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Proton neutron antisymmetric wavefunction

  1. Feb 14, 2009 #1
    I was told that if there is a wavefunction containing a proton and a neutron that it must be anti-symmetric under exchange of a proton and a neutron. I am having trouble understanding this. The short handwavy explaination is that they are indistinguishable via the color force.

    What bothers me is that they are disinguishable particles. Why should there be a requirement that an exchange introduces a minus sign? Since the particles are distinguishable, why would the exchange even give a state related to the first one ... or is that a hint that this applies for specific situations only?

    Any help you can provide would be appreciated.
  2. jcsd
  3. Feb 14, 2009 #2
    Is it not just merely that the person who told you that was assuming isospin symmetry ?
  4. Feb 14, 2009 #3


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    I'll try. If you don't introduce isospin (It can be introduced without requiring isospin symmetry.), then the p and n are distinguishable and need not be antisymmetized.
    With isospin, the full wave function: (Space)(spin)(isospin) must be antisymmetric.
  5. Feb 15, 2009 #4
    I assume that is what was meant by the neutron and proton are "indistinguishable via the color force". But I am fuzzy on how this is "elevated" to a symetry.

    For example, the [itex]e^+[/itex] and [itex]\pi^+[/tex] have the same quantum numbers for the electric force, but that doesn't lead to any requirement on the symmetry of a wavefunction containing both (does it?).

    Also, since the proton and neutron are distinguishable, I don't really understand why exchanging them would lead to any relavently related situation. Imagine a hydrogen atom and a neutron separated by meters... exchange the neutron and proton in the wavefunction for this system ... why would they be correlated at all? Or do they have to be close enough for the strong force to be "pertinent to the situation" for this anti-symmetry condition to occur?
  6. Feb 15, 2009 #5


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    But a proton and neutron are indisinguishable when you introduce isopspin as clem and humanino said. If you introduce isospin, they are spin configurations in isopspin space.

    Proton is spin up of the nucleon in isospin space and the neutron is spin down of the nucleon. I.e. compare this with an electron which can have spin up or spin down in spin-space.
  7. Feb 15, 2009 #6
    This is the (at least one) part that I need explained. For they still have other quantum numbers which distinguish them (electric charge being the obvious one).

    Consider the hydrogen atom in the ground state. If the electron is found to be spin up, the proton is spin down. Their spins are correlated. But they have other quantum numbers which are not. What does this have to do with whether the overall wavefunction is anti-symmetric to exchange?

    Based on everyone's answers, I'm missing some really fundemental starting point here.
  8. Feb 15, 2009 #7


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    You can't compare the electron and the proton in an hydrogen atom. They are bound by the electromagnetic force. Isospin of nucleons are for the STRONG force. Also the proton have mass of 1840 times the electron, so that is one more thing which you need to take into account. The most fundemental reason is that electron and proton are not even close to be viewed as two faces of the same particle. A proton is a composite hadron and the electron is an elementary lepton...

    Now coming back to hadronic isospin. The reason for why you introduce it is that under the STRONG force, the nucleon and the proton is the same particle; the Nucleon. The strong force is charge independent. Nuclei are held together by the strong force and thus you need to consider the symmetries involved there.

    Study the deuteron (the only bound 2 nucleon state) for instance, the charge of the proton is irrelevant for the system, what matters is isospin and regular spin.

    This pattern then propagates to all nuclei, but then you need to do corrections for the repulsive electro-force if you want to compare spectra.

    One famous case is the 14C, 14N, 14O example (http://books.google.com/books?id=0ERX65ocQIQC&printsec=frontcover&dq=povh+nuclei&hl=sv#PPA22,M1 ) page 22. Make corrections for the increasing electro-repulsing energy and the levels will be the same. N have more states since more states are allowed from this isospin symmetry consideration.
  9. Feb 15, 2009 #8
    I used that as an example to try to show where I am coming from. You are saying that one quantum number is correlated between two particles ... and then, from my point of view, appear to be jumping to saying that therefore the wavefunction must be anti-symmetric to exchange of these two particles. There is a huge gulf there that I don't understand how to traverse. What let's you make this leap? From your above statements, it seems to somehow rely on the fact that the proton and neutron have similar masses. Why does it matter how many of their parameters are similar, when clearly others allow them to so easily be distinguished? If I exchange all protons for neutrons in the world, I'd surely notice ... this doesn't seem like a real symmetry of the hamiltonian.

    There's something very very basic here that I don't understand, which is apparently implied/assumed.

    If the neutron and proton are the same particle according to the strong force, then why can we have a bound proton-neutron state but not a bound neutron-neutron state. Or maybe I'm misunderstanding what you mean by "the same particle".
  10. Feb 15, 2009 #9


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    Isospin is for particles with same masses, so ... (is also called isobaric spin, bar is latin for mass i think..). You have no reason to think that the electron and the proton is the same particle is some space.. isospin is here for the STRONG force. You can do this for the pions, etc.

