Spin Measurements

A spin measurement is a measurement of a chosen spin component. For spin-$1/2$, measuring along a unit axis $\hat{\mathbf n}$ means measuring

$$S_{\hat n} = \hat{\mathbf n}\cdot\mathbf S = \frac{\hbar}{2}\hat{\mathbf n}\cdot\boldsymbol\sigma.$$

The possible outcomes are

$$\pm\frac{\hbar}{2}.$$

The axis is part of the measurement specification. The invariant $S^2$ has the fixed value $3\hbar^2/4$, but it does not specify a spin projection. To predict the two outcomes $\pm\hbar/2$, one must name the measured component, such as $S_z$, $S_x$, or $S_{\hat n}$.

This page applies the projective measurement formalism to spin. The physical Stern–Gerlach analyzer model is treated in Stern–Gerlach Revisited, and the general measurement rules are treated in Projective Measurement.

Measurement Along z

In the standard $S_z$ basis,

$$\lvert\uparrow\rangle=\lvert+z\rangle, \qquad \lvert\downarrow\rangle=\lvert-z\rangle,$$

with

$$S_z\lvert\uparrow\rangle = \frac{\hbar}{2}\lvert\uparrow\rangle, \qquad S_z\lvert\downarrow\rangle = -\frac{\hbar}{2}\lvert\downarrow\rangle.$$

The corresponding projectors are

$$P_{+z} = \lvert\uparrow\rangle\langle\uparrow\rvert = \frac12(I+\sigma_z),$$

and

$$P_{-z} = \lvert\downarrow\rangle\langle\downarrow\rvert = \frac12(I-\sigma_z).$$

For a normalized state

$$\lvert\psi\rangle = \alpha\lvert\uparrow\rangle +\beta\lvert\downarrow\rangle,$$

the Born probabilities are

$$p(+z)=\lvert\alpha\rvert^2, \qquad p(-z)=\lvert\beta\rvert^2.$$

These probabilities describe the specified measurement. The formula alone neither assigns pre-existing values to all spin components nor rules out every hidden-variable model; such claims require additional assumptions and experiments.

Arbitrary-Axis Projectors

Let

$$\hat{\mathbf n} = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$$

be a unit vector. The spin component along $\hat{\mathbf n}$ is

$$S_{\hat n} = \frac{\hbar}{2}\hat{\mathbf n}\cdot\boldsymbol\sigma.$$

Because

$$(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I,$$

the eigenvalues of $\hat{\mathbf n}\cdot\boldsymbol\sigma$ are $\pm1$, and the eigenvalues of $S_{\hat n}$ are $\pm\hbar/2$.

The projectors are

$$P_\pm^{(\hat n)} = \frac12 \left( I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma \right).$$

They obey the projective-measurement identities

$$P_+^{(\hat n)}+P_-^{(\hat n)}=I,$$ $$P_\pm^{(\hat n)}P_\pm^{(\hat n)} = P_\pm^{(\hat n)},$$

and

$$P_+^{(\hat n)}P_-^{(\hat n)}=0.$$

Thus $\{P_+^{(\hat n)},P_-^{(\hat n)}\}$ is the two-outcome projective measurement associated with the spin component $S_{\hat n}$.

Probabilities from Spinors

For a pure state $\lvert\psi\rangle$, the probability of outcome $\pm$ along $\hat{\mathbf n}$ is

$$p_\pm = \langle\psi|P_\pm^{(\hat n)}|\psi\rangle.$$

If the state is the $+\hat{\mathbf a}$ eigenstate, where $\gamma$ is the angle between $\hat{\mathbf a}$ and $\hat{\mathbf n}$, then

$$p_+ = \frac12 \left( 1+\hat{\mathbf a}\cdot\hat{\mathbf n} \right) = \cos^2\frac{\gamma}{2},$$

and

$$p_- = \sin^2\frac{\gamma}{2}.$$

For example, $|+z\rangle$ measured along $x$ has $\gamma=\pi/2$, so both outcomes have probability $1/2$.

