Noether in field theory — conserved currents and the stress–energy tensor

Wherever a continuous symmetry lives, a conserved current $j^\mu$ follows — a first encounter with the U(1) 4-current of a complex scalar and the stress–energy tensor $T^{\mu\nu}$ generated by spacetime translations.

Opening

Noether’s theorem made its first appearance in Analytical Mechanics I as a one-line statement: if the Lagrangian is invariant under a continuous transformation, a conserved quantity exists. Back then we saw the particle-mechanics flavour — time translation gives energy, spatial translation gives momentum, rotation gives angular momentum. In this chapter we rewrite the same theorem on infinitely many degrees of freedom, i.e. on a field $\phi$. The output is no longer a single conserved number but a conserved current $j^\mu$ (with $\mu$ a 4-spacetime index) spread over spacetime. By the end of the chapter you will be able to derive, with your own pen, the two slogans that open every QFT textbook: U(1) symmetry implies charge conservation, and spacetime translation invariance implies energy–momentum conservation. Throughout the chapter we use natural units $c = 1$ and, consistent with chapter 6, the Minkowski metric $\eta_{\mu\nu} = \mathrm{diag}(+, -, -, -)$.

Main 1 — Noether for fields, statement and one-line proof

For a field $\phi(x)$ (with $x = (t, \vec x)$) the action is the spacetime integral of a Lagrangian density $\mathcal{L}(\phi, \partial_\mu \phi)$. Suppose under an infinitesimal transformation $\phi \to \phi + \epsilon\, \delta\phi$ the Lagrangian changes by a total divergence, i.e. $\delta \mathcal{L} = \epsilon\, \partial_\mu K^\mu$ for some $K^\mu$. The variation then leaves the action invariant up to a boundary term, so the transformation is a symmetry. The associated conserved current is

$$j^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\, \delta\phi - K^\mu.$$

On any solution of the Euler–Lagrange (EL) equations, $\partial_\mu j^\mu = 0$. The corresponding conserved charge is

$$Q = \int d^3 x\, j^0$$

and satisfies $\partial_t Q = 0$.

One-line proof. Expand $\delta \mathcal{L} = (\partial \mathcal{L}/\partial \phi)\,\delta\phi + (\partial \mathcal{L}/\partial(\partial_\mu \phi))\,\partial_\mu \delta\phi$ and use the Leibniz rule to rewrite

$$\delta \mathcal{L} = \partial_\mu\!\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\,\delta\phi\right) + \left[\frac{\partial \mathcal{L}}{\partial \phi} - \partial_\mu\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\right]\delta\phi.$$

The bracket on the right is exactly the EL equation, so it vanishes on solutions. The remaining identity reads $\partial_\mu K^\mu = \partial_\mu(\partial \mathcal{L}/\partial(\partial_\mu \phi)\,\delta\phi)$; moving everything to one side gives $\partial_\mu j^\mu = 0$ for the $j^\mu$ above. That single rearrangement is the whole ladder from “symmetry” to “conserved current” in field theory.

Main 2 — U(1) example: complex scalar and the 4-current

For a complex scalar field $\phi(x) \in \mathbb{C}$ the simplest Lorentz-invariant Lagrangian is

$$\mathcal{L} = \partial^\mu \phi^* \partial_\mu \phi - m^2 \phi^* \phi$$

where $m$ is the field’s mass (dimension length$^{-1}$ in natural units). This Lagrangian is exactly invariant under the phase rotation $\phi \to e^{i\alpha}\phi$, $\phi^* \to e^{-i\alpha}\phi^*$ for any real $\alpha$, since both $|\phi|^2$ and $\partial^\mu \phi^* \partial_\mu \phi$ are $\alpha$-independent. The group of these one-parameter rotations is called U(1).

