Srednicki 2.17. How does metric act on Levi Cevita symbol?

[LAHLH]

Sorry to be asking again so soon, the help yesterday was great. I'm now trying to reach 2.17 from the generator commutation relation:

[M^{\mu\nu}, M^{\rho\sigma}]=i\hbar(g^{\mu\rho}M^{\nu\sigma}-g^{\nu\rho}M^{\mu\sigma})+... 2.16

J is defined by its components as J_i=\frac{1}{2}\epsilon_{ijk}M^{jk}

I'm trying to work out the commutation relation [J_i, J_l]

I've started this and plugged in the expression for the J components and used the generator commutation relation. I came to a point where I had:

[J_i, J_l]= \frac{1}{4} i\hbar \epsilon_{ijk}\epsilon_{lmn} ( g^{jm}M^{kn}-g^{km}M^{jn}-g^{jn}M^{km}+g^{kn}M^{jm} )

I paused at this point and wondered how to get rid of the metrics, can I just raise the the relevant index on each Levi Cevita symbol? (I'm not sure as I've heard this isn't a standard tensor but a tensor density). I blindly assumed I could act with the metrics in this way, and raised the Levi Cevita symbol indices, and I ended up with (after relabelling lots of dummy indices, using the antisymmetry of the M, and using the fact that single index swaps of the Levi Cevita pick up a minus sign (although not sure if this is legal between a raised and lowered index which I did do):

[J_i, J_l]=i \hbar \epsilon_i^{.m}_k \epsilon_{lmn}M^{kn}

The expression I actually want (Equation 2.17) is [J_i, J_l]=i \hbar \epsilon_{ilk}J_k=\frac{1}{2}i \hbar \epsilon_{ilk}\epsilon_{kmn}M^{mn}

Cheers for any help at all

 

[Jimmy Snyder]

[J_i, J_l]= \frac{1}{4} i\hbar \epsilon_{ijk}\epsilon_{lmn} ( g^{jm}M^{kn}-g^{km}M^{jn}-g^{jn}M^{km}+g^{kn}M^{jm} )

I solved this by concentrating on non-zero terms. For instance:
if i = l, then j = m => k = n, otherwise the product of the epsilons will be zero. But then in the 4 terms gM, whenever g is not zero, M is zero and so [J_i, J_l] is zero, as is ih\epsilon_{iik}J^k.

If i \ne l, then consider, for instance, i = 1, l = 2. There are only 4 assignments of j, k, m, and n that leave the product of the epsilons non-zero. In each of these 4 cases, the contribution is -i\hbar M^{21} so the sum of 4 of them, divided by 4 is -i\hbar M^{21}. A simple computation shows that i\hbar\epsilon_{12k}J^k = -i\hbar M^{21}

 

[LAHLH]

I solved this by concentrating on non-zero terms. For instance:
if i = l, then j = m => k = n, otherwise the product of the epsilons will be zero. But then in the 4 terms gM, whenever g is not zero, M is zero and so [J_i, J_l] is zero, as is ih\epsilon_{iik}J^k.

If i \ne l, then consider, for instance, i = 1, l = 2. There are only 4 assignments of j, k, m, and n that leave the product of the epsilons non-zero. In each of these 4 cases, the contribution is -i\hbar M^{21} so the sum of 4 of them, divided by 4 is -i\hbar M^{21}. A simple computation shows that i\hbar\epsilon_{12k}J^k = -i\hbar M^{21}

Thanks for your reply Jimmy Snyder, I greatly appreciate it. I can certainty now see the two things are equal because of your post, but I still feel one should be able to derive explicitly the result? I thought it would be just some identity of the Levi Cevita symbols I was missing or something? I guess one way would be to explicitly expand the levi cevita product is terms of deltas, but maybe there is a much simpler way?
Also strictly to contract the last index with the one on the J, shouldn't it be raised so we can employ Einstein summation? unlike how Srednicki has it

 

[ismaili]

Thanks for your reply Jimmy Snyder, I greatly appreciate it. I can certainty now see the two things are equal because of your post, but I still feel one should be able to derive explicitly the result? I thought it would be just some identity of the Levi Cevita symbols I was missing or something? I guess one way would be to explicitly expand the levi cevita product is terms of deltas, but maybe there is a much simpler way?
Also strictly to contract the last index with the one on the J, shouldn't it be raised so we can employ Einstein summation? unlike how Srednicki has it

(1) Yes. The metric tensor can raise or lower the indices of the Levi-Civita symbols. Even if you were dealing with a non-tensor, the raise-lower-rule can be applied. These rules are independent of the transformation laws to new coordinate systems.

(2) Yes. You can expand the Levi Civita symbol in terms of Kronecker delta's. And the result is much simpler than you thought. For example, for the first term in your stuck equation,
<br /> \rightarrow \frac{i\hbar}{4}\epsilon_{ijk}\epsilon_{\ell}{}^j{}_{n}M^{kn}<br /> = \frac{i\hbar}{4}(\delta_{i\ell}\delta_{kn} - \delta_{in}\delta_{k\ell})M^{kn}<br /> = \frac{i\hbar}{4}(M^{k}{}_k\delta_{i\ell} - M^{\ell i})<br />
Now, you can see that the other three terms have similar contributions, this results in,
<br /> [J_i, J_\ell] = i\hbar M^{i\ell}<br />
Note that, from the definition of the operators J_i = \frac{1}{2}\epsilon_{ijk}M^{jk}
One could write,
<br /> M^{\ell m} = \frac{1}{2}\left(\delta^{\ell}_j\delta^m_k - \delta^{\ell}_k\delta^{m}_j \right)M^{jk}<br /> = \frac{1}{2}\epsilon^{i\ell m}\epsilon_{ijk}M^{jk}<br /> = \epsilon^{i\ell m}J_i<br />
This is exactly what you want.

By the way, because of the indices of the rotation generators J_i are defined in the Euclidean space, there is no difference in writing J_i as J^i. This is why Srednicki write the final expression in that way.

 

[LAHLH]

Thanks so much ismaili, your post has helped me a tonne.