Symmetries and Conservation Laws
Noether's theorems, conserved currents, and the energy-momentum tensor
1.11 What Is a Symmetry of the Action?
A symmetry is a transformation of the fields (and possibly coordinates) that leaves the equations of motion unchanged. In the Lagrangian framework, the precise requirement is weaker than “the action is invariant” — we only need the action to change by at most a boundary term.
Definition (Symmetry of the Action)
A continuous family of field transformations $\phi \to \phi + \epsilon\,\Delta\phi$ is a symmetry if and only if the Lagrangian density transforms as:
for some 4-vector $K^\mu$. If $K^\mu = 0$ the symmetry is strict; otherwise the Lagrangian changes by a total divergence that does not affect the equations of motion.
Infinitesimal Transformations
We always work infinitesimally. Write the general transformation as:
where the index a runs over all field species. Common cases:
| Symmetry | $\Delta\phi$ | Parameters | Type |
|---|---|---|---|
| Spacetime translation | $-a^\nu\partial_\nu\phi$ | $a^\mu$ (4 params) | Spacetime |
| Lorentz transformation | $-\tfrac{1}{2}\omega^{\rho\sigma}(\Sigma_{\rho\sigma}\phi + x_\rho\partial_\sigma\phi - x_\sigma\partial_\rho\phi)$ | $\omega^{\mu\nu}$ (6 params) | Spacetime |
| U(1) phase | $i\phi$ | $\alpha$ (1 param) | Internal |
| SU(N) | $iT^a\phi$ | $\alpha^a$ ($N^2-1$ params) | Internal |
Important Distinction
Internal symmetries act only on the fields ($\delta x^\mu = 0$). Spacetime symmetries also shift coordinates ($\delta x^\mu \neq 0$). Noether's theorem handles both, but the current formula looks different in each case.
1.12 Noether's First Theorem — Full Proof
Theorem (Emmy Noether, 1918)
If the action $S = \int d^4x\,\mathcal{L}(\phi_a, \partial_\mu\phi_a)$ is invariant (up to a boundary term) under a continuous transformation with $r$ parameters, then there exist $r$ conserved currents $j^\mu_{(\alpha)}$ satisfying:
valid on shell (i.e., when the fields satisfy the Euler-Lagrange equations).
Step-by-Step Proof
Step 1. Start from the action and compute $\delta S$ under$\phi_a \to \phi_a + \epsilon\,\Delta\phi_a$:
Step 2. For a global symmetry, $\epsilon$ is constant, so$\partial_\mu\epsilon = 0$. Use the product rule on the second term:
Step 3. Rewrite using a total derivative identity:
Step 4. Substituting back:
Step 5. On shell (using E-L equations), the first term vanishes. Since the transformation is a symmetry,$\delta\mathcal{L} = \epsilon\,\partial_\mu K^\mu$. Therefore:
Step 6. Define the Noether current:
Then $\partial_\mu j^\mu = 0$ on shell. $\;\square$
Common Pitfall: The $K^\mu$ Term
Many textbooks write $j^\mu = \pi^\mu_a \Delta\phi_a$ and drop the $K^\mu$ term. This is only correct for strict symmetries where $\delta\mathcal{L} = 0$(such as internal symmetries of the Klein-Gordon Lagrangian). For spacetime translations, the Lagrangian changes as $\delta\mathcal{L} = \epsilon\,\partial_\mu(\mathcal{L}\,a^\mu)$, so $K^\mu = \mathcal{L}\,a^\mu \neq 0$. Forgetting this gives a wrong energy-momentum tensor!
Conserved Charge
The continuity equation $\partial_\mu j^\mu = 0$ implies a conserved charge:
Proof of conservation:
where we used $\partial_\mu j^\mu = 0$ and Gauss's theorem. The boundary term vanishes if the fields fall off fast enough at spatial infinity (which is ensured by finite energy conditions).
