Quantum Field Theory Course

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Part I, Chapter 2 | Page 1 of 6

Noether's Theorem

Symmetries and conserved currents in field theory

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Lecture 2: Symmetries and Conservation Laws - MIT 8.323

Noether's theorem and conserved currents in field theory (MIT QFT Course)

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2.1 The Power of Symmetry

Noether's theorem (1918) is one of the most profound results in theoretical physics. It establishes a deep connection between continuous symmetries of the action and conserved quantities.

Noether's Theorem (Statement)

For every continuous symmetry of the action, there exists a corresponding conserved current Jμsatisfying the continuity equation:

$$\partial_\mu J^\mu = 0$$

Examples of Symmetries

  • Time translation invariance → Energy conservation
  • Spatial translation invariance → Momentum conservation
  • Rotation invariance → Angular momentum conservation
  • U(1) phase rotation → Electric charge conservation
  • Lorentz invariance → Conservation of angular momentum tensor

This theorem tells us that conservation laws are not independent principles—they are consequences of symmetries.

2.2 Infinitesimal Symmetry Transformations

Consider an infinitesimal transformation of spacetime coordinates and fields:

$$x^\mu \to x'^\mu = x^\mu + \delta x^\mu$$
$$\phi(x) \to \phi'(x') = \phi(x) + \delta \phi(x)$$

where δxμ and δφ are infinitesimal. This transformation is a symmetry if the action is invariant (up to boundary terms that vanish):

$$\delta S = S[\phi'] - S[\phi] = 0$$

Total Variation

The total variation of the field has two parts:

$$\Delta \phi = \phi'(x') - \phi(x) = \phi'(x') - \phi(x') + \phi(x') - \phi(x)$$

Define:

  • Functional variation: Change in field value at same spacetime point
    $$\delta \phi(x) = \phi'(x) - \phi(x)$$
  • Coordinate change: Due to coordinate transformation
    $$\phi(x') - \phi(x) = \delta x^\mu \partial_\mu \phi$$

Therefore, the total variation is:

$$\Delta \phi = \delta \phi + \delta x^\mu \partial_\mu \phi$$

2.3 Derivation of the Conserved Current

The action is:

$$S = \int d^4x \, \mathcal{L}(\phi, \partial_\mu \phi)$$

Under the transformation, the Lagrangian density changes as:

$$\mathcal{L}(x) \to \mathcal{L}'(x') = \mathcal{L}(x) + \delta \mathcal{L}$$

Variation of Action

The change in action has two sources:

$$\delta S = \int d^4x \left[\frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\partial_\mu(\delta \phi)\right] + \int d^4x \, \partial_\mu(\mathcal{L} \delta x^\mu)$$

The second term is a total derivative (boundary term). Using the Euler-Lagrange equations:

$$\frac{\partial \mathcal{L}}{\partial \phi} = \partial_\mu\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\right)$$

we can write:

$$\delta S = \int d^4x \, \partial_\mu\left[\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi + \mathcal{L} \delta x^\mu\right]$$

Noether Current

If the transformation is a symmetry (δS = 0), and the boundary terms vanish, then:

$$\partial_\mu J^\mu = 0$$

where the Noether current is:

$$\boxed{J^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi + \mathcal{L} \delta x^\mu}$$

For internal symmetries (δxμ = 0), this simplifies to:

$$J^\mu = \pi^\mu \delta \phi = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi$$

2.4 Conserved Charge

The continuity equation ∂μJμ = 0 implies a conserved charge (or Noether charge):

$$Q = \int d^3x \, J^0(x)$$

Proof of Conservation

Taking the time derivative:

$$\frac{dQ}{dt} = \int d^3x \, \partial_t J^0 = \int d^3x \, \partial_t J^0$$

Using the continuity equation:

$$\partial_t J^0 + \nabla \cdot \vec{J} = 0 \quad \Rightarrow \quad \partial_t J^0 = -\nabla \cdot \vec{J}$$

Therefore:

$$\frac{dQ}{dt} = -\int d^3x \, \nabla \cdot \vec{J} = -\int_{\partial V} \vec{J} \cdot d\vec{A}$$

If the current vanishes at spatial infinity (|J| → 0 as |x| → ∞), then:

$$\boxed{\frac{dQ}{dt} = 0}$$

The charge Q is conserved (time-independent).

Key Concepts (Page 1)

  • • Continuous symmetries → conserved currents (Noether's theorem)
  • • Conserved current: ∂μJμ = 0
  • • Noether current: Jμ = (∂$\mathcal{L}$/∂(∂μφ))δφ + $\mathcal{L}$δxμ
  • • Conserved charge: Q = ∫d³x J0, dQ/dt = 0
  • • Internal symmetries: δxμ = 0, spacetime symmetries: δxμ ≠ 0
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