Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Pauli exclusion principle

  1. Dec 9, 2007 #1
    I've just been reading my notes on quantum mechanics and got to the Pauli exclusion principle. The way it was explained makes no sense to me. It says "no two identical fermions can be in the same quantum state". Surely there are some restrictions to the applicability of this statement? It doesnt make sense to say that no two electrons in the entire universe can occupy the same quantum state. Any help would be much appreciated.
  2. jcsd
  3. Dec 9, 2007 #2
    No. There are no restrictions on the exclusion principle. But, don't forget that states may be distinguished by localization. That is, an electron in the ground state of a Hydrogen atom in the sun is not in the same state as an electron in the ground state of a Hydrogen atom on Mars.
  4. Dec 9, 2007 #3
    Surely in that case two electrons in an atom can have the same quantum numbers provided they don't occupy the same spatial location? Where is the distinction between two electrons in an atom or two on different planets if there is no restrictions on the applicability of the Pauli exclusion principle?
  5. Dec 9, 2007 #4
    The distinction is that what is relevant is the localization of the wavefunction. The wavefunction of an electron in an atom is localized to the vicinity of and centered on that atom. It's shape is entirely determined once you know the quantum numbers of its state. So, two electrons in the same state of the same atom would necessarily have the same localization. Unless this is the case, the electrons are simply not in the same state.
  6. Dec 10, 2007 #5
    It didn't make sense to me either. The claim that the wave function must be antisymmetric in the exchange of electrons, that means psi(x1,x2) = -psi(x2,x1), is mathematically well defined claim. The claim "electrons cannot be in the same state" seems to be a special case of this. It only makes sense if you can assume that the electrons feel the identical potential.

    For example, I wouldn't know what it would mean for electrons to be a in a different or in the same state, if we didn't ignore the interactions of the electrons themselves.
  7. Dec 10, 2007 #6
    If what is relevant is the localisation of the wavefunction, then why not choose the system to be two atoms, or a galaxy? Then the wavefunction would include all electrons in the system, and no two electrons in the system could occupy the same quantum state.
  8. Dec 10, 2007 #7
    Ok i think i understand better now. The last post doesn't make sense since all the electrons are in different spatial positions, whereas the wavefunction of a single atom specifies the spatial position exactly wrt the nucleus.
  9. Dec 11, 2007 #8
    Ok so if there is a wavefunction for an electron which depends on angle as well as distance from the nucleus, then surely a rotation by an arbitrary angle will give another quantum state, since there is nothing special about the coordinate system chosen. Wouldn't this allow more than one electron to occupy a state with the same quantum numbers?
  10. Dec 14, 2007 #9
    Can nobody answer the last question? I don't think there is anything wrong with the question. Given the earlier responses in the thread it seems an obvious question to ask.
  11. Dec 14, 2007 #10


    User Avatar
    Staff Emeritus
    Science Advisor

    I thought the Pauli exclusion principle applied to electrons within a given atom. Clearly all atoms of a given element and in the ground state have the same electron configuration. Basically, what PEP means is that no two or more electrons in a given atom will have the same set of quantum numbers.
  12. Dec 14, 2007 #11
    Thanks Astronuc, I thought that might be the case too, but I'm unsure why the PEP would only apply to atoms. I was told earlier in the thread that it applies to every identical fermion in the universe, with position being a factor, but then I haven't had my above question on that explained yet.
  13. Dec 14, 2007 #12
    If you rotate by an arbitrary value in [itex]\phi[/itex], you will, at worst, add an overall phase to the wavefunction. This can be absorbed into the normalization; so, it makes no physical difference. If, however, you rotate in [itex]\theta[/itex], you are effectively changing your z-axis; so, you're transforming into a superposition state that mixes values of m. In neither case will an arbitrary rotation allow you to end up with a state orthogonal to your starting one. So, no, such a rotation will not allow you to fit arbitrarily many electrons into a given energy level.
  14. Dec 15, 2007 #13


    User Avatar
    Science Advisor

    Indeed, the Pauli Exclusion Principle applies to two or more electrons or two or more protons or two or more of any of the same brand of fermions. Let's work with just electrons -- representative Fermions if you will. The basic stricture of the PEP is: any multiple-electron state must be antisymmetric under the interchange of any two electrons. This necessarily prevents two electrons from being in the same state -- for example, suppose we are interested in the atomic structure of Helium, and we'll use hydrogen wave functions as our basis. A one-electric state is characterized by a complete set of quantum numbers, and for hydrogen we'll use the standard set: orbit number, n, orbital and spin angular momenta L , S =1/2; Lz and Sz, the angular projections on the z-axis. The Helium state with 2 s state electrons (2s for short) with 2Sz=1 or -1. This state is impossible according to the Pauli Principle. It is symmetric under electron interchange.

