Question about the magnetic dipole moment of the electron

[georgenewman]

Let's say that an electron is moving with constant velocity through a magnetic field. If the magnitude of the field is the same everywhere in space and if it is perpendicular to the velocity of the particle everywhere, the electron will follow a circular path. We all know that. What I need to know is, in what way does the direction of the magnetic dipole moment of the electron change as it moves along this circular path?

 

[tim_lou]

I suppose you are talking about the intrinsic magnetic momentum of the electron. I'll talk about spin instead of magnetic moment since they are proportional anyway.

In QM, the spin of the electron will oscillate with some phase, like the following:
[tex]a e^{i\mu_B B/\hbar }\left| \uparrow\right>+ b e^{-i\mu_B B/\hbar}\left| \downarrow\right>[/tex]

I think (not too sure) the electron will go through photoemission and end up in the ground state.

 

[astrokreng]

As the electron moves along the circular path, the direction of its magnetic dipole moment will remain constant. This is because the magnetic dipole moment of an electron is defined as the product of its charge and the distance between its center of mass and its axis of rotation. Since the electron is moving with constant velocity, the distance between its center of mass and its axis of rotation remains the same, and therefore the magnetic dipole moment remains constant. However, the orientation of the magnetic dipole moment with respect to the magnetic field may change as the electron moves along the circular path. This is because the magnetic field is always perpendicular to the velocity of the electron, and the direction of the magnetic dipole moment is defined as being along the axis of rotation. Therefore, as the electron moves along the circular path, the direction of its magnetic dipole moment may change relative to the direction of the magnetic field, but the magnitude will remain constant.