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beginner · Physics · Spin & the Stern–Gerlach Experiment

Measuring Spin Along x

The SxS_x, SyS_y, and SzS_z operators for a spin-12\tfrac{1}{2} particle are given by the Pauli matrices:

Si=2σi,i{x,y,z}.S_i = \frac{\hbar}{2}\,\sigma_i, \qquad i \in \{x, y, z\}.

The SzS_z eigenstates |{\uparrow}\rangle and |{\downarrow}\rangle form a convenient basis. When a particle is in one of these states and we measure SxS_x instead, the result is no longer certain — the SzS_z eigenstates are not eigenstates of SxS_x.

Eigenstates of SxS_x

The SxS_x operator has matrix representation

Sx=2(0110).S_x = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0\end{pmatrix}.

Solving Sxv=λvS_x |v\rangle = \lambda |v\rangle gives eigenvalues λ=±/2\lambda = \pm\hbar/2 and, after normalisation, the eigenstates expressed in the zz-basis:

+x=12(+),x=12().|{+}\rangle_x = \frac{1}{\sqrt{2}}\bigl(|{\uparrow}\rangle + |{\downarrow}\rangle\bigr), \qquad |{-}\rangle_x = \frac{1}{\sqrt{2}}\bigl(|{\uparrow}\rangle - |{\downarrow}\rangle\bigr).

You can verify these directly: acting SxS_x on +x|{+}\rangle_x gives 2+x\tfrac{\hbar}{2}|{+}\rangle_x, and on x|{-}\rangle_x gives 2x-\tfrac{\hbar}{2}|{-}\rangle_x.

Predicting x-measurement outcomes

Suppose the particle has been prepared in |{\uparrow}\rangle (spin-up along zz). Invert the two relations above to express |{\uparrow}\rangle in the SxS_x eigenbasis:

=12+x+12x.|{\uparrow}\rangle = \frac{1}{\sqrt{2}}\,|{+}\rangle_x + \frac{1}{\sqrt{2}}\,|{-}\rangle_x.

By the Born rule, the probability of each outcome is the squared modulus of the corresponding amplitude:

P ⁣(Sx=+2)=122=12,P ⁣(Sx=2)=122=12.P\!\left(S_x = +\tfrac{\hbar}{2}\right) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}, \qquad P\!\left(S_x = -\tfrac{\hbar}{2}\right) = \left|\frac{1}{\sqrt{2}}\right|^2 = \frac{1}{2}.

Both outcomes are equally likely — a 50:5050{:}50 split. This is the defining signature of a state that is an eigenstate of one spin component being measured along a perpendicular component.

Bloch sphere picture

On the Bloch sphere, |{\uparrow}\rangle sits at the north pole and |{\downarrow}\rangle at the south pole, while +x|{+}\rangle_x points along +x+x and x|{-}\rangle_x along x-x. Measuring SxS_x for a state on the zz-axis is geometrically equivalent to projecting a vector that is perpendicular to the measurement axis — the two projections are equal, explaining the 50:50 outcome directly.

Try it

This is a numerical exercise — your code should return a number. A spin-12\tfrac{1}{2} particle is prepared in z|{\uparrow}\rangle_z and a Stern–Gerlach device is oriented along xx. Compute the probability of obtaining Sx=+/2S_x = +\hbar/2.

Run your code to see the quantum state.

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