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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:

    [tex]
    \begin{array}{cll}
    \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)
    \end{array}
    [/tex]
     
    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

    selfAdjoint

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    [tex]e^{i\frac{\pi}{4}}[/tex]?
     
  10. Nov 2, 2006 #9
    And do you mean

    [tex]
    \langle x_1',x_2' | \hat{S} | x_1,x_2 \rangle =
    \langle x_1',x_2' | x_2,x_1 \rangle =
    \delta_{x_1'x_2,x_2'x_1}
    [/tex]
     
    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

    Galileo

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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

    Galileo

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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.

    proof:

    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.

    Cheers!
     
  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 !

    Careful
     
  22. Nov 10, 2006 #21
    Well, it's not that difficult. If they are flying in two different directions, then their momenta will have opposite signs. Therefore they will be in different eigenstates of momentum, and then they are obviously in different states.
     
  23. Nov 10, 2006 #22
    Euh that is not the point. In the above people tried to derive bose and fermi statistics from projective invariance of any two particle state under the swapping operation. The issue I tried to raise is; when do you call two particles indistinguishable and consequently when does the statistics apply ? You simply say, well both particles are in different states, but in a two particle wave, you cannot even speak about the state of a single particle; that is the message of EPR (at least in the standard QM formalism).

    Moreover, if they are in true eigenstates of the momentum, then the latter probabilities are ill defined.

    For example, two electrons bound to different nuclei spaced by a distance of a few micron cannot be expected to satisfy the above criterion (antisymmetrization of state). Even the wavefunction of two electrons in different shells of one atom should not anti-symmetrize by this argument. I am not even sure if it would apply exactly to two electrons in the same shell if all radiative corrections are taken into account.


    Careful
     
    Last edited: Nov 11, 2006
  24. Nov 10, 2006 #23
    That's fair enough: I must admit, I didn't really read the rest of the thread.

    As far as I know, two electrons bound to different nuclei do not satisfy the principle: even if they have the same [itex]n,l,j,m_j[/itex], their total state vector is specified by the position of the center of mass of each atom as well: and these will be different.

    Why should they? They have different values of [itex]n, l[/itex].
     
  25. Nov 11, 2006 #24
    Right, now prove that they even do not approximately form an antisymmetric state starting from fermionic free QFT where you know that creation operators anti-commute and as such always form antisymmetric wave functions. The issue is that I would for sure not expect such anti symmetrization properties to hold for free particles, which are clearly distinguishable.


    Since you seem to accept this so easily, you might now want to show us why we should believe in an entangled state where we have a left mover and a right mover ; both particles being clearly distinguishable, but the wave function still being antisymmetric (of course I am aware that you will appeal to conservation of spin , but nothing says this needs to be the case since we do not control the mechanism for creating an entangled state AFAIK (and only total angular momentum needs to be conserved).).
     
    Last edited: Nov 11, 2006
  26. Nov 11, 2006 #25
    There is a theory called the spi-statistics theorem, which proves that half integer spin particles cannot occupy the same state. If you generate greens function from an intrinsic spin half theory like the Dirac equation, then if 2 particles occupy the same state then the greens functions will equal zero. Since Greens functions are related to the density which is related to wave functions then this give the antisymmetry condition on the wavefunction. This is however a function of dimensionality. In theories with even space dimensions we are able to have states which are neither totally symmetric nor antisymmetric. This is related to parastatistics wgich was first Elucidated by H. S. Green in 1953 (Phys. Rev 1953,90,270).

    You can find a 'proof' of the spin statistics theorem in Scwabl, 'Advanced Quantum Mechanics'
     
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