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 if you have something to share.

CC-BY-SA 4.0 Jonny Evans.