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Enforced symmetry in multi-electron problems

  1. Apr 3, 2010 #1

    LeonhardEuler

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    I've been thinking about how the requirement of anti-symmetry of the wavefunction is introduced in multi-electron problems and I am left puzzled over some aspects of it.

    Various types of symmetry come automatically in classical physics. If you are studying water flowing in a cylindrical pipe, you know, if you think about it, that the velocity will not depend on [tex]\theta[/tex]. If you don't realize this, it's just an inconvenience and you will get the same unique solution when you solve the Navier-Stokes equations.

    Apparently, this is not how it works in QM. I say this because antisymmetry of the wavefunction is introduced as an independent postulate from the Schrödinger equation in all the books I've read. Treating the Schrödinger equation as a boundary value problem like any other, it seems the solution(s) must be either symmetric or antisymmetric under interchange of electrons due to their indistinguishably. My understanding is that we must restrict ourselves to looking at the antisymmetric ones.

    Now for what I find confusing. The variational principle is often stated by saying that the expectation of the energy for any wavefunction will be greater than or equal to the expectation for the ground state, with equality holding only for the ground state itself. This makes no reference to antisymmetry. Either the lowest energy wavefunction is antisymmetric without imposing any restriction, making the restriction unnecessary, or it is not antisymmetric, meaning either the variational principle or antisymmetry of the wavefunction is false.

    I looked at a proof of the variational principle in a book I have, and it depends on expanding the arbitrary wavefunction as a linear combination of eigenfunctions of the Hamiltonian. Perhaps the way out of this is that the books I have just have an imprecise statement of the variational principle and it should be amended to say the "expectation of the energy for any antisymmetric wavefunction is greater etc...". But this is still a little unsatisfactory to me, because how do we know any antisymmetric function can be expanded in just the antisymmetric eigenfunctions of the Hamiltonian? Certainly antisymmetric functions can have expansions containing terms that are not themselves antisymmetric, as I can think of examples.
    (e.g. f(x,y) = sin(x) - sin(y), break the 2 sin's into Taylor expansions, but when defining [tex]\phi_n[/tex]'s, mix and match different order terms, like

    [tex]\phi_1 = x[/tex], [tex]\phi_2 = -\frac{x^3}{6} - y[/tex] [tex]\phi_3 = \frac{x^5}{120}+\frac{y^3}{6}[/tex] etc.)

    Another bothersome thing is that the idea of the distinction between fermions and bosons seems a little artificial and unable to handle generality. By this I mean, consider the following scenario: It is discovered that there are actually two types of electrons that vary just extremely minutely in their mass or charge. If you have a helium atom with two types of electrons, you no longer expect perfect antisymmetry, but you expect the result to be extremely similar to the identical electron case because the situations are so similar. How do you actually express the way a fermion is different from a boson in this case, since there will now be neither symmetry nor antisymmetry? Is the difference really how they behave in cases of exact symmetry, or is there something more fundamental, or more physical to explain the difference?
     
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  3. Apr 3, 2010 #2

    SpectraCat

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    Wavefunctions are only required to be antisymmetric with respect to exchange of *fermionic* particles ... they must be symmetric with respect to exchange of bosonic particles. So, if you are dealing with electronic wavefunctions, then it is correct that the overall wavefunction must be antisymmetric with respect to electron exchange.

    Regarding the variational principle, I guess I don't see what your issue is. If you are dealing with fermions .. all of the wavefunctions must be anti-symmetric, not just the ground state. So there doesn't seem to be any contradiction there.

    Perhaps what is confusing you is that this problem only arises for systems with more than one particle. One particle wavefunctions, like the harmonic oscillator or H-atom solutions don't have exchange symmetry .. since there is only ever one particle. When you have multiple fermionic particles to deal with, then the overall wavefunction is the product of the spatial part and the spin part. If one is symmetric with respect to exchange, then the other has to be anti-symmetric, and vice versa. Read about excited states of the helium atom in your textbook, that is where the most thorough and clear explanation of this usually appears.

    Regarding your last question .. it is a good one! I think the scenario you postulate is unlikely in the extreme, but your analysis about it is correct. If electrons turned out to be distinguishable in any way, then all of the symmetry properties that we have to deal with in the solutions would be null and void. The math would get a whole lot easier actually, because we wouldn't have to worry about anti-symmetrizing our solutions .. bye-bye Slater determinants. On the other hand, physics and chemistry would get a lot hard to understand, since all of a sudden we would lose the ability to assume that all atoms of a given element are identical. For example, there would be 3, or perhaps 4 types of helium atoms: those where both electrons were of one type, and those having one electron of each time, which might or might not be distinguishable depending on if one could tell the difference between the spin-up and spin-down states of the "new" electrons.

    So, physics does really require that the electrons and other fundamental particles be indistinguishable from one another.
     
  4. Apr 3, 2010 #3

    LeonhardEuler

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    The issue I'm having is that the variational principle and the Schrödinger equation make no reference to the fact that we're dealing with fermions. Just think of it as a pure math problem. There is one function out of all possible functions that minimizes the expectation of the Hamiltonian. Either that function is always antisymmetric for Hamiltonians of the multi-electron molecule type, or it isn't. If it always is, then there is no need for a separate postulate for the Fermi exclusion principle. If it sometimes isn't, then either the Fermi exclusion principle is false, or the variational principle is.
     
  5. Apr 3, 2010 #4

    Vanadium 50

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    The spin-statistics theorem is the thing that is enforcing this symmetry. It is a consequence of relativistic quantum mechanics. The Schroedinger equation, however, is a non-relativistic equation. So you're left with two choices: add on spin-statistics in an ad hoc way, or give up entirely on NRQM.

    The former is more useful.
     
  6. Apr 3, 2010 #5

    SpectraCat

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    The variational principle isn't quite as all-powerful as you are making it. For example, it is subject to the symmetry of the problem under consideration. That is why it works for fermions, or bosons, or vibrational states of molecules with a C2v point-group, or whatever. From a mathematical point of view, the symmetry is determined by the Hamiltonian and the trial function that you choose, and in some cases, the values of the variational parameters (i.e. if you allowed anti-symmetric basis functions, but set their initial coefficients to zero). So if you make a "pathologically bad" choice, such as forcing the wrong symmetry on the system, then you will get a mathematically sensible, but physically nonsensical answer.

    So, in a sense you are correct about the mathematical point of view .. you can construct a trial function however you like, and the variational principle will guarantee that you get the lowest available answer within the symmetry constraints of the problem. Whether or not that answer is physically relevant depends on the characteristics of your trial function. If you are solving problems with multiple electrons, then it probably makes sense to express your answer in terms of one or more Slater determinants, since those will enforce the exchange antisymmetry that we *know* is correct from the physical analysis of the system. If you choose a different variational basis, then you may have some additional work to do to create the appropriately anti-symmetrized versions of the answers.

    So, hopefully this helps explain why both the mathematical theorem we call the variational principle and the quantum mechanical postulate we call the Pauli exclusion principle are independent, useful concepts in quantum mechanics.

    One final point is that the Pauli-exclusion principle is only required in non-relativistic QM ... when relativistic effects are considered, then the spin-symmetry is properly included in fundamental equations (it is not in the SE).

    EDIT: I see Vanadium 50 already beat me to the punch while I was typing.
     
  7. Apr 4, 2010 #6
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