# Commutator relations

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,

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

Related Advanced Physics Homework Help News on Phys.org
fzero
Homework Helper
Gold Member
That's the right way. Do you know how to use the $$\epsilon$$ symbol to write $$(Q\times p)_\alpha$$ in terms of components?

Gold Member
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.

fzero
Homework Helper
Gold Member
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.

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

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

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

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 dont know how to write the beta cross product in terms of components..which indices can I reuse from cross product alpha?

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

vela
Staff Emeritus
Homework Helper
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.

Gold Member
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
Staff Emeritus
Homework Helper
Gold Member
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

vela
Staff Emeritus
Homework Helper
For the initial summations, only the indices on the Levi-Civita symbols will change.

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

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

vela
Staff Emeritus
Homework Helper
Yes.

vela
Staff Emeritus
Homework Helper
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.

Gold Member
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:
vela
Staff Emeritus
Homework Helper
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.

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

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:
Gold Member
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!

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