Subtlety: On-Shell vs Off-Shell
The Noether current is conserved on shell only — i.e., when the fields obey the equations of motion. In the quantum theory, this becomes the statement that Noether's current is conserved inside correlation functions (Ward-Takahashi identities), up to contact terms.
1.13 Example 1: U(1) Phase Symmetry → Charge Conservation
Consider the complex Klein-Gordon field:
Step 1: Identify the Symmetry
Global U(1) phase rotation:
Infinitesimally: $\Delta\phi = i\phi$, $\Delta\phi^* = -i\phi^*$. The Lagrangian is strictly invariant ($K^\mu = 0$).
Step 2: Compute the Current
Step 3: Verify Conservation
Check $\partial_\mu j^\mu = 0$ on shell:
Step 4: The Conserved Charge
where $\pi = \dot\phi^*$ is the conjugate momentum. After quantization, $Q$ becomes the electric charge operator: particles contribute $+1$ and antiparticles $-1$.
1.14 Example 2: Translations → Energy-Momentum Tensor
This is where most students get tripped up. The spacetime translation is a coordinate transformation, so the Lagrangian is not strictly invariant — it changes by a total derivative. Let's be very careful.
Step 1: The Translation
Under $x^\mu \to x^\mu + a^\mu$ with constant $a^\mu$:
So $\Delta\phi = a^\nu\partial_\nu\phi$. But wait — this is the total change. We need the functional change at the same point, which is$\delta\phi(x) = \phi'(x) - \phi(x) = -a^\nu\partial_\nu\phi(x)$.
Step 2: Change in Lagrangian
Since $\mathcal{L}$ depends on $x$ only through the fields (no explicit $x$-dependence):
This is a total derivative! So $K^\mu = a^\nu\delta^\mu_\nu\,\mathcal{L} = a^\mu\mathcal{L}$. The Lagrangian is not invariant — but it changes by a boundary term, so it's still a symmetry.
Step 3: Build the Noether Current
Using $\Delta\phi_a = a^\nu\partial_\nu\phi_a$ in the general formula (with the minus sign for functional variation):
Factor out $a^\nu$:
Since $a^\nu$ is arbitrary, conservation $\partial_\mu j^\mu = 0$for all $a^\nu$ gives:
This is the canonical energy-momentum tensor.
Step 4: Identifying the Conserved Quantities
The four conserved charges are:
$\nu = 0$: Energy
$\nu = i$: Momentum
Common Pitfall: Sign Conventions
The sign and index placement of $T^{\mu\nu}$ varies between textbooks. Peskin-Schroeder defines $T^{\mu\nu} = \pi^{\mu}_a\partial^\nu\phi_a - g^{\mu\nu}\mathcal{L}$(mostly-minus metric). Weinberg uses mostly-plus. Always check which convention your source uses before comparing results!
1.15 Example 3: Lorentz Invariance → Angular Momentum
Under an infinitesimal Lorentz transformation $x^\mu \to x^\mu + \omega^\mu{}_\nu x^\nu$with antisymmetric $\omega^{\mu\nu} = -\omega^{\nu\mu}$:
For a scalar field
Applying Noether's theorem gives the conserved current:
The 6 conserved charges (one for each independent $\omega^{\rho\sigma}$):
Spatial components $M^{ij}$
The angular momentum: $L^k = \tfrac{1}{2}\epsilon^{ijk}M^{ij}$
Mixed components $M^{0i}$
Center-of-energy motion: $M^{0i} = tP^i - \int d^3x\, x^i T^{00}$
For fields with spin
A spinor field $\psi$ transforms as $\delta\psi = -\frac{i}{4}\omega_{\rho\sigma}\sigma^{\rho\sigma}\psi - \omega^\mu{}_\nu x^\nu\partial_\mu\psi$where $\sigma^{\rho\sigma} = \frac{i}{2}[\gamma^\rho, \gamma^\sigma]$. This adds an intrinsic spin term:
Total angular momentum = orbital + spin. This is why spin emerges naturally from Lorentz symmetry in QFT.
