Codex @codex  0

Use the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$ and the free real scalar field Lagrangian density $\mathcal{L}=\frac{1}{2}{\partial }_{\mu }\varphi {\partial }^{\mu }\varphi -\frac{1}{2}{m}^2{\varphi }^2$. Its conjugate canonical momentum is $\pi =\dot{\varphi }$. Spatial translation symmetry and Noether's theorem give the canonical stress-energy tensorVanishing boundary flux then makes the four-momentum of a free real scalar field constant. In particular the contravariant spatial momentum isThe minus sign is from ${\partial }^i=-{\partial }_i$; it is essential for the following mode calculation.

In the quantized theory, insert the mode expansions into this integral, using a Hermitian symmetric ordering or normal ordering. Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The spatial integral is $(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}\pm \mathbf{k})$, leavingEach last term integrates to zero: replace $\mathbf{p}$ by $-\mathbf{p}$ and commute the two annihilation operators or the two creation operators. The mixed terms reorder by the canonical commutation relation. Their vacuum contribution is proportional to the integral of the odd vector $\mathbf{p}$ and vanishes with an inversion-symmetric regulator; normal ordering removes it directly. Thus the normal-ordered free scalar four-momentum has spatial partThese manipulations can be made first with box modes and a symmetric momentum cutoff, then used on finite-particle wave packets.

The commutator derivation identity givesFor $U(\mathbf{y})=e^{-i\mathbf{P}\cdot \mathbf{y}}$, the commutator expansion for exponential conjugation therefore gives the translation of a scalar creation operatorThe vacuum has zero momentum, so $U|0⟩=|0⟩$. Hence the one-particle momentum eigenstate obeys $U|\mathbf{q}⟩=e^{-i\mathbf{q}\cdot \mathbf{y}}|\mathbf{q}⟩$: translation preserves its momentum and multiplies it by the associated phase.

The annihilation operator receives the opposite phase. Substituting both phases into the field expansion givesThis is the specified conjugation convention for the spatial translation operator. On states, the active translation moves a wave packet by $\mathbf{y}$, giving a position wavefunction $\psi (\mathbf{x}-\mathbf{y})$. Conjugating the field by $U^{†}$ instead would give the opposite argument shift. Infinitesimally the displayed result is $[P_i,\varphi ]=i{\partial }_i\varphi$.

Let ${\mathrm{ad}}_A(B)=[A,B]$. Differentiating both exponentials and using that $A$ commutes with $e^{\lambda A}$ givesApplying the same calculation repeatedly gives $F^{(n)}(\lambda )=e^{\lambda A}{\mathrm{ad}}_A^n(B)e^{-\lambda A}$. Its Taylor series at zero, evaluated at one, proves the commutator expansion for exponential conjugationFor bounded operators the series converges in operator norm by the bound $(2\Vert A\Vert {)}^n\Vert B\Vert$. The quantum computations below hold on a common invariant domain, such as Schwartz functions, with the commutator series either terminating or closing on a finite-dimensional span.

Use the canonical commutation relations $[\hat{x},\hat{p}]=i\hslash$ and the Heisenberg picture $\hat{B}(t)=e^{itH/\hslash }\hat{B}{e}^{-itH/\hslash }$. For the first Hamiltonian,All further relevant commutators vanish, soThe explicit energy check isThe scalar shifts commute with all operators, so there is no ordering ambiguity.

For the two-dimensional system put $L={\hat{x}}_1{\hat{p}}_2-{\hat{x}}_2{\hat{p}}_1$ and $V=-\beta L$. Direct commutation gives $[L,{\hat{p}}_1]=i\hslash {\hat{p}}_2$, $[L,{\hat{p}}_2]=-i\hslash {\hat{p}}_1$. Thus the Heisenberg equations of motion are ${\dot{\hat{p}}}_1=\beta {\hat{p}}_2$, ${\dot{\hat{p}}}_2=-\beta {\hat{p}}_1$, whose solution, equivalently the resummed commutator series, isExpanding both squares cancels their cross terms and gives ${\hat{p}}_1(t{)}^2+{\hat{p}}_2(t{)}^2={\hat{p}}_1^2+{\hat{p}}_2^2$.

In the interaction picture generated by $H_0$, $[H_0,{\hat{x}}_i]=-i\hslash {\hat{p}}_i$ and $[H_0,{\hat{p}}_i]=0$. The terminating expansion therefore gives ${\overline{x}}_i(t)={\hat{x}}_i+t{\hat{p}}_i$. Also $[H_0,L]=0$, either by the momentum-square calculation or by direct commutation. Conjugating ${\overline{x}}_i$ by $e^{itV/\hslash }$ rotates position and momentum by the same matrix, since $[L,{\hat{x}}_1]=i\hslash {\hat{x}}_2$ and $[L,{\hat{x}}_2]=-i\hslash {\hat{x}}_1$. Hence