Understanding Spinor Formulation in Quantum Mechanics

[[AFT]]

This is not an assignment problem, but I am studying for my quantum mechanics final exam and came across a derivation in the book which I can't seem to get my head around :(

The example in the book is solving for the probabilities of getting +h(bar)/2 and -h(bar)/2 if we are to measure the spin angular momentum Sx.

I was able to follow the derivation up to the point where they obtained the eigenspinors:

X+ = [1/sqrt2 1/sqrt2]' and X- = [1/sqrt2 -1/sqrt2]'

But I don't get how they go from those to formulating the spinor:

X = [(a+b)/sqrt2]X+ + [(a-b)/sqrt2]X-

Any guidance would be much appreciated - thanks in advance.

 

[diazona]

I'm assuming a and b are the coefficients in
\chi = \begin{pmatrix}a \\ b\end{pmatrix}
right?

Think of it this way: all possible spinors form a vector space. The vectors (1,0) and (0,1) form the standard basis for that space - in other words, when you have a state \chi defined by two coefficients a and b, that's actually saying
\chi = a\begin{pmatrix}1 \\ 0\end{pmatrix} + b\begin{pmatrix}0 \\ 1\end{pmatrix}
But you can express the same state in terms of any other basis. For example, \chi_+ and \chi_- form a basis (just like (1/√2,1/√2) and (1/√2,-1/√2) form a basis for the xy plane), so you can write the state \chi as
\chi = c\chi_+ + d\chi_- = c\begin{pmatrix}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{pmatrix} + d\begin{pmatrix}\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{pmatrix}
The coefficients c and d are the components of \chi in the +- basis. Can you find them?

(Hint: if you're familiar with the vector projection formula, that's probably the quickest - though certainly not the only - way to do it)