Sping Matrices and Commutation Relations

[Rahmuss]

Homework Statement

Check that the spin matrices (Equations 4.145 and 4.147) obey the fundamental commutation relations for angular momentum, Equation 4.134.

Homework Equations

Eq. 4.147a --> S_{x} = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}

Eq. 4.147b --> S_{y} = \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}

Eq. 4.145 --> S_{z} = \frac{\hbar}{2}\begin{pmatrix}1 & 0 \\ 0 & -1 \end{pmatrix}

\left[ S_{x}, S_{y}\right] = i\hbar S_{z}

\left[ S_{y}, S_{z}\right] = i\hbar S_{x}

\left[ S_{z}, S_{x}\right] = i\hbar S_{y}

The Attempt at a Solution

Well, from Eq. 4.147 I have S_{x} = \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} and also S_{y} = \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix}

So \left[ S_{x}, S_{y} \right] = i\hbar S_{z} \Rightarrow

\left[ \frac{\hbar}{2}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix}, \frac{\hbar}{2}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \right] \Rightarrow

\frac{\hbar^{2}}{4}\begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} - \frac{\hbar^{2}}{4}\begin{pmatrix}0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix}0 & 1 \\ 1 & 0 \end{pmatrix} \Rightarrow

\frac{\hbar^{2}}{4}\begin{pmatrix}i & 0 \\ 0 & -i \end{pmatrix} - \frac{\hbar^{2}}{4}\begin{pmatrix}-i & 0 \\ 0 & i \end{pmatrix} = \frac{\hbar^{2}}{4}\begin{pmatrix}2i & 0 \\ 0 & -2i \end{pmatrix}

It doesn't quite seem right to me because the answer shows:

\left[ S_{x}, S_{y}\right] = i\hbar S_{z}

But I can't see how I'd get that from where I'm going.

 

[Dick]

Your answer IS i*hbar*S_z.

 

[Rahmuss]

Err... hey, you're right. :D

This idiot thanks you.