# Noether's theorem in field theory

[2017-11-24 Fri]

## Energy and momentum in scalar field theory

Let $$\mathcal{F}$$ denote the space of fields $$\phi\colon\mathbf{R}^n\to\mathbf{R}$$. The phase space for the classical theory of scalar fields is the cotangent bundle $$T^*\mathcal{F}$$. Since $$\mathcal{F}$$ is just a vector space whose elements are field configurations $$\phi$$, its cotangent fibre is also a vector space of functions $$\pi$$ (technically, this should probably be a space of distributions, as it's the dual of $$\mathcal{F}$$; we will ignore this issue). If $$A$$ is a vector on $$T^*\mathcal{F}$$ then it has a component $$A_\phi$$ in the $$\mathcal{F}$$-directions and a component $$A_\pi$$ in the cotangent directions. We can think of $$A_\phi$$ and $$A_\pi$$ as functions. The symplectic structure $$\Omega$$ on $$T^*\mathcal{F}$$ is given by $\Omega(A,B)=\int d^nx(A_\phi(x)B_\pi(x)-A_\pi(x)B_\phi(x)).$ Given a function $$H\colon T^*\mathcal{F}\to\mathbf{R}$$, we can write down a corresponding Hamiltonian vector field $$A$$; if $$\delta H(B)$$ denotes the directional (functional) derivative of $$H$$ in the $$B$$-direction then $$A$$ is defined by Hamilton's equations $\Omega(A,B)=-\delta H(B),$ in other words $\delta H(B_\phi)=\int d^nx A_\pi B_\phi,\quad\delta H(B_\pi)=-\int d^nx A_\phi B_\pi.$

Consider the function $$P_i(\phi,\pi)=\int d^nx\pi(x)\partial_i\phi(x)$$. We have $\delta P_i(B)=\int d^nx(B_\pi(x)\partial_i\phi(x)+\pi(x)\partial_i(B_\phi(x))),$ so the Hamiltonian vector field has components $$A_\phi(x)=\partial_i\phi(x)$$ and $$A_\pi(x)=\partial_i\pi(x)$$. We can integrate this Hamiltonian vector field to find a Hamiltonian $$\mathbf{R}$$-action on $$T^*\mathcal{F}$$. Indeed, consider the $$\mathbf{R}$$-action on $$\mathcal{F}$$ given by sending a function $$\phi$$ to $$\phi_t(x)=\phi(x+te_i)$$ (where $$e_i$$ is the $$i^{th}$$ basis vector in $$\mathbf{R}^n$$). To first order, this action is $\phi_t(x)=\phi(x)+t\partial_i\phi(x)+\mathcal{O}(t^2),$ so the vector field generating this action is $$\partial_i\phi$$. A group action on $$\mathcal{F}$$ induces a canonical group action on $$T^*\mathcal{F}$$ by symplectomorphisms and the corresponding vector field on $$T^*\mathcal{F}$$ is $$(\partial_i\phi,\partial_i\pi)$$, which is precisely the Hamiltonian vector field $$A$$ associated to $$P_i$$. In other words, $$P_i$$ is the momentum associated to the $$\mathbf{R}$$-action which changes functions (fields) by translating their argument in the $$e_i$$-direction.

Consider the function $$H(\phi,\pi)=\frac{1}{2}\int d^nx(\pi^2+|\nabla\phi|^2+m^2\phi^2)$$. We have \begin{align*} \delta H(B)&=\int d^nx(\pi B_pi+\nabla\phi\cdot \nabla B_\phi+m^2\phi B_\phi)\\ &=\int d^nx(\pi B_\pi+(m^2\phi-\nabla^2\phi)B_\phi), \end{align*} so the Hamiltonian vector field $$A=(A_\phi,A_\pi)$$ is $$(\pi,\nabla^2\phi-m^2\phi)$$. The flow of this vector field is given by $$(\phi(t,x),\pi(t,x))$$ satisfying $\dot{\phi}=\pi,\qquad\ddot{\phi}=\dot{\pi}=\nabla^2\phi-m^2\phi,$ in other words, by a solution of the Klein-Gordon equation $\ddot{\phi}-\nabla^2\phi=-m^2\phi.$

