# Magnetic field of a Spin-1/2-Particle

Suppose we have a Spin-1/2-Particle with no charge, like a Silver Atom, fixed at the origin. The magnetic dipole moment is $\mathbf{\mu}=\gamma\mathbf{S}$, where $\gamma$ ist the gyromagnetic ration and $\mathbf{S}$ is the spin angular momentum.
The magnetic moment creates the magnetic field:
$$\mathbf{B(r)}=\frac{\mu_0}{4\pi}(\frac{3\mathbf{r}(\mathbf{\mu}\cdot\mathbf{r})}{r^5}-\frac{\mathbf{\mu}}{r^3})$$
Further suppose we have a charged, spin 0 particle, like a Silver-Ion, at the position $z\mathbf{\hat{k}}$ with the velocity $v\mathbf{\hat{i}}$. Also suppose that the particle is heavy enough to be treated like a classical object, The magnetic field at the position of the second particle in $y$ and $z$ direction is:
$$B_z=\frac{\mu_0\gamma}{4\pi z^3}(3S_z-S_z)=\frac{\mu_0\gamma}{2\pi z^3}S_z$$
$$B_y=-\frac{\mu_0\gamma}{4\pi z^3}S_y$$
Now the second particle will experience a force in the $y$-Direction proportional to $B_z$ and a force in the $z$-Direction proportional to $B_y$. Now if we let this particle hit a screen we can measure the deflection in y- and in z-direction. From this, we can infer $B_y$. and $B_z$ and therefore $S_y$ and $S_z$. But the uncertainty principle does not allow for $S_y$ and $S_z$ to be measured simultaneously. So where is the mistake?

I'm not sure if the approximation to treat the second particle classically is valid, but it is the same approximation commonly used for calculations in the Stern-Gerlach Experiment. If it is not valid, could you please provide a qualitative description how the probability of finding the second particle on a specific spot on the screen depends on the spin of the first particle?

Thank you for your help :)

$$B_y=-\frac{\mu_0\gamma}{4\pi z^3}S_y$$