Normalised eigenspinors and eigenvalues of the spin operator

[says]

Homework Statement

Find the normalised eigenspinors and eigenvalues of the spin operator S_y for a spin 1⁄2 particle

If X₊ and X_- represent the normalised eigenspinors of the operator S_y, show that X₊ and X_- are orthogonal.

Homework Equations

det | S_y - λI | = 0
S_y = $ħ/2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}$

The Attempt at a Solution

$det | ħ/2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}$ - $\begin{bmatrix} λ & 0 \\ 0 & λ \\ \end{bmatrix} | = 0$

skipping a few steps here but the eigenvalues = ±ħ/2

normalised eigenspinors

$ħ/2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \begin{bmatrix} a \\ b \\ \end{bmatrix} = ± ħ/2 \begin{bmatrix} a \\ b \\ \end{bmatrix}$

$ħ/2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ γ \\ \end{bmatrix} = ħ/2 \begin{bmatrix} 1 \\ γ \\ \end{bmatrix}$

$\begin{bmatrix} -iγ \\ i \\ \end{bmatrix} = \begin{bmatrix} 1 \\ γ \\ \end{bmatrix}$

γ = i ⇒ X₊ = 1/√2 $\begin{bmatrix} 1 \\ i \\ \end{bmatrix}$

$ħ/2 \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix} \begin{bmatrix} 1 \\ γ \\ \end{bmatrix} = -ħ/2 \begin{bmatrix} 1 \\ γ \\ \end{bmatrix}$

$\begin{bmatrix} -iγ \\ i \\ \end{bmatrix} = \begin{bmatrix} -1 \\ -γ \\ \end{bmatrix}$

γ = -i ⇒ X_- = 1/√2 $\begin{bmatrix} 1 \\ -i \\ \end{bmatrix}$

Eigenvalues of the spin operator S_y = ±ħ/2

Normalised eigenspinors =
X₊ = 1/√2 $\begin{bmatrix} 1 \\ i \\ \end{bmatrix}$

and

X_- = 1/√2 $\begin{bmatrix} 1 \\ -i \\ \end{bmatrix}$I've got the eigenvalues and normalised eigenspinors but I'm not sure how to show the eigenspinors are orthogonal.

 

[blue_leaf77]

I've got the eigenvalues and normalised eigenspinors but I'm not sure how to show the eigenspinors are orthogonal.

They are orthogonal if $X_-^\dagger X_+ = 0$ (or alternatively $X_+^\dagger X_- = 0$).

 

[says]

## X^† = (X^T)*

where T: transpose of matrix
*: complex conjugate

Thanks for that. I calculated it and can see that it equals 0.

I also found that eigenfunctions of hermitian operators are orthogonal. So I figured if S_y is hermitian then the eigenfunctions of S_y will be orthogonal. I essentially did the same thing.

If S_y = S_y^† then S_y is hermitian and it's eigenfunctions / eigenspinors are orthogonal.