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Does exchange of identical particles lead to new state?

  1. Aug 18, 2015 #1
    Does an exchange of two identifical particles (electrons) lead to a new microscopic state?
     
  2. jcsd
  3. Aug 18, 2015 #2

    Avodyne

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    No. That's what it means for the particles to be identical!
     
  4. Aug 18, 2015 #3
    Exchange of identical fermions changes the sign of the state, but its properties remain the same.,
     
    Last edited by a moderator: Aug 18, 2015
  5. Aug 19, 2015 #4
    Jilang's correct. Exchanging identical fermions multiplies the wavefunction by -1. This leads to the Pauli exclusion principle.
    You are free to exchange identical bosons at will - nothing changes.
     
  6. Aug 19, 2015 #5
    Sorry... But I guess I'm confused by the term "identical" with reference to fermions. Does that imply the same fermion type, with all the same physical attributes with the same location/momentum?
     
  7. Aug 19, 2015 #6
    "Identical" here means fermion type. This means, for example, if you exchange two electrons then the wavefunction of the system of those two electrons is multiplied by -1.
    The electrons could, therefore, not have been in a completely indistinguishable initial state (as you say, same location/momentum) because if they were absolutely indistinguishable then nothing could possibly change when they are exchanged. Therefore you get the Pauli exclusion principle: two electrons can never be in exactly the same state.

    Important to make the distinction between "identical" (i.e., same type) and "indistinguishable".
     
    Last edited: Aug 19, 2015
  8. Aug 20, 2015 #7

    Jano L.

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    Two things are identical when they have the same properties yet they are two separate objects (at different places).

    In physics, identical particles are particles that have the same constant properties (same mass, charge, etc.) but not some other variables (state, like position and momentum). It does not mean "their exchange does not lead to new state" as Avodyne suggests.

    In Newtonian mechanics, exchange of two identical particles means the two particles change their states:

    First, particle 1 has position ##\mathbf r## and momentum ##\mathbf p## and particle 2 has position ##\mathbf r'## and momentum ##\mathbf p'##.

    Then, particle 1 has position ##\mathbf r'## and momentum ##\mathbf p'## and particle 2 has position ##\mathbf r## and momentum ##\mathbf p##. That's what the exchange means. So unless ##\mathbf r = \mathbf r'## and ##\mathbf p'=\mathbf p##, the exchange means the system gets to a new state.

    In quantum theory, it is not standard to ascribe particles positions and momenta; not even ascribe them separate states. What one works instead with is global "quantum state" identified with set of functions of two positions ##\mathbf r_1, \mathbf r_2## of the two particles; the functions in the set may differ from each other by a factor ##e^{i\alpha}## with ##\alpha## being any constant real number; they all represent the same quantum state. "Exchange of particles" in this scheme is nothing like the above change of states; it is merely interchanging the arguments in the function:

    First, we have function ##\psi(\mathbf r_1,\mathbf r_2) = \phi(\mathbf r_1,\mathbf r_2)##.
    Then, we have function ##\psi(\mathbf r_1,\mathbf r_2) = \phi(\mathbf r_2,\mathbf r_1)##

    So, if no ##\alpha## could make the expression ##\phi(\mathbf r_2,\mathbf r_1) = e^{i\alpha} \phi(\mathbf r_1,\mathbf r_2)## true, the exchange would change the quantum state.

    At this point a new assumption is often introduced; that only those functions ##\phi## that satisfy the last relation are admissible in quantum theory. There are two possibilities how this can happen, ##e^{i\alpha} =1## and ##e^{i\alpha} =-1##. The first possibility means the functions are symmetric, the second means the functions are anti-symmetric.

    This assumption is called assumption of indistinguishability, because inspecting the function ##\phi## that satisfies the mentioned relation, we cannot distinguish the two particles; the state is the same whether we put ##\mathbf r_1## or ##\mathbf r_2## the first into the list of arguments ; the quantum state is the same.

     
  9. Aug 20, 2015 #8
    Rather than saying, "their exchange leads to a new state" it may be more precise to say, "the possible outcomes (measurement results) of this state include certain pairs of outcomes where the location of the particles are mutually interchanged. The complex amplitudes for the mutually exchanged pair of outcomes are equal (bosons) or opposite in sign (fermions)"
     
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