Probabilities from the Bloch Vector

For a spin-$1/2$ density operator,

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right),$$

where $\lvert\mathbf r\rvert\leq1$, the measurement probabilities are

$$p_\pm = \operatorname{Tr} \left( \rho P_\pm^{(\hat n)} \right) = \frac12 \left( 1\pm\mathbf r\cdot\hat{\mathbf n} \right).$$

This is the compact Bloch-sphere form of the Born rule. The component of the Bloch vector along the analyzer axis controls the bias; transverse components affect other axes but not this outcome probability directly.

The expectation value is

$$\langle S_{\hat n}\rangle = \frac{\hbar}{2}\mathbf r\cdot\hat{\mathbf n}.$$

Thus a state can have zero expectation value along an axis while still giving two possible outcomes $\pm\hbar/2$.

State Update

If the outcome $+$ along $\hat{\mathbf n}$ occurs in an ideal selective projective measurement, a pure state updates as

$$\lvert\psi\rangle \longrightarrow \frac{ P_+^{(\hat n)}\lvert\psi\rangle }{ \sqrt{p_+} }.$$

Similarly, outcome $-$ gives

$$\lvert\psi\rangle \longrightarrow \frac{ P_-^{(\hat n)}\lvert\psi\rangle }{ \sqrt{p_-} }.$$

For density operators, the update is

$$\rho \longrightarrow \frac{ P_\pm^{(\hat n)}\rho P_\pm^{(\hat n)} }{ \operatorname{Tr} \left( \rho P_\pm^{(\hat n)} \right) }.$$

These conditional formulas apply only when the outcome has nonzero probability. They describe the ideal repeatable (Lüders) measurement; a device with the same outcome probabilities can apply an additional transformation afterward. This is a conditional state assignment after a specified outcome. It is not a full microscopic model of a detector.

If this ideal measurement is performed but its recorded outcome is disregarded, the nonselective update is

$$\rho \longrightarrow P_+^{(\hat n)}\rho P_+^{(\hat n)} + P_-^{(\hat n)}\rho P_-^{(\hat n)}.$$

In Bloch-vector language, this keeps only the component of $\mathbf r$ along $\hat{\mathbf n}$:

$$\mathbf r \longrightarrow (\mathbf r\cdot\hat{\mathbf n})\hat{\mathbf n}.$$

The measurement removes coherence between the two eigenspaces of $S_{\hat n}$.

Noncommuting Axes

Spin measurements along different axes are generally incompatible. For two unit vectors $\hat{\mathbf a}$ and $\hat{\mathbf b}$,

$$[S_{\hat a},S_{\hat b}] = i\hbar S_{\hat{\mathbf a}\times\hat{\mathbf b}}.$$

Here $S_{\mathbf v}=\mathbf v\cdot\mathbf S$ also denotes a nonunit vector $\mathbf v$; the cross product has length $\sin\gamma$. If the axes are neither parallel nor antiparallel, the spin-$1/2$ commutator is nonzero. A state prepared with definite $S_z$ is therefore not generally definite in $S_x$.

For spin-$1/2$,

$$\lvert+z\rangle = \frac{1}{\sqrt2} \left( \lvert+x\rangle+\lvert-x\rangle \right).$$

An $x$ measurement of a $+z$ state gives equal probabilities. If one outcome is selected, the post-measurement state is an $x$ eigenstate, and a later $z$ measurement again has equal probabilities. This is the spin version of the general Sequential Measurements rule.

Repeated and Sequential Measurements

Repeated measurement along the same axis is deterministic in the ideal projective model. If a measurement of $S_{\hat n}$ gives $+$ and is immediately repeated, the second measurement gives $+$ with probability $1$ because the state has been updated into the $+\hat{\mathbf n}$ eigenspace.

Changing the axis changes the measurement. A useful three-step example is:

1. prepare $|+z\rangle$;
2. measure $S_x$ and select $+x$;
3. measure $S_z$ again.

