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Why are nuclear excited states always eigenstates of parity?

  1. Jan 28, 2012 #1
    QM says that states which are simultaneous eigenstates of two commuting observables are allowed. If you don't have such states to start with you can construct them with the Gramm-Schmidt orthogonalization procedure.

    Consider the excited states of a nucleus. (They can be considered eigenstates of the Hamiltionian even though their lifetime is not infinite). Since the parity operator commutes with the Hamiltonian (ignore weak interactions), QM says simultaneous eigenstates of the Hamiltonian and of parity can be constructed.

    But tables of nuclear excited states determined experimentally are always labeled with their spin and parity. That is, experimentally measured energy eigenstates are also eigenstates of parity. Why? QM only says you can construct such states; it doesn't say that states occurring in nature must already be simultaneous eigenstates.

    It should be perfectly permissible to have nuclear excited states that are Hamiltonian eigenstates but are a linear combination of parity eigenstates. Why do we never see such states? If they do exist, but experimentalists like to assign quantum numbers, there should be in the tables degenerate states, one with positive parity, one with negative parity, with the same energy.

    One might think one could say the same thing about energy and angular momentum. Why are energy states always also eigenstates of angular momentum? But in this case, the Hamiltonian is a function of angular momentum, so states of different angular momentum have different energy and the question does not arise. That is not the case with parity.
  2. jcsd
  3. Jan 28, 2012 #2

    Simon Bridge

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    Welcome to LQ:

    In an atom there are is coupling between energy, angular momentum, spin, parity etc. which makes a difference to the total energy of the state. Some combinations are more tightly bound than others. It is useful to distinguish them anyway.

    But you should know that already - just look up the derivation of the quantum states and you'll see.

    Note: "why" is not a good question in science - ask "how" or "what" and leave the "why" questions to philosophers.

    You sound a lot like someone else new here...
  4. Jan 28, 2012 #3
    Since the question referred only to the nucleus I doubt atomic structure plays any role. If there is any atomic electron effect, only the inner K shell electrons should play a role, since their's is the only wave function that overlaps the nucleus. And since their wave function is isotropic, they should have no parity effect.

    But I may see the resolution: If the nucleus conforms to the shell model, its nucleons are in atomic-like orbits, with wave functions that are spherical harmonics (angular momentum eigenstates). And each of those spherical harmonics has a specific parity. So maybe a nuclear excited state is a parity eigenstate just as an atomic excited state is. If the nucleus does not conform to the shell model then I still don't understand.

    The question "Why" is not always philosophical. In the present context the question makes perfectly good physics sense.

    Thanks for your reply and for welcoming me to LQ. But: what is "LQ"?
  5. Jan 29, 2012 #4

    Simon Bridge

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    I don't believe I mentioned atomic structure at all.
    In atomic nuclei the spin-orbit coupling - for eg - is much stronger than for atomic electrons.

    Yes: this is a shell-model description. Nucleons are confined by their aggregate potential, which is central, so they have all the same sorts of terms you see in atomic structure.

    LQ - is shorthand for another forum I am active on "Linux Questions" ... I meant to say: "welcome to PF" :)
  6. Jan 29, 2012 #5
    OK, thanks. I believe the nuclear shell model does not apply to all nuclei. When it doesn't, I'm still a bit amazed that the nuclear state is an eigenstate of parity. I don't see any reason why it should be.
  7. Jan 29, 2012 #6


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    ncarron, It's the other way around. Shell model or not, states of opposite parity will necessarily have very different wavefunctions, and with a system as complex as a nucleus, nondegeneracy is what you would expect. The degeneracy of such states would be an extraordinary coincidence and require explanation.
  8. Jan 29, 2012 #7

    Simon Bridge

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    IIRC: Even with something as simple as an infinite square well, each energy eigenstate has a single parity (they alternate in parity).

    I thought states of opposite parity must have different wavefunctions because that is what "parity" is telling us. I thought the surprise came from the realization that they also have different energy eigenstates as well.

    But shouldn't it be clear(ish)? An even parity state must have an antinode at the origin, and the odd parity state must have a node. (Or the other way around - I also have trouble with my own parity.) It follows that the probability of finding a particle close to the origin when it is in an even-state is higher than when it is in an odd state ... this gives a picture of the even-parity particle hanging around closer to the center than it's odd brother... so it seems reasonable to expect it to be more tightly bound.

    So why would OP expect to find parity and energy to be independent?

    Perhaps this is due to a slight misreading of the part about simultaneous eigenstates ... correct me if mistaken: it means that parity and energy can be measured successively without destroying the result of the previous measurement. Does not mean that parity is independent of the energy state.

    Now I wonder which nuclei the shell model does not apply to?
    It is, after all, just a model.
  9. Jan 29, 2012 #8
    The parity eigenstate arguments in this discussion all make sense if the nucleus is a composite object made of "elementary" nucleons. Then there's a spatial (and spin) wave function, and all the comments make sense. That is no doubt a very good approximation.
    Then as a general rule it looks like any composite system can be considered a bound state of structureless elementary particles (quarks, in the final analysis). Each elementary constituent has its own intrinsic parity, and the spatial wavefunction (perhaps a spherical harmonic) has its own parity. Together they determine the parity of the composite system. An energy eigenstate will almost certainly also be an eigenstate of parity. Is that a generally valid way to look at it for an arbitrary quantum system?

    Simon Bridge: Perhaps you could share what IIRC and OP stand for. Thanks.
  10. Jan 30, 2012 #9

    Simon Bridge

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    there are infinite hydrogen states but only one occupied.
    the hamiltonian itself has even parity.

    the wavefunction is associated with the state, the state may or may not be associated with a particle ... also any wavefunction can be represented as a sum of other wavefunctions. I think you are reading too much into the representation. When we say a quantum object has spin, say, we don't mean to say that it is literally a little top rotating about some axis. We mean that it's wavefunction has angular-momentum terms in the math.

    You'll find spin described in terms of a rotational symmetry.

    It may well be that a quantum-described object has a real physical extent that makes sense and actual rotating going on so it can have angular momentum in the intuitive sense ... but as you progress you will find yourself giving up these ideas in favor of ideas about symmetry.

    Take all these models with a grain of salt.

    IIRC: If I Recall Correctly
    OP: Original Post(er).
    It's been a while since I've done this routinely and there is a more knowledgeable person responding.
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