Why the magnetic moment is zero for occupied levels?

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SUMMARY

The discussion centers on the concept of magnetic moments in quantum mechanics, specifically addressing why filled energy levels with an even number of electrons result in a net magnetic moment of zero. Participants clarify that for a closed subshell, all magnetic quantum numbers (m) are occupied, leading to a total angular momentum (L) of zero. The conversation emphasizes that the total angular momentum vector does not point in a specific direction when it is zero, and the wavefunction of a closed subshell is spherically symmetric, confirming the zero magnetic moment. Additionally, the uncertainty principle is discussed in relation to angular momentum components.

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hokhani
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I can not get convinced why for energy level, filled with an even number of electrons, the net magnetic moment is zero!
If we have a level with the quantum number l, we have two electrons with opposite spins in it. Also magnetic moment is proportional to Jtotal where J_t= J_1+J_2(1and 2 refer to electrons). Moreover J=L+S. If magnetic moment were to be zero, the J_t must be zero. How can we show that?
 
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Isn't the correct statement that the magnetic moment is zero only for an atom whose subshells are completely filled. That is, electrons occupy all m values for a given L, so that Σm = 0
 
Bill_K said:
Isn't the correct statement that the magnetic moment is zero only for an atom whose subshells are completely filled. That is, electrons occupy all m values for a given L, so that Σm = 0
OK, But if so(Σm = 0) we can only deduce that the z-component of L is zero not the L itself!
 
hokhani said:
OK, But if so(Σm = 0) we can only deduce that the z-component of L is zero not the L itself!
Along every z-axis. A closed subshell is spherically symmetric, i.e. its total L is zero.
 
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Bill_K said:
Along every z-axis. A closed subshell is spherically symmetric, i.e. its total L is zero.
Thank you very much. It just remains another question:
By the above statement, do you mean that in all the spherically symmetric systems, If we have a z-component, then we have x and y component? In this case, the uncertainty principle would be invalid!
 
hokhani said:
Thank you very much. It just remains another question:
By the above statement, do you mean that in all the spherically symmetric systems, If we have a z-component, then we have x and y component? In this case, the uncertainty principle would be invalid!

Well if you have 0 total angular momentum which way is the J vector pointing? If you can tell which way it's pointing, how do you know it's projection on the (x,y,z) basis?

Granted the quantum angular momentum vector really doesn't "point" in a specific direction when if is non-zero, but you see my point.
 
hokhani said:
By the above statement, do you mean that in all the spherically symmetric systems, If we have a z-component, then we have x and y component? In this case, the uncertainty principle would be invalid!
No, of course not, that is not at all what I said. A closed subshell contains 2(2l+1) electrons, an electron pair occupying every possible value of m. The wavefunction is a Slater determinant, and is spherically symmetric, i.e. invariant under rotations. Not just the potential is spherically symmetric, the total wavefunction is spherically symmetric. This means the total L value is zero.

EDIT: Are you thinking that [Lx, Ly] = iħ Lz is an example of the uncertainty principle? The uncertainty principle gives the commutator of variables that are canonical conjugates, which Lx and Ly are not.
 
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