Commutator with Tensor Notation

In summary, In Liboff's quantum mechanics textbook, the commutator \left[Lx^2,Ly^2\right] is given and is equal to \left[Ly^2,Lz^2\right] using the index notation. However, when trying to solve for the components of L using this notation, it's not helpful to use the summation convention. Instead, you need to use the commutation relations for the components of L.
  • #1
Septim
167
6
Greetings,

I would like to find the commutator [itex]\left[Lx^2,Ly^2\right][/itex] and prove that
[itex]\left[Lx^2,Ly^2\right][/itex]=[itex]\left[Ly^2,Lz^2\right][/itex]=[itex]\left[Lz^2,Lx^2\right][/itex] 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.
 
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  • #3
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?
 
  • #4
Septim said:
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. :devil: )
 
  • #5
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.
 
  • #6
if you really want to invoke index notation then note that for example Li2 can be written as
Li2ijkrjpkεilmrlpm=(δjlδkmjmδkl)rjpkrlpm
after it you will have to use the first quantized form for momentum operators.
 
  • #7
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.
 
  • #8
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=xijxij-xijxji=(r22+r.∇)-{3r.∇+(r.∇)2-(r.∇)}=r22-r.∇-(r.∇)2,which ultimately gives the L2.you can see it will be of no use to determine the components separately.
 
  • #9
andrien said:
if you really want to invoke index notation then note that for example Li2 can be written as
Li2ijkrjpkεilmrlpm=(δjlδkmjmδkl)rjpkrlpm
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.
 
  • #10
Fredrik said:
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.
Li2 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 Lx2 really poses problem because indices should not be summed even if it appears twice.So it is better not use index notation.
 
  • #11
andrien said:
Li2 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.
 
  • #12
Okay thanks I also think I cannot make any contractions so tensor notation is not that suitable for this problem at hand.
 

1. What is a commutator in tensor notation?

A commutator in tensor notation is a mathematical operation that measures the degree of non-commutativity between two tensors. It is represented by the symbol [A,B] and is calculated by taking the difference between the products AB and BA.

2. How is a commutator calculated in tensor notation?

A commutator in tensor notation is calculated by using the Einstein summation convention. This involves summing over repeated indices in the tensors, multiplying the corresponding components, and then subtracting the results to obtain the final value of the commutator.

3. What is the physical significance of a commutator in tensor notation?

The commutator in tensor notation is used to quantify how two physical quantities behave under different mathematical operations. It helps determine if these quantities can be measured simultaneously or if their order affects the final result. It is particularly useful in quantum mechanics and relativity.

4. Can a commutator in tensor notation be zero?

Yes, a commutator in tensor notation can be zero. This indicates that the two tensors commute, meaning their order does not affect the final result. This is often the case for classical physical quantities, but in quantum mechanics, a zero commutator can also have other implications.

5. How is a commutator used in physics?

A commutator in tensor notation is used in various areas of physics, including quantum mechanics, relativity, and classical mechanics. It helps determine the non-commutativity of physical quantities and plays a crucial role in understanding the behavior of particles, fields, and other physical systems.

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