The final $S_z$ measurement gives

$$p(+z)=p(-z)=\frac12.$$

The intermediate $S_x$ selection prepares a new state. It does not merely reveal a value while leaving the original $S_z$ certainty intact.

Connection to Stern–Gerlach

An ideal Stern–Gerlach analyzer oriented along $\hat{\mathbf n}$ is the standard physical picture for the measurement $\{P_+^{(\hat n)},P_-^{(\hat n)}\}$. The apparatus correlates the two spin projectors with two spatial output paths.

For how magnetic forces separate the output paths, and where the ideal measurement approximation enters, see Stern–Gerlach Revisited.

Common Mistakes

• Saying “measure spin” without specifying the axis.
• Using $|\alpha|^2$ and $|\beta|^2$ for every axis instead of only the chosen basis.
• Forgetting that $P_\pm^{(\hat n)}$ changes when the analyzer axis changes.
• Treating the post-measurement state as the same as the pre-measurement state after a selective outcome.
• Assuming that definite $S_z$ implies definite $S_x$.
• Confusing expectation value $\langle S_{\hat n}\rangle$ with a single-shot outcome.
• Treating the projective update rule as a complete physical detector model.

Exercises

1. A state is $\lvert\psi\rangle=(\lvert\uparrow\rangle+i\lvert\downarrow\rangle)/\sqrt2$. Find the probabilities for an $S_z$ measurement.

Solution

In the $S_z$ basis,

$$\alpha=\frac{1}{\sqrt2}, \qquad \beta=\frac{i}{\sqrt2}.$$

Therefore

$$p(+z)=\lvert\alpha\rvert^2=\frac12, \qquad p(-z)=\lvert\beta\rvert^2=\frac12.$$

2. A spin starts in $|+z\rangle$. It is measured along an axis $\hat{\mathbf n}$ making angle $\theta$ with $z$, and the outcome is discarded. Find the probability of $+z$ in a final measurement.

Solution

Initially $\mathbf r=\hat{\mathbf z}$. The nonselective intermediate measurement gives $\mathbf r'=\cos\theta\,\hat{\mathbf n}$, so

$$P(+z)=\frac{1+\mathbf r'\cdot\hat{\mathbf z}}{2} =\frac{1+\cos^2\theta}{2}.$$

It is one for parallel or antiparallel analyzer axes and $1/2$ for a perpendicular axis. Discarding the outcome does not undo the measurement disturbance.

3. Verify that $P_\pm^{(\hat n)}=(I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma)/2$ are orthogonal projectors.

Solution

Use

$$(\hat{\mathbf n}\cdot\boldsymbol\sigma)^2=I.$$

Then

$$\left( P_\pm^{(\hat n)} \right)^2 = \frac14 \left( I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma \right)^2 = \frac12 \left( I\pm\hat{\mathbf n}\cdot\boldsymbol\sigma \right) = P_\pm^{(\hat n)}.$$

Also,

$$P_+^{(\hat n)}P_-^{(\hat n)} = \frac14 \left[ I- (\hat{\mathbf n}\cdot\boldsymbol\sigma)^2 \right] = 0.$$

4. A spin is prepared in $|+z\rangle$, then $S_x$ is measured and the $+x$ outcome is selected. What are the probabilities for a later $S_z$ measurement?

Solution

After selecting $+x$, the state is $|+x\rangle$. In the $S_z$ basis,

$$|+x\rangle = \frac{1}{\sqrt2} \left( |+z\rangle+|-z\rangle \right).$$

Therefore the later $S_z$ probabilities are

$$p(+z)=\frac12, \qquad p(-z)=\frac12.$$

References

• C. Cohen-Tannoudji, B. Diu, and F. Laloë, Quantum Mechanics, Wiley, 1977.
• M. A. Nielsen and I. L. Chuang, Quantum Computation and Quantum Information, Cambridge University Press, 2010.
• A. Peres, Quantum Theory: Concepts and Methods, Kluwer, 1995.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.