Past exam of the theoretical physics major option course of the University of Oxford / 2022 / 5

At fixed time, the coordinate field and the canonical momentum are independent variables. Their functional derivatives satisfyPutting $F=Q$, $G={\varphi }_a(\mathbf{x})$ in the Poisson bracket definition therefore givesThe superscript on the species label $a$ in the printed denominator has no spacetime-metric significance; with the ordinary species Kronecker metric it is the same expression.

A scalar field has $\delta {\varphi }_a=0$ for a pure change of spacetime argument. With $\delta x^k={ϵ}^k$ and $\delta x^0=0$, its variation at a fixed argument is $\bar{\delta }{\varphi }_a=-{ϵ}^k{\partial }_k{\varphi }_a$. The supplied Noether charge becomesSince $\delta Q/\delta {\pi }_a=-{ϵ}^k{\partial }_k{\varphi }_a$, solution and antisymmetry of the Poisson bracket giveCanonical quantization replaces the Poisson bracket by a commutator divided by $i\hslash$:With this sign convention one may use $U=e^{-i\hat{Q}/\hslash }$ and ${\hat{\varphi }}_a^{\prime }=U^{†}{\hat{\varphi }}_{a}U$. Because $Q=-{ϵ}^k{P}_k$, the negative shift of the field argument follows with the required sign.

For $\delta x^0={ϵ}^0$ the Noether charge is $Q=-{ϵ}^{0}H$, with the Hamiltonian density $\mathcal{H}={\pi }_a{v}_a-\mathcal{L}(\varphi ,v,\nabla \varphi )$ and $v_a={\dot{\varphi }}_a$. Invertibility of the momentum-velocity relation allows $v_a=v_a(\varphi ,\pi ,\nabla \varphi )$. When taking a functional derivative at fixed $\varphi$, the implicit dependence of $v$ cancels in the Legendre transform:It follows thatThis proves the same generator relation for time translations. In a constrained theory the assumed inversion fails, so the appropriate reduced variables or Dirac brackets are needed instead. That is precisely the issue encountered for the Proca field in Question 4.

For a first-derivative Lagrangian density with no explicit spacetime dependence, introduce the canonical energy-momentum tensorDirect differentiation giveson solutions of the Euler-Lagrange equations. The four Noether currents are the four columns of this tensor, whereas their spatial integrals are the conserved four-momentum components:For $\mu =0$ this is the Hamiltonian; for a spatial lower index it is $P_k=\int {\pi }_a{\partial }_k{\varphi }_a$. Under the paper's passive-translation convention, the corresponding generator is $Q=-{ϵ}^{\mu }{P}_{\mu }$. Conservation of the integrated quantities also uses the stated vanishing boundary flux.

For the spatial momenta, integration by parts givesTheir Poisson bracket is thereforeFor the mixed case, $P_0=H$ and the spatial generator acts on both canonical fields by differentiation: $\{{\varphi }_a,P_i\}={\partial }_i{\varphi }_a$, $\{{\pi }_a,P_i\}={\partial }_i{\pi }_a$. Hence, for a translation-invariant local Hamiltonian density,Finally $\{H,H{\}}_{\mathrm{PB}}=0$ by antisymmetry. Together these proveThe proof requires commuting derivatives and no surviving boundary terms. It expresses the Abelian translation subgroup of Poincaré symmetry at the level of its canonical generators.

Write $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2{\omega }_p]$, $p^{\mu }=({\omega }_p,\mathbf{p})$ and ${\omega }_p=\sqrt{{\mathbf{p}}^2+m^2}$. The Fourier expansion in the source is compatible with the oscillator canonical commutation relationsFor the free real scalar field, $\pi =\dot{\varphi }$ andSubstitution of the Fourier expansion produces spatial Dirac delta functions. Terms with two annihilation operators or two creation operators cancel: in $P_0$ their coefficient is $-{\omega }_p^2+{\mathbf{p}}^2+m^2=0$, and in $P_i$ their integral is odd under $\mathbf{p}\mapsto -\mathbf{p}$. The two mixed terms combine, with normal ordering removing the vacuum constant, to giveIn particular, $P_i$ contains the lower component $p_i=-{\mathbf{p}}_i$ in this Minkowski metric convention.

To check commutativity without manipulating a square of a Dirac delta function, define $N_f=\int d{\Pi }_{p}f(p)a_p^{†}{a}_p$. The oscillator algebra gives $[N_f,a_q]=-f(q)a_q$ and $[N_f,a_q^{†}]=f(q)a_q^{†}$. Thus $[N_f,a_q^{†}{a}_q]=0$ and $[N_f,N_g]=0$. Taking $f=p_{\mu }$, $g=p_{\nu }$ provesThese identities hold on the usual dense domain of finite-particle states, or can first be proved in a finite box and then continued to the continuum.