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I read that according to QM the electron in the ground state orbital produces no magnetic field, if that is true, can you explain why so and in what states there is a magnetic field and how that is produced? Why the difference?

Anyway, I am interested in learning how the classical model calculated it, since I read different values in different papers: for example hyperphysics says its value is .4 T, elesewhere I found .5 and .6, and If I use the classical formula I get .125 T

Does 0.4 Tesla refers to the exact value of the magnitude of the magnetic field od 1S state?

  • Orbitals don't "produce" magnetic field, only electrons. Where your original statement about ground states come from? – Greg Oct 25 '16 at 17:45
  • The hyperphysics page is referring to the 2p state, not the ground state. – DavePhD Oct 26 '16 at 14:48
  • @DavePhD,thanks, do you know how they got .4 T and what would the value of ground state be? is it 1/8 T? –  Oct 27 '16 at 05:31
  • @user104372 hyperphysics means that given that the splitting between the two different 2p spin-orbit states is 0.000045 eV, and considering that this is due to the energy of the states being +/-
    μB, where μ is the magnetic moment of the electron, B is 0.4T. For the ground state, 1s, there is no orbital momentum. The magnetic moment of the electron is only in the field of the proton, which is a much weaker field. The magnetic field would be equation 77 here: http://www.pha.jhu.edu/~rt19/hydro/node9.html     – DavePhD Oct 27 '16 at 20:11

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Hyperphysics means that the energy difference between the 2p J=3/2 state and the 2p J=1/2 state is as if the magnetic moment of the electron were aligned with and against a magnetic field of 0.4T.

The energy of the two states, relative to the ground state, is (according to NIST)

2p J=3/2 : 10.19885142904 eV
2p J=1/2 : 10.19880606470 eV

The difference is 0.00004538204 eV or $7.271004 \times 10^{-24}$ J

The electron magnetic moment is $−9.28476462 \times 10^{-24}$ J/T

So the energies are as if they are raised and lowered by being in a field of 0.3915556 T.

For the ground state, 1s, there is no orbital angular momentum. The 1s state is only split by interaction with the proton. This is sometimes referred to as hyperfine structure. The energy difference is $5.87433 \times 10^{-6}$ eV.

DavePhD
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    @tvb orbitals that are not s-orbitals have angular momentum and this creates a magnetic field. A classical formula for the magnetic field is equation 42 here http://www.pha.jhu.edu/~rt19/hydro/node4.html#SECTION00031000000000000000 , but that is as if there are circular orbits, which is not true – DavePhD Oct 29 '16 at 12:42
  • Thanks, Dave, very helpful answer and comment. Can you briefly say why s-orbital have no L while other do have, or should I ask a separate question? –  Oct 30 '16 at 08:42
  • @tvb "s" means l=0, "p" means l=1, "d" means l=2, etc. Also, the angular momentum operators involve derivatives with respect to angular coordinates, while the wavefunctions for s orbitals are independent of angle, so such derivatives are zero. For more see http://www2.ph.ed.ac.uk/~ldeldebb/docs/QM/lect8.pdf especially at the bottom of page 78. – DavePhD Oct 30 '16 at 12:58