    We are only looking at the STRONG force, and from the strong force point of view, they are the same particle (the nucleon) but different spin components in isospin space. The strong force can't "see" the Electric charge. You need to calm down a bit I think. What textbook's do you have? The important issue that needs to be introduced is that when you discuss the nucleon-nucleon force, electric charge must be forgotten totally. The total hamiltonian on the system will indeed depend on charge, but you can decompose the Hamiltonian into the EM part and the Strong part. Recall the picture I asked you to study? Neglecting the electro-part of the Hamiltonian, the isospin symmetry is obvious-> for the strong part of the hamiltonian, the electric charge doesn't matter.

    The neutron-neutron is not bound since the nucleon-nucleon force is isospin dependent. The pn can have I = 1 or 0, but the nn have I = 1 ...
    Last edited: Feb 15, 2009
  11. Feb 15, 2009 #10


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    Just pretend you see everything from the perspective as the strong force does: Ignore electric charge of particles.
  12. Feb 15, 2009 #11
    While the strong force can't "see" the electric charge, the Hamiltonian can. So why does the fact that some of the quantum numbers are the same, lead to a requirement for the total wavefunction under exchange?

    You seem to place strong importance on the fact that the mass is the same, but don't care at all that the charge is different. Does the strong force "see" the gravitational charge (ie. mass)?

    You are starting to repeat a lot of what you say, which means I'm probably asking really dumb questions here. I just don't see what allows that leap over the chasm. I've tried to read some on wikipedia, and they mention many of the same things you say, but they still don't explain why having a few quantum numbers the same leads to global requirements on the wavefunction despite the particles being distinguishable.

    I don't even understand in what sense this is even a symmetry (for as mentioned, if we switched all the neutrons for protons and vice versa, we'd surely notice). And if we ignore the electric charge, all the remaining quantum numbers are the same, yet they are still distinguishable (p-p can form a short lived bound state, while n-n has no such bound state) ... so in what sense is there any symmetry between the two? What do you even mean by "symmetry" here, since it is not a symmetry of the hamiltonian ... maybe that is where my confusion lies.

    Huh? Is there some other requirement on n-n wavefunctions that is not on p-n or p-p wavefunctions?
  13. Feb 15, 2009 #12


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    Which hamiltonian?! You are talking about an object which is not given.

    We are only discussing the STRONG part of the hamiltonian.

    So switch off the electric charge, the proton and the neutron are indistinguishable. They are the spin up and spin down in isospin space of the nucleon. the p-p and the n-n system are the same if you switch off the electric charge.

    the n-n system (and the p-p) can only have I = 1, and the nucleon-nucleon force can only bind nucleons which are in I = 0 states... very simple. you know how to construct spin wave functions of 2 spin 1/2 particles yes?

    You want to do too many things in one step, that is the problem I guess. I can't help you more. Isospin is that you switch off the electric charge, but still want to have it turned on, that is why you fail in understanding I think.

    The p-p forming a semi bound state is just one way to interpret the p+p -> d formation. One can argue that one has semibound n-n states in halo nuclei also...
  14. Feb 15, 2009 #13
    Okay, so the approximation is that we truncate the hamiltonian to only have the strong force. In this approximation, the proton and neutron have the same quantum numbers with almost the same mass so there is nothing really distinguishing them in this hamiltonian, and since they are spin 1/2 particles, they must be anti-symmetric to exchange.

    Is this correct so far?

    Maybe I interpretted wrong, but I thought people were suggesting this can be "extended" to say the proton and neutron must be anti-symmetric to exchange in the full hamiltonian (electric force included) somehow. Is that not the case? That is how it was presented to me, and confused me, and led me on this journey. If that is not the case, I guess that answers the question.

    However, if indeed this "symmetry" leads to some requirement that a wavefunction containing a proton and neutron must be anti-symmetric to exchange ... I don't understand what let's us say that because the color part is anti-symmetric, that the whole thing must be anti-symmetric.

    Does that describe my question/confusion a little better?

    I'm sorry I'm having so much trouble explaining myself here.

    Yes, I do understand that at least.

    Okay, so Isospin in only useful when the color force dominates? So if I want to look at say the proton - neutron effective force, at distances larger than say 10 nucleon widths, then the electromagnetic force interactions dominate and isospin is no longer a 'useful symmetry' and furthermore the wavefunction need no longer be anti-symmetric to exchange of the two?

    Am I at least getting a little closer?
  15. Feb 15, 2009 #14


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    Yes, that is correct, we neglect the EM charge/interaction, only consider the strong part of the Hamiltonian.

    It depends on how the Hamiltonian looks, in the case of deuteron, electromagnetic force is not even included. In the many-body case, there are many way's how to construct states, there is a discussion in e.g. the book of K Heyde (the nuclear shell model).

    There is no color force between proton and neutron, only the residual, which is called the Nuclear force.. that was a side remark.

    Isospin is always useful, since you can do coulomb corrections afterwards.. but in your example, yes, maybe one can put it in that way.
  16. Jun 24, 2009 #15
    I can understand what confused you. Perhaps it has been said before me, but the thing here is that the Isospin symmetry people talk about in this context is an approximate symmetry. It is a good symmetry when you neglect the electroweak force. Then when you include corrections your wavefunction might not be completely antisymmetric to the n-p exchange, but it will be to zeroth order.
    It is important that the mass is roughly equal because you cannot neglect the kinetic part of the Hamiltonian, which contains the mass. Again, the masses being roughly equal means it is a good approximate symmetry.
    I hope this helps..

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