In Griffiths' Introduction to Quantum Mechanics, he briefly introduces the Stern-Gerlach experiment with the following example.

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The magnetic field is $$ \mathbf{B} = -\alpha x \mathbf{\hat{x}} + \left(B_0 + \alpha z\right)\mathbf{\hat{z}}, $$ where $\alpha$ is a small deviation from uniform magnetic field, and an $x$ component is added to satisfy $\nabla \cdot \mathbf{B} = 0$.

Using $$ \mathbf{F} = \nabla\left(\mathbf{\mu} \cdot \mathbf{B}\right) $$ and $$ \mathbf{\mu} = \gamma \mathbf{S}, $$ where $\gamma$ is the gyromagnetic ratio, we find that (averaging over time, the influence of $S_x$ is equal to zero due to Larmor precession) $$ \mathbf{F} = \gamma\alpha S_z \mathbf{\hat{z}} = \frac{\gamma\alpha\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \mathbf{\hat{z}} . $$ Now I don't get what it means for the component of a vector to be a matrix, or if it makes sense at all.


1 Answer 1


What you wrote down is the operator for z-projection of the force, using the operator for angular momentum.

The force itself is still a scalar, where it's magnitude is the expectation value, e.g., $\langle\uparrow|\hat{\mathbf{F}}|\uparrow\rangle = \frac{\gamma \alpha \hbar}{2}$.

  • $\begingroup$ I suppose $\newcommand{\ket}[1]{|#1\rangle} \ket{\uparrow}$ means the spin-up state with respect to the $z$ axis? $\endgroup$ Commented Jun 12, 2023 at 16:39
  • 1
    $\begingroup$ Yes. In the representation you use, it would correspond to the $[1, 0]^T$ state vector. $\endgroup$
    – TomP
    Commented Jun 12, 2023 at 22:58

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