So there are two unit vectors $\hat{m}$ and $\hat{n}$ with arbitrary directions in 3D space.

There is a spin operator along a particular direction in space, say that of $\hat{r}$, is:

$\sigma_r= \begin{bmatrix} r_z & r_x-ir_y \\ r_x+ir_y & -r_z \end{bmatrix}$

..which has the eigenvalues +1,-1.

Here is the question:

Suppose that a spin is prepared so that $\sigma_m = +1$. The apparatus is then rotated to the $\hat{n}$ direction and $\sigma_n$ is measured. What is the probability that the result is $+1$?

Now I've tried this using spherical coordinates, and Cartesian coordinates but it always ends up in a huge spaghetti of algebra. I have a feeling I'm doing something subtly wrong or missing some fundamental facts.

Here is what my procedure of attempt was:

I find the eigenvector (the possible state) of $\sigma_m$ corresponding to the eigenvalue +1. I get the normalized vector:

$|m_+> \ = \frac{1}{\sqrt{2(1-m_z)}} \begin{bmatrix} m_x-im_y \\ 1-m_z \end{bmatrix}$

Similarly, I find the same for the $\sigma_n$ operator:

$|n_+> \ = \frac{1}{\sqrt{2(1-n_z)}} \begin{bmatrix} n_x-in_y \\ 1-n_z \end{bmatrix}$

Then, for the probability I simply find (or try to..) $P = |<n_+|m_+>|^2$. The answer given is $\cos^2(\rho/2)$ where $cos(\rho)=\hat{n}.\hat{m}$.

Am I attempting this correctly?


1 Answer 1


If you want to reduce the "spaghetti of algebra" you can reorient the coordinates. If $\hat{z}'=\hat{m}$ then in spherical coordinates you have $$ \sigma_\hat{m}= \left[ \begin{array}{cc} \cos(\theta) & \sin(\theta) \\ \sin(\theta) & -\cos(\theta) \\ \end{array} \right] $$ where $\cos(\theta)=\hat{n} \cdot \hat{m}$. It's easier to find the eigenvectors of this matrix using: $$ \sin(\theta/2)= \sin(\theta-\theta/2)= \sin(\theta)\cos(\theta/2)-\cos(\theta)\sin(\theta/2) $$ and the analogous equation for cosine.

Therefore $$ |n_{+}> = \left[ \begin{array}{c} \cos(\theta/2) \\ \sin(\theta/2) \\ \end{array} \right] $$ and finally: $$ |<m_+|n_+>|^2=\cos^2(\theta/2) $$

  • $\begingroup$ Thank you for your answer. Effectively, that is re-orienting our 3D coordinates such that m and n lie in the same plane formed by a pair of new axis - am I right? Also, what does the z' mean? $\endgroup$ Commented Nov 10, 2015 at 20:12
  • $\begingroup$ You're right, that way you have less variables. z' mean the z axis of the new coordinate system that way there is no azimuthal angle in the equations. $\endgroup$
    – Hu Al
    Commented Nov 11, 2015 at 1:27

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