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McIntyre Quantum mechanics while describing Stern-Gerlach experiment

The results of the experiment suggest an interaction between a neutral particle and a magnetic field. We expect such an interaction if the particle possesses a magnetic moment $\boldsymbol{\mu}$. The potential energy of this interaction is $E=-\mu \cdot \mathbf{B}$, which results in a force $\mathbf{F}=\nabla(\boldsymbol{\mu} \cdot \mathbf{B})$. In theStern-Gerlach experiment, the magnetic field gradient is primarily in the z-direction, and the resulting -component of the force is $$ \begin{aligned} F_{z} &=\frac{\partial}{\partial z}(\boldsymbol{\mu} \cdot \mathbf{B}) \\ & \cong\mu_{z} \frac{\partial B_{z}}{\partial z} \end{aligned} $$

Why wasn't $\boldsymbol{\mu}$ differentiated in $$ \begin{aligned} F_{z} &=\frac{\partial}{\partial z}(\boldsymbol{\mu} \cdot \mathbf{B}) \\ & \cong \mu_{z} \frac{\partial B_{z}}{\partial z}~? \end{aligned} $$ Won't $\boldsymbol{\mu}$ depend on $z$ as the particle travels through space?

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    $\begingroup$ The magnetc moment of the electron is a constant ${\boldsymbol \mu}= ge/2m_{\rm electron}{\bf S}$. $\endgroup$
    – mike stone
    Commented Dec 24, 2021 at 15:32
  • $\begingroup$ Thank you sir. It's the same about the atom? Also is the magnetic moment you wrote from classical electrodynamics? $\endgroup$
    – Kashmiri
    Commented Dec 24, 2021 at 16:08
  • $\begingroup$ No. Not classical. It's a quantum effect. $\endgroup$
    – mike stone
    Commented Dec 24, 2021 at 18:23

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$$\mathbf{F}=\nabla(\boldsymbol{\mu} \cdot \mathbf{B})$$

is a slightly facetious way of putting it. The electron's magnetic moment $\mu$ is a constant and independent from other quantities in the Stern-Gerlach experiment, so it's the same as writing:

$$F_z=\mu \frac{\partial B_z}{\partial z}$$

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  • $\begingroup$ Thank you. In the stern gerlach we've atoms/molecules instead of electrons, does the atoms/molecules have a fixed magnetic moment vector? Also is this result i.e a fixed magnetic moment from classical electrodynamics? $\endgroup$
    – Kashmiri
    Commented Dec 24, 2021 at 16:23
  • $\begingroup$ Stern Gerlach uses silver atoms, which have a lone electron (look up the electron configuration of $\text{Ag}$) So each atom has a net MM, of one electron. $\endgroup$
    – Gert
    Commented Dec 24, 2021 at 17:16

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