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Pauli paramagnetism - signs are driving me crazy

  1. Apr 2, 2009 #1
    Hi all,

    I am looking into the discussions of Pauli paramagnetism (arising from free neutral fermions with spin 1/2) in chapter 11.6 of K. Huang's Stat. Mech. (II ed) and in chapter 31 of Ashcroft and Mermin's Solid State Physics.
    It seems to me that these books do not agree on signs.

    So in this discussion electrons are neutral, i.e. they interact with a (constant) magnetic field only through their dipole moment. The interaction energy should be [tex]-\vec{\mu}\cdot \vec{H}[/tex], i.e. the dipole wants to be aligned with the field.
    But

    [tex]
    \vec{\mu} = g_s \frac{q}{2 m c} \vec{s} = -g_s \frac{e}{2 m c} \vec{s}
    [/tex]

    so that the interaction energy becomes

    [tex]
    g_s \frac{e}{2 m c} \vec{s} \cdot \vec{H} = \pm \mu_B H
    [/tex]

    for an electron of spin [tex]\pm[/tex] (recalling that the g factor of the electron is basically 2). That is the energy is raised for spin up and decreased for spin down.

    Now this is what A&M say.

    On the other hand, it seems to me that K. Huang says the opposite when writing the Hamiltonian for a single dipole ("neutral electron") as

    [tex]
    {\cal H} = \frac{p^2}{2 m}-\mu_B \vec{\sigma} \cdot \vec{H}
    [/tex]


    Could it be that Huang is using the intrinsic dipole moment for a particle of charge e instead of -e? Then he gets the correct result because his definition (11.112) of magnetization agrees with this assumption:
    [tex]
    {\cal M} = \frac{\mu_B}{V}(N_{+}-N_{-})
    [/tex]
    which differs from A&M's (31.55) by a sign.

    Does this make sense to you?
    Sorry about all this fuss, but all the story here is about signs, you know, paramagnetism vs diamagnetism, and the sign discrepancy bothers me a little.

    Anyway, the bottom line seems to be that one gets paramagnetism independent of the sign in the definition of the dipole moment. That sort of makes sense, since magnetization here is the average dipole moment.

    Thanks

    F
     
  2. jcsd
  3. Apr 2, 2009 #2

    alxm

    User Avatar
    Science Advisor

    Well:
    [tex]\vec{\sigma}\cdot\vec{H}=\pm H[/tex]

    Because the electron's spin/magnetic moment can only be oriented in two ways; 'with' or 'against' the field. So the sign doesn't really matter, it's just a question of how you define [tex]\vec{\sigma}[/tex] (or [tex]\vec{s}[/tex]), so the sign doesn't really matter, as long as you're consistent in your definition you'll get the correct result.
     
  4. Apr 2, 2009 #3
    People are often sloppy in defining the charge of an electron. It's either e or -e, depending on author. I've had lecturers who've switched half-way through a chapter and never noticed...
     
  5. Apr 2, 2009 #4
    Yes, I see that. You mean that you do not need to really need to connect the intrinsic magnetic dipole moment of a particle with its charge and spin. You just need to know that you have an integer number of possible orientations (two in this case).

    In some sense Huang defines the "positive" spin so that it is oriented along the field. Hence he has to set [tex]M \propto (n_+ + n_-)[/tex]. With these choices everything is ok.

    A&M use the first definition of the dipole moment I gave, so that, owing to the negative charge of the electron, the positive spin is oriented against the field. That's why the field increases its energy, in this case.
    But again, since the magnetic dipole moment is opposite to the spin, it must be [tex]M \propto (n_- + n_+)[/tex].

    Yes, everything figures, provided one keeps the same notation along the calculation (I'm sure the next time I'll think about this I'll get it mixed up :grumpy:).

    Anyway I find Huang's notation a bit confusing, because in eq (11.99) one might think he's using -e for the charge of the electron in the Lorentz force (minimal substitution) and e in the relation between spin and dipole moment.

    But, as we discuss, the latter is in some sense arbitrary.

    Cheers

    F
     
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