# Verifying Commutator Relations for $\vec{J}=\vec{Q}\times \vec{p}$

• kreil
In summary: You can use the property of commutators to simplify the equation. Do the following:\begin{align*}Q_mp_j&=Q_pm_j\\Q_ip_n&=Q_pi_n\end{align*}After that, you can use the sifting property to combine the terms into a single equation.In summary, the first equation is cross product of two vectors, and the second equation is the sum of the products of the vectors with the delta function.
kreil
Gold Member

## Homework Statement

Verify the following commutation relations using $$\vec J = \vec Q \times \vec p$$ and $$[Q_{\alpha},p_{\beta}]=i \delta_{\alpha \beta} I$$

1. $$[J_{\alpha}, J_{\beta}]=i \epsilon_{\alpha \beta \gamma} J_{\gamma}$$

2. $$[J_{\alpha}, p_{\beta}]=i \epsilon_{\alpha \beta \gamma} p_{\gamma}$$

3. $$[J_{\alpha}, G_{\beta}]=i \epsilon_{\alpha \beta \gamma} G_{\gamma}$$

## Homework Equations

note epsilon is 1 when alpha beta gamma are in permutable order, -1 when they are not, and 0 if any are equal.

## The Attempt at a Solution

Diving right in on the first one,

$$[J_{\alpha},J_{\beta}]=[(Q \times p)_{\alpha}, (Q \times p)_{\beta}] = (Q \times p)_{\alpha}(Q \times p)_{\beta}-(Q \times p)_{\beta}(Q \times p)_{\alpha}=?$$

That's the right way. Do you know how to use the $$\epsilon$$ symbol to write $$(Q\times p)_\alpha$$ in terms of components?

fzero said:
That's the right way. Do you know how to use the $$\epsilon$$ symbol to write $$(Q\times p)_\alpha$$ in terms of components?

No, I don't.

Well you can easily verify that
$$(Q\times p)_\alpha = \epsilon_{\alpha\beta\gamma} Q_\beta p_\gamma .$$
It should make things a lot easier.

Thanks a lot, that helps enormously. Would this mean that

$$(Q\times p)_\beta = \epsilon_{\beta\gamma\alpha} Q_\gamma p_\alpha$$

$$(Q \times p)_{\alpha}(Q \times p)_{\beta}-(Q \times p)_{\beta}(Q \times p)_{\alpha}=\epsilon_{\alpha\beta\gamma}Q_{\beta}p_{\gamma}(Q \times p)_{\beta}-(Q \times p)_{\beta}\epsilon_{\alpha\beta\gamma}Q_{\beta}p_{\gamma}$$

I can see the end of the problem in sight, since I know that

$$[Q_{\alpha},p_{\beta}]=i \delta_{\alpha\beta}I$$

I just don't know how to write the beta cross product in terms of components..which indices can I reuse from cross product alpha?

Last edited:
I'm still having trouble with this. Can anyone help?

You're overusing the index β. It might be easier to keep track if you use regular letters for the indices you're summing over, so

\begin{align*} (Q \times P)_\alpha &= \epsilon_{\alpha i j}Q_i p_j \\ (Q \times P)_\beta &= \epsilon_{\beta m n}Q_m p_n \\ (Q \times P)_\alpha(Q \times P)_\beta &= \epsilon_{\alpha i j}Q_i p_j\epsilon_{\beta m n}Q_m p_n \end{align*}

It may be a bit less tedious if you keep everything in the commutator, e.g.,

$$[\epsilon_{\alpha i j}Q_i p_j, \epsilon_{\beta m n}Q_m p_n]$$

and use the properties of commutators you should hopefully be familiar with.

After applying the identities

[A,BC] = [A,B]C + B[A,C]

[AB,C] = A[B,C] + [A,C]B

I arrived at the following expression

$$i\epsilon_{\alpha i j}\epsilon_{\beta nm} \hat 1 \left ( Q_mp_j\delta_{in}-Q_ip_n \delta_{jm} \right )$$

What assumptions can I make to proceed from here?

vela said:
You need to use the delta functions to evaluate some of the summations.

This is what is giving me trouble. I've read and reread the levi-civita WiKi page for a while now (since my book did an awful job of defining the symbol).

