Does a free electron have an orbital magnetic moment?

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SUMMARY

The discussion clarifies that a free electron does not possess an orbital magnetic moment due to the absence of orbital motion around an atom. The total magnetic moment of an electron is the sum of its orbital magnetic moment and spin magnetic moment. The orbital angular momentum operator, defined as L = x × p = -iħ x × ∇, is applicable to electrons, but for free electrons, it does not contribute to the magnetic moment since they do not orbit. The discussion also highlights the relevance of spherical symmetry in the context of angular momentum conservation.

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Edge5
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I know that total magnetic moment of an electron (I am not sure if it is magnetic moment of electron or atom, please clarify this) is sum of magnetic moment caused by orbital motion and spin angular momentum.

So,
Total magnetic moment = Orbital magnetic moment + spin magnetic moment

Do I have orbital magnetic moment for a free electron? (I think not, because electron doesn't orbit around an atom so it should only have spin angular momentum.-In my opinion I mean)
 
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The orbital angular momentum operator is defined as
<br /> \mathbf{L} = \mathbf{x} \times \mathbf{p} = - i \hbar \, \mathbf{x} \times \mathbf{\nabla}<br />
This is a well defined quantity for any particle whose Hilbert space has a basis in position space, so certainly for an electron. Note that it is also coordinate-dependent.

Orbital angular momentum is most useful when you have spherical symmetry so that it is a conserved operator, [H,\mathbf{L}] = 0. This is true for the free particle, but it's sort of a simple consequence of the fact that linear momentum is conserved. Most often, people parametrize solutions of the free particle in terms of linear momentum, but you can certainly use angular momentum instead!
 
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Yes, you can show that ##Y_{lm}(\phi,\theta)J_m(kr)## solves the free particle Schrödinger equation, where ##Y_{lm}## are spherical harmonics and ##J_m## is a Bessel function.
 

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