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Magnetic dipole in magnetic field

  1. Sep 9, 2012 #1
    Hi

    The energy U of a magnetic dipole in an external magnetic field is given by
    [tex]
    U = -\mu \cdot B
    [/tex]
    so the energy is zero when they are perpendicular and maximal when they are antiparallel. This makes very good sense intuitively. Quantum-mechanically we have that
    [tex]
    \mu = -m_Fg_F \mu_B
    [/tex]
    so U becomes
    [tex]
    U = \mu_Bg_Fm_FB,
    [/tex]
    which is just the Zeeman shift of an atom. My questions is on how these two different scenarios - quantum and classical - are related.

    The first relation states that the particle gains energy due to the torque exerted on it by B. However a Zeeman shift of an atom is - how I have understood it - basically not related to that the atom gains enegy. It just means that its internal levels are shifted. So it is not intuitive to me how the magnetic field "imparts" energy onto the particle in the second relation.

    I hope my question is clear.

    Best,
    Niles.
     
  2. jcsd
  3. Sep 9, 2012 #2

    mfb

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    Staff: Mentor

    Only if it can rotate in that direction. The transition between the corresponding quantum mechanical energy levels is not really a rotation, but has a similar effect: it is a change of µ, which changes the energy.
     
  4. Sep 9, 2012 #3
    Thanks. When you say that it is a change of μ, then you are referring to that mF is changed in
    [tex]
    \mu = -m_Fg_F \mu_B
    [/tex]
    ?

    Best,
    Niles.
     
  5. Sep 9, 2012 #4

    mfb

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    Staff: Mentor

    As the other two are constant... right ;).
     
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