How to evaluate commutator with angular momentum? [closed]

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I need to evaluate the commutator $[\hat,\hat_z]$. I believe the $L_z$ is referring to the angular momentum operator which is: $L_z = xp_y - yp_x$ using this relationship i end up with: $[x,L_z] = x(xp_y - yp_x)-(xp_y - yp_x)x$ my next step is substituting in for the p operator but i still dont get anywhere. any suggestions.

3,350 17 17 silver badges 23 23 bronze badges asked Mar 2, 2013 at 20:15 quantum savant quantum savant 35 2 2 gold badges 2 2 silver badges 9 9 bronze badges

$\begingroup$ Word to the wise: you've posted three relatively simple quantum problems in a short span of time. You really should attempt to struggle to solve these yourself before getting help. You'll never really get good at this stuff otherwise. $\endgroup$

Commented Mar 2, 2013 at 20:23

$\begingroup$ if you saw these sheets of paper ive been scribbling on for the past 2 hours i think you would agree i have worked relatively hard on them. The fact the someone would show me the solution would teach me far more then what ive learned trying to do it on my own. Thanks for the advice though @joshphysics. Also im in 3rd year chemistry, whereas you are a grad student - something that may be simple to you may very well be challenging for me, if you care to help I would greatly appreciate it. $\endgroup$

Commented Mar 2, 2013 at 20:31 $\begingroup$ Well that's fair enough. $\endgroup$ Commented Mar 2, 2013 at 20:53

2 Answers 2

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I'll help you with a general solution. First, you can write the angular momentum as $L_=\epsilon_x_p_$. Where $e_$ is the Levi-Civita symbol. Now write a general commutator as $[x_,L_]$ where $l,i$ run from 1 to 3. With this you have

Now, you use the following identity for commutator $[A,BC]=[A,B]C+B[A,C]$. With this in mind, you get

Now, the commutator $[x_,x_]$ is zero and $[x_,p_]$ is equal to $i\hbar\delta_$. So, you are left with

From this result you can find your particular commutator.

answered Mar 2, 2013 at 21:01 nijankowski nijankowski 2,168 14 14 silver badges 21 21 bronze badges

$\begingroup$ You're correct but your post might be too advanced for OP - just a thought. $\endgroup$

Commented Mar 2, 2013 at 21:29 $\begingroup$

Usually I find it easiest to evaluate commutators without resorting to an explicit (position or momentum space) representation where the operators are represented by differential operators on a function space.

In order to evaluate commutators without these representations, we use the so-called canonical commutation relations (CCRs) $$ [x_i,p_j] = i\hbar \,\delta_, \qquad [x_i, x_j]=0,\qquad [p_i, p_j]=0 $$ Now, in order to evaluate and angular momentum commutator, we do precisely as you suggested using the expression $$ L_z = x p_y - y p_x $$ and we use the CCRs \begin [x, L_z] &= [x, xp_y-yp_x]\\ &= [x,xp_y] - [x,yp_x]\\ &= x[x,p_y]+[x,x]p_y-y[x,p_x]-[x,y]p_x \\ &= -i\hbar y \end In the last step, only the third term was non-vanishing because of the CCRs. I have also used the fact that the commutator is linear in both of its arguments, $$ [aA+bB,C] = a[A,C] + b[B,C], \qquad [A,bB + cC] = b[A,B] + c[A,C] $$ where $a,b,c$ are numbers and $A,B,C$ are operators, and the following commutator identity that you'll find useful in general: $$ [AB,C] = A[B,C] + [A,C]B $$