I'm trying to overcome some misunderstanding that I have in Noether's theorem.

There is formula in David Gross's Lectures on QFT for Noether's theorem:

$$J^\mu_\alpha=\mathcal{L}X^\mu_\alpha+\Pi^\mu_i\left[\Psi_{i\alpha}-\partial_\nu\phi_i X^i_\alpha\right]$$ $J^\mu_\alpha$ - conserved vector current, $\Pi^\mu_i$-canonical momentum, $X^\mu_\alpha$ and $\Psi_{i\alpha}$ are generators of symmetry which are defined by $$x'^\mu=x^\mu+X^\mu_\alpha \omega^\alpha,$$ $$\phi'(x'^\mu)=\phi(x^\mu)+\Psi_{i\alpha} \omega^\alpha,$$

Particularly I'm working out Lorentz symmetry:

$$x'^\mu=\Lambda^\mu_\nu x^\nu$$

Infinitesimal version of this transformation is: $\Lambda^\mu_\nu=\delta^\mu_\nu+\omega^\mu_\nu$, where $\omega^\mu_\nu=-\omega^\nu_\mu$ is antisymmetric.

So, my problem that I dont understand why (they write this expression in many sources, not only in Gross) $$X^{\mu,\alpha}_\beta=\delta^\mu_\alpha x^\beta-\delta^\mu_\beta x^\alpha$$.

If I'm calculating $X^{\mu,\alpha}_\beta\omega^\beta_\alpha$, I get $2\omega^\mu_\nu x^\nu$, not the $\omega^\mu_\nu x^\nu$. Also I've done this calculation: $$f(x'^\mu)=f(x^\mu+\omega^\mu_\alpha x^\alpha)\approx f(x^\mu)+\omega_{\alpha \mu}x^\alpha \partial^\mu f = f(x^\mu) + \frac 1 2 \omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)f $$

If $f=x^\mu$ (coordinate function), then $$X^{\mu,\alpha}_\beta=\frac 12\left(\delta^\mu_\alpha x^\beta-\delta^\mu_\beta x^\alpha\right).$$

What exactly am I doing wrong?


2 Answers 2


I think this calculation is the answer for your question:

$$\sum_{\alpha,\mu}\frac 1 2 \omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\mu<\alpha}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\alpha<\mu}\omega_{\mu\alpha } \left(x^\alpha \partial^\mu-x^\mu \partial^\alpha \right)=\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)+\frac 1 2 \sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)=\sum_{\alpha<\mu}\omega_{\alpha \mu} \left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)$$

Where I

  1. Splited sum in "$\alpha<\mu$" and "$\mu<\alpha$" parts.
  2. Changed $\alpha \leftrightarrow \mu$
  3. Used antisymmetric property of $\omega_{\alpha \mu}$ and $\left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)$

This is one of the few times when using the Einstein convention isn't helpful: recall that in Noether's theorem you need to sum over the independent parameters of your transformation, which is usually trivial but, for Lorentz transformations, you need to be careful because of the antisymmetry of $\omega_{\mu\nu}$, which implies that you have only 6 generators.

When you sum over $\alpha,\mu$ using the Einstein convention, you're summing twelve terms (and double counting the contribution of each generator). To have a sum that takes into account that you only have six, you need to sum using, say, $\alpha>\mu$, and that will imply not using the Einstein convention for a while. If you do that you don't get the factor if 1/2.


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