I Clarification on a statement about spin 1/2 z component related to x and y

[Sparky_]

Actually, I take that back; you didn't do it entirely correctly in post #22, you just got lucky.

The correct part of what you did in post #22 is this equation:

$$
\frac{\hbar}{2} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix} = \lambda \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
$$

where I have put back the factors of $1 / \sqrt{2}$ on each side to make it clear exactly what we are doing, namely, writing the eigenvalue equation for $\hat{S}_x$ and $| x_+ \rangle$. But this gives two equations for $\lambda$, not one: one equation for the upper component of the RHS, the second for the lower component of the RHS. Those equations are

$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} \left( 0 \times 1 + 1 \times 1 \right) = \lambda \times \frac{1}{\sqrt{2}} \times 1
$$

$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} \left( 1 \times 1 + 0 \times 1 \right) = \lambda \times \frac{1}{\sqrt{2}} \times 1
$$

Both of these equations give the same value for $\lambda$, namely $\hbar / 2$.

Now do the same with the column vector

$$
\frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}
$$

shoot, (wasn't being careful) , should have looked up the actual matrix operation -

$$
\frac{1}{\sqrt{2}}\begin{pmatrix}
0&\frac{\hbar}{2}\\ \frac{\hbar}{2}&0\end{pmatrix} \begin{pmatrix}
1\\-1\end{pmatrix} = \lambda \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix}
$$
2 equations:
$$
\frac{\hbar}{2} \frac{1}{\sqrt{2}} = \frac{\lambda}{\sqrt{2}} \\
-\frac{\hbar}{2} \frac{1}{\sqrt{2}} = -\frac{\lambda}{\sqrt{2}} \\

\lambda = \frac{\hbar}{2}
$$

 

[PeterDonis]

2 equations

You have the signs wrong. The eigenvalue should be $- \hbar / 2$.

 

[Sparky_]

crap, went too fast

(I did on scratch paper earlier the "characteristic equation" deal for finding eigenvalues - the determinant resulting from subtracting the eigenvalues times the identity matrix and setting to zero ...

there I got $$ \pm \frac{\hbar}{2} $$

I should have caught that. I knew there was a $$-\frac{\hbar}{2}$$

corrected:
$$
\frac{1}{\sqrt{2}}\begin{pmatrix} 0&\frac{\hbar}{2}\\ \frac{\hbar}{2}&0\end{pmatrix} \begin{pmatrix} 1\\-1\end{pmatrix} = \lambda \frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ -1 \end{bmatrix} \\

-\frac{\hbar}{2} \frac{1}{\sqrt{2}} = \frac{\lambda}{\sqrt{2}} \\
\frac{\hbar}{2} \frac{1}{\sqrt{2}} = -\frac{\lambda}{\sqrt{2}} \\

\lambda = -\frac{\hbar}{2}
$$