It is generally assumed that there is no limit on how many bosons are allowed to occupy the same quantum mechanical state. However, almost every boson encountered in every-day physics is not a fundamental particle (with the photon being the most prominent exception). They are instead composed of a number of fermions, which can not occupy the same state.

Is it possible for more than one of these composite bosons to be in the same state even though their constituents are not allowed to be in the same state? If the answer is "yes", how does this not contradict the more fundamental viewpoint considering fermions?


4 Answers 4


This is a nice puzzle--- but the answer is simple: the composite bosons can occupy the same state when the state is spatially delocalized on a scale larger than the scale of the wavefunction of the fermions inside, but they feel a repulsive force which prevents them from being at the same spatial point, so that they cannot sit at the same point at the same time. The potential energy of this force is always greater than the excitation energy of the composite system, so if you force the bosons to sit at the same point, you will excite one of them, so that the composing fermions are no longer in the same state, and the two particles become distinguishable. The scale for this effective repulsion is the decay-length of the wavefunction of the composing fermions, and this repulsion is what leads matter to feel hard.

The reason you haven't heard this is somewhat political--- there are people who say that the exclusion principle is not the cause of the repulsive contact forces in ordinary matter, that this force is electrostatic, and despite this being ridiculously false, nobody wants to get into the mud and argue with them. So people don't explain the fermionic exclusion principle forces properly.

If you have a two-fermion composite which is net bosonic, like a H atom with a proton nucleus and spin-polarized electron, when you bring the H-atoms close, the energy of the electronic ground state is the effective Hamiltonian potential energy for the nuclei. When the nuclei are close enough so that the electronic wavefunctions have appreciable overlap, you get a strong repulsion. You can see that this repulsion is pure Pauli, because if the electrons have opposite spins, you don't get repulsion at short distances, you get attraction, and the result is that you form an H2 molecule of the two H atoms.

You can see this exclusion force emerge in an exactly solvable toy model. Consider a 1d line with two attractive unit delta function pontetials at positions a and -a, each with a fermion attached in the ground state. Each one has an independent ground state wavefunction that has the shape $exp(-|x|)$, but when the two are together at separation 2a, the two states are deformed, and the ground state energy for the fermions goes up. The effect is quadratic in the separation, because the ground state (one fermion) goes down in energy, and the first excited state goes up in energy, and to leading order in perturbations, the two are cancelling when both states are occupied. To next leading order, the effect is positive potential energy, a repulsion. This potential is the effective potnetial of the two delta functions when you make them dynamical instead of fixed.

The maximum value of the repulsive potential in this model is exactly where the model breaks down, which is at a=1. At this point, the ground state is exp(-2x) to the left of -1, constat between the two delta functions, then exp(2x) to the right, with energy -2, and the first excited state is constant to the left of -1, a straight line from -1 to 1, and constant past 1, with energy 0. The result is a net energy of -1 unit. This is half the binding energy of the two separated delta functions, which is -2.

This effect is the exclusion repulsion, and it reconciles the fermionic substructure with the net bosonic behavior of the particle. You can only see the substructure when the wavefunction of the boson is concentrated enough to have appreciable overlap on the scale of the composing fermion wavefunctions, and this is why you need high energies to probe the compositeness of the Higgs (or for that matter, the alpha particle). To get the wavefunctions to sit at the same point to this accuracy, you need to localize them at high energy.

  • 1
    $\begingroup$ You state that "if you force the bosons to sit at the same point, you will excite one of them". There is the problem I am having. When the two bosons are in the same state, they are on the same point, in the sense that their wavefunctions are the same and therefore overlap entirely. It all comes down to this: If the wavefunctions of two bosons are the same, the wavefunctions of their constituents should pairwise be the same, which is impossible. What am I missing? $\endgroup$
    – Friedrich
    Dec 3, 2012 at 0:10
  • $\begingroup$ @Friedrich: Not exactly--- the wavefunction is only exactly the same for two bosons if they are non-interacting. If they have a repulsion at ultra-short distances, the wavefunction is entangled so that it is very close to the product of the wavefunction with itself at most points, but the entanglement zeros out the diagonal (so they aren't ever at the same point). This is a property of higher dimensional multi-particle wavefunctions. The best definition for the condensate wavefunction by giving the effective field for the boson a VEV, and this only produces a product state for a free field. $\endgroup$
    – Ron Maimon
    Dec 3, 2012 at 3:26
  • $\begingroup$ Ron, how can the volume of the wavefunction of a composite boson be larger than the volume of the wavefunction of its fermions? $\endgroup$
    – lurscher
    Dec 4, 2012 at 19:44
  • $\begingroup$ @lurscher: The volume of the wavefunction of the fermions is in the dimension of the relative coordinate, while the volume of the composite is in the center of mass coordinate. $\endgroup$
    – Ron Maimon
    Dec 4, 2012 at 23:37
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    $\begingroup$ It is one the best answers posted to the PSE ever. $\endgroup$
    – peterh
    Mar 10, 2018 at 17:20

Yes, they can, an experimental example of that is Bose-Einstein Condensate of fermions. And that is possible because actually they will have the same wave function, in sense that nature no more capable of making any distinguish between them.

Regarding everyday life, actually saying that it is bosonic is just a formal statement, in sense that Pauli exclusion principle not working here, but that not because the composite things are bosons, but because in everyday life there is almost nothing in the same state, because accomplishing that is very hard, and if done, it will reproduce Bose-Einstein condensate.


I find that a useful way to think of this is to consider the momenta of the individual fermions. Let's say a neutron and a proton comprise a deuteron, and we have two deuterons, or two deuterium atoms, in the same state. Are the neutrons in the same state? Well, no - and one way to see it is that if we could look ''inside'' the deuteron to see the neutron, it's banging about like crazy in there - it has kinetic energy on the order of the binding energy of the neutron, MeV. To make it even more specific, if we measured the momentum of one neutron, and then the other, we'd be very, very likely to get wildly different values. The earlier posts discussing the necessary short range repulsive potential are also correct - but to me the individual momenta are an easier handle on this rather counterintuitive state.


Yes. Examples include the superfluid state at $0K$ of integer spin 0 boson He-4 which itself is composed of several fermions.

Other effective composite "bosons" include Cooper pairs of fermions which makes their spin integer thus Pauli Exclusion Principle no longer applies to them. This makes Fermionic Condensate possible.


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