The infinitesimal form (take $\alpha \to 0$) is $\delta\phi = i\phi$, $\delta\phi^* = -i\phi^*$. Because the Lagrangian is strictly invariant we have $K^\mu = 0$. Plugging into the formula of Main 1 gives

$$j^\mu = i\bigl(\phi^* \partial^\mu \phi - \phi\, \partial^\mu \phi^*\bigr).$$

This single line is the seed of what QFT calls the “charge 4-current.” On a plane-wave solution $\phi(x) = A\, e^{-i(\omega t - \vec k \cdot \vec x)}$ (with $A$ the complex amplitude, $\omega$ the angular frequency, $\vec k$ the wavevector, and the dispersion relation $\omega^2 = \vec k^2 + m^2$),

$$j^0 = 2\omega |A|^2, \qquad \vec j = 2\vec k\, |A|^2.$$

So the 4-current $j^\mu$ is directly proportional to the 4-momentum $p^\mu = (\omega, \vec k)$. The takeaway is that the non-relativistic “probability current $\rho \vec v$” of Schrödinger theory, lifted to a relativistic field, packages into precisely these four components. A subtle sign warning: $j^0$ can be either positive or negative, so it cannot serve as a probability density directly — QFT eventually reinterprets it as the charge density of particles minus antiparticles.

Main 3 — Spacetime translations and the stress–energy tensor

Spacetime translations $x^\mu \to x^\mu + \epsilon^\mu$ are also symmetries of any Poincaré-invariant $\mathcal{L}$. Because the parameter has four components ($\mu = 0, 1, 2, 3$), four conserved currents emerge simultaneously and naturally bundle into a rank-2 tensor $T^{\mu\nu}$. Applying the Main 1 formula carefully (with $\delta\phi = \partial^\nu \phi$ and $K^\mu = \eta^{\mu\nu}\mathcal{L}$) yields

$$T^{\mu\nu} = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\,\partial^\nu \phi - \eta^{\mu\nu}\mathcal{L}$$

with $\partial_\mu T^{\mu\nu} = 0$ holding for every $\nu = 0, 1, 2, 3$ on solutions. This object is the (canonical) stress–energy tensor.

The physical content of each component is the following. $T^{00}$ is the energy density. $T^{0i}$ is the momentum density (equivalently, the energy flux divided by $c^2$, which equals it in natural units). $T^{ij}$ is the rate at which the $i$-component of momentum flows in the $j$-direction — the stress. The four conservation laws split into $\partial_\mu T^{\mu 0} = 0$, which is energy conservation, and $\partial_\mu T^{\mu i} = 0$, which is momentum conservation. The very same $T^{\mu\nu}$ appears on the right-hand side of Einstein’s field equation in general relativity; a formal derivation is deferred to the next chapter.

In Python

# Verify that the U(1) conserved current j^0 on a plane wave equals 2 omega |A|^2,
# then check that the charge Q = ∫ j^0 dx is independent of time.
import numpy as np

A     = 0.7
omega = 2.0
k     = 1.6
m     = 1.2          # omega^2 - k^2 = 4.0 - 2.56 = 1.44 = m^2 -> OK

x  = np.linspace(-10.0, 10.0, 200)
dx = x[1] - x[0]
ts = [0.0, 0.5, 1.0]

expected = 2.0 * omega * abs(A)**2     # = 2.8
print(f"theory j^0 = 2 omega |A|^2 = {expected:.4f}")

Q_history = []
for t in ts:
    phase  = -(omega * t - k * x)
    phi    = A * np.exp(1j * phase)         # complex plane wave
    # Analytic time derivative is d phi / dt = -i*omega*phi.
    # We compute it numerically with a small dt to mimic measurement.
    dt     = 1e-4
    phi_p  = A * np.exp(1j * (-(omega*(t+dt) - k*x)))
    dphidt = (phi_p - phi) / dt
    j0     = 1j * (np.conj(phi) * dphidt - phi * np.conj(dphidt))
    j0     = j0.real                         # imaginary part is numerical zero
    Q      = np.trapezoid(j0, x)
    print(f"t={t:.1f}: mean j^0 = {j0.mean():.4f}, std = {j0.std():.2e}, Q = {Q:.4f}")
    Q_history.append(Q)

print(f"Q drift = {max(Q_history) - min(Q_history):.2e}  (conserved if <= ~1e-3)")

If the mean stays near $2.8$, the standard deviation is at numerical-noise level, and the time drift of $Q$ sits below $10^{-3}$, then the theoretical identity has survived intact on the discrete grid.

To the next chapter

Chapter 9: From classical to quantum lifts the conserved current $j^\mu$ and the stress–energy tensor $T^{\mu\nu}$ obtained here to operators in the quantisation procedure, and shows how the conserved charge $Q$ is reinterpreted as a particle-number operator. The fact that Noether’s theorem is one of the sturdiest bridges between classical and quantum physics should be entirely tangible by the end of that chapter.