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

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.

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 $$\cos^2 \frac{\theta}{2}$$.