Angular Momentum commutation relationships

[ognik]

It seems to be implied, but I can't 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 [L_x,L_y] = i(h-bar)L_z, using only position and momentum operators...<<mentor note: originally posted in homework forum, template removed>>

 

[Orodruin]

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

 

[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 YP_zZP_x - the last 4 could cancel the 1st 4 out - but only if it was OK to change the order - like ZP_xYP_z (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]

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]

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.

 

[ognik]

Thanks all