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Electron spin and the Pauli Exclusion Principle.

  1. May 16, 2012 #1
    How is it that only 1 spin up and 1 spin down electron are allowed in an atom even though there is no measurement to collapse the state function?
     
  2. jcsd
  3. May 17, 2012 #2

    tom.stoer

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    That is not the case.

    You typically chose a basis in a Hilbert space. One possibility is |+1/2> and |-1/2> w.r.t. to the z-direction; but all other directions are allowed as well to define a basis.

    In addition in an atom with more than one electron (like He² with total spin S=0) it is not true that the "first electron has spin +1/2" and the "second one has spin -1/2" w.r.t. to z. Instead the two electrons are in an entangled state. An ansatz taking antisymmetrization into account is the Slater determinant.

    Of course one may chose the z-direction to define the basis; but the state is independent from this choice.
     
  4. May 17, 2012 #3
    Then how does the third electron 'know' that it can't have spin n,l,m.s = 1,0,0,+1/2 (s w.r.t z)? As you just said yourself, this state is unoccupied.
     
  5. May 17, 2012 #4

    tom.stoer

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    I am only saying that you cannot distinguish between "the first" and "the second" electron. And you should not say that "one electron has spin +1/2 w.r.t. z" whereas "the other one has spin -1/2 w.r.t. z"; that's not wrong but misleading. Both spins couple to S=0. You don't have to mention the z-axis in order to specify the singulet state S=0.

    The two states

    [tex]|1s,\uparrow_z\rangle|1s,\downarrow_z\rangle - |1s,\downarrow_z\rangle|1s,\uparrow_z\rangle[/tex]

    and

    [tex]|1s,\uparrow_x\rangle|1s,\downarrow_x\rangle - |1s,\downarrow_x\rangle|1s,\uparrow_x\rangle[/tex]

    are identical w.r.t. to total spin S.
     
    Last edited: May 18, 2012
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