Spin-½ measurement along a rotated axis (Pauli-matrix derivation)


Spin-½ measurement along a rotated axis (Pauli-matrix derivation)

We will show explicitly—using Pauli matrices—why a spin-½ measurement along any direction still yields exactly two outcomes, and compute the probabilities when the measurement axis is rotated by 30^\circ from the original x-axis.

1) Pauli matrices and spin operators

    \[ \sigma_x = \begin{pmatrix} 0 & 1 \\[4pt] 1 & 0 \end{pmatrix},\quad \sigma_y = \begin{pmatrix} 0 & -i \\[4pt] i & 0 \end{pmatrix},\quad \sigma_z = \begin{pmatrix} 1 & 0 \\[4pt] 0 & -1 \end{pmatrix}. \]

The spin operator along a unit direction \hat n is

    \[ \hat S_{\hat n} \;=\; \frac{\hbar}{2}\,(\vec\sigma\!\cdot\!\hat n)\,, \quad \text{with}\quad \vec\sigma\!\cdot\!\hat n \;=\; n_x \sigma_x + n_y \sigma_y + n_z \sigma_z. \]

For any direction \hat n, the eigenvalues of \hat S_{\hat n} are always \pm \frac{\hbar}{2}. Thus there are always two outcomes, independent of the axis.

2) Choose a new axis rotated by 30^\circ from the x-axis

Let the original measurement axis be \hat x. Rotate the apparatus by an angle \theta=30^\circ toward \hat z about the y-axis. The new unit vector is

    \[ \hat n' \;=\; (\cos\theta,\,0,\,\sin\theta). \]

Then

    \[ \vec\sigma\!\cdot\!\hat n' \;=\; \cos\theta\,\sigma_x \;+\; \sin\theta\,\sigma_z \;=\; \begin{pmatrix} \sin\theta & \cos\theta \\[4pt] \cos\theta & -\sin\theta \end{pmatrix}. \]

For \theta=30^\circ with \cos\theta=\tfrac{\sqrt{3}}{2} and \sin\theta=\tfrac{1}{2},

    \[ \vec\sigma\!\cdot\!\hat n' \;=\; \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\[6pt] \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}, \qquad \hat S_{\hat n'} \;=\; \frac{\hbar}{2}\,\vec\sigma\!\cdot\!\hat n'. \]

3) Eigenvalues (two outcomes only) and eigenvectors

The characteristic equation of \vec\sigma\!\cdot\!\hat n' is

    \[ \det\big(\vec\sigma\!\cdot\!\hat n' - \lambda I\big) = \lambda^2 - 1 = 0 \;\;\Rightarrow\;\; \lambda=\pm 1. \]

Therefore the measurement outcomes for \hat S_{\hat n'} are always \pm \frac{\hbar}{2}.

A corresponding +1 eigenvector (for general \theta) can be taken as

    \[ |+\hat n'\rangle \;=\; \begin{pmatrix}\cos\frac{\theta}{2}\\[4pt]\sin\frac{\theta}{2}\end{pmatrix} \quad\text{and}\quad |-\hat n'\rangle \;=\; \begin{pmatrix}-\sin\frac{\theta}{2}\\[4pt]\cos\frac{\theta}{2}\end{pmatrix}, \]

when we work in the \sigma_z (i.e., |+\!z\rangle, |-\!z\rangle) basis and choose azimuth \phi=0 (the x\!-\!z plane).

4) Start in |+\!x\rangle and measure along the rotated axis \hat n'

In the \sigma_z basis, the \sigma_x eigenstates are

    \[ |+\!x\rangle \;=\; \frac{1}{\sqrt{2}} \begin{pmatrix}1\\[2pt]1\end{pmatrix}, \qquad |-\!x\rangle \;=\; \frac{1}{\sqrt{2}} \begin{pmatrix}1\\[2pt]-1\end{pmatrix}. \]

If the system is prepared in |+\!x\rangle and we measure \hat S_{\hat n'}, the probability of obtaining +\frac{\hbar}{2} is

    \[ P_{+}(\hat n' \mid +x) \;=\; \big|\langle +\hat n' \,|\, +x \rangle\big|^2 \;=\; \cos^2\!\frac{\theta}{2}, \]

and P_{-}(\hat n' \mid +x) = \sin^2\!\frac{\theta}{2}. This can be derived either by the explicit overlap of the spinors above, or by the Bloch-vector identity

    \[ P_{+}(\hat n' \mid \hat a) \;=\; \frac{1+\hat a\!\cdot\!\hat n'}{2}, \quad\text{with}\;\; \hat a=\hat x \;\Rightarrow\; P_{+}=\frac{1+\cos\theta}{2}=\cos^2\!\frac{\theta}{2}. \]

5) Plug in \theta=30^\circ

    \[ \cos\frac{\theta}{2}=\cos 15^\circ =\sqrt{\frac{1+\cos30^\circ}{2}} =\sqrt{\frac{1+\tfrac{\sqrt{3}}{2}}{2}} =\sqrt{\frac{2+\sqrt{3}}{4}} \quad\Rightarrow\quad \cos^2 15^\circ=\frac{2+\sqrt{3}}{4}\approx 0.933. \]

    \[ \sin^2 15^\circ=\frac{2-\sqrt{3}}{4}\approx 0.067. \]

Therefore, for a 30^\circ rotation of the measurement axis relative to x, a state prepared as |+\!x\rangle yields

  • Outcome +\frac{\hbar}{2} along \hat n' with probability \displaystyle \frac{2+\sqrt{3}}{4}\approx 93.3\%.
  • Outcome -\frac{\hbar}{2} along \hat n' with probability \displaystyle \frac{2-\sqrt{3}}{4}\approx 6.7\%.

6) What is special about the chosen direction?

Physically, no direction is special: quantum spin for a spin-½ particle always has two eigenvalues along any axis. What is special is that the experimentalist’s chosen direction defines the observable:

    \[ \hat S_{\hat n} \;=\; \frac{\hbar}{2}\,(\vec\sigma\!\cdot\!\hat n), \]

and distinct directions correspond to non-commuting operators (e.g. [\hat S_x,\hat S_z]\neq 0). Changing the apparatus orientation rotates the measurement basis; it does not change the two-valued spectrum.

7) One-line summary

Rotate the experiment by any angle you like (e.g. 30^\circ): the spin-½ measurement still has outcomes \pm \tfrac{\hbar}{2}; only the probabilities change according to \cos^2(\theta/2) and \sin^2(\theta/2).