This problem is closely related to Problem 7 in this problem set from David Tong's QFT course: http://www.damtp.cam.ac.uk/user/tong/qft/oh1.pdf

So I am studying a Klein-Gordon field $\phi$ with Lagrangian $\mathcal{L}=\frac{1}{2} ( \partial_{\mu} \phi )(\partial^{\nu} \phi) - \frac{1}{2} m \phi^{2}$.

I'm dealing with the consequences of Noether's Theorem as a result of the symmetry $x^{\mu} \to x^{\mu} + \omega^{\mu}_{\ \nu} x^{\nu}$, where $\omega^{\mu}_{\ \nu}$ is an infinitesimal anti-symmetric tensor (overall, this is essentially an infinitesimal Lorentz transformation). This gives me a field variation $\delta \phi = - \omega^{\mu}_{\ \nu} x^{\nu} (\partial_{\mu} \phi)$.

I have been able to show that $\delta \mathcal{L} = \partial_{\mu}( - \omega^{\mu}_{\ \nu} x^{\nu} \mathcal{L} ) := \partial_{\mu} F^{\mu} $. So using the proof from Neother's theorem, I know I can construct the following conserved current:

\begin{align}j^{\lambda} &= \frac{ \partial \mathcal{L} }{ \partial( \partial_{\lambda} \phi ) } \delta \phi - F^{\lambda} \\ &= - \omega^{\mu}_{\ \nu} x^{\nu} \frac{ \partial \mathcal{L} }{ \partial( \partial_{\lambda} \phi ) } ( \partial_{\mu} \phi ) + \omega^{\lambda}_{\ \nu} x^{\nu} \mathcal{L} \\ &= - \omega^{\mu}_{\ \nu} x^{\nu} \left( \frac{ \partial \mathcal{L} }{ \partial( \partial_{\lambda} \phi ) } ( \partial_{\mu} \phi ) - \delta^{\lambda}_{\ \nu} \mathcal{L} \right) \\ &= - \omega^{\mu}_{\ \nu} x^{\nu} T_{\ \nu}^{\lambda}\end{align}

where $T$ is the energy momentum tensor.

This is the answer for the conserved current in the above link. I'm okay with everything so far.

$\ $

My Question:

I am somehow supposed to take the above and come up with the conserved current:

$$(j^{\lambda})^{\nu \mu} = x^{\nu} T^{\mu \lambda} - x^{\mu} T^{\nu \lambda}$$

How do I do this? How is it that I suddenly have the two extra ''$\mu\nu$'' components? My guess is that I somehow have to "peel off" the $\omega_{\mu\nu}$ from the above $j^{\lambda}$, and use the anti-symmetry of $\omega$, but I don't know how to do this.

Also: Supposedly, there are six of the above currents $(j^{\lambda})^{\nu\mu}$... Where does this number come from? How is that the case?


1 Answer 1


Your guess is right, your infinitesimal term $\omega_{\mu,\nu}$ is a tensor, with $\frac{4(4-1)}{2}$ component (anti-symmetry) so you should have the same number of conserved currents. Remember that you can choose anything you want for your tensor $\omega_{\mu,\nu}$ while it is antisymmetric.

I would advise you to first put $j^\lambda$ in antisymmetric form, then work component by component.

[For the number $6 = \frac{4(4-1)}{2}$, you could just count the number of independant component of an antisymmetric matrix of size 4, there's $0,1$,$0,2$,$0,3$,$1,2$,$1,3$,$2,3$, the rest is either zero or the opposite of one of those. The general formula for a $n\times n$ antisymmetric matrix just happens to be $\frac{n(n-1)}{2}$]


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