Commutation relations of angular momentum with position, momentum.

In summary, the conversation discusses the use of position and momentum operators in proving equations involving commutators. The attempt at a solution involves the use of tensor algebra and the importance of respecting laws and conventions. The confusion arises from the lack of summation convention in the problem. The solution requires understanding of cartesian vectors/tensors and the application of the identity/equality in a specific manner.
  • #1
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Homework Statement


Using the position space representation, prove that:
[itex] \left[L_i, x_j\right] = i\hbar\epsilon_{ijk}x_k [/itex].
Similarly for [itex] \left[L_i, p_j\right] [/itex].

Homework Equations


Presumably, [itex] L_i = \epsilon_{ijk}x_jp_k [/itex].
[itex] \left[x_i, p_j\right] = i\hbar\delta_{ij} [/itex].

The Attempt at a Solution


[itex] \left[L_i, x_j\right] = \epsilon_{ijk}\left[x_jp_k, x_j\right]
= \epsilon_{ijk}\left(x_jp_kx_j - x_jx_jp_k\right)
= \epsilon_{ijk}x_j\left(p_kx_j - x_jp_k\right)
= \epsilon_{ijk}x_j\left[p_k, x_j\right]
= -i\hbar\epsilon_{ijk}x_j\delta_{jk} [/itex]
which is where I become confused - it seems to me that the right hand side is always zero (if the Kronecker delta is nonzero, the Levi-Civita symbol is zero, and vice-versa).

Any help is much appreciated.
 
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  • #2
You should respect the laws of tensor algebra. Don't use the same index 3 times in a tensor product.

So [itex] [L_i,x_j]=\epsilon_{ikl}[x_k p_l, x_j] = ... [/itex]
 
  • #3
I can't edit my post ? Ok...

It's not true that epsilon times (in a tensor way) delta =0. This happens iff there's a double contracted tensor product among them.
 
  • #4
That's the thing: when I asked my TA, he said that we are not to assume the summation convention. That's the source of my confusion - I have no idea what this question even means without the summation convention.

(I'm literally the only person in the course who knows what a tensor is).
 
  • #5
You need the summation convention and know a little about cartesian vectors/tensors, if you're approaching the identity/equality from the all-components-at-the-same-time perspective
 

1. What are the commutation relations of angular momentum with position and momentum?

The commutation relations of angular momentum with position and momentum are given by:
[Lx, x] = i,
[Ly, y] = i,
[Lz, z] = i,
[Lx, px] = 0,
[Ly, py] = 0,
[Lz, pz] = 0,
where L represents the angular momentum operators, x, y, and z represent the position operators, and px, py, and pz represent the momentum operators.

2. What is the significance of the commutation relations of angular momentum with position and momentum?

The commutation relations of angular momentum with position and momentum are important because they describe the fundamental quantum mechanical behavior of these physical quantities. They demonstrate that the position and momentum operators do not commute, meaning that they cannot be simultaneously measured with arbitrary precision. This is one of the key principles of Heisenberg's uncertainty principle.

3. How do the commutation relations of angular momentum with position and momentum relate to the quantization of angular momentum?

The commutation relations of angular momentum with position and momentum are a crucial component in the quantization of angular momentum. They show that the angular momentum operators have discrete eigenvalues, rather than continuous values, which is a key feature of quantization in quantum mechanics. These discrete eigenvalues correspond to the possible states of a quantum system with angular momentum.

4. Can the commutation relations of angular momentum with position and momentum be generalized to higher dimensions?

Yes, the commutation relations of angular momentum with position and momentum can be generalized to higher dimensions. In three-dimensional space, the angular momentum operators have three components (x, y, and z), and the position and momentum operators also have three components each. The commutation relations hold for each of these components separately, and they can also be extended to higher dimensions with more components.

5. How do the commutation relations of angular momentum with position and momentum affect the behavior of quantum systems?

The commutation relations of angular momentum with position and momentum play a crucial role in determining the behavior of quantum systems. They impose restrictions on the simultaneous measurement of position and momentum, leading to the uncertainty principle. They also affect the energy levels and allowed transitions in atoms and other quantum systems, as well as the behavior of particles in magnetic fields. Understanding these commutation relations is essential for understanding the behavior of quantum systems at the atomic and subatomic level.

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