Magnetic moment of an silver atom

Question

Image from Sakurai, Modern quantum mechanics.

Question: Why $\partial_z(\mu\cdot B)\simeq\mu_z \partial_z B_z$? it's because $\mu_z$ is constant with respect to the $z$-axis? thanks.

Answer 1

Everywhere along the path of the flying silver atoms the magnetic field $\mathbf{B}$ is pointing in $z$ direction, i.e. the components $B_x$ and $B_y$ are zero.

_{(image from Lexikon der Physik - Stern-Gerlach-Experiment)}

Therefore the dot product simplifies to $$\mathbf{\mu}\cdot\mathbf{B}=\mu_xB_x+\mu_yB_y+\mu_zB_z =\mu_zB_z.$$

The magnetic moment $\mathbf{\mu}$ is a property of the atom. Its magnitude $|\mu|$ is constant and characteristic for the atom. Its direction can change because an external magnetic field $\mathbf{B}$ exerts a torque $\mathbf{\tau}=\mathbf{\mu}\times\mathbf{B}$ on the magnetic moment. So when the magnetic moment $\mathbf{\mu}$ is aligned with the field $\mathbf{B}$ (like here in the Stern-Gerlach experiment), then there is no torque, and hence $\mathbf{\mu}$ stays constant.

Hence we finally have $$\frac{\partial}{\partial z}(\mu\cdot\mathbf{B}) =\frac{\partial}{\partial z}(\mu_zB_z) =\mu_z\frac{\partial B_z}{\partial z}$$

Answer 2

In the expression $\partial_z(\mu \cdot B)$, $\mu$ is a constant, so it's the same as $\mu \cdot \partial_z B$. And since it says we ignore all other components of $B$, the only term that contributes is the one containing $B_z$, namely $\mu_z \partial_z B_z$.

Moreover, even in situations where $\mu$ is a function of position (e.g. when computing the force density exerted upon a magnetized material) the formula $F = \nabla(\mu \cdot B)$ is still supposed to be interpreted to mean that $\mu$ is treated as a constant by the $\nabla$. To be more precise, it should say something like $F(x) = \left[\nabla'(\mu(x) \cdot B(x'))\right]_{x'=x}$.