Noether's theorem in field theory
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. \(\dot{G}_\alpha=\{H,G_\alpha\}\).