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Identical particles

  1. Nov 1, 2006 #1
    When discussing identical particles, it is said that a wave function must be either symmetric or antisymmetric wrt exchange of particles. Could we not have, in general,
    [tex] \psi(r_1,r_2)=e^{i\phi}\psi(r_2,r_1)[/tex]
    since the overall factor is physically meaningless?
  2. jcsd
  3. Nov 1, 2006 #2
    Is this like bosons, fermions and anyons?
  4. Nov 1, 2006 #3
    I've never heard of an anyon. I found a brief section on it on wiki. Can you provide any details?
  5. Nov 1, 2006 #4
    The symmetry or antisymmetry drops out of field theory in three dimensions. If you restrict yourself to two dimensions, you no longer have a strict exclusion principle, and flipping the two particles yields whatever phase you want with respect to the original state vector. These are anyons.
  6. Nov 1, 2006 #5
    Let the operator for swapping the particles be [itex]\hat{S}[/itex] in the position representation. Now apply it twice to our wavefunction:

    \hat{S}^2\psi(r_1,r_2) = \hat{S}\psi(r_2,r_1)=\psi(r_1,r_2) &\Rightarrow \hat{S}^2 = 1 \\
    &\Rightarrow \hat{S} = \pm 1 \\
    &\Rightarrow \psi(r_1,r_2)=\pm\psi(r_2,r_1)
    Last edited: Nov 1, 2006
  7. Nov 1, 2006 #6
    [tex] S^2 =1[/tex] does not imply [tex] S=\pm 1[/tex]. All this implies is [tex] S=S^{-1}[/tex].

    And I believe you just proved that all states are either symmetric or antisymmetric. What about the state |0>|1> ?

    In fact, S must have some zero's on the diagonal in the position basis:
    [tex] <x,y|S|x,y>=<x,y|y,x> = \delta_{xy}[/tex]

    Am I mistaken?
  8. Nov 2, 2006 #7
    Can you think of an example where

    [tex]\hat{S}^2 = 1 \mbox { but } \hat{S} \ne \pm 1[/tex]
  9. Nov 2, 2006 #8


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  10. Nov 2, 2006 #9
    And do you mean

    \langle x_1',x_2' | \hat{S} | x_1,x_2 \rangle =
    \langle x_1',x_2' | x_2,x_1 \rangle =
    Last edited: Nov 2, 2006
  11. Nov 2, 2006 #10
    [tex]\hat{S}=e^{i\frac{\pi}{4}} \Rightarrow \hat{S}^2 = e^{i\frac{\pi}{2}} = i \ne 1[/tex]
  12. Nov 2, 2006 #11


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    You can conclude from S^2=I (and that S is Hermitian) that the eigenvalues are [itex]\pm 1[/itex] (take the Pauli spin operators for example). If the particles are identical, then H must treat them on equal footing. It just means that [H,S]=0, so you can find a complete set of eigenstates common to H which are either symmetric or antisymmetric.
  13. Nov 2, 2006 #12
    Yes, a diagonal matrix with 1 and -1's on the diagonal.
  14. Nov 2, 2006 #13
    Where do anyons come into all this? What you say here shows that all particles can be thought as a superposition of fermionic and bosonic states.

    Also, why would we expect S to be hermitian?
    Last edited: Nov 2, 2006
  15. Nov 2, 2006 #14


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    They don't, since that wasn't my intention.

    You can prove S is Hermitian quite easily.

    I don't think you can prove the wavefunction must be (anti)-symmetric from QM. The fact that the states are required to be either completely symmetric or antisymmetric under the exchange of two identical particles is a postulate.
  16. Nov 2, 2006 #15
    Is that true?

    I know that if you derive the (relativistic) Dirac equation the Pauli spin matrices fall right out. So I can see how the antisymmetric nature comes about for fermions. As for bosons...i have no idea.
  17. Nov 2, 2006 #16
    Because S commutes with respect to the Hamiltonian, and since the wavefuction is an eigenfunction of H it will be an eigenfunction of S also - the eigenvalue being the (experimentally measurable) spin. Whichever way to want to look at it, S must be hermitian.


    If H is hermitian then:

    [tex]H^\dagger = H[/tex]

    and if S and H commute:

    [tex][H,S] = 0[/tex]

    [tex]HS - SH = 0[/tex]

    [tex]HS = SH[/tex]

    [tex](HS)^\dagger = (SH)^\dagger[/tex]

    [tex]S^\dagger H^\dagger = H^\dagger S^\dagger[/tex]

    [tex]S^\dagger H = H S^\dagger[/tex]

    but since [H,S] = [S,H] = 0 then SH=HS and so it must be true that

    [tex]S^\dagger = S[/tex]

    and so S must be hermitian
    Last edited: Nov 2, 2006
  18. Nov 2, 2006 #17
    First: Requierment that permuting particles leaves physical state intact leads to conclusion that representation of permutational grouop must act as multiplication with [tex]e^{i\phi}[/tex]. Even, if you take that in form of some general unitary operator you will finally end with reducible operator whose irreducible subspaces are 1D, and hence his action is reduced to simple scalar multiplication by phase factor. Then, application of same permutation again must be identical operator, thus giving only two possible 1D representations, +1 and -1.
    Second: take a system which is in state [tex]|\phi>=|s>+|a>[/tex], where [tex]|s>[/tex] is from symmetric subspace of permutational group representation and [tex]|a>[/tex] is from antisymmetric part. Attack this state with some permutation and you will have [tex]D(S)|\phi>=|s>-|a>[/tex], which is not colinear with original state, and this is not allowed by assumption of identical particles. Thus all systems must be in either symmetric or antisymmetric state.
  19. Nov 2, 2006 #18
    I think I am only becoming more confused.

    Anyons do not exist?
  20. Nov 10, 2006 #19
    The answer is here: M. D. Girardeau "Permutation Symmetry of Many-Particle Wave Functions", Phys. Rev. 139, 500 (1965).

    In short, the phase [tex]\phi[/tex] can depend on positions in 1D, but in 3D the only
    possibilities are 0 and [tex]\pi[/tex].

    Beautiful paper, anyway.

  21. Nov 10, 2006 #20
    As a general note, you have to be VERY careful when applying the credo that two particles can be considered to be indistinguishable. For example, when I have two electrons 1 and 2, each flying off in two different directions, say to the left and to the right ; then clearly the probability that the first electron is at my left and the second at my right is much bigger than the first being at my right and the second at my left. In other words flipping the positions might change the physical state !

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