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

[kreil]

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,

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

is this the right way to go about this? should i be using the jacobi identity instead?

 

[fzero]

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

 

[kreil]

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]

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.

 

[kreil]

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

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

 

[kreil]

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

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?

 

[kreil]

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

 

[vela]

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.

 

[kreil]

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]

You need to use the delta functions to evaluate some of the summations. After you do that, you'll be able to contract the Levi-Civita symbols:

http://en.wikipedia.org/wiki/Levi-Civita_symbol

 

[kreil]

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]

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

 

[kreil]

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]

Yes.

 

[vela]

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.

 

[kreil]

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..

 

[vela]

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.

 

[kreil]

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

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?

 

[kreil]

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]

Great!

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

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