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Hydrogen orbits, magnetic field, orbit-spin interaction

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    In different (scientific) sites I find different values for MF of H-1 ranging from 0.3 to 0.6 Tesla. Isn't there a generally acknowledged value?

    what about H-2 and H-3 ?

    3. The attempt at a solution
    using the general formula I get 1/ 8 tesla, that formula it not valid?

    1. The problem statement, all variables and given/known data

    can the axis of electron spin be oriented in any direction,(without external influence)an be
    or it can be only anti-parallel to orbit axis?
    (again: in many (scientific) applets the show axis at a tilt)
    Last edited: Jul 18, 2011
  2. jcsd
  3. Jul 20, 2011 #2


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    To start off with, lets assume the nucleus makes no contribution to spin. In reality, it will, but lets hope it is effects are shielded by the electron or its effects are comparatively small.

    So now we just have the one electron. Generally, it can have many possible values for its orbital angular momentum. But usually it will fall into the lowest energy state. This state has zero orbital angular momentum.
    So now, we just have the intrinsic spin of the electron, which is hbar/2. This gives the electron an electronic spin magnetic moment of about (plus or minus) 1.4*(10^-5) eV/T.
    So when there is an external magnetic field of one Tesla applied to the atom, an energy difference of twice the number mentioned will separate the two states.
    They usually talk about the magnetic moment of the electron because it is the energy difference which is usually measured (not the magnetic field itself). If you wanted to know about the magnetic field created by the atom, I don't know the answer, that's not something I've come across in basic quantum mechanics. If I had to guess, I'd say there isn't a nice, classical picture of magnetic field lines flowing through space. I suppose QED is the proper description for a magnetic field in quantum mechanics.

    Even the idea of the electron's spin axis is a bit tricky. It can never be perfectly defined in a single direction, but you can define the component of spin in a given direction. (which is what happens when we apply an external magnetic field). So then you can imagine the axis of the electron's spin to lie somewhere on a cone of possible axis.

    For a hydrogen atom in its ground state, when no external magnetic field is applied, the electron's spin will be equally as likely to be in any direction.

    But when the electron is orbiting around the nucleus, there is another effect I have not yet mentioned. It is spin-orbit coupling, which can be roughly explained as an effective magnetic field at the electron, which means the different spin states will have different associated energy levels.
  4. Jul 20, 2011 #3
    1)If I got it right, you don't know it but it is known
    Can't we calc it by the rule of MF in a loop?
    Once I found the value for H-1 can I automatically deduce the other values (H-2,H-3......)

    2, 3)That is my problem:
    inside the orbit we a strong magnetic field of say 0.4 T. No external influence !

    shouldn't the e-axis be always antiparallel to MFaxis, that is, rotating like a wheel on a road? (of course it may wobble a bit.)

    3)it shouldn't be any direction, because it takes energy ,to deflect an e- from the antiparallel direction
  5. Jul 20, 2011 #4


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    1) What is usually important is the magnetic moment of the electron. This tells us the energy difference of different electron states. The magnetic moment due to intrinsic spin cannot be calculated by classical rules (although the rules are similar in some ways). If the atom is not hydrogen, approximation can be used to find the energy levels.
    If you want to know the magnetic field created by the electron's intrinsic spin, this implies you want to find out what effect this has on other particles. But such experiments are a lot more tricky (I think) which is why its not taught as one of the basics of quantum physics. As I said before, I guess quantum electrodynamics has the answer. But keep in mind that the quantum version of electrodynamics is very different from the classical version.
    2,3) The axis of orbital angular momentum is also not defined precisely. Quantum mechanics does not allow it. Even if we do have a precisely defined axis (i.e. in the case of an external magnetic field), the electron's spin cannot take an exact direction because this state is not allowed by quantum mechanics.
    3)In the ground state, the electron has no orbital angular momentum. its not moving, so there is no spin-orbit interaction.
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