The apparatus used in the Stern Gerlach experiment is somewhat like the one shown below with magnetic pole pieces of different shapes producing an inhomogeneous magnetic field.

But what would have happened if we had used symmetrical pole peices which produced a near-homogeneous magnetic field? Basically, why not use a homogeneous field?

Stern Gerlach Diagram


In the Stern-Gerlach experiment you want the atoms to be deflected depending on the direction of their magnetic dipole moment.

But you get a net force on the magnetic dipole moment only if the magnetic field is non-homogenous.

enter image description here
(image from "Force and Torque on a Magnetic Dipole" (page 26))

If the magnetic field would be homogenous, then the forces ($d\vec{F}_1$ and $d\vec{F}_2$ in the image above) would excactly cancel each other, and the net force would be zero.


@Thomas Fritsch uploaded a nice picture that provides intuition about the dynamics of the situation. I would just like to add that the force $\textbf{F}$ exerted on this infinitesimal loop pictured (which is how we model a single silver atom in the Stern-Gerlach experiment), is classically given by: $$\textbf{F} := \nabla \left( \boldsymbol{\mu} \cdot \textbf{B}\right) \tag{A}$$ where $\boldsymbol{\mu}$ is the silver atom's magnetic dipole moment and is a constant vector (that is, it doesn't exhibit spatial variation). You can easily observe that had $\textbf{B}$ been a constant vector, we would have $$\textbf{F} := \nabla \left( \boldsymbol{\mu} \cdot \textbf{B}\right) \tag{B} = \left(\boldsymbol{\mu} \cdot \nabla\right) \cdot \textbf{B} = \mu_x \frac{\partial B_x}{\partial x} + \mu_y \frac{\partial B_y}{\partial y} + \mu_z \frac{\partial B_z}{\partial z} = \textbf{0}$$ Thus, the magnetic field has to be inhomogeneous if we wish to observe a noticeable deflection in the atom's trajectory and thus investigate the distribution of $\boldsymbol{\mu}$ for an ensemble of silver atoms.

$$\bbox[4px,border:1px solid black]{\textbf{An additional note on Equation $(A)$}}$$ Instead of equation $(A)$, some authors simply write: $$F_z = \mu_z \left(\frac{\partial B_z}{\partial z}\right) \hat{\textbf{z}}$$ The simplification occurs by the realization that the average force in the $x$ and $y$ directions vanishes, as a result of spin precession on the same plane.

Image taken from page 399, Principles of Quantum Mechanics, Shankar (Image taken from page 399, Principles of Quantum Mechanics, Shankar)


The simple answer is that the spins are magnetic dipoles, not monopoles. A monopole will feel a force in any field. But a dipole needs a spatially varying field, because otherwise each pole of the dipole would feel an equal and opposite force (and the net force would be zero).


The potential energy $\text{U}$ of a magnetic dipole moment $\textbf{m}$ in a magnetic field $\textbf{B}$ is $$\text{U} = - \textbf{m} \cdot \textbf{B}$$

The force on an object in a potential field is in general given by the negative gradient of the potential

$$\textbf{F} = - \nabla\text{U}$$

Which in this case gives

$$\textbf{F} = - \nabla\left(-\textbf{m} \cdot \textbf{B}\right) = \left(\textbf{m}\cdot\nabla\right)\textbf{B} + \textbf{m}\times\left(\nabla\times\textbf{B}\right)$$

Where the cross product term is generally zero unless there's a current density (Ampere's law), and I've already dropped the $\nabla\times\textbf{m}$ term (as it's always zero), leaving $\textbf{F} = \left(\textbf{m}\cdot\nabla\right) \textbf{B}$. For a homogeneous field, $\nabla\cdot\textbf{B} = 0$ everywhere, so there is no force, thus an inhomogeneous field is required to provide a force to separate the atoms according to their dipole moments (spins).

  • $\begingroup$ A friendly suggestion: do add a parenthesis after the grad operator; that is: $$\textbf{F} = \left(\textbf{m} \cdot \nabla \right) \cdot \textbf{B}$$ I am also a little curious to what the expression $\nabla \textbf{B}$ evaluates to. $\endgroup$ – Andreas Mastronikolis Nov 26 '20 at 19:47
  • 2
    $\begingroup$ @AndreasMastronikolis ahh yeah I looked at that for a bit and felt uncomfortable before I hit post, been too long since I've actually done real vector calc. Need to un-bold that $U$ as well $\endgroup$ – llama Nov 26 '20 at 19:49
  • $\begingroup$ Another friendly suggestion: Drop the dot after the parenthesis after the grad operator, as in: $$\textbf{F} = \left(\textbf{m} \cdot \nabla \right) \textbf{B}$$ The $\textbf{m}\cdot\nabla$ is a scalar, so it can’t be dotted with a vector. $\endgroup$ – Gilbert Nov 27 '20 at 20:38

While the "how" has been well described, the "why" has not been explicitly stated:

The reason for using an inhomogeneous field is so that the device takes a pure state and entangles the spatial part of the wave function with the spin wave function.


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