Calculating Comutator of H and 1/r∙r

[jarvinen]

I need to calculate [itex] \left[ H , \ \frac{1}{r} \mathbf{r} \right] [/itex].

My initial idea is to use [itex] \left[ H, \ U \mathbf{r} \right] = \left[ H, \ U \right] \mathbf{r} + U \left[ H , \ \mathbf{r} \right] [/itex].

Then clearly [itex] \left[H , \ U \right] \psi = \frac{ - \hbar ^{2} }{2m} \left( \nabla ^{2} \left( U \psi \right) - U \nabla ^{2} \psi \right) = 0 [/itex] in this case (as U = 1/r).

So we only have [itex] \left[ H, \ \mathbf{r} \right] [/itex] to consider.

I am only interested in the first component so I do [itex] \left[ H, \ r \right] = \frac{- \hbar ^{2}}{2m} \frac{2}{r} [/itex].

Is this correct? I ask because it seems out-of-place in the rest of the question (which I can add later if the above is right).

 

[kuruman]

I don't think you can do it this way. Why not find

[tex]\left[ H,\frac{x}{r} \right][/tex]

and then do a cyclic permutation to get the other two commutators?

 

[jarvinen]

I don't think you can do it this way. Why not find

[tex]\left[ H,\frac{x}{r} \right][/tex]

and then do a cyclic permutation to get the other two commutators?

Thanks for the response. I am a bit unsure about what you mean by the 'cyclic permutation' and this whole idea of using 'r' has been troubling me from the start of this question.

 

[kuruman]

In this context "cyclic permutation" means that the system doesn't know the alphabet. In other words, x is indistinguishable from y which is indistinguishable from z which is indistinguishable from x. So, if you find the commutator [H, x/r], then the commutator [H, y/r] will be the same with x → y, y → z, z → x. Similarly for [H, z/r].

What do you mean by "this whole idea of using 'r' has been troubling me from the start of this question."?

r = [x²+y²+z²]^1/2.

Put this in the denominator and see what you get.

 

[jarvinen]

OK, but surely if we have that [itex] H = \frac{- \hbar ^{2}}{2m} \nabla ^{2} + U [/itex] then it is clear that the 'U' part will commute with things like x/r, then clearly one is considering something like [itex] \nabla ^{2} \left( \frac{x \psi}{r} \right) - \frac{x}{r} \nabla ^{2} \left( \psi \right) [/itex]? This is obvious if we split the first Laplacian up.

Edit: This seems to give me that [itex] \left[ H , \ \frac{x}{r} \right] = \frac{ - \hbar ^{2}}{2m} \left( \nabla ^{2} \left( \frac{x}{r} \right) \right) = 0 [/itex].

 

[kuruman]

*** On edit ***

Exactly how do you get zero?

 

[jarvinen]

OK, I think that [itex] \mathbf{0} [/itex] was what I wanted the answer to be. This is all from a larger question.

I have shown that [itex] \left[ H , \ \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p} \right] = [/itex] [itex] \mathbf{r} \left( \mathbf{G} \cdot \mathbf{p} + \mathbf{p} \cdot \mathbf{G} \right) - \left( \left( \mathbf{r} \cdot \mathbf{G} \right) \mathbf{p} + \mathbf{p} \left( \mathbf{r} \cdot \mathbf{G} \right) \right) - i \hbar \mathbf{G} [/itex]

where [itex] \mathbf{G} = \left[ H , \ \mathbf{p} \right] [/itex]

I now need to work out [itex] \left[ H , \ \mathbf{K} \right] [/itex]

where [itex] \mathbf{K} = \frac{1}{2m} \left( \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p} \right) + U(r) \mathbf{r} [/itex]

(So here we have a potential of U that depends upon r alone).

I have many pages of working that I have spend many hours on however it is all not working. I need to calculate [itex] \mathbf{G} [/itex] and be totally sure of it.

I was thinking that I could calculate the r-coordinate (in spherical polars) for this commutator, clearly the [itex] \nabla ^{2} [/itex] part of the H will commute and thus I am left with [itex] \mathbf{G} _{r} = i \hbar \left( \frac{ d U }{dr} \right) [/itex].

