From spinor to ket space: Equivalents eigen equations

[LCSphysicist]

Homework Statement

Hello. I am having a little of trouble to understand how do we go from one equation involving spinors, to the "same equations in ket language".

Relevant Equations

.

"$\sigma . n X = 1*X$"
to
"$S. n| S. n; +\rangle = \frac{h}{4\pi}| S .n; +\rangle$"​

X is a spinor
n is any unitary vector
sigma are the pauli matrices $(\sigma 0, \sigma x,\sigma y,\sigma z)$
S is the spin vector.

It was claimed that both equations are equivalent, but i couldn't see why.

 

[jambaugh]

I'm a little bit confused by your notation (the periods specifically. do they represent subscripts?).
But I believe the only substantive difference in the two equations is that S is the spin operator i.e. the physical observable which generates rotations around some axis while $\sigma$ is a Pauli spin matrix which is proportional to the representation of the spin operator in the spinor representation of the rotation group.
J_n \to S_n= \frac{\hbar}{2} \sigma(and $h = 2\pi \hbar$.)

So you could just as aptly have written $S_.n X = \frac{h}{4\pi} X$.

So for example in the spinor representation (in the spin-z operator's eigen-basis):
J_z \to S_z = \frac{\hbar}{2}\sigma_z=\frac{\hbar}{2}\left(\begin{array}{rr} 1 & 0\\0 & -1\end{array}\right) while in the vector representation (say of a massive boson) you have:
J_z \to \hbar\left(\begin{array}{rr}1 &0 &0\\ 0 & 0 & 0 \\ 0 & 0 & -1\end{array}\right)
Here the $J_z$ is the physically interpreted operator representing the observable for z-component of spin in any representation. $S_z$ is (I believe in most texts) specifically its spinor representation and the half of Plank's constant factor is factored out to give the more purely mathematical Pauli spin matrix.