    Why the PEP? The reason is a bit arcane, and is best seen within the context of QFT. In order to guarantee that all non-interacting particles have positive energy (0 is allowed), the particles involved must either obey canonical commutation rules or canonical anti-commutation rules. The proof can be found in many texts, and, probably, in Google, or other internet venues.

    And, of course, the major triumph of the PEP is the explanation of why the Periodic Table is as it is.. Atomic shell structures would be otherwise if not for the PEP.
    Reilly Atkinson

    Note: the PEP applies to all electrons. However, if you include all these other electrons, you will find their influence is small, due mainly to distance.
  15. Mar 15, 2008 #14
    I know this thread is old, but I have brought it back up because the answers I got seem inconsistent with new things i have learned. You say that the reason the PEP doesnt apply to electrons in separate atoms is that the occupy different positions in space. If this is the case then how can it be applied to electrons in a fermi gas, as is done in statistical mechanics?
  16. Mar 15, 2008 #15


    User Avatar
    Science Advisor
    Homework Helper

    … indistinguishable … that's what you are! … or was it you? …

    Hi madness! :smile:

    Isn't it simply that, for the Pauli exclusion principle, the electrons have to be indistinguishable?

    Two electrons in the same shell and with the same spin round the same atom are indistinguishable.

    Two electrons in the same shell and with the same spin round different atoms aren't.

    (erm … I don't know anything about fermi gases … :redface:)
  17. Mar 15, 2008 #16
    two electrons are clearly indistinguishable no matter what. how do you tell the difference between them?

    if you have a seperable potential in an atom, ie the electron electron interaction is ignored. then the pauli exclusion principle holds, the idea of fermions and bosons is that one is symmetric and one anti-symmetric, now what does this condition mean?

    well (use y instead of psi) y=y1y2-y2y1

    now if the two electrons have the same quantum numbers that means y1=y2

    hence y=0 and this combination doesn't exist, hence the PEP. this holds as long as you can seperate your hamiltonian into something that looks like H1+H2=H
    if you can't do that then the PEP would have no reason to hold as you can't use seperation of variables.

    unless there is some stronger form which holds for non-seperable hamiltonians?
  18. Mar 16, 2008 #17
    would that make electrons distinguishable , in principle?
  19. Mar 16, 2008 #18


    User Avatar
    Science Advisor
    Homework Helper

    … I have PET electrons which I keep in separate boxes …

    I give them names, and then follow their careers in the electronic media! :smile:

    Unless they're hanging out with their buddies in the p-shell, in which case I can't because… they're indistinguishable!

    (I think it has something to do with being in the same Hilbert space … :confused:)
  20. Mar 16, 2008 #19
    It may be to do with their indistinguishability. Electrons are only (at least in my stat mech course) considered indistinguishable when they are not localised. For example atoms with fixed positions in a crystal are distinguishable but in a gas are not. Perhaps electrons in a single atom are indistinguishable but electrons in seperate atoms are distinguishable. I would be interested to see if this is provable from QM.
  21. Mar 18, 2008 #20
    I think you've misconstrued what I said some months back. My whole point was that the Pauli exclusion principle does apply between any two electrons. But, that the equivalent energy levels in two atoms are not the same state, which can be understood by recognizing that they have different localization. If two states have different representations in the position basis, then they must be different states.

    As has been pointed out by several people on this thread, the Pauli exclusion principle can be seen as a consequence of the necessity that all multi-electron states be antisymmetric under the exchange of any pair of electrons. If two electrons were in identical states, the combined wavefunction would necessarily be symmetric under exchange of these two electrons, since it wouldn't be possible to tell what change had been made. The only way for the wavefunction to be simultaneously symmetric and antisymmetric under the exchange is for it to be identically 0.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook