# Angular Momentum commutation relationships

It seems to be implied, but I cant find it explicitly - the order in which linear operators are applied makes a difference. IE given linear operators A,B then AB is NOT necessarily the same as BA ? I thought it was only with rotation operators that the order made a difference?

I noticed this while looking at text that showed [Lx,Ly] = i(h-bar)Lz, using only position and momentum operators...

<<mentor note: originally posted in homework forum, template removed>>

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Orodruin
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In general, linear operators will not commute. Another common example is the position and momentum operators.

Thanks - of course, not a clever question when I am studying commutation relationships... But now I might see what was bothering me (I think) - the text expands [Lx,Ly] in terms of position and momentum operators, you get 8 terms like YPzZPx - the last 4 could cancel the 1st 4 out - but only if it was OK to change the order - like ZPxYPz (which is the 1st of the last 4, to complete the example). So are they OK in assuming that the operators don't commute - in order to prove that other operators don't commute?

Orodruin
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So are they OK in assuming that the operators don't commute - in order to prove that other operators don't commute?

Yes. You can easily derive the commutation relations for ##P_i## with ##X_i## using the position basis representation ##P_i \to -i\partial_i## and ##X_i \to x^i## and their action on any wave function ##\psi(x)##.

ChrisVer
Gold Member
the order in which linear operators are applied makes a difference.

If they don't commute yes... if they commute, no...
If they commute, you have to be careful when changing the order -> new terms can be brought in.
For example if I have $x p_x$ and I want for some calculation to rewrite it in $p_x x$ (because it would come handy) I would have to use the relation:
$[x, p_x ] = x p_x - p_x x= i \hbar \Rightarrow x p_x = p_x x + i \hbar$ and that's with what you change $x p_x$.

Thanks all