Angle between spins

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
[tex]|\alpha'>=\cos \frac{\theta}{2}|\alpha>+\sin \frac{\theta}{2}|\beta>[/tex]
[tex]|\beta'>=\sin \frac{\theta}{2}|\alpha>-\cos \frac{\theta}{2}|\beta>[/tex]
Why?

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 m_z = -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 m_a along any other axis a, and the transformation from one basis to another is a unitary transformation,

|m_a> = Σ|m_z><m_z|D^(j)(α,β,γ)|m_a>

where D^(j)(α,β,γ) is a unitary operator whose matrix elements <m_z|D^(j)(α,β,γ)|m_a> 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,

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

Amok

Nvm, I had misunderstood the question.

matematikuvol

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

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.

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?

Amok

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

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).

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) .

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?