum mechanics. Here we have a many body wavefunction $\psi(r_1, r_2, r_3,\cdots,r_N)$ describing a fixed number of particles of given types, and which may or may not obey any symmetry properties.

Now let's imagine the system is two electrons with wavefunction $\psi(r_1, r_2)$. Fermi statistics tell us that $\psi(r_1, r_2)=-\psi(r_2,r_1)$. An arbitrary function of two coordinates won't do, though for any $f(r_1,r_2)$ we can make a valid state by $\psi \sim f(r_1,r_2)-f(r_2,r_1)$. The symmetric part $f(r_1,r_2)+f(r_2,r_1)$ is not a valid state. So what we are doing is decomposing the Hilbert space of all two particle wavefunctions into two parts: $\mathcal{H}_2=\mathcal{H}_- \oplus \mathcal{H}_+$ and cutting off the symmetric part $\mathcal{H}_+$, identifying the physical Hilbert space as $\mathcal{H}_{phys}=\mathcal{H}_-$. For $\geq 3$ particle systems we cut off all of the states with mixed symmetries as well, keeping only the totally antisymmetric subspace. Importantly states like $\psi(r_1,r_2)=\chi(r_1)\chi(r_2)$ don't exist because they are not antisymmetric and they can't be made antisymmetric. It's not that such states are just inaccesible, they simply don't exist.

Here is the key point: this can only possibly work for truly identical particles. What I mean is that exchange symmetry must really be a symmetry: the particle exchange operators must all commute with the Hamiltonian! Otherwise time evolution starting from an antisymmetric state would generate mixed symmetry states that are unphysical. This shows that fermions/bosons must be identical particles in every way. The proof going the other way, i.e. that identical particles must be either bosons or fermions is a consequence of the fact that two states differing by an exchange operation are physically indistinguishable, hence must correspond to the same ray in the physical Hilbert space. Thus exchanges take a ray to itself, i.e. change the state by at most a phase factor $\exp(i\phi)$, and the only representations of the permutation group as phase factors are the trivial (symmetric) representation (i.e., bosons) and the $\mathbb{Z}_2$ (antisymmetric) representation (i.e. fermions).

Now let's look at your situation. Imagine we have a system of an electron and a doppelganger, with $\psi(r_1; r_2)$ where the first argument refers to the electron and the second to the doppelganger. Now there is no reason for the state to obey any symmetry property. Certainly you can make an antisymmetric state, but you can make a symmetric state as well and the time evolution doesn't preserve any particular symmetry. Even if you can't practically distinguish them in your detector $\psi(r_1; r_2)$ and $\psi(r_2; r_1)$ are physically different states, hence they are orthogonal rays in Hilbert space. So the previous argument falls apart. The Hilbert space still breaks into smaller representations of the permutation group but the Hamiltonian doesn't preserve the subspaces: it generates evolution between them. If the exchange symmetry is almost good then the evolution between subspaces will be slower than evolution within subspaces.

Importantly the ground states $\chi_{e,d}$ for the electron and the doppelganger let you make $\chi_e(r_1)\chi_d(r_2)$, $\chi_e(r_2)\chi_d(r_1)$, $\chi_e(r_1)\chi_d(r_2)+\chi_e(r_2)\chi_d(r_1)$ and maybe even $\chi_e(r_1)\chi_d(r_2)-\chi_e(r_2)\chi_d(r_1)$! (The antisymmetric combination might not exist if the ground state wavefunctions $\chi_e = \chi_d$, which is the case for the infinite well but not a general potential. You may think this wavefunction is numerically small if the functions $\chi_e \approx \chi_d$, but you always normalize the many body states so this is no issue. The point is it's not always identically zero.) So there is no obstruction at all to the system dropping down the ground state. Moreover, since the properties of the electron and the doppelganger are very similar the transition rates from the excited states to the ground state will be similar for the two particles. So the transition takes place quickly. There is no Pauli blocking. It might take a long time for the symmetry character of the state to change significantly (because the difference in masses is small), but this doesn't matter since the system can drop into any combination of $\chi_e(r_1)\chi_d(r_2)$ and $\chi_e(r_2)\chi_d(r_1)$.

The only exception is if nature conspires to always produce states in a symmetry subspace that doesn't include the ground state, in which case you have to wait for the state to evolve into a different symmetry subspace which may take a while. But this is a conspiracy that has no good reason to happen. In particular it depends on the shape of the ground state wavefunctions for the electron and doppleganger which can be engineered to be different, so one could always design a new experiment if the one you tried doesn't work fast enough.

  • $\begingroup$ Nice explanation, one thing confuses me: "...it depends on the shape of the ground state wavefunctions for the electron and doppleganger". So if the shape is very similar (which would be necessary for my definition of "almost identical") you would see partial Pauli blocking? $\endgroup$ Sep 26, 2013 at 15:52
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    $\begingroup$ I was trying to evoke the situation where the ground state has a particular symmetry which is not shared by the initial state (actually, being an energy eigenstate of a system where exchanges don't commute with $H$, the ground state can't have a definite symmetry, but if the exchange is almost exact then the "spill over" of the ground state into other symmetry sectors can be small) then direct transitions are strongly suppressed and you have to wait for the system to evolve into another sector which includes the ground state. $\endgroup$
    – Michael
    Sep 26, 2013 at 16:13
  • $\begingroup$ Thanks. So it seems my reasoning is correct. Just to clarify: An "almost exact" exchange is similar to saying that the eigenstates are "almost" the same? And only transitions that would be forbidden if they were truly identical fermions would be suppressed/slowed down? $\endgroup$ Sep 26, 2013 at 16:32
  • $\begingroup$ 1) "Almost exact" means that the commutator of an exchange operator $\Pi$ with the Hamiltonian $[\Pi,H]$ is "small." It's a loose idea since it's really smallness of an operator, but you could say the commutator has small eigenvalues on a subspace of interest compared to the energies of those states. Something like that. 2) No. There are other possibilities, for instance a symmetric initial state decaying to an almost antisymmetric final state. Or initial and final states with different classes of mixed symmetries. $\endgroup$
    – Michael
    Sep 26, 2013 at 16:50
  • $\begingroup$ Ok thanks. I am not a specialist in QM, so your translation from math to intuition is very helpful. (2): I should have said: transitions that preserve symmetry or antisymmetry are not slowed down (antisymmetry is preserved in identical fermionic transitions, of course), but transitions "between" these "worlds" are suppressed. $\endgroup$ Sep 26, 2013 at 18:30

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