Two hydrogen atoms and Pauli's exclusion principle

In summary: The wavefunction for the H2 molecule is a valid Schr¨dinger equation, and it has the following three properties:-It is symmetric under exchange of electrons and protons-It is anti-symmetric under exchange of protons-It remains unchanged under exchange of both electrons and protons
  • #1
TL;DR Summary
Paradox regarding H atoms being bosons but protons and electrons being fermions
Hello, I recently came across the following (apparent, I hope) paradox: suppose we have two H atoms. Now, a hydrogen atom is made up of one proton and one electron (fermions), so it is a boson. Then one could have two hydrogen atoms which are in the exact same state (including position). This should be extremely unstable because of the electromagnetic repulsion between nuclei, but, as I understand it, it is possible, in principle.

But if the atoms are in the same state, then so are the protons and the electrons of each atom, but this is a contradiction, since protons and electrons are fermions, which obey the Pauli's exclusion principle. What is happening?

Thank you in advance!
 
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  • #2
josemi_guelll said:
if the atoms are in the same state, then so are the protons and the electrons of each atom,
No, they're not. This is a case where one has to be very careful when looking at how QM models systems.

When we model a single H atom as being a boson, we are ignoring its internal degrees of freedom. Physically, this is only justified under circumstances where those internal degrees of freedom do not affect the observed behavior. So we can model a collection of H atoms as forming a Bose-Einstein condensate (which is the kind of boson state you are describing, where all of the bosons are "in the same state"), but this is only justified as long as we can ignore the fact that the H atoms are actually made up of a pair of fermions. This works on length scales that are large compared with the "sizes" of fermions themselves, so that the Pauli exclusion principle does not come into play. Roughly speaking, this means that we can treat the H atoms as bosons only on length scales significantly larger than the size of the atoms.
 
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  • #3
PeterDonis said:
No, they're not. This is a case where one has to be very careful when looking at how QM models systems.

When we model a single H atom as being a boson, we are ignoring its internal degrees of freedom. Physically, this is only justified under circumstances where those internal degrees of freedom do not affect the observed behavior. So we can model a collection of H atoms as forming a Bose-Einstein condensate (which is the kind of boson state you are describing, where all of the bosons are "in the same state"), but this is only justified as long as we can ignore the fact that the H atoms are actually made up of a pair of fermions. This works on length scales that are large compared with the "sizes" of fermions themselves, so that the Pauli exclusion principle does not come into play. Roughly speaking, this means that we can treat the H atoms as bosons only on length scales significantly larger than the size of the atoms.
Oh, thank you so much! That was enlightening.

So, my doubt is now regarding spin-statistics theorem. The spin of the H atom can be 0 or 1, that is, an integer, and I believe this is not an approximation. But if this system can only be considered a boson when we are dealing with lengths scales much larger than its size, does that mean the spin-statistics theorem only strictly applies to elementary particles, and not composite ones, like this case?
 
  • #4
josemi_guelll said:
The spin of the H atom can be 0 or 1, that is, an integer, and I believe this is not an approximation.
Yes.

josemi_guelll said:
if this system can only be considered a boson when we are dealing with lengths scales much larger than its size, does that mean the spin-statistics theorem only strictly applies to elementary particles, and not composite ones, like this case?
No. It means that if you have a collection of H atoms, and you only do things to them that don't involve their internal degrees of freedom, then they will show bosonic statistics. (Being able to form a Bose-Einstein condensate is an example of showing bosonic statistics.) And that is consistent with the spin-statistics theorem, because, as you note, H atoms have integral spin (see above).

But if you do something to the H atoms that does involve their internal degrees of freedom--for example, if you ionize them--then you will not see bosonic statistics.
 
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  • #5
Back up a second:

If I have a valid wavefunction for the H2 molecule, what are its properties? If I swap the electrons, the wavefunction must change sign. If I swap the protons, the wavefunction must change sign, If I swap both the wavefunction remains unchanged: two sign flips cancel. This is the same as swapping the two H atoms, which are bosons.
 
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  • #6
Vanadium 50 said:
Back up a second:

If I have a valid wavefunction for the H2 molecule, what are its properties? If I swap the electrons, the wavefunction must change sign. If I swap the protons, the wavefunction must change sign, If I swap both the wavefunction remains unchanged: two sign flips cancel. This is the same as swapping the two H atoms, which are bosons.
Yes, that makes sense.