So assuming I am summing over all i,j,m,n in the above expression, how do I apply the sifting property of the delta, i.e.

$$\sum_{i=- \infty}^{\infty} a_i \delta_{ij}=a_j$$

given that the indices of the Q and p in each term don't match that of the delta

For the initial summations, only the indices on the Levi-Civita symbols will change.

I'm not sure I follow..I think this 'shorthand' notation is causing me problems. Is this expression correct?

$$i\sum_i \sum_j \sum_m \sum_n \epsilon_{\alpha i j}\epsilon_{\beta mn} \hat 1 \left ( Q_mp_j\delta_{in}-Q_ip_n \delta_{jm} \right )$$

Yes.

No, you can only sum over either i or n for the first term and only over either j or m for the second term because the respective Kronecker deltas only multiply one term.

OK so in that case we would get

edit:
$$i\sum_j\sum_m \sum_n \epsilon_{\alpha n j}\epsilon_{\beta mn} Q_mp_j - i\sum_i\sum_m \sum_n \epsilon_{\alpha im}\epsilon_{\beta mn} Q_ip_n$$

where i summed the i's in the first term and the j's in the second. I have no clue what to do next..

Last edited:
That's better. Now refer to the Wiki page to see how to deal with the Levi-Civita symbols by turning them into combinations of Kronecker deltas.

$$\epsilon_{\alpha n j}\epsilon_{\beta mn} = \left | \begin{array}{ccc} \delta_{\alpha\beta} & \delta_{\alpha m} & \delta_{\alpha n} \\ \delta_{n\beta} & \delta_{nm} & \delta_{nn} \\ \delta_{j\beta} & \delta_{jm} & \delta_{jn} \end{array} \right |$$

and similarly,

$$\epsilon_{\alpha i m}\epsilon_{\beta mn} = \left | \begin{array}{ccc} \delta_{\alpha\beta} & \delta_{\alpha m} & \delta_{\alpha n} \\ \delta_{i \beta} & \delta_{im} & \delta_{in} \\ \delta_{m \beta} & \delta_{mm} & \delta_{mn} \end{array} \right |$$

I left the determinants in matrix form for simplicity (there are now 12 total terms)

do i now go through term by term and compute the i and j sums?

Last edited:
Thank you so much for your patience! I got the right answer for (a) after exhaustively evaluating all the sums, and then (b) and (c) were quite easy. I understand this much better now!

Great!

Just wanted to point out that because the Levi-Civita symbols have an index in common, you can write, for example,

$$\epsilon_{\alpha nj}\epsilon_{\beta mn} = -\epsilon_{\alpha jn}\epsilon_{\beta mn} = -(\delta_{\alpha \beta}\delta_{jm} - \delta_{\alpha m}\delta_{\beta j})$$

## 1. What is the significance of verifying commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$?

The commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$ are important because they describe the way in which the operators representing angular momentum ($\vec{J}$), position ($\vec{Q}$), and momentum ($\vec{p}$) interact with each other. This helps us to understand how these physical quantities behave in quantum systems.

## 2. How do you verify commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$?

To verify the commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$, we use the definition of the commutator and the fundamental commutation relations for position and momentum. We also use the properties of the cross product to simplify the calculations.

## 3. What are the applications of verifying commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$?

The commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$ have many applications in quantum mechanics. They are used to derive the quantization of angular momentum in quantum systems, as well as to understand the symmetries and conservation laws in quantum mechanics. They also play a crucial role in the study of quantum systems with rotational symmetry.

## 4. Are there any experimental methods for verifying commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$?

Yes, there are experimental methods for verifying the commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$. One example is the Stern-Gerlach experiment, which demonstrates the quantization of angular momentum and the commutation relations between position and momentum. Other experiments involve measuring the properties of spin and orbital angular momentum in quantum systems.

## 5. What happens if the commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$ are not satisfied?

If the commutator relations for $\vec{J}=\vec{Q}\times \vec{p}$ are not satisfied, it means that the operators representing angular momentum, position, and momentum do not behave in a consistent way. This can lead to inconsistencies in the mathematical description of the system and may indicate a flaw in the underlying theory. Therefore, verifying these commutator relations is an important step in ensuring the validity of the quantum mechanical framework.

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