Spinors and eigenspinors confusion

In summary: So$$\hat{S}_z \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} \hbar/2 & 0 \\ 0 & -\hbar/2 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} (\hbar/2) a \\ (-\hbar/2) b \end{pmatrix} = \begin{pmatrix} (\hbar/2) a \\ (0) b \end{pmatrix}.$$And you see that ##a## is the component of the eigenvector corresponding to the ##+\hbar/2
  • #1
happyparticle
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TL;DR Summary
Spinors and eigenspinors of ##S_z## and ##S_x## for a spin 1/2
Hi,
While studying the spin 1/2, I'm facing some confusions about the spinors and the eigenspinors.

I understand that ##\chi = \begin{bmatrix}a \\ b \end{bmatrix}## is the spinor with ##\chi_+ = \begin{bmatrix}1 \\ 0 \end{bmatrix}## and ##\chi_-= \begin{bmatrix}0 \\ 1 \end{bmatrix}## the eigenspinors.

However, in my book I have ##\chi_+ = \begin{bmatrix}1 \\ 0 \end{bmatrix}, (eigenvalue + \frac{\hbar}{2})## which I'm not sure to understand.
Furthermore, to find the eigenvalue of ##S_x##, Griffith uses ##\begin{bmatrix}-\lambda & \frac{\hbar}{2}\\ \frac{\hbar}{2} & -\lambda \end{bmatrix} = 0##, but I don't see where this expression come from?
Finally, he says that ##\chi = (\frac{a+b}{\sqrt{2}} \chi_+ + \frac{a-b}{\sqrt{2}} \chi_-)##, which I suposse ##\frac{a+b}{\sqrt{2}} = a ## and ##\frac{a-b}{\sqrt{2}} = b##, again I'm not sure how he get those expressions for a and b.
 
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  • #2
happyparticle said:
TL;DR Summary: Spinors and eigenspinors of ##S_z## and ##S_x## for a spin 1/2

Hi,
While studying the spin 1/2, I'm facing some confusions about the spinors and the eigenspinors.

I understand that ##\chi = \begin{bmatrix}a \\ b \end{bmatrix}## is the spinor with ##\chi_+ = \begin{bmatrix}1 \\ 0 \end{bmatrix}## and ##\chi_-= \begin{bmatrix}0 \\ 1 \end{bmatrix}## the eigenspinors.

However, in my book I have ##\chi_+ = \begin{bmatrix}1 \\ 0 \end{bmatrix}, (eigenvalue + \frac{\hbar}{2})## which I'm not sure to understand.
Furthermore, to find the eigenvalue of ##S_x##, Griffith uses ##\begin{bmatrix}-\lambda & \frac{\hbar}{2}\\ \frac{\hbar}{2} & -\lambda \end{bmatrix} = 0##, but I don't see where this expression come from?
Finally, he says that ##\chi = (\frac{a+b}{\sqrt{2}} \chi_+ + \frac{a-b}{\sqrt{2}} \chi_-)##, which I suposse ##\frac{a+b}{\sqrt{2}} = a ## and ##\frac{a-b}{\sqrt{2}} = b##, again I'm not sure how he get those expressions for a and b.
You are close. To find the eigenvalues we take the determinant of ##S_x - I \lambda## and set it equal to 0:
##\left | \dfrac{\hbar}{2} \left ( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right ) - \left ( \begin{matrix} \lambda & 0 \\ 0 & \lambda \end{matrix} \right ) \right | = 0##

The last statement is going to take a bit more work and I'm going to try to be a bit psychic because you didn't say where this came from.

The +x eigenvector written in the z basis is
## \dfrac{1}{\sqrt{2}} \left ( \begin{matrix} 1 \\ 1 \end{matrix} \right )##

and the -x eigenvector is written as
## \dfrac{1}{\sqrt{2}} \left ( \begin{matrix} 1 \\ -1 \end{matrix} \right )##

So, I believe the question is: If ##\chi = \left ( \begin{matrix} a \\ b \end{matrix} \right )## is written in the x basis, how can we rewrite this in the z basis?

-Dan
 
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  • #3
topsquark said:
So, I believe the question is: If ##\chi = \left ( \begin{matrix} a \\ b \end{matrix} \right )## is written in the x basis, how can we rewrite this in the z basis?
I'm not totally sure, but I think ##\chi## is in x basis.
Since the expression is ##
\chi = (\frac{a+b}{\sqrt{2}} \chi_+^{x} + \frac{a-b}{\sqrt{2}} \chi_-^{x})##

I'm still not sure to understand the expression ##
\chi_+ = \begin{bmatrix}1 \\ 0 \end{bmatrix}, (eigenvalue + \frac{\hbar}{2})
##
Is ##\chi_+ = \begin{bmatrix}1 \\ 0 \end{bmatrix},## and/or ##(eigenvalue + \frac{\hbar}{2})## ?

Griffith says, if you mesure ##S_x##, the probability of getting ##+\hbar/2## is ##(1/2) |a+b|^2##.
There should be a link with the expression above, but I don't see why he says the probability of getting ##+ \hbar/2##
 
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  • #4
If you work in a basis of eigenvectors of ##\hat{s}_3##, ##|s_3 \rangle##, with the eigenvalues ##s_3 \in \{\hbar/2,-\hbar/2 \}##, then you can write any spin vector as
$$|\chi \rangle=a |\hbar/2 \rangle + b |\hbar/2 \rangle.$$
Then there is a one-to-one mapping between the vectors and their components wrt. that basis
$$\chi \rangle \mapsto \begin{pmatrix} a \\ b \end{pmatrix}.$$
Then the eigenvectors are orthonormal, i.e., ##a=\langle \hbar/2 |\chi \rangle=\chi_{\hbar/2}## and ##b=\langle \hbar/2|\chi \rangle=\chi_{-\hbar/2}##.
Any operator acting on the spinors then has a one-to-one map to ##2 \times 2## matrices:
$$\hat{A} |\chi \rangle=\sum_{s_{31},s_{32}} |s_{31} \rangle \langle s_{31}|\hat{A} s_{32} \rangle \langle{s_{32}}|\chi \rangle.$$
The matrix elements are
$$A_{s_{31} s_{32}}=\langle s_{31}|\hat{A} s_{32} \rangle \langle{s_{32}}$$
and the corresponding matrix
$$\hat{A}=\begin{pmatrix} A_{\hbar/2,\hbar/2} & A_{\hbar/2,-\hbar/2} \\ A_{-\hbar/2,\hbar} & A_{-\hbar/2,-\hbar/2} \end{pmatrix}.$$
And the the components of ##\hat{A} |\chi \rangle## are obviously given by ##\hat{A} \chi##.
 
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