Commutator with Tensor Notation

[Septim]

Greetings,

I would like to find the commutator \left[Lx^2,Ly^2\right] and prove that
\left[Lx^2,Ly^2\right]=\left[Ly^2,Lz^2\right]=\left[Lz^2,Lx^2\right] I infer from the cyclic appearance of the indices that using the index notation would be much more compact and insightful to solve the problem. However due to summation convention I do not know how to write a component squared instead of the whole vector squared. What is the remedy? Any help suggestion is welcome.

 

[atyy]

Is the index really squared? Not like http://quantummechanics.ucsd.edu/ph130a/130_notes/node214.html , where the operator is squared?

 

[Septim]

Thanks for the reply. Yes, I am sure the index is squared the question is from Liboff Introduction to Quantum Mechanics 4th edition, and it is given as I stated. I do not know the physical insight behind this question but I want to prove that identity using index notation. How can I do so?

 

[strangerep]

Thanks for the reply. Yes, I am sure the index is squared the question is from Liboff Introduction to Quantum Mechanics 4th edition, and it is given as I stated. I do not know the physical insight behind this question but I want to prove that identity using index notation. How can I do so?

I don't have a copy of Liboff, but I guess the $L$'s are so(3) generators? If so, you don't need index notation. Just apply the Leibniz product rule for commutators, i.e., $[a,bc] = b[a,c] + [a,b]c$, to the first commutator $[L_x^2, L_y^2]$ twice. Then cyclically permute x,y,z to get another expression and use the so(3) commutation relations to show it's the same as the original.

If you need more help than this, don't forget to show your work -- like in the homework forums.

(Actually, this probably should be in the homework forums. )

 

[Septim]

I do not have a solid background in the mathematical aspect of the quantum theory so I may not say that if the operator is a so(3) generator or not. I applied the same approach as you illustrated to the problem and it is tractable that way but I insist on using the index notation.

 

[andrien]

if you really want to invoke index notation then note that for example L_i² can be written as
L_i²=ε_ijkr_jp_kε_ilmr_lp_m=(δ_jlδ_km-δ_jmδ_kl)r_jp_kr_lp_m
after it you will have to use the first quantized form for momentum operators.

 

[Septim]

andrien I think your suggestion also sums over all components of angular momentum from the way you expand the product of two Levi Civita Tensors.

 

[andrien]

yes,that is what I wanted from you to recognise.It is the total angular momentum square and so you will get the famous operator L² from it i.e.
L²=x_i∂_jx_i∂_j-x_i∂_jx_j∂_i=(r²∇²+r.∇)-{3r.∇+(r.∇)²-(r.∇)}=r²∇²-r.∇-(r.∇)²,which ultimately gives the L².you can see it will be of no use to determine the components separately.

 

[Fredrik]

if you really want to invoke index notation then note that for example L_i² can be written as
L_i²=ε_ijkr_jp_kε_ilmr_lp_m=(δ_jlδ_km-δ_jmδ_kl)r_jp_kr_lp_m
after it you will have to use the first quantized form for momentum operators.

The first equality isn't true. The expression after the first equality is equal to $L_iL_i$, which is equal to $\mathbf L^2$, not $L_i^2$.

There's no need to use the definition of L in terms of r and p here. It's sufficient to use the commutation relations for the components of L. There's also no need to use tensor index notation, because what strangerep said completely solves the problem. However, if you really want to use tensor index notation for some reason, then you need to start with $[L_i^2,L_j^2]$. Ah, I see the problem. When you use the rule [AB,C]=A[B,C]+[A,C]B, the same index will appear twice, and you don't want this to imply summation. This just means that you can't use the summation convention in this problem.

So if you insist on using the index notation, you need to start by saying that in this calculation, you're not using the summation convention. Then if you need to sum over some index, you just write a summation sigma.

 

[andrien]

The first equality isn't true. The expression after the first equality is equal to $L_iL_i$, which is equal to $\mathbf L^2$, not $L_i^2$.

There's no need to use the definition of L in terms of r and p here. It's sufficient to use the commutation relations for the components of L. There's also no need to use tensor index notation, because what strangerep said completely solves the problem. However, if you really want to use tensor index notation for some reason, then you need to start with $[L_i^2,L_j^2]$. Ah, I see the problem. When you use the rule [AB,C]=A[B,C]+[A,C]B, the same index will appear twice, and you don't want this to imply summation. This just means that you can't use the summation convention in this problem.

So if you insist on using the index notation, you need to start by saying that in this calculation, you're not using the summation convention. Then if you need to sum over some index, you just write a summation sigma.

L_i² means LiLi where the summation is understood (no use of metric tensor) which is equal to L².It should not pose any ambiguity.Also with the components of say L_x² really poses problem because indices should not be summed even if it appears twice.So it is better not use index notation.

 

[Fredrik]

L_i² means LiLi where the summation is understood

This is where we disagree. Obviously, if you just write $L_iL_i$ without any comments, then it should be interpreted as $\mathbf L^2$ because of the summation convention, but $L_i^2$ simply means $L_1{}^2$ if i=1, $L_2{}^2$ if i=2 and $L_3{}^2$ if i=3. That's why you can't rewrite $L_i^2$ as $L_iL_i$ without saying that we're not using the summation convention.

I obviously can't rule out that there's some book that uses the notation $L_i^2$ to mean $L_iL_i$, but I would consider that a bad notation, because an index that doesn't explicitly appear twice shouldn't be summed over. Also, we already have the notation $\mathbf L^2$ for $L_iL_i$. We don't need another.

 

[Septim]

Okay thanks I also think I cannot make any contractions so tensor notation is not that suitable for this problem at hand.