The symmetry connected to Baryon/Lepton Number conservation is, as far as I understand, global U(1) symmetry (which is called here global gauge invariance).

Does anyone know of an explicit computation of this, using Noether etc.?

Any idea, link or book recommendation would be much appreciated


1 Answer 1


"Derivation" of Baryon Number Conservation -

Consider the QCD Lagrangian (density)

$$\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu - m)\psi - \frac{1}{4}G^a_{\mu\nu}G_a^{\mu\nu}$$

where the symbols have their usual meaning.

This is invariant under $U(1)$, which is nothing but a multiplication of $\psi$ by a global phase factor $e^{i \theta}$. This is because $\bar \psi$ picks up a corresponding factor with an opposite phase, and the two cancel each other, in the first term.

To make use of the Noether theorem, rewriting this as $$\psi'(x) = e^{i \theta} \ \psi(x) \approx \psi(x) + i\theta \psi(x)$$ to first order (small $\theta$, as in infinitesimal transformations). This $U(1)$ invariance form is a very special case of invariance under global transformations of the type $$\psi'(x) = e^{i \theta^a \Gamma_a} \psi(x) \approx \psi(x) + i\theta^a \Gamma_a \psi(x)$$ where $\Gamma_a$ refers to the generators of the unitary group acting on the quark field $\psi$.

Using the Noether's theorem for the general case, one has to go through the usual steps: writing the change in the action, $$\delta S \equiv \int d^4 x \delta \mathcal{L} = 0$$ then expanding this $\delta \mathcal{L}$ in terms of a changes in $\psi$ and those in $\partial_{\mu} \psi$, then integrating by parts to arrive at the conserved Noether current $$J_a^{\mu}(x) = \bar \psi (x) \gamma^{\mu} \Gamma_a \psi (x)$$ which satisfies $$\partial_{\mu} J_a^{\mu}(x) = 0$$ The same may be cast into the form of a charge $$Q_a = \int d^3 x J_a^0 (x) $$ with $$\frac{dQ_a}{dt} = \int d^3 x \frac{\partial J_a^0 (x)}{\partial t} = - \int d^3 x \nabla \cdot J_a = 0$$ provided the currents decay sufficiently rapidly at the spatial infinity. That condition is anyways employed even in electrodynamics in the deriving total charge conservation, using an identical argument.

Thus, the zeroth component of this Noether current, integrated over 3-space (hence giving a charge), is globally conserved. That's a powerful result, for the general case.

Returning back to the issue of interest - $U(1)$ invariance. Making use of the general result, you may notice that in this case, there is no index $a$ and the sole gamma is identity (correlating the two equations). Thus, substituting in the general case, our conserved current reads $$J_{\mu} = \bar \psi \gamma_{\mu} \psi$$ and the conserved charge reads $$Q = \int d^3 x \bar \psi \gamma_0 \psi$$ which can be rewritten as $$Q = \int d^3 x \psi^{\dagger} \psi$$ by using the well known properties of gamma matrices.

Notice at this point two things -

1) We have used invariance arguments, which can always accommodate a multiplicative factor, since that won't change the invariance.

2) $\psi^{\dagger} \psi$ is a number density. (You may want to recall the definition of number operator in e.g. harmonic oscillator in basic QM, and extrapolate.)

Thus, taking QCD Lite for example, where the quarks of interest are $u$, $d$ and $s$, any baryon would be composed of these, and we can always normalize the conserved charge by a factor of $1/3$, one for each quark flavor. Also, the number density integrated over 3-space would give a number. Hence, putting these two together, we can write as the conserved charge $$B = \frac{1}{3} \int d^3 x \ \psi^{\dagger} \psi$$ (where $\psi$ covers the three flavors), which can be interpreted as, (gulp) Baryon Number.

Thus, baryon number can be interpreted as the conserved charge corresponding to $U(1)$ invariance of the QCD Lagrangian.

  • $\begingroup$ Noether theorem is for classical field, do we need to go to quantum theory? $\endgroup$
    – user26143
    Sep 26, 2014 at 4:48
  • $\begingroup$ @user26143 - I'm not sure I understand. Noether's theorem would give the conserved currents corresponding to every differential symmetry of the action $S$. This ${\mathcal L}$ is of course of QuantumCD, but whether or not we quantize this ${\mathcal L}$ doesn't appear to enter anywhere in the statement/application of the theorem. But, there may be another derivation based on the quantum analog of these, the Ward-Takahashi identity. I've never seen it done, but I'll be (contd.) $\endgroup$
    – 299792458
    Sep 26, 2014 at 5:10
  • $\begingroup$ (contd.) happy to learn how it is done. :) $\endgroup$
    – 299792458
    Sep 26, 2014 at 5:11
  • $\begingroup$ I am not sure either... since the derivation of Noether theorem does not involve quantization procedure, I do not know if quantization matters. I am happy to see further comment :) $\endgroup$
    – user26143
    Sep 26, 2014 at 5:14
  • $\begingroup$ 6 years too late but I can potentially chime in. Quantisation matters a bit. Often it is the case that a classical symmetry results in a quantum symmetry, however, there is the possibility of quantum corrections that break the conservation law, however so far I havent come accross any examples of this (although I almost definitely wil at some point) $\endgroup$ Jan 24, 2021 at 20:27

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