Lorentz symmetry and Noether's theorem

Question

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?

Answer 1

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 $\left(x^\mu \partial^\alpha-x^\alpha \partial^\mu \right)$

Answer 2

Your expression is wrong and you got things mixed up somewhere. With the notation you used, it should be: $$J^μ_α = \mathcal{L} X^μ_α + Π^{μi} \left[Ψ_{iα} - ∂_νφ_i X^ν_α\right],$$ where the respective infinitesimal transforms are: $$Δφ_i = Ψ_{iα} ω^α,\quad Δx^μ = X^μ_α ω^α,$$ and $Π^{μi}$ is the Lagrangian derivative $$Π^{μi} = \frac{∂\mathcal{L}}{∂\left({∂_μφ_i}\right)},$$ not the canonical stress tensor density. The canonical stress tensor density, in your notation, which I'll denote here by $Π^μ_ν$, is $$Π^μ_ν = Π^{μi} ∂_νφ_i - δ^μ_ν \mathcal{L},$$ so that you could write $J^μ_α$, just as well, as: $$J^μ_α = Π^{μi} Ψ_{iα} - Π^μ_ν X^ν_α;$$ i.e. $$J^μ = J^μ_α ω^α = Π^{μi} Δφ_i - Π^μ_ν Δx^ν.$$ It's best to keep it as $J^μ$, without the decomposition.

For your application, $$Δφ_i = 0, \quad Δx^ν = ω^ν_ρ x^ρ.$$ Then $$J^μ = -Π^μ_ν Δx^ν = -Π^μ_ν ω^ν_ρ x^ρ.$$ The transform coefficients have the property: $$ω^ν_ρ = g_{μρ} ω^{νμ}, \quad ω^{νμ} = -ω^{μν},$$ where $g_{μρ}$ are the coefficients of a non-degenerate metric. Equivalently, $$g_{μρ} ω^ρ_ν + g_{νρ} ω^ρ_μ = 0,$$ or just $$ω_{νμ} = -ω_{μν}, \quad ω_{μν} = g_{μρ} ω^ρ_ν,$$ if you use the metric to lower the indexes, instead. The transform is defined as one that preserves the metric.

Thus, upon substitution: $$J^μ = -Π^μ_ν ω^ν_ρ x^ρ = -Π^μ_ν ω^{νσ} g_{ρσ} x^ρ.$$ Since $ω^{νσ} = -ω^{σν}$, then it can be anti-symmetrized: $$J^μ = \frac12 ω^{νσ} x^ρ \left(Π^μ_σ g_{ρν} - Π^μ_ν g_{ρσ}\right).$$ If you lower the coordinates' indices, $x_σ = g_{ρσ}x^ρ$, then you can also write this as: $$J^μ = \frac12 ω^{νσ} \left(x_ν Π^μ_σ - x_σ Π^μ_ν\right) = \frac12 ω^{νσ} J^μ_{νσ},\quad J^μ_{νσ} = x_ν Π^μ_σ - x_σ Π^μ_ν.$$

If, instead, you do it with the indices in the opposite positions, then it will be: $$J^μ = \frac12 ω_{νσ} \left(x^ν Π^{μσ} - x^σ Π^{μν}\right) = \frac12 ω_{νσ} J^{μνσ},\quad J^{μνσ} = x^ν Π^{μσ} - x^σ Π^{μν},$$ where $Π^{μν} = Π^μ_ρ g^{ρν}$.

The same anti-symmetrization applies to the transform $Δx^μ$: $$Δx^μ = ω^μ_ν x^ν = ω^{ρσ} δ^μ_ρ g_{σν} x^ν = \frac12 ω^{ρσ} \left(δ^μ_ρ g_{σν} - δ^μ_σ g_{ρν}\right) x^ν.$$ or: $$Δx^μ = \frac12 ω^{αβ} X_{αβμ}^ν x^ν, \quad X_{αβμ}^ν = δ^μ_α g_{βν} - δ^μ_β g_{αν}.$$ If you swap the $β$ index, then you'll get: $$Δx^μ = \frac12 ω^α_β X_{αμ}^{βν} x^ν, \quad X_{αμ}^{βν} = δ^μ_α δ^β_ν - g^{βμ} g_{αν},$$ where $g^{βν}$ are the coefficients of the inverse of the metric.

Answer 3

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.