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GR0177 #43
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Quantum Mechanics}Commutators

Write it out and recall the commutator identities and .

Thus , using the identities supplied by ETS.  Alternate Solutions
 QM3202008-11-05 20:50:33 You can also do it by brute force: Using the given commutation relations, this is equal to Rearranging and canceling gives choice (D). Note: This was supposed to be an alternate solution, so I've reposted it as such. If someone could tell me how to delete my original (duplicate) comment, below, that would be helpful. :)Reply to this comment QM320
2008-11-05 20:50:33
You can also do it by brute force:

Using the given commutation relations, this is equal to

Rearranging and canceling gives choice (D).

Note: This was supposed to be an alternate solution, so I've reposted it as such. If someone could tell me how to delete my original (duplicate) comment, below, that would be helpful. :)
 thisguy2009-10-23 15:16:39 There's a small typo. [LxLy,Lz] = LxLyLz - LzLxLy. Not LxLyLz - LzLyLx as you wrote.
 hybridusmanus2010-06-15 18:06:16 Why can't re-associate the commutations after expansion? rnrni.e. rn[LL,L] = rnLLL - LLL =rnL(LL) - (LL)L = rnLihL - ihLL = rnih(L - L)rnrndoes this not always work?
 walczyk2011-04-07 08:51:16 Just wanted to clarify what the typo was and why it is significant. We cannot commute these orbital momentum operators willy nilly, without adding extra terms and whatnot so its a delicate procedure you could say. So beginning we have . Let's follow the original approach and try to substitute something for and . Quickly write out , rearrange to get . You can keep going with it if you'd like. I'm curious what happens if you try to do something different.. , so we've got . QM320
2008-11-05 20:49:13
You can also do it by brute force:

Using the given commutation relations, this is equal to

Rearranging and canceling gives choice (D). Anastomosis
2008-04-10 14:20:54
A quick way to do this:
As the commutator is ,
Use the right hand rule to point first in the +X+Y (diagonal) direction, and then curl up to the Z direction. Your thumb will be pointing in the +X-Y direction, so thus choice (D) bootstrap
2007-04-06 20:15:33
The question have it setup as [AB,C] which is [AB,C] = A[B,C] + [A,C]B. Its very similar to what is posted..      LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$