# Homework Help: Linear algebra - Anti-Commutation Relations

1. Jul 13, 2010

### SpiffyEh

1. The problem statement, all variables and given/known data
Show that {x_i, x_j} = 2*y_ij* I for i = 1; 2; 3 and j = 1; 2; 3.
where
y_ij: N x N -> {0,1}, such that y_ij = {1, if i = j ; 0, if i not = j

2. Relevant equations

3. The attempt at a solution

I'm confused about exactly what i'm supposed to do here. Do i do all the combinations between i and j or what? Can someone please show me an example or two so that I can do the rest. Thank you

2. Jul 13, 2010

### vela

Staff Emeritus
Yes, you need to show it for all combinations of i and j.

How are the xi's defined?

3. Jul 13, 2010

### SpiffyEh

x_1 =
[0 1
1 0]

x_2 =
[ 0 -i
i 0]

x_3 =
[1 0
0 -1]

could you show me one of them just so that i know i'm doing them correctly please.

Last edited: Jul 13, 2010
4. Jul 13, 2010

### vela

Staff Emeritus
The easiest thing to do is just calculate the six possibilities. For example, for x1 and x2, you want to calculate

\begin{align*} \{x_1,x_2\} & = x_1 x_2 + x_2 x_1 \\ & = \begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}0 & -i \\ i & 0\end{bmatrix} + \begin{bmatrix}0 & -i \\ i & 0\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix} \end{align*}

5. Jul 13, 2010

### SpiffyEh

I calculate that then do the 2*y_ij* I and compare? how do i know {x1, x2} = x1x2 + x2x1?

Last edited: Jul 13, 2010
6. Jul 14, 2010

### vela

Staff Emeritus
Yes.
That's the definition of the anticommutator.

7. Jul 14, 2010

### SpiffyEh

oh, ok. I wish we had been told that. But it makes sense.

I also have one similar to that

Commutation Relations. Show that [x_i; x_j ] = 2$$\sqrt{-1}$$$$\sum$$e_ijk * x_k
from k= 1 to 3
for i = 1,2,3 and j= 1,2,3

where e_ijk = { 1 if(i,j,k) is (1,2,3),(2,3,1),(3,1,2)
-1 if (i,j,k) is (3,2,1),(1,3,2) or (2,1,3)
0 if i= j or j=k or k=1

For this do i basically do the same thing? is the relationship the same for the x1x2 + x2x1? And also say i do x1 and x2, i'm not sure what to choose for k

8. Jul 14, 2010

### vela

Staff Emeritus
The commutator is defined as [A,B]=AB-BA.

The righthand side of the equation is

$$2\sqrt{-1}\sum_{k=1}^3 \epsilon_{ijk}x_k$$

You're summing over k. You don't get to choose it.

Suppose i=1 and j=2, then $\epsilon_{ijk}=0$ if k=1 or k=2, so the only term that will survive is the k=3 term. So you want to show

$$[x_1,x_2] = 2\sqrt{-1}x_3$$

Last edited: Jul 14, 2010
9. Jul 14, 2010

### SpiffyEh

oh! ok that makes sense. Thank you so much