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QM angular momentum question

  1. Oct 25, 2007 #1
    The following is NOT a homework problem.I have difficulty in understanding something is a book quoted below:

    After developing the concept of quantum mechanical angular momentum
    [ I mean the relation L_z=m (h') where h' means h/2π
    http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/qangm.html] [Broken]

    a book says:

    "The quantum mechanical origin of these strange restrictions lies in the require-
    ment that if either the particle or the laboratory is turned through a complete
    rotation around any axis,the observed situation will be the same as before the
    rotation.Because observables are related to the square of the wavefunction,the
    wavefunction must turn into either plus or minus itself under a rotation by 2π
    radians.Its sign remains unchanged if the angular momentum around the rotation
    axis is an integer multiple of h(i.e.,forbosons)but changes if the angular momen-
    tum around the rotation axis is a half-integer multiple of h (i.e.,forfermions).
    Because of this difference in sign under 2π rotations,bosons and fermions each
    obey a different type of quantum statistics"

    I cannot exactly follow the book here.How can exp[i 2π] result in - of the same wave function?

    can anyone please explain?
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Oct 25, 2007 #2


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    If [itex]|\psi\rangle[/itex] is the state before a rotation, then you can show that after a rotation by an angle [itex]\vec{\theta}[/itex] the state is given by


    If you perform a rotation by 2*pi, then by single-valuedness of the wavefunction, you must get back the same state UP TO A PHASE. In three dimensions you can show that this phase must be an overall sign (+ or -).

    Now for a spin-1/2 particle


    so due to the 1/2, a rotation of 2*pi (in the z-direction, say) will give an overall phase of [itex]e^{i\pi\sigma_3}|\psi\rangle=-|\psi\rangle[/itex]. Of course, there was nothing special about the z-direction, so this argument holds irregardless of the choice of axis.
  4. Oct 25, 2007 #3

    Hans de Vries

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    Gold Member

    It does... In any way you rotate it, after 360 degrees it is in a state
    which negatively interferes with parts of the wave function which
    didn't (Say in a two split experiment) If you turn the laboratory
    instead of the electron then the electron does not change, contrary
    to what your book says, however, all other electrons in the laboratory
    do change.

    In principle, any object with spin angular moment senses rotations
    of its spin axis, while an object with spin precession senses rotations
    it makes about any arbitrary axis. So, presumably, there is a mechanism
    at work which changes the state of the electron (fermion) during rotation
    until the state is inverted after a rotation of 360 degrees.

    The spin of an electron should be understood as a continuous spin
    density distribution spread out over its wave function. Each point
    represents an infinitesimal small point spin and magnetic moment.
    Molecular modeling software uses this to calculate the total magnetic
    field for instance to determine the magnetic properties of a material.


    Compare spin 1/2 particles with the spin 1 electromagnetic field:

    Polarized Light:
    Polarized light has two independent components: Horizontal and
    Vertical under an angle of 90 degrees. Horizontal polarized light
    does not interfere with vertical polarized light. Light which is rotated
    180 degrees interferes negatively with parts of the wave which are
    not rotated.

    An electron has two independent components: Spin up and spin
    down which are under an angle of 180 degrees. Spin up electrons
    do not interfere with spin down electrons. An electron wave
    function which is rotated 360 degrees interferes negatively with
    parts of the wave which are not rotated.

    Regards, Hans
    Last edited: Oct 25, 2007
  5. Oct 25, 2007 #4
    I suppose this requires that "Because observables are related to the square of the wavefunction",right?

    Hans de Vries, I fear I could not follow you.However,I have to read a couple of more times...

    thanks to both of you.

  6. Oct 25, 2007 #5


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    yeah, that's the idea.
  7. Oct 25, 2007 #6
    OK,thank you...
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