Question about the magnetic dipole moment of the electron

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SUMMARY

The discussion focuses on the behavior of an electron's magnetic dipole moment as it moves through a uniform magnetic field. It is established that the electron follows a circular path due to the perpendicular relationship between its velocity and the magnetic field. The intrinsic magnetic momentum, related to the electron's spin, oscillates with a phase described by the equation a e^{i\mu_B B/\hbar }\left| \uparrow\right>+ b e^{-i\mu_B B/\hbar}\left| \downarrow\right>. The conversation also touches on the potential for the electron to undergo photoemission, leading to a transition to the ground state.

PREREQUISITES
  • Quantum Mechanics (QM) fundamentals
  • Understanding of magnetic dipole moments
  • Knowledge of electron spin and its representation
  • Familiarity with photoemission processes
NEXT STEPS
  • Study the principles of magnetic dipole moments in Quantum Mechanics
  • Explore the mathematical representation of electron spin states
  • Investigate the effects of uniform magnetic fields on charged particles
  • Learn about the mechanisms and implications of photoemission in quantum systems
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Physicists, students of Quantum Mechanics, and researchers interested in the behavior of electrons in magnetic fields will benefit from this discussion.

georgenewman
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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?
 
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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.
 

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