For the composite wavefunction, I would write (up to normalization):
$$\psi=\psi_1\psi_2\psi_3\psi_4-\psi_3\psi_2\psi_1\psi_4-\psi_1\psi_4\psi_3\psi_2+\psi_3\psi_4\psi_1\psi_2$$

(to be understood as tensor product of proton space x electron space x proton space x electron space)

Then you get what you said, swapping only both protons or both electrons flip the sign of ##\psi##, and swapping them at the same time doesn't do anything.

But if both electrons and both protons are in the same state, that is, ##\psi_1=\psi_3## and ##\psi_2=\psi_4##, then the resulting wavefunction is 0, in apparent contradiction with the H atom being a boson. That was my original question. If I understood the first answer correctly, the solution to this is that the H atom only behaves like a boson when you can ignore its internal structure (for example, if you work with larger scales), which is not the case when trying to put them in the same state.I wonder then if you can arrive at the same explanation using wavefunction theory and permutations, perhaps, using the wavefunction ##\psi## above. Because I don't see how.
 
  • #7
josemi_guelll said:
But if both electrons and both protons are in the same state,
But you can't. That won't flip the sign under interchange.

This is a better way to think about it than two stes of rules, one "internal" and one "external". The spin statistics theorem always applies. But that means you can't just assume any arbitrary wavefunction is allowed. It has to have the right symmetry properties under every single exchange. A C-12 atom has 6 protons, 6 neutrons and 6 electrons, or (15)(15)(15)=3375 pairs. A valid wavefunction needs to flip sign under all 3375 of these exchanges.
 
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  • #8
To piggyback on V50's answer, the error you are making is in assuming that the total wave function can be written as a product of single-particle wave functions. This is not necessarily the case, and doesn't work here.
 
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  • #9
Then I don’t know what to do :( I have been told that a many particle system can be described using the tensor product of each Hilbert space, and that when these particles are identical, I should (anti)symmetrize the wavefunction. So I thought I could describe this atom in this way. Does anyone know any bibliography where this is discussed? I started the thread because I couldn’t find anything related to it, but I’m afraid I am more lost than before. Thank you once again :)
 
  • #10
The product space, however, doesn't only consist of tensor products of vectors but of superpositions of such products!
 
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  • #11
vanhees71 said:
The product space, however, doesn't only consist of tensor products of vectors but of superpositions of such products!
I’m aware, that’s why I wrote a linear combination of ##\psi_i\psi_j\psi_k\psi_l##, permutating the indexes
 
  • #12
josemi_guelll said:
Then I don’t know what to do :( I have been told that a many particle system can be described using the tensor product of each Hilbert space, and that when these particles are identical, I should (anti)symmetrize the wavefunction. So I thought I could describe this atom in this way. Does anyone know any bibliography where this is discussed? I started the thread because I couldn’t find anything related to it, but I’m afraid I am more lost than before. Thank you once again :)
I'm not sure what exactly you are after. But the concept appear when one wants to treat molecules. For instance, the spin of the two electrons in H2 cannot be written (always) as the product of the individual spin states, as the eigenstates are the singlet and triplet states. Same thing for the two protons, as they are also spin-1/2 particles. Then the spatial wave function must also possess a definite symmetry with respect the interchange of particles, and so on.

Consider a harmonic trapping potential in 3D. You can have two hydrogen atoms (bosons), each in their own internal ground state, occupying the ground state of the harmonic potential. You can then either treat each atom as a separate (composite) particle, and not worry about the details of the state of individual electrons and protons, and you can write the full state as the product of the state of the individual atoms (which will be a good approximation). If you want to consider the individual electrons and protons, then you have to do as for molecules, and have proper identical-particle functions for fermions, and the result is no longer separable.
 
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  • #13
PeterDonis said:
But if you do something to the H atoms that does involve their internal degrees of freedom--for example, if you ionize them--then you will not see bosonic statistics.
Perhaps... one could argue that the strength of interactions between internal degrees of freedom play a stronger role in determining evident boson/fermion behavior, and the magnitude of interactions depends strongly on screening. For example, if the Coulomb interaction were attractive rather than repulsive, you could easily situate two "H" atoms on top of each other by transferring the antisymmetry to the spin degrees of freedom. The simplest way to eliminate the bosonic character of a bunch of (actual) H atoms would be to compress them all into a volume of order ##a^3##, or to move some electrons away from their associated protons (that is, to ionize them as you suggest) but in principle one could still observe Bose statistics among a subset of quasi-particles analogous to Cooper pairs in BCS theory (especially if the ionized H atoms became conductive.)
 
