# The generic spinor as a linear combination

1. Mar 28, 2008

### nowits

I have trouble understanding the concept of spin (spin 1/2 in this case). In Introduction to Quantum Mechanics Griffiths states that "the generic spinor X can be expressed as a linear combination of [eigenvectors of the spin component Sx]
$$\chi = \frac{a+b}{\sqrt{2}}\chi^{(x)}_+ + \frac{a-b}{\sqrt{2}}\chi^{(x)}_-$$

What I don't understand (among other things) is why Sy and Sz don't count. Because there are only to directions, up and down? (If so, how do I know which of Sx,Sy and Sz to choose?)

Another thing is that I'm told that a spin in the direction (x,y,z) is defined as $$\textbf{S}=\frac{\hbar}{2}(x\sigma _x +y\sigma _y +z\sigma _z)$$. What exactly is this and how does it relate to the spinor? It is a matrix but a spinor (that's also supposed to describe the spin state) is a vector.

2. Mar 28, 2008

### siddharth

Well, Sy and Sz do count. Point is, $$\chi^{(x)}_+$$ and $$\chi^{(x)}_-$$ are eigenspinors of the $$S_x$$ operator. These two eigenspinors span the 1/2 spin states, so any general state can be written as a linear combination of the two.

If you write the general spin state in terms of the eigenspinors of the $$S_z$$ operator,

$$\chi_+ = \left( \begin{array}{c} 1 \\ 0 \end{array} \right)$$

$$\chi_-=\left( \begin{array}{c} 0 \\ 1 \end{array} \right)$$

as

$$\chi = a \chi_+ + b \chi_-$$, this general spin state can also be written in the basis of the eigenspinors of the $$S_x$$ operator as

$$\chi = \frac{a+b}{\sqrt{2}}\chi^{(x)}_+ + \frac{a-b}{\sqrt{2}}\chi^{(x)}_-$$

S is not the spin state, but the operator to measure the spin in the required direction. In this notation, x,y,z are components of a unit vector in the required direction. $$\sigma _x, \sigma _y, \sigma _z$$ are the pauli matrices.

Last edited: Mar 28, 2008
3. Mar 28, 2008

### nowits

Ok, many thanks. Though I'm still not sure if I understand this. I mean it's quite unclear to me how these are used.

Edit: The problem is the following:
We have vectors a=(a1,a2,a3) and b=(b1,b2,b3)
Express the spin state |+>b as a linear combination of the normalized eigenvectors of the spin operator Sa and show that the probabilities to measure +/- hbar/2 are $$cos^2 (\theta/2)$$ and$$sin^2 (\theta/2)$$ where theta is the angle between a and b.

I know the eigenvectors of Sa but can't get past that.

Below is everything I have so far.
-------

How do I calculate the eigenvalue of S?

$$\textbf{S}=\frac{\hbar}{2}(x \sigma _x + y \sigma _y + z \sigma _z)= \frac{\hbar}{2}\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)$$
So the eigenvalues would be..?
$$\left| \begin{array} {cc}\frac{\hbar}{2}z-\lambda & \frac{\hbar}{2}(x-iy) \\ \frac{\hbar}{2}(x+iy) & -\frac{\hbar}{2}z-\lambda \end{array}\right| \Rightarrow \lambda = ^+_- \frac{\hbar}{2}$$
because (x,y,z) is a unit vector.
If I'm to find the eigenvector (of the positive eigenvalue):
$$\textbf{Sv}=\frac{\hbar}{2}\textbf{v}\Rightarrow\frac{\hbar}{2}\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)\textbf{v}=\frac{\hbar}{2}\textbf{v}\Rightarrow\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)\textbf{v}=\textbf{v}\Rightarrow\left( \begin{array}{cc} z & x -iy \\ x +iy & -z \end{array} \right)\left( \begin{array}{c} u \\ v \end{array} \right)=\left( \begin{array}{c} u \\ v \end{array} \right)$$
I get two equations:
$$zu+(x-iy)v=u \ \ ,\ \ (x+iy)u-zv=v$$
Rearranging these:
$$(z-1)u+(x-iy)v=0 \ \ = \ \ (x+iy)u-(z+1)v=0$$
And again:
$$(z-1)u-(x+iy)u=(-x+iy)v-(z+1)v\Rightarrow (-x-iy+z-1)u=(-x+iy-z-1)v\Rightarrow u=-x+iy-z-1 \ \ ,\ \ v=-x-iy+z-1$$
So the eigenvector would be
$$\left( \begin{array}{c} -x+iy-z-1 \\ -x-iy+z-1 \end{array}\right)$$
And for the negative eigenvalue
$$\left( \begin{array}{c} -x+iy-z+1 \\ -x-iy+z+1 \end{array}\right)$$
These seem to work if I insert them into
$$\textbf{Sv}=^+_-\frac{\hbar}{2}\textbf{v}$$
Normalizing the eigenvectors I get:
$$\chi_+=\frac{1}{2\sqrt{x+1}}\left( \begin{array}{c} -x+iy-z-1 \\ -x-iy+z-1 \end{array}\right)$$
$$\chi_- =\frac{1}{2\sqrt{-x+1}}\left( \begin{array}{c} -x+iy-z+1 \\ -x-iy+z+1 \end{array}\right)$$
But I don't actually understand what I've just done. Is this physically reasonable?

How do I actually use these to get some generic spinor?
$$\chi = \frac{a+b}{2\sqrt{x+1}}\chi _+ + \frac{a-b}{2\sqrt{-x+1}}\chi _-\ \ ?$$

And to make things even more laborious, I'd have to know how to express the spin state of some other vector (i,j,k) as a linear combination of the previous eigenvectors. I know that when vectors m = (x,y,z) and n = (i,j,k) are unit vectors, then there's a relation $$m \cdot n =cos\ \theta$$
The problem is I don't understand how I can apply that relation to the spinors. If theta is the angle between m and n, how can I apply it to figure how the eigenvectors in the direction m relate to the eigenvectors in the direction n? If the eigenvectors X+ and X- are the eigenvectors of the spin operator S in the direction m, how can I use them to get the spinor in the direction n?

Last edited: Mar 28, 2008
4. Jun 7, 2008

### gn0m0n

I'd also like to see this topic's explanation developed, especially with an eye to the physical meaning of the operators and matrices... Anyone help us out?