Undergrad Spinors and eigenspinors confusion

[happyparticle]

TL;DR

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.

 

[topsquark]

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

 

[happyparticle]

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 $+ \hbar/2$

 

[vanhees71]

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 $\hat{A} \chi$.