Setup: eigenstates of $\sigma_y$#

The Pauli matrix $\sigma_y = \begin{pmatrix}0 & -i \\ i & 0\end{pmatrix}$ has eigenvalues $\pm 1$, corresponding to spin $\pm\hbar/2$. Its normalized eigenstates are:

$|\uparrow_y\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\i\end{pmatrix}, \qquad |\downarrow_y\rangle = \frac{1}{\sqrt{2}}\begin{pmatrix}1\\-i\end{pmatrix}$

Our state $|\Psi\rangle = \begin{pmatrix}\alpha\\\beta\end{pmatrix}$ is written in the $S_z$ eigenbasis, i.e. $|\Psi\rangle = \alpha|\uparrow_z\rangle + \beta|\downarrow_z\rangle$. To find spin-$y$ probabilities we project onto the $\sigma_y$ eigenstates.

Probability of $s_y = +\hbar/2$#

By the Born rule, $P(s_y = +\tfrac{\hbar}{2}) = |\langle\uparrow_y |\Psi\rangle|^2$: the squared overlap with the eigenstate, with no operator in between. The bra is the conjugate transpose, $\langle\uparrow_y| = \tfrac{1}{\sqrt2}\begin{pmatrix}1 & -i\end{pmatrix}$:

$P(s_y = +) = \frac{1}{2}\left|\begin{pmatrix}1 & -i\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}\right|^2 = \frac{1}{2}|\alpha - i\beta|^2$

(A common slip is to write $|\langle\uparrow_y|S_y|\Psi\rangle|^2$ instead. Here it would give $\tfrac{\hbar^2}{4}$ times the right answer, or exactly the right answer if you use $\sigma_y$. That’s only because $\langle\uparrow_y|\sigma_y = +\langle\uparrow_y|$ and $\langle\downarrow_y|\sigma_y = -\langle\downarrow_y|$, and it isn’t the Born rule.)

Probability of $s_y = -\hbar/2$#

Similarly, with $\langle\downarrow_y| = \tfrac{1}{\sqrt2}\begin{pmatrix}1 & i\end{pmatrix}$:

$P(s_y = -) = \frac{1}{2}\left|\begin{pmatrix}1 & i\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}\right|^2 = \frac{1}{2}|\alpha + i\beta|^2$

Sanity check: $P(+) + P(-) = \frac{1}{2}(|\alpha - i\beta|^2 + |\alpha + i\beta|^2) = \frac{1}{2}(2|\alpha|^2 + 2|\beta|^2) = 1$ ✓

Probabilities along $z$#

For $\sigma_z = \begin{pmatrix}1&0\\0&-1\end{pmatrix}$ with eigenstates $|\uparrow_z\rangle = \begin{pmatrix}1\\0\end{pmatrix}$, $|\downarrow_z\rangle = \begin{pmatrix}0\\1\end{pmatrix}$:

$P(s_z = +) = \left|\begin{pmatrix}1&0\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}\right|^2 = |\alpha|^2$

$P(s_z = -) = \left|\begin{pmatrix}0&1\end{pmatrix}\begin{pmatrix}\alpha\\\beta\end{pmatrix}\right|^2 = |\beta|^2$

Which is just the Born rule applied directly — reassuringly, $\alpha$ and $\beta$ are exactly the probability amplitudes for $s_z = \pm\hbar/2$.

Verification with known states#