## Internal symmetries

Consider a free complex scalar field $$\psi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2)$$. The Hamiltonian which describes the time evolution of this system is $H=\int d^nx\left(|\pi|^2+\nabla\psi^*\cdot\nabla\psi+m^2|\psi|^2\right).$ Writing this out in terms of $$\phi_1$$ and $$\phi_2$$ we get the Hamiltonan for two free Klein-Gordon fields with mass $$m$$: $H=\sum_{k=1}^2\frac{1}{2}\int d^nx\left(\pi_k^2+|\nabla\phi_k|^2+m^2\phi_k^2\right).$ Here, $$\pi_1$$ and $$\pi_2$$ are the conjugate momenta to $$\phi_1$$ and $$\phi_2$$. One slightly confusing aspect of this is that $$\pi=\frac{1}{\sqrt{2}}(\pi_1-i\pi_2)$$.

Consider the circle action on the space of fields where $$e^{i\alpha}\in U(1)$$ acts on $$\psi$$ to produce $$e^{i\alpha}\psi$$; equivalently, $$e^{i\alpha}$$ acts on $$(\phi_1,\phi_2)$$ to produce $$(\phi_1\cos\alpha-\phi_2\sin\alpha, \phi_1\sin\alpha+\phi_2\cos\alpha)$$. This action induces an action by symplectomorphisms on $$T^*\mathcal{F}$$: $\left(\begin{array}{c}\phi_1 \\ \phi_2 \\ \pi_1 \\\pi_2\end{array}\right) \stackrel{e^{i\alpha}}{\to} \left(\begin{array}{c}\phi_1\cos\alpha-\phi_2\sin\alpha \\ \pi_1\sin\alpha+\pi_2\cos\alpha \\ \pi_1\cos\alpha-\pi_2\sin\alpha \\ \pi_1\sin\alpha+\pi_2\cos\alpha\end{array}\right).$ The infinitesimal action (i.e. the vector field on $$T^*\mathcal{F}$$ generating this action) is $V=(-\phi_2,\phi_1,-\pi_2,\pi_1).$

This vector field $$V$$ is the Hamiltonian vector field associated to the Hamiltonian $G=\int d^nx\left(\pi_1\phi_2-\pi_2\phi_1\right).$ In terms of $$\psi=\frac{1}{\sqrt{2}}(\phi_1+i\phi_2)$$ and $$\pi=\frac{1}{\sqrt{2}}(\pi_1-i\pi_2)$$, this is $$G=-i\int d^nx\left(\pi\psi-\pi^*\psi^*\right)$$, or, using the fact that under the Hamiltonian flow of $$H$$ we have $$\pi=\dot{\psi}^*$$, $$G=i\int d^nx\left(\dot{\psi}\psi^*-\dot{\psi}^*\psi\right)$$. If you are a physicist, this is the usual expression for the conserved charge associated with the $$S^1$$-action (modulo a sign?).
We have \begin{align*} \delta G(W)&=\int d^nx\left(-\pi_2W_{\phi_1}+\pi_1W_{\phi_2} +\phi_2W_{\pi_1}-\phi_1W_{\pi_2}\right)\\ &=-\Omega(V,W), \end{align*} so $$V$$ is the Hamiltonian vector field associated to $$G$$.

## Noether currents

Up till now, everything we have said is just like in the case of finite-dimensional Hamiltonian systems: when we have a symplectic phase space, Hamiltonian functions give rise to Hamiltonian symmetries of the phase space and vice versa. Now we come to a phenomenon which only really makes sense in the field theory setting: the idea of Noether currents.

We start with a Hamiltonian $$H=\int d^nx\mathcal{H}$$ which generates the time-evolution of our system. Suppose we have another function $$G=\int d^nx\mathcal{G}$$ which generates a Hamiltonian flow that commutes with $$H$$, in other words $$\{H,G\}=0$$. An example would be the free complex scalar field $$\psi$$ with its free Hamiltonian and the circle action rotating the phase of $$\psi$$. The claim is that, not only does the time derivative of $$G$$ vanish along the flow of $$H$$, but the time derivative of $$\mathcal{G}$$ is equal to the divergence of a vector field $$J$$ on space. This is a stronger assertion: if you integrate $$\dot{\mathcal{G}}=\nabla\cdot J$$ over space then you get $$\dot{G}=0$$ because the integral of a divergence is zero. Moreover, you can figure out what the vector field $$J$$ is using the time-honoured trick of integrating by parts. The 4-vector $$(\mathcal{G},J)$$ is then called the Noether current associated to the symmetry. It is not uniquely determined, since we can always add a gradient vector field to $$J$$ without changing the property that $$\nabla\cdot J=\dot{\mathcal{G}}$$. Nonetheless, there's usually a sensible-looking choice.

