# Homework Help: Simple matrix algebra

1. Jul 24, 2007

### Jimmy Snyder

1. The problem statement, all variables and given/known data
This is from Ryder's QFT book, second ed. page 37. At the bottom of the page it says that the commutation relations (eqn 2.68?) are satisfied by:
$K = \pm i\frac{\sigma}{2}$
However, I do not find this to be so. What am I missing?

2. Relevant equations
Here is one of the commutation relations that I think he means.
$[K_x,K_y] = -iJ_z$

3. The attempt at a solution
Using $K = i\frac{\sigma}{2}$, I get:
$[K_x,K_y] = [i\frac{\sigma_x}{2},i\frac{\sigma_y}{2}] = \frac{-1}{4}[\sigma_x,\sigma_y] = -\frac{1}{2}\sigma_z = iK_z \neq -iJ_z$

Last edited: Jul 24, 2007
2. Jul 24, 2007

### Dick

Isn't there an i in the commutation relations of the Pauli matrices as well?

3. Jul 24, 2007

### olgranpappy

yep,
$$\left[\sigma_j,\sigma_k\right]=2i\epsilon_{jkl}\sigma_l$$

4. Jul 24, 2007

### Jimmy Snyder

Thanks Dick. Here is the corrected attempt. I still don't get the right commutation relation.

$[K_x,K_y] = [i\frac{\sigma_x}{2},i\frac{\sigma_y}{2}] = \frac{-1}{4}[\sigma_x,\sigma_y] = -i\frac{\sigma_z}{2} = -K_z \neq -iJ_z$

5. Jul 24, 2007

### olgranpappy

yes, you do get the right relation. Ryder is talking about (2-component) Pauli spinors for which $$J_z=\frac{\sigma_z}{2}$$.

Look at equation (2.74). That is a boost and a rotation of a 2-component spinor where the rotation generator is
$$\frac{\vec \sigma}{2}$$ and the boost generator is $$i\frac{\vec \sigma}{2}$$.

6. Jul 25, 2007

### Jimmy Snyder

Thanks olgranpappy, your reply is what I needed. If I make the substitutions $$K = i\frac{\sigma}{2}$$ and $$J = \frac{\sigma}{2}$$, then I get:

$$[K_x,K_y] = [i\frac{\sigma_x}{2},i\frac{\sigma_y}{2}] = -i\frac{\sigma_z}{2} = -iJ_z$$ just as in (2.68)

I have also verified the other relations in (2.68). I came close to solving it this morning as I was driving to work. It occured to me that there might be a typo in the book and that the author meant J instead of K in eqn (2.69). If I had followed that thought a while longer, I might have come up with the solution on my own. Thanks again for your help.

7. Jul 25, 2007

### olgranpappy

no problem. cheers.