# How to evaluate commutator with angular momentum? [closed]

I need to evaluate the commutator $[\hat{x},\hat{L}_z]$. 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???

-

## closed as too localized by David Z♦Mar 3 '13 at 2:24

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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. – joshphysics Mar 2 '13 at 20:23
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. – quantum savant Mar 2 '13 at 20:31
Well that's fair enough. – joshphysics Mar 2 '13 at 20:53

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) $$[x_i,p_j] = i\hbar \,\delta_{ij}, \qquad [x_i, x_j]=0,\qquad [p_i, p_j]=0$$ 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{align} [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{align} 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: $$[AB,C] = A[B,C] + [A,C]B$$

-
I appreciate you putting the solution up. I am a little unclear of how you went from the second line to the third line (i will continue to look at your commutator identities and hopefully come to the conclusion on my own). However, can you explain why the third term in your last step is the only nonvanishing term? because you say [x,p]=ih... so why would your first term, in your last step, disappear? also what property makes the last term disappear? sorry if this is a stupid question im just trying to understand it fully. – quantum savant Mar 2 '13 at 21:16
@quantumsavant: Josh has used the product rule for commutators, written at the bottom of this post. Try to prove it yourself, and similarly work out $[A,BC] = ??$. As for the 2nd question: $x$ only has a nonzero commutator with $p_x$, not with $p_y$ etc. That's because of the $\delta_{ij}$ in above. $[x,y] = 0$ because a) they are both position operators, so their commutator always vanishes, or more formally b) it follows from the identity $[x_i,x_j] = 0$, where $x_1 = x,$ $x_2 = y.$ – Vibert Mar 2 '13 at 21:28
@NijankowskiV. thank you for the explanation i think i finally got it! appreciate all the help – quantum savant Mar 2 '13 at 21:34
@josh I'm not sure if you knew about this but our homework policy says that you shouldn't post complete answers to homework-like questions, so please don't give away so much in the future. (Ordinarily I would delete this, but quantum savant has already seen it so there's not much point.) – David Z Mar 3 '13 at 2:26

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

$[x_{l},L_{i}]=[x_{l},\epsilon_{ijk}x_{j}p_{k}]$

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

$[x_{l},\epsilon_{ijk}x_{j}p_{k}]=\epsilon_{ijk}[x_{l},x_{j}]p_{k}+\epsilon_{ijk}x_{j}[x_{l},p_{k}]$

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

$[x_{l},\epsilon_{ijk}x_{j}p_{k}]=[x_{l},L_{i}]=i\hbar\epsilon_{ijk}x_{j}\delta_{lk}=i\hbar\epsilon_{ijl}x_{j}$

From this result you can find your particular commutator.

-
You're correct but your post might be too advanced for OP - just a thought. – Vibert Mar 2 '13 at 21:29