Why do the eigenvectors of a spin observable not align with the direction of the spin?

Question

The italicized section below is referring to chapter 3 of "Quantum Mechanics: The Theoretical Minimum" by Leonard Susskind and Art Friedman.

It is written that the operator that corresponds to measuring spin in the $\hat{n}$ direction is $\sigma_\hat{n} = n_x X + n_y Y + n_z Z$. The book gives an example where $\hat{n}$ lies in the z-x plane, $$\hat{n} = \begin{pmatrix}\sin(\theta) \\ 0\\ \cos(\theta)\end{pmatrix}$$ where $\theta$ is the angle between $\hat{n}$ and the $z$ axis. In matrix form, this gives: $$ \sigma_\hat{n}= \begin{pmatrix} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \end{pmatrix} $$ with eigenvalues and eigenvectors $[1, \big(\begin{smallmatrix} \cos(\theta/2) \\ \sin(\theta/2) \end{smallmatrix}\big)]$ and $[-1, \big(\begin{smallmatrix} -\sin(\theta/2) \\ \cos(\theta/2) \end{smallmatrix}\big)]$.

So, if a state, $|\psi\rangle$, is pointing in the direction $\hat{n}$, it has an angle of $\theta$ from the Z axis, then it's state vector in the Z basis should be $|\psi\rangle = \big(\begin{smallmatrix} \cos(\theta) \\ \sin(\theta) \end{smallmatrix}\big) = \hat{n}$.

Why do the eigenvectors of $\sigma_n$ not align with the vector for $\hat{n}$? My thinking is that measuring the spin in the direction $\hat{n}$ should leave a state either in $\hat{n}$ or a state orthogonal to $\hat{n}$.

My guess is that I'm making a mental error going from the Cartesian system to Bloch Sphere system.

Answer 1

The key idea is to distinguish between the physical space and the Hilbert space.

The observable $\sigma_{\hat n}$ is associated with the physical space, here modeled as $\mathbb{R}^3$. Consequently, $\sigma_{\hat n}$ has three real components $n_x, n_y$ and $n_z$.

The state $|\psi\rangle$ lives in the Hilbert space, here modeled as $\mathbb{C}^2$. Consequently, $|\psi\rangle$ has two complex components. Note that the eigenvectors you found, e.g. $(\cos\frac{\theta}{2}, \sin\frac{\theta}{2})$ happen to have real components, but this is purely accidental. They would have non-zero imaginary part if you considered an observable with non-zero $n_y$.

Since $|\psi\rangle$ does not live in the physical space, it does not make sense to ask which physical direction it points along. However, it does make sense to ask which direction $\sigma_{\hat n}$ points along (and the answer is $\hat n$). This is related to the physical significance of the components of vectors in the two spaces.

The physical significance of the components of $\sigma_{\hat n}$ lies in specifying the direction along which the spin is measured. It can be thought of as a setting of a measurement device. By contrast, the physical significance of the components of $|\psi\rangle$ lies in determining the probabilities of various measurement outcomes. For example, since the first component of your first eigenvector is $\cos \frac{\theta}{2}$ we know that if we were to measure $\sigma_z$ we would obtain the $+1$ or "spin up" result with probability $\cos^2 \frac{\theta}{2}$.