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Quantum, commutation, suffix

by Chronos000
Tags: commutation, suffix
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Chronos000
#1
Apr12-11, 10:04 AM
P: 80
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
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tiny-tim
#2
Apr12-11, 10:33 AM
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Hi Chronos000!

(try using the X2 and X2 icons just above the Reply box )
Quote Quote by Chronos000 View Post
ih(bar) Xk ekjl Xl + ih(bar) ekjl Xk Xl
That's just 2ih(bar) Xk ekjl Xl.

Maybe the second one should be ih(bar) Xk eljk Xl ?
Chronos000
#3
Apr12-11, 03:26 PM
P: 80
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?

vela
#4
Apr12-11, 04:04 PM
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Quantum, commutation, suffix

That is equal to 0 because XoXm is symmetric while ejmo is antisymmetric.
Chronos000
#5
Apr12-11, 04:19 PM
P: 80
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
vela
#6
Apr12-11, 04:25 PM
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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.
tiny-tim
#7
Apr12-11, 04:56 PM
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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)
Chronos000
#8
Apr12-11, 05:09 PM
P: 80
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.
tiny-tim
#9
Apr12-11, 05:15 PM
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Quote Quote by Chronos000 View Post
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?
That's right

I only did that to prove that it works, but you can just say "it's obvious!"
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.
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).
Chronos000
#10
Apr12-11, 05:22 PM
P: 80
thanks I'm pretty happy with this now


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