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  • #14
Fermions obey the Pauli principle by definition, and it's not due to any interactions but because of the realization of the symmetry under exchange of indistinguishable particles such that the state flips sign under exchange of any pair of fermions.
 
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  • #15
vanhees71 said:
Fermions obey the Pauli principle by definition, and it's not due to any interactions but because of the realization of the symmetry under exchange of indistinguishable particles such that the state flips sign under exchange of any pair of fermions.
But Hydrogen atoms aren't fermions.
 
  • #16
Couchyam said:
But Hydrogen atoms aren't fermions.
No, but they're also not bosons. They are quantum systems consisting of two interacting fermions and that can be approximated as a single bosonic particle as long as we avoid interactions that probe too deeply.
 
  • #17
Nugatory said:
No, but they're also not bosons. They are quantum systems consisting of two interacting fermions and that can be approximated as a single bosonic particle as long as we avoid interactions that probe too deeply.
Thank you for that tactfully administered dose of reality. Now I see where you and @vanhees71 are coming from (after all, H3 is an unstable molecule), and that my earlier contribution to this thread could be more than just a little bit misleading.

There is still a (admittedly somewhat trivial or contrived) sense in which hydrogen atoms can be thought of as bosons, however, which might be easier to see with second quantization. Consider a many body wave function with an equal number of electrons and protons. This can be expanded (for example, by rearranging operators in lexicographical order of their coordinate/spin values and relabeling ##j##-indices accordingly) as
\begin{align*}
\int_{(\mathbb R_e^{3}\times\{\pm\}_e)^N/S_N}\int_{(\mathbb R_p^{3}\times\{\pm\}_p)^N/S_N} d^{3N}x_ed^{3N}x_p \Psi_N(\vec x_e, \vec x_p, \vec \sigma_e, \vec\sigma_p)\prod_{j=1}^N \psi^\dagger_e(x_{e,j},\sigma_{e,j})\psi^\dagger_p(x_{p,j},\sigma_{p,j})|0\rangle
\end{align*}
(where the integral over the 'quotient space' omits points where coordinates coincide.) This is pretty standard, and explains the correspondence between 2nd quantization and 1st quantization in many-body theory for Fermions (EDIT: see eq. 21.58 from [1]; here a quotient space is used for the integral instead of the usual normalization factor of ##\frac{1}{\sqrt{N!}}##, while the wave function ##\Psi_N## can still be defined outside of the integration bounds through antisymmetry.)

On a configuration by configuration basis, we associate electron coordinates with the nearest unpaired proton, starting with the electron closest to a proton. There will be some configurations where the closest unpaired electron-proton isn't well-defined, but those configurations are on a set of measure zero (e.g. an infinitely thin hypersurface) and can be ignored.
After pairing the electrons with protons in this way, reorder the product ##\Pi_j\psi_{e,j}^\dagger\psi_{p,j}^\dagger## so that electrons are paired with their corresponding proton, and relabel coordinates (and change the bounds of integration) accordingly.
When this has been done, each pair ##\psi^\dagger_e(x_{e,j},\sigma_{e,j})\psi^\dagger_p(x_{p_j},\sigma_{p,j})## can (again in a somewhat trivial or contrived way) be thought of as a joint "H-atom" creation operator ##\phi_H^\dagger(x_{e,j},x_{p,j},\sigma_{e,j},\sigma_{p,j})## over a 6-dimensional + 2##\times## spin-1/2 d.o.f. configuration space. Claim: pairs of spin-1/2 fermions over a 3-dimensional space can be recast as (individual) spin-1 or spin-0 bosons over a 6-dimensional space.

These are very special bosons, in that they satisfy additional constraints beyond those of "normal" garden-variety 6D+2##\times##spin-1/2 bosons, but I'm at least 90% sure they are still bosons, even if they don't look anything like a real-world, well-adjusted, healthy-boundary-setting hydrogen atom, and can be constructed from arbitrary fermionic states (with equal numbers of electrons and protons.)

EDIT:
References:
[1] Merzbacher, E., Quantum Mechanics, 3rd Ed.
 
Last edited:
  • #18
Couchyam said:
This is pretty standard
Can you give a reference?
 

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