Normalised eigenspinors and eigenvalues of the spin operator

  • #1
594
12

Homework Statement


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

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

Homework Equations


det | Sy - λI | = 0
Sy = ## 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 Sy = ±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.
 

Answers and Replies

  • #2
blue_leaf77
Science Advisor
Homework Helper
2,637
785
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##).
 
  • #3
594
12
## X = (XT)*

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 Sy is hermitian then the eigenfunctions of Sy will be orthogonal. I essentially did the same thing.

If Sy = Sy then Sy is hermitian and it's eigenfunctions / eigenspinors are orthogonal.
 

Related Threads on Normalised eigenspinors and eigenvalues of the spin operator

Replies
6
Views
3K
  • Last Post
Replies
6
Views
719
Replies
15
Views
2K
Replies
2
Views
761
Replies
0
Views
2K
Replies
1
Views
272
  • Last Post
Replies
6
Views
5K
Top