Where Did I Go Wrong? Troubleshooting a Commutator Relation

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In summary, the conversation is discussing a commutator relation and the Dirac Hamiltonian equation. The problem is to show the commutator relation [H_D,L_k] = -i\epsilon_{kln}\alpha_lp_n, where H_D is the Dirac Hamiltonian and L_k is the angular momentum component. The conversation also addresses some discrepancies in notation and suggests using the correct notation for the angular momentum components in order to find the correct solution.
  • #1
Spinny
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Hi, I've got a commutator relation I'm trying to figure out here. I don't know what I'm doing wrong, but I don't seem to be able to get it right, so hopefully someone can help me through it.

Anyway, here's the problem. We're given the Dirac Hamiltonian [tex]H_D = \alpha_j p_j + \beta m[/tex], where [tex]p_j = -i\nabla_j[/tex] and the angular momentum components [tex]L_k = i\epsilon_{kln}x_lp_n[/tex]. We are then going to show the commutator relation

[tex] [H_D,L_k] = -i\epsilon_{kln}\alpha_lp_n[/tex]

Here's what I've got so far:

[tex][H_D,L,k]\psi = (H_DL_k-L_kH_D)\psi[/tex]

[tex]=(\alpha_j p_j+\beta m)i\epsilon_{kln}x_lp_n\psi -i\epsilon_{kln}x_lp_n(\alpha_jp_j+\beta m)\psi[/tex]

[tex]=\alpha_j p_j i\epsilon_{kln}x_lp_n\psi +\beta mi\epsilon_{kln}x_lp_n\psi -i\epsilon_{kln}x_lp_n\alpha_jp_j\psi-i\epsilon_{kln}x_lp_n\beta m\psi[/tex]

So far, (hopefully) so good. Now, as far as I can see, the second and the last part cancel, so we're left with

[tex][H_D,L_k]\psi = \alpha_j p_j i\epsilon_{kln}x_lp_n\psi - i\epsilon_{kln}x_lp_n\alpha_jp_j\psi[/tex]

Rewriting the first part we get

[tex]i\epsilon_{kln}p_n\alpha_j p_j(x_l\psi)[/tex]

and knowing that [tex]p_j[/tex] is a differential operator, we use the product rule, and get

[tex]i\epsilon_{kln}p_n(\psi \alpha_j p_j x_l + x_l\alpha_jp_j \psi)[/tex]

We then have that [tex]p_j x_l = -i[/tex] for j = l, and 0 for j != l. Thus

[tex]i\epsilon_{kln}p_n(-i\alpha_l\psi + x_l\alpha_jp_j\psi)= \epsilon_{kln}p_n\alpha_l \psi+i\epsilon_{kln}p_nx_l\alpha_jp_j\psi[/tex]

Putting this back in we get

[tex][H_D,L_k]\psi = \epsilon_{kln}p_n\alpha_l \psi+i\epsilon_{kln}p_nx_l\alpha_jp_j\psi- i\epsilon_{kln}x_lp_n\alpha_jp_j\psi[/tex]

But now the last two parts cancel, and we're left with

[tex][H_D,L_k]\psi = \epsilon_{kln}p_n\alpha_l \psi[/tex]

This is almost what I was supposed to get, only the factor -i is missing. It seems so close, so hopefully I'm not way off, but, where did I go wrong?!?
 
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  • #2
One observation (not a solutoin to your problem, but i'll grt some paper for that in a second), is that you unnecessarily include the beta m in you calcs.

if H=a_jp_j+bm, then commutating anything with H is the same as commutating it with just a_jp_j since bm is a constant and commutes with everything (it is a central element).

you also don't need to include the phi either.

just in case you want to write this out in latex again you can save yourself some bother.
 
  • #3
I think you have you angular momentum operator slightly wrong (i'm not an expert here) but it ought to be

[tex]L_k =-i\epsilon_{lkn}x_k\partial_n[/tex]

this changes everything in your post by a factor if -i as required I think, though I've not gotten down to checking the signs with any great enthusiasm. (i'm a pure mathematician, and so if it's true up to minus signs [we call this mod 2] then it's true full stop.)
 
  • #4
matt grime said:
I think you have you angular momentum operator slightly wrong (i'm not an expert here) but it ought to be

[tex]L_k =-i\epsilon_{lkn}x_k\partial_n[/tex]

this changes everything in your post by a factor if -i as required I think, though I've not gotten down to checking the signs with any great enthusiasm. (i'm a pure mathematician, and so if it's true up to minus signs [we call this mod 2] then it's true full stop.)

Thanks for your insight. I'm not able to see, however, how it is the ang. mom. that's incorrect. The -i still comes from the mom. in the hamiltonian and cancel the i in the ang. mom.

Anyway, I'm still having trouble finding out where that ang. mom. actually came from, I haven't been able to find a similar anywhere yet (at least not one I recognize as the one given in this problem). If I ever get in the mood again, I'll review the problem with your insight in mind.
 
  • #5
Spinny said:
Thanks for your insight. I'm not able to see, however, how it is the ang. mom. that's incorrect. The -i still comes from the mom. in the hamiltonian and cancel the i in the ang. mom.

Anyway, I'm still having trouble finding out where that ang. mom. actually came from, I haven't been able to find a similar anywhere yet (at least not one I recognize as the one given in this problem). If I ever get in the mood again, I'll review the problem with your insight in mind.
The angular momentm components are incorrectly written. They should either be written without the "i" (when writing in terms of momentum components) or as Matt says (on plugging in for the mom. components).

Also note that the non-rel KE term is written using Einstein notation, so it really is the sum of all three products (ie : [itex] \sum_{j=k,l,n} \alpha _j p_j [/itex] ).
 
  • #6
Spinny said:
Thanks for your insight. I'm not able to see, however, how it is the ang. mom. that's incorrect. The -i still comes from the mom. in the hamiltonian and cancel the i in the ang. mom.

Anyway, I'm still having trouble finding out where that ang. mom. actually came from, I haven't been able to find a similar anywhere yet (at least not one I recognize as the one given in this problem). If I ever get in the mood again, I'll review the problem with your insight in mind.


The problem is that you're mixing notations. Angular momentum can be with or without the i, but that changes it from a nabla to a p or back again. So, you're using one convention in your working out but the solution is written with the other convention, that is why they don't agree. At least, that is my analysis: i checked this relation and that seems to be the problem.
 

1. What is a troublesome commutator?

A troublesome commutator is a mechanical component found in some electric motors and generators. It consists of a series of copper segments connected to the armature, which rotates within a magnetic field. The commutator is responsible for converting the alternating current produced by the armature into direct current that can be used to power devices.

2. What causes a commutator to become troublesome?

There are several factors that can cause a commutator to become troublesome. These include wear and tear from normal use, buildup of dirt and debris, and improper maintenance. Over time, the surface of the commutator may become worn or uneven, leading to poor contact with the brushes and resulting in problems such as sparking, overheating, and reduced efficiency.

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