The key point which allows us to find these currents is the well-known fact in symplectic geometry that if you have Hamiltonian functions $$A$$ and $$B$$ (with Hamiltonian vector fields $$V$$ and $$W$$), you let $$\phi_t$$ be the Hamiltonian flow associated to $$A$$, and you consider the function $$B(\phi_t(x))$$ then $$\dot{B}=\{A,B\}=\Omega(V,W)$$. Armed with this fact, we will study the simplest example.

Let's revisit the example of the charged scalar field. Let $$\alpha(x)$$ be a function on space, and consider the modified Hamiltonian $G_\alpha=\int d^nx\alpha(x)\left(\pi_1\phi_2-\pi_2\phi_1\right).$ This no longer commutes with the Hamiltonian $$H$$. It generates the circle action which sends the field $$\psi(x)$$ to the field $$e^{i\alpha(x)}\psi(x)$$ (i.e. the phase shift depends on position). The corresponding Hamiltonian vector field is $V_\alpha=(-\phi_2\alpha,\phi_1\alpha,-\pi_2\alpha,\pi_1\alpha).$ Since $$G_\alpha$$ and $$H$$ do not commute, we can ask how $$G_\alpha$$ evolves along the flow of $$H$$. From the previous paragraph we know that $\dot{G}_\alpha=\{H,G_\alpha\}=\Omega(X,V_\alpha),$ where $$X=(\pi_1,\pi_2,\nabla^2\phi_1-m^2\phi_1,\nabla^2\phi_2-m^2\phi_2)$$ is the Hamiltonian vector field associated to $$H$$. In other words: \begin{align*} \dot{G}_\alpha&=\int d^nx\left(\pi_1(-\pi_2\alpha)+\pi_2(\pi_1\alpha)\right.\\ & \left.+(\nabla^2\phi_1-m^2\phi_1)\phi_2\alpha-(\nabla^2\phi_2-m^2\phi_2)\phi_1\alpha\right)\\ &=\int d^nx\alpha\left(\phi_2\nabla^2\phi_1-\phi_1\nabla^2\phi_2\right) \end{align*} If there were no $$\alpha$$ here then we could integrate by parts and everything would cancel; indeed, that is why $$G$$ is a conserved quantity (when $$\alpha=1$$). Note that both sides are given by an integral: $\int d^nx\alpha\dot{\mathcal{G}}=\int d^nx\alpha\left(\phi_2\nabla^2\phi_1-\phi_1\nabla^2\phi_2\right),$ and $$\alpha$$ is completely arbitrary. Therefore the integrands themselves must agree: $\dot{\mathcal{G}}=\phi_2\nabla^2\phi_1-\phi_1\nabla^2\phi_2.$ But the quantity on the right is equal to $$\nabla\cdot J$$ with $$J=\phi_2\nabla\phi_1-\phi_1\nabla\phi_2$$. Therefore we end up with $\dot{\mathcal{G}}=\nabla\cdot J,$ as required. The 4-vector $$j=(\mathcal{G},J)$$ is then $j^\mu=\phi_2\partial^{\mu}\phi_1-\phi_1\partial^{\mu}\phi_2,$ which is what I meant by sensible-looking''. The equation $$\dot{G}=\nabla\cdot J$$ can be written in a more Lorentz-invariant-looking way as $\partial_\mu j^\mu=0.$

As an exercise, show that playing the same trick with the free real scalar field $$\phi$$ and the Hamiltonian $$H_\alpha=\frac{1}{2}\int d^nx\alpha(\pi^2+|\nabla\phi|^2+m^2\phi^2)$$ yields $$\pi(x)\nabla\phi(x)$$ as the Noether current; this is the integrand of the field momentum we studied first of all.

This trick only makes sense in field theory, where our Hamiltonian functions have the form $$\int d^nx\mathcal{G}$$: otherwise we cannot insert a cheeky factor of $$\alpha$$. It also relies heavily on the equations of motion for $$G_\alpha$$, i.e. $$\dot{G}_\alpha=\{H,G_\alpha\}$$.

Comments, corrections and contributions are very welcome; please drop me an email at j.d.evans at lancaster.ac.uk if you have something to share.

CC-BY-SA 4.0 Jonny Evans.