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Angular momentum of an electron

  1. May 31, 2005 #1
    Hi,

    Say you have an electron in the hydrogen atom. Can this electron be in a state that is a superposition of the usual energyeigenstates (which are also eigenstates of L_z) AND have different expectation values for L_x and L_y? Are x and y symmetric in the sense that their expectation values are always equal in this situation?
     
  2. jcsd
  3. May 31, 2005 #2
    You can't measure the expectation value of L_z and L_y simultaneously cause of the uncertainty. You can only measure total angular momentum and L_y or total angu. mom. and L_z precisely.
     
  4. May 31, 2005 #3

    dextercioby

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    You can measure them all (5) with arbitrary precision,if the quantum system is in a state with [itex] l=0 [/itex].

    Daniel.
     
  5. Jun 1, 2005 #4
    No no, I think you misunderstand. The EXPECTATION value is not the result of a single measurement. This is what I mean: Can you have a particle in a state, which is a superposition of eigenstates of L_z, but which - at some point in time (say t=0) - has different EXPECTATION values for L_x and L_y.

    The reason I'm asking is that I found just that in my exam the other day. The assignment was this:

    In this problem we consider a hydrogen atom whose normalized stationary states is denoted [tex]|nlm\rangle[/tex], where n is the main quantum number and l and m is the quantum numbers belonging to L^2 and L_z, respectively. Consider a hydrogen atom described by the state:
    [tex]|\psi\rangle=\frac1{\sqrt2}(|210\rangle + |211\rangle)[/tex]​

    Answer these questions:

    1. Blah blah
    2. Blah blah
    3. Blah blah
    4. Calculate the expectation values of L_x and L_y in the state |psi>.

    I got different answers for <L_x> and <L_y> (h/sqrt(2) and 0, respectively). Is this possible? When you choose z as reference axis the theory seems to treat x and y equally.
     
  6. Jun 1, 2005 #5

    dextercioby

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    There's no problem whatsoever.The two operators are different (see their expressions as a function of the ladder operators),so it's obvious to get different expectation values.

    Daniel.
     
  7. Jun 2, 2005 #6
    Ok, thanks.
     
  8. Jun 2, 2005 #7
    That's right, but your specific state doesn't treat x and y equally. So, as dextercioby already pointed, there's no contradiction if you find different expectation values.
     
  9. Jun 4, 2005 #8
    Choosing z as the reference axis doesn't necessarily treat x and y equally, it's
    just a convention. Only m=0 states have symmetry about the z-axis. So <L_x> = <L_y> (= 0 I think) whenever m= 0, but since [tex] |211\rangle [/tex] is part of
    the wavefunction, it doesn't have the symmetry.
     
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