Last question for this series. I started answering this one and got stuck. Again, this subject is new to me, so forgive me if I miss obvious stuff.. I'm getting used to the format and the method.

 

I don't want to copy an answer, of course. I would appreciate an explanation and/or a way to continue so I can go on to other similar problems and also understand how to solve these type of questions in the future.

 

Thanks!

 

Problem 3:

 

The Hamiltonian for three-dimensional system is of the form:

[math]H=\frac{p_{i}p_{i}}{2m}+V[/math]

 

where V depends only on r. Obtain the commutator [math] [x_{k}p_{k},H][/math]

 

--

 

Here's what I have so far:

 

[math] [x_{k}p_{k},H] = [x_{k}p_{k},\frac{p_{i}p_{i}}{2m}+V] =[/math]

 

[math] [x_{k}p_{k},\frac{p_{i}p_{i}}{2m}] + [x_{k}p_{k}, V] =[/math]

 

[math]\left( x_{k}p_{k}\frac{p_{i}p_{i}}{2m} - \frac{p_{i}p_{i}}{2m}x_{k}p_{k} \right) + \left( x_{k}p_{k} V - V x_{k}p_{k} \right) = [/math]

 

Now, in the left part, I recognize that I might be able to rebuild the commutator from the p's... I tried:

 

[math]\frac{1}{2m} \left( x_{k} \left( p_{k}p_{i}p_{i} \right) - p_{i}p_{i}x_{k}p_{k} \right) [/math]

I can switch between x_k and p_k and get the remainder -ihbar:

[math]\frac{1}{2m} \left( x_{k} \left( p_{k}p_{i}p_{i} \right) + i\hbar\left( p_{i}p_{i}p_{k} \right)x_{k} \right) [/math]

 

These two can be switched for the commutator, but I have an extra factor of ihbar in there... meh... where do I go from here, and am I even in the right direction?

Also, the next question is exactly the same, only V® is defined so that r carries all three dimensions of x. So I think that this means this current question treats r without much relation to x (otherwise why the distinction in the next question?) and hence, V® has no x, and is commutable with both p and x and can be neglected.

 

Am I right? Help.

 

Thanks!

 

~moo

If [A,B]=C, then AB can be rewritten as C + BA. My point here is: if you rewrite a product using a commutator relation, then it would be replaced by two addends, not gain a factor ([math]i\hbar[/math]) as it seems in your case.

Yikes! I see it. Okay, I think I have an idea of how to solve it now. I'll try and post my solution.

 

Thanks