1.16 Example 4: SU(2) Isospin → 3 Conserved Charges
Consider a doublet of real scalar fields $\vec\phi = (\phi_1, \phi_2, \phi_3)$ with O(3)-invariant Lagrangian:
Under infinitesimal rotation $\phi_a \to \phi_a + \epsilon^b(\epsilon_{bac})\phi_c$(here $\epsilon_{bac}$ is the Levi-Civita symbol, $\epsilon^b$ is the rotation parameter):
The three conserved currents ($b = 1,2,3$):
The three conserved charges $Q_b = \int d^3x\,j^0_{(b)}$ are the components of isospin. In particle physics, this becomes the SU(2) weak isospin when applied to the electroweak sector.
1.17 The Belinfante Improvement
The canonical $T^{\mu\nu}$ has a problem: it is not symmetric in general. For a vector field or spinor field, $T^{\mu\nu} \neq T^{\nu\mu}$. But the energy-momentum tensor that couples to gravity (in General Relativity) must be symmetric.
The Ambiguity
If $\partial_\mu j^\mu = 0$, then adding a “superpotential” preserves conservation:
where $B^{\rho\mu\nu} = -B^{\mu\rho\nu}$ is antisymmetric in the first two indices. Then $\partial_\mu\Theta^{\mu\nu} = \partial_\mu T^{\mu\nu} + \underbrace{\partial_\mu\partial_\rho B^{\rho\mu\nu}}_{= 0\text{ (antisymmetry)}} = 0$.
Belinfante's Procedure
Choose $B^{\rho\mu\nu}$ to absorb the spin contribution$S^{\mu\rho\sigma}$ and make the result symmetric:
The resulting Belinfante tensor $\Theta^{\mu\nu}$ is:
- Symmetric: $\Theta^{\mu\nu} = \Theta^{\nu\mu}$
- Conserved: $\partial_\mu\Theta^{\mu\nu} = 0$
- Gauge invariant (for gauge theories)
- Equals $\frac{2}{\sqrt{-g}}\frac{\delta S_{\text{matter}}}{\delta g_{\mu\nu}}$ — the same tensor that appears in Einstein's equations!
Why This Matters
For scalar fields, $T^{\mu\nu}$ is already symmetric (no spin). For the electromagnetic field, the canonical $T^{\mu\nu}$ is not gauge-invariant and not symmetric. The Belinfante procedure fixes both problems and gives the standard$\Theta^{\mu\nu} = F^{\mu\alpha}F^\nu{}_\alpha - \frac{1}{4}g^{\mu\nu}F^2$.
1.18 Noether's Second Theorem (Gauge Symmetries)
Emmy Noether actually proved two theorems. The second is less well-known but equally important:
Noether's Second Theorem
If the action is invariant under a local (gauge) symmetry — where the transformation parameters are arbitrary functions of spacetime — then the equations of motion are not independent. They satisfy certain identities (rather than giving conservation laws).
The Key Difference
| Feature | First Theorem (Global) | Second Theorem (Local/Gauge) |
|---|---|---|
| Parameters | Constants $\epsilon^a$ | Functions $\epsilon^a(x)$ |
| Result | Conserved currents | Identities between E-L equations |
| Consequence | Conservation laws | Equations are redundant (gauge freedom) |
| Example | U(1) global → charge | U(1) local → $\partial_\mu F^{\mu\nu} = J^\nu$ and $\partial_\nu J^\nu = 0$ are linked |
Example: Electromagnetism
The Maxwell Lagrangian is invariant under $A_\mu \to A_\mu + \partial_\mu\alpha(x)$for arbitrary $\alpha(x)$. Noether's second theorem tells us the four E-L equations $\partial_\mu F^{\mu\nu} = 0$ are not independent — they satisfy the identity:
This means one of the four Maxwell equations is redundant (only 3 are independent), reflecting the gauge freedom of choosing $\alpha(x)$. The system is underdetermined without a gauge-fixing condition.
