2.2. Noether’s Theorem

Noether’s theorem deals with continuous symmetry.

2.2.1. Noether’s Theorem of Particles

2.2.2. Noether’s Theorem of Fields

Suppose we have a continuous transformation, which is internal, that transforms the fields according to

\[\phi_i (x^\mu) \to \phi_i (x^\mu) + \delta \phi_i(x^\mu).\]

For convenience, we explicity write the variation \(\delta \phi_i(x^\mu)\) as a continuous quantity \(\alpha\), i.e.,

\[\delta \phi_i(x^\mu) = \alpha \Delta\phi(x^\mu).\]

Noether’s theorem states that if this continuous preserves the Lagrangian, we can define conserved Noether current thus conserved charge.

Planning the Proof

1. Write down the variation of Lagrangian.
2. Combine the terms and apply the Euler-Lagrangian equation.
3. If the Lagrangian is invariant under such a continuous tranformation, blablabla.

The variation of Lagrangian is

(1)\[\delta \mathcal L(\phi_i, \dot \phi_i) = \frac{\partial \mathcal L}{\partial \phi_i} \delta \phi_i + \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \delta(\phi_{i,\mu}).\]

We know that the variation and partial derivative can be exchanged, such that

\[\delta(\phi_{i,\mu}) = \partial_\mu (\delta \phi_i).\]

We rewrite the second term in Eq. (1),

\[\begin{split}&\frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \delta(\phi_{i,\mu}) \\ = & \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \partial_\mu( \delta \phi_{i} ) \\ = & \partial_\mu\left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \delta \phi_{i} \right) - \delta \phi_{i} \partial_\mu \left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \right).\end{split}\]

Plug it back into Eq. (1), we have

\[\begin{split}\delta \mathcal L(\phi_i, \dot \phi_i) =& \frac{\partial \mathcal L}{\partial \phi_i} \delta \phi_i + \partial_\mu\left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \delta \phi_{i} \right) - \delta \phi_{i} \partial_\mu \left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \right) \\ = & \delta \phi_i \left( \frac{\partial \mathcal L}{\partial \phi_i} - \partial_\mu \left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \right) \right) + \partial_\mu\left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \delta \phi_{i} \right).\end{split}\]

The first term is zero by Euler-Lagrangian equation. Thus

\[\delta \mathcal L(\phi_i, \dot \phi_i) =\partial_\mu\left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \delta \phi_{i} \right).\]

Now we impose the condition that the Lagrangian is invariant under such a continuous transformation, so that \(\delta \mathcal L = 0\).

(2)\[\partial_\mu\left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \delta \phi_{i} \right) = 0.\]

Eq. (2) defines the constant of motion. Put the definition of \(\delta \phi_i\) back in,

(3)\[\alpha \partial_\mu\left( \frac{\partial \mathcal L}{\partial \phi_{i,\mu}} \Delta \phi_{i} \right) = 0.\]

2.2.3. Examples of Noether Current

2.2.3.1. Global Phase Transformation

For the Lagrangian

\[\mathcal L = \partial^\mu \phi^* \partial_\mu \phi - m^2 \phi^* \phi,\]

that leads to Klein-Gordon equation, a transformation

\[\begin{split}\phi \to e^{i\alpha}\phi,\\ \phi^* \to e^{-i\alpha}\phi^*,\end{split}\]

will not change the scalar particle Lagrangian.

The corresponding Noether current is defined by

\[\partial_\mu j^\mu = 0,\]

where

\[j^\mu = -i(\phi^* \partial^\mu \phi - \phi\partial^\mu \phi^*).\]

Along with the current we find the conserved charge

\[Q = \int d^3 x j^0,\]

which satisfies

\[\frac{\partial Q}{\partial t}= 0.\]

Proof

Here is the proof.

2.2.3.2. Space-time Translation

For arbitary Lagrangian \(\mathcal L(x^\mu)\) which is space-time dependent, we can calculate the action

\[S = \int d^4x \mathcal L.\]

If the action is invariant under space-time translation

\[x^\mu\to x^\mu + \alpha a^\mu,\]

we find the conserved current to be the energy-momentum tensor \(T^{\mu\nu}\)

\[T^{\mu\nu} = \frac{\partial \mathcal L}{\partial (\partial_\mu\phi)}\partial^\nu \phi - g^{\mu\nu} \mathcal L.\]

The corresponding conservation equation is

\[\partial_\mu T^{\mu\nu} = 0,\]

which defines the four charges

\[Q^\mu = \int d^3 T^{\mu\nu} .\]

Proof Energy-momentum Tensor as Noether Current

QED.

For the Lagrangian

\[\mathcal L = \frac{1}{2} \partial^\mu \phi \partial_\mu \phi - \frac{1}{2} m^2 \phi^2,\]

one can easily prove that the corresponding energy-momentum tensor is

\[T^{\mu\nu} = \partial^\mu \phi\partial^\nu \phi - g^{\mu\nu} \mathcal L.\]

Derivation of Energy-momentum for Real Scalar Lagrangian

QED.

The 00 component is in fact the Hamiltonian density \(\mathcal H\).

Prove that \(T^{00}=\mathcal H\)

Calculate \(T^{00}\),

\[\begin{split}T^{00} =& \partial^0 \phi \partial^0\phi - \mathcal L \\ =& \frac{1}{2} ( \partial^0\phi\partial^0\phi + \partial^i \phi \partial^i\phi + m^2\phi^2 ).\end{split}\]

Notice that the Hamiltonian density is

\[\mathcal H = \Pi \partial^0 \phi - \mathcal L,\]

where

\[\Pi = \frac{\partial \mathcal L}{\partial (\partial^0 \phi)} = \partial^0\phi.\]

Plug in the momentum we find

\[\mathcal H = \partial^0\phi\partial^0\phi - \mathcal L = T^{00}.\]

Dialation and Noether Current

Dilation can be written as

\[\begin{split}x_\mu \to & a x^\mu,\\ \phi \to & a^{-1} \phi.\end{split}\]

The Noether current corresponding to such transformation is

\[j_{\mathrm D}^{\mu} = T^{\mu\rho}x_\rho + \frac{1}{2} \partial^\mu \phi^2.\]

Notice that Lagrangian

\[\mathcal L = \frac{1}{2} \partial_\mu \phi \partial^\mu \phi - \frac{1}{4\,! } \lambda \phi^4,\]

which is \(\phi^4\) theory, is invariant under dilation.

2.2.4. References and Notes