Angular Momentum commutation relationships

1. Apr 23, 2015

ognik

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>>

Last edited by a moderator: Apr 24, 2015
2. Apr 24, 2015

Orodruin

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

3. Apr 24, 2015

ognik

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?

4. Apr 24, 2015

Orodruin

Staff Emeritus
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 $\psi(x)$.

5. Apr 24, 2015

ChrisVer

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 $x p_x$.

6. Apr 28, 2015

Thanks all