# Angle between spins

1. Oct 23, 2012

### matematikuvol

If $|\alpha>$ is spin up, and $|\beta>$ is spin down. Then if angle between those spins and some other up and down spin is $\theta$, then
$$|\alpha'>=\cos \frac{\theta}{2}|\alpha>+\sin \frac{\theta}{2}|\beta>$$
$$|\beta'>=\sin \frac{\theta}{2}|\alpha>-\cos \frac{\theta}{2}|\beta>$$
Why?

2. Oct 23, 2012

### Bill_K

For each value j of angular momentum there are 2j+1 linearly independent states. For example these can be taken as the states with spin projection mz = -j,... +j along the z axis. They form a basis in a 2j+1-dimensional space. We can just as well take for a basis the states with projection ma along any other axis a, and the transformation from one basis to another is a unitary transformation,

|ma> = Σ|mz><mz|D(j)(α,β,γ)|ma>

where D(j)(α,β,γ) is a unitary operator whose matrix elements <mz|D(j)(α,β,γ)|ma> are called the rotation matrix. An arbitrary rotation in three dimensions requires three Euler angles α,β,γ to describe.

For spin 1/2 the space is two-dimensional, just spin up and spin down. The simplest rotation from the z axis to some other axis a is through an angle θ directly down a line of longitude, and the rotation matrix is (almost!) what you have written,

$$\left(\begin{array}{cc}cos θ/2&sin θ/2\\-sin θ/2&cos θ/2\end{array}\right)$$

3. Oct 23, 2012

### Amok

Nvm, I had misunderstood the question.

4. Oct 23, 2012

### matematikuvol

But why you get $\frac{\theta}{2}$ in matrix if you rotate for angle $\theta$?

5. Oct 23, 2012

### Amok

Wait, the state-space is 2-dimensional, but isn't this problem making reference to a rotation in the real space where this spin-1/2 particle is? I mean, there's a z-axis.

6. Oct 23, 2012

### tom.stoer

simple question: are we talking about the angle between the two axes in 3-dim. position space or about the angles between two spin states in 2-dim spin. space?

7. Oct 23, 2012

### Amok

Yes, that's what I meant. It's not really that clear from the question.

8. Oct 23, 2012

### Bill_K

Tom, All I have tried to do is give a simple answer to a simple question. Please let's not throw confusion at it. Especially since you know how a spinor transforms inside out, forwards and backwards. The appearance of the half angle in the rotation matrix is a result of the mapping between SO(3) and SU(2).

9. Oct 24, 2012

### andrien

I think it has to do with a direct spin character i.e. it is true for spin 1/2 something like
exp(imθ) .

10. Oct 27, 2012

### matematikuvol

Ok if I don't know that. I have some up spin. How to get up spin which is rotate for angle $\theta$ from that spin. Can I use Pauli matrices and spherical coordinates and get that result?