# Quantum, commutation, suffix

1. Apr 12, 2011

### Chronos000

1. The problem statement, all variables and given/known data

The question asks for [Xi^2, Lj]

I can get to the line:

ih(bar) Xk ekjl Xl + ih(bar) ekjl Xk Xl

this line must be zero but I don't see how it is.

It looks like an expansion of a commutation between Xk and Xl but not quite right

2. Apr 12, 2011

### tiny-tim

Hi Chronos000!

(try using the X2 and X2 icons just above the Reply box )
That's just 2ih(bar) Xk ekjl Xl.

Maybe the second one should be ih(bar) Xk eljk Xl ?

3. Apr 12, 2011

### Chronos000

I done it again

got ih Xo ejmo Xm + ih ejmo Xm Xo

can I switch the order of the last X operators and put a negative in front? which would give me a commutation

why did u say the previous solution was 2ih(bar) Xk ekjl Xl when the order was different in the last expression?

4. Apr 12, 2011

### vela

Staff Emeritus
That is equal to 0 because XoXm is symmetric while ejmo is antisymmetric.

5. Apr 12, 2011

### Chronos000

I'm sure I've seen something like this before but I really don't have a clue how you can see which is symmetric or not

6. Apr 12, 2011

### vela

Staff Emeritus
XoXm is symmetric because you can swap the subscripts without changing anything, i.e. XoXm=XmXo. The tensor ejmo on the other hand is antisymmetric because it changes sign when you exchange a pair of subscripts, i.e. ejmo=-ejom.

7. Apr 12, 2011

### tiny-tim

Hi Chronos000!

sorry, I wasn't thinking straight earlier

vela is right, ejmo is antisymmetric, so if you multiply it by something symmetric (in 2 of the 3 indices in ejmo), you get 0

(when you do all the adding, for example you add ej12X1X2 and ej21X2X1 … the latter is -ej12X2X1 = -ej12X1X2)

8. Apr 12, 2011

### Chronos000

thanks for your replies. So my expression is just 0 + 0. I don't do anything like you have done in your final comment tiny-tim?

Are you allowed to change the position of the epsilon yes? as it does really "act" on anything but just tells you what the cross product is.

9. Apr 12, 2011

### tiny-tim

That's right …

I only did that to prove that it works, but you can just say "it's obvious!"
You can change the position of any whole thing …

what you can't do is change the position of indices (unless they're symmetric indices, in which case you can, or unless they're anti-symmetric indices, in which case you can if you multiply by -1).

10. Apr 12, 2011

### Chronos000

thanks I'm pretty happy with this now