1.19 Common Pitfalls and Subtleties
1. Discrete Symmetries Don't Give Currents
Noether's theorem applies only to continuous symmetries. Discrete symmetries (P, C, T) give multiplicative quantum numbers (±1), not additive conserved charges with currents. There is no “parity current.”
2. The Current Is Not Unique
You can always add a “trivially conserved” term:
Then $\partial_\mu j'^\mu = \partial_\mu j^\mu + \underbrace{\partial_\mu\partial_\nu\Sigma^{\mu\nu}}_0 = 0$. The charge $Q = \int d^3x\,j^0$ is unchanged (Gauss's theorem).
3. Spontaneously Broken Symmetries
If a symmetry is spontaneously broken (the vacuum is not invariant), Noether's theorem still gives a conserved current. But the charge $Q$ does not annihilate the vacuum: $Q|0\rangle \neq 0$. Instead, the broken symmetry gives rise to Goldstone bosons (massless excitations). The current is still conserved — but it creates Goldstone particles from the vacuum.
4. Anomalies Break Classical Symmetries
A symmetry of the classical action may fail after quantization — this is called an anomaly. The most famous example: the axial U(1) symmetry of massless QCD is broken by the chiral anomaly:
The classical conservation law is violated by quantum effects (loop diagrams). Anomalies have profound physical consequences ($\pi^0 \to \gamma\gamma$ decay).
5. Confusing Total vs Functional Variation
For spacetime symmetries, the total variation $\Delta\phi = \phi'(x') - \phi(x)$and the functional variation $\delta\phi = \phi'(x) - \phi(x)$differ by $\delta x^\mu\partial_\mu\phi$. Mixing these up is the #1 source of sign errors in computing $T^{\mu\nu}$. For internal symmetries, $\delta x^\mu = 0$so both variations coincide.
1.20 Ward-Takahashi Identities (Preview)
In the quantum theory, Noether's theorem becomes the Ward-Takahashi identity. The classical equation $\partial_\mu j^\mu = 0$ is promoted to:
The right-hand side consists of contact terms — they arise when the current insertion coincides with a field operator. These identities:
- Constrain the structure of Feynman diagrams
- Ensure the photon remains massless in QED (Ward identity)
- Prove renormalizability of gauge theories
- For the energy-momentum tensor: constrain graviton couplings
The QED Ward Identity
For QED with the U(1) current $j^\mu = \bar\psi\gamma^\mu\psi$:
where $\mathcal{M}^\mu$ is any amplitude with an external photon of momentum $q$. This ensures $Z_1 = Z_2$ (vertex and wavefunction renormalizations are equal), and that the photon has no mass counterterm.
Summary: Symmetry → Conservation Law
| Symmetry | Transformation | Conserved Current | Conserved Charge |
|---|---|---|---|
| Time translation | $t \to t + \epsilon$ | $T^{0\mu}$ | Energy $H$ |
| Space translation | $x^i \to x^i + a^i$ | $T^{i\mu}$ | Momentum $P^i$ |
| Rotations | $x^i \to R^i{}_j x^j$ | $\mathcal{M}^{0ij}$ | Angular momentum $L^k$ |
| Boosts | $x^0 \to x^0 + \omega^{0i}x_i$ | $\mathcal{M}^{00i}$ | Center of energy |
| U(1) phase | $\phi \to e^{i\alpha}\phi$ | $i(\phi^*\partial^\mu\phi - \text{c.c.})$ | Electric charge $Q$ |
| SU(N) | $\phi \to e^{i\alpha^a T^a}\phi$ | $j^{\mu a} = \bar\phi\gamma^\mu T^a\phi$ | Color / isospin |