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Electron Magnetic Moment

  1. Mar 2, 2008 #1
    So far I have found two completely different values of the electron magnetic moment and i need help to know in which cases these are applicable:
    They both refer (i think) to values of the particle as such, without orbital contribution. Just the spin contribution to the magnetic moment

    the magnetic moment of the electron is 1.73 Bohr magnetons

    the magnetic moment of the electron is 1.0011596521931 Bohr magnetons
    http://www.isis.rl.ac.uk/neutronSites/index.htm?content_area=/neutronSites/constants.htm&side_nav=/neutronSites/neutronsitesSideNav.htm&" [Broken]
    You have to look up : (Electron magnetic moment in Bohr magnetons)

    the first one has to do with the fact that according to wave mechanics the magnitude of the spin angular momentum is not a multiple of the quantum number s but sqrt[s(s+1)].
    It is the value normally used to predict magnetic moments in transition-metal coordination chemistry.

    The second one is the value experimentally measured and appears everywhere.

    Are they both correct?
    When can we use which?

    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Mar 2, 2008 #2


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    Where do you see that value on that site?!? I have never seen the electron magnetic moment given as that value.
    Last edited by a moderator: Apr 23, 2017
  4. Mar 3, 2008 #3
    It is at the top of the page
  5. Mar 3, 2008 #4


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    The mu ans s at the top of the page are vectors.
    g is defined n terms of their z components, where the \sqrt{s(s+1)} does not enter.
    Parts of that website are confusing.
    The "S-G Experiment" part seems to imply that the experiment was performed on electrons, which wouldn't work. Other parts mention that it was done on neutral atoms, and then some atomic physics theory was used to infer the spin of the electron.
    The S-G expt is not accurate enough to find the difference of mu_e from 1.
  6. Mar 3, 2008 #5
    Yes, they are vectors.
    g is a scalar quantity, and is called the electron gyromagnetic ratio or spin g-factor, i.e., the ratio between the MAGNITUDES of the magnetic spin angular momentum to the spin angular momentum ans is very close to two.

    Actually the very accurate value of 1.0011596521931 Bohr magnetons is exactly half of the very accurate value given for g.

    However there are two things I do not understand:

    Why do they assume the magnitude of the spin angular momentum (Total) to be 1/2 hbar instead of sqrt[s(s+1)] hbar, which is the real value?. 1/2 hbar is just the z-component of the total spin angular momentum!!

    The magnetic moment that can be measured (using a SQUID magnetometer) in any transition-metal compound with one unpaired electron and with no orbital contribution can be calculated using the SPIN-ONLY contribution to the magnetic moment (mu = g sqrt[s(s+1)] Bohr magnetons) with very good agreement with the experiment, for instance in the case of Ti +3.
    How can this be explained?
    Last edited by a moderator: Apr 23, 2017
  7. Mar 3, 2008 #6
  8. Mar 3, 2008 #7


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    Those two articles are confusing, at best.
    The gyromagnetic ratio is always (if correctly) defined in terms of the z component of spin.
    The square root factor never enters in a correct definition of the GR.
    The Lande g factor mentioned in one of the articles as to do with the vector addition of
    L and 2S in an atomic state, and is not the same as the g factor for an isolated electron.
  9. Mar 10, 2008 #8
    ok! After racking my brain i got the solution to the riddle.
    Experimentally measured electron magnetic moment:
    In order to measure the magnetic moment of a free electron a magnetic field must be applied and only the z-coordinate of the magnetic moment is measured. The question of how we measure experimentally the total magnetic moment of the electron is pointless, because the Larmor precession frequency cancels out the x and y contributions. Then, the z-component of the electronic magnetic moment is the measured value and amounts to g/2 Bohr magnetons.

    the magnetic moment of the electron is 1.73 Bohr magnetons:
    However, there are other indirect techniques that allow to calculate the magnitude of the whole spin vector, that gives rise to the above-mentioned z-component when a magnetic field is applied. The Curie Law for paramagnetic materials is an example of such techniques. Then 1.73 Bohr magnetons is the total magnitude of the spin magnetic moment vector for one single electron mu = g sqrt[s(s+1)] when the magnetic field tends to 0.

    Thanks pam
  10. Mar 10, 2008 #9


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    You have to learn to start with the statement "The mm of the electron is NOT 1.73 Bohr magnetons." I have seen that number nowhere, but in your posts.
    You get sqrt(3) by misinterpreting that web page.
    You have to understand the difference in QM between the operator [tex]{\vec s}[/tex]
    and the operator [tex]s^2[/tex]. Because the 3 components of [tex]{\vec s}[/tex]
    don't commute, you can't use your classical "intuition". g is defined only in terms of
    [tex]\mu_z=g s_z[/tex]. That should leave no confusion.
  11. Mar 18, 2008 #10

    Hans de Vries

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    pam is right, the magnetic moment of the electron is 0.00115965218085
    Bohr magnetrons. This corresponds with the z-component of the spin and
    not with the total spin.

    The total magnetic moment corresponding to the total spin, which axis is
    rotated over ~54.735 degrees with the z-component is considered to
    be averaged out over all directions with only the z-component remaining.

    Interestingly, This purely QM picture can be derived from classical magneto-
    statics by simply requiring that the charge and spin-densities associated
    with the wave-function do not lead to infinite energies.

    Abandoning the classical point-charge resolves the infinite electrostatic
    energy of the classical electron. The Pauli-Weisskopff interpretation spreads
    out the charge and spin continuously over the wave-function.

    However, this still does not resolve the infinite magneto-static energy because
    the magneto-static dipole force increases with the fourth order if the distance

    [tex]\vec{F}_r\ =\ \frac{3\mu_o\mu_e^2}{4\pi r^4}\left[\ \left(1 -
    3\cos^2(\theta) \right)\mbox{\Large $\hat{r}$}\ \right][/tex]

    Compared with the electrostatic force which goes only with [itex]1/r^2[/itex]. This is
    too fast and leads to an infinite magnetostatic energy except when the angle
    theta is 54.735 degrees, where the term between brakets becomes zero and
    the force is neither attractive nor repulsive.

    So classical magnetostatics defines the angle [itex]\theta[/itex] between the spin at any
    certain point of the wave function, and the local average magnetic moment
    corresponding to the z-component. It does not define the angle [itex]\phi[/itex] which is
    undefined and averaged out. This is exactly the quantum mechanical picture.

    Regards, Hans
    Last edited: Mar 18, 2008
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