 

[jarvinen]

*** On edit ***

Exactly how do you get zero?

[itex] \nabla ^{2} \left( \frac{x}{r} \right) = \frac{1}{r} \nabla ^{2} x + x \nabla ^{2} \frac{1}{r} = 0 [/itex]?

 

[kuruman]

The product rule works for first derivatives, not second.

 

[jarvinen]

The product rule works for first derivatives, not second.

Indeed. I did the differentiation and got [itex] -2xr^{-3} [/itex] however this could be incorrect. I was just trying to see if taking a specific example would help me with the longer question.

I need to calculate [itex] \left[ H , \ U(r) \mathbf{r} \right] [/itex] in general case of U and am somewhat unsure about how to do this.

 

[jarvinen]

My attempt to calculate [itex] \left[ H , \ U(r) r \right] [/itex] is as follows. Clearly we are just having to calculate [itex] \nabla ^{2} \left( U r \psi \right) - Ur \nabla ^{2} \psi [/itex], we drop the 'r' because U is just U(r) so we can absorb it into U (later do a substitution).

Then note that [itex] \nabla ^{2} \left( U \psi \right) = U \nabla ^{2} \psi + 2 \nabla U \cdot \nabla \psi + \psi \nabla ^{2} U [/itex]

Hence consider that [itex] \left[ H , \ U \right] = \frac{ - \hbar ^{2}}{2m} \left( \nabla ^{2} U + \frac{ dU}{dr} \frac{ \partial }{ \partial r} \right) [/itex].

I get that [itex] \left[ H, \ Ur \right] = \frac{ - \hbar ^{2}}{2m} \left( \left( \frac{2}{r} U + 4 \frac{dU}{dr} + r \frac{ d^{2} U}{d r^{2}} \right) + 2 \left( U + r \frac{dU}{dr} \right) \frac{ \partial }{ \partial r } \right) [/itex].

However it seems that this would not follow well because if one has that [itex] r \frac{dU}{dr} + U = 0 [/itex] then the above is just 0 and this is satisfied by the example I gave in the original post.

 

[kuruman]

My attempt to calculate [itex] \left[ H , \ U(r) r \right] [/itex] is as follows. Clearly we are just having to calculate [itex] \nabla ^{2} \left( U r \psi \right) - Ur \nabla ^{2} \psi [/itex], we drop the 'r' because U is just U(r) so we can absorb it into U (later do a substitution).

You can do that if you are looking for the commutator [H, U(r)r].
You cannot do that if you are looking for the commutator [H,U(r)r]

In the latter, r is a vector and "three commutators in one" are involved, namely [H, U(r)x], [H, U(r)y] and [H, U(r)z].

Also, what is this business with the cross products and what relation do they bear to the problem? What is the original question and how did you get to the point of needing the commutator that you posted?

 

[jarvinen]

The question is just to calculate the commutator [itex] \left[ H , \ \mathbf{K} \right] [/itex] where [itex] \mathbf{K} = \frac{1}{2m} \left( \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p} \right) + U(r) \mathbf{r} [/itex].

The reason I wanted to do [H,Ur] was because I wanted to consider [itex] \mathbf{r} = \left( r, \ \theta , \ \phi \right) [/itex].

 

[kuruman]

I see. So you want to calculate three commutators of the form [H, U(r)x], etc. for y and z and then see if you can deduce a vector out of the result. Perhaps you can use the identity [A,BC] = [A,B]C+B[A,C] to simplify your task.

 

[jarvinen]

I have a result for the commutator of [itex] \left[ H , \ \mathbf{p} \times \mathbf{L} - \mathbf{L} \times \mathbf{p} \right] [/itex] which is almost certainly true (the question was "show that ..." so unless there is an error then this is fine) and thus I am calculating the last part.

The reason I am tempted to use the spherical polars is because of the U(r) and thus all of the grads will be simple. Is it not possible to use the spherical polar coordinates?