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Let $A=Dy(x)$ be the Jacobian matrix and $K(y)=H(x(y))$. The chain rule gives ${\nabla }_{x}H=A^T{\nabla }_{y}K$. Therefore the Hamilton equations transform to $\dot{y}=AJ{A}^T{\nabla }_{y}K$, with $J=\left(\begin{array}{cc}0& -I\\ I& 0\end{array}\right)$. They have the required canonical form for every Hamiltonian exactly whenNecessity follows by choosing Hamiltonians whose gradients take arbitrary values at a point; sufficiency follows from the displayed transformed equation. This is the symplectic matrix condition for a canonical transformation, and it also forces $A$ to be invertible.

For the given transformation, $A=\left(\begin{array}{cc}1& 2\\ -1/4& 1/2\end{array}\right)$ has determinant one. For any $2\times 2$ matrix, $AJ{A}^T=(\det A)J$, so the proposed transformation is canonical.

With Hamilton's equations ${\dot{q}}_j=H_{p_j}$ and ${\dot{p}}_j=-H_{q_j}$, the chain rule givesFor the vortex problem, combine the two ordered terms for each unordered pair, writing $D_{ij}=(p_i-p_j{)}^2+(q_i-q_j{)}^2$ and $H=-(\kappa /2){\sum }_{i<j}\log D_{ij}$. Since $F_{q_j}=2q_j$, $F_{p_j}=2p_j$, its pairwise contribution to the Poisson bracket isEvery pair cancels, so $[F,H]=0$ and ${\sum }_j(q_j^2+p_j^2)$ is conserved along any collision-free solution. The logarithmic Hamiltonian is undefined at coinciding vortices, so the derivation applies on its natural domain.

Use the maximizing form of the Pontryagin maximum principle. For a cost-minimization problem there are a nonnegative multiplier ${\lambda }_0$ and a costate $\psi$, not both identically zero, with Hamiltonian $\mathcal{H}=\psi b-{\lambda }_{0}c$. The optimal control maximizes $\mathcal{H}$, the costate obeys $\dot{\psi }=-{\mathcal{H}}_x$, and free terminal time at the fixed terminal state zero gives $\mathcal{H}(\tau )={\lambda }_0{C}^{\prime }(\tau )$. There is no fixed terminal costate value since the terminal state is fixed. In the normal case scale ${\lambda }_0=1$.

For the energy problem $b=-u$, $c=up(t)$, $C=-R$, so $\psi$ is constant and the normal Hamiltonian is $u(p_{\ast }-p(t))$, where $p_{\ast }=-\psi$. Maximization gives full power for $p(t)<p_{\ast }$ and zero power for $p(t)>p_{\ast }$; either value, or a fractional value, is allowed on a price tie. On the final active arc, the free-time condition givesThis is the threshold energy scheduling with a completion reward rule. In this example the normal case is justified: an abnormal constant costate would require either no processing or processing at all times, and free-time transversality would then force that nonzero costate to vanish.

For $p_{\ast }=1/4$, processing starts at $t=1/2$. Energy $1/2$ is completed at $\tau =1$, before the price rises again. Its cost and reward areAny threshold carrying completion into day two has $\tau \ge 2$ and reward at most $1/3$. Since energy cost is nonnegative, $J\ge -1/3>-11/24$, so it is suboptimal.

For a global check, even any completion time $\tau \ge 5/4$ has reward at most $4/9<11/24$ and cannot beat the displayed candidate. Also $\tau \ge 1/2$ since power is bounded by one. At a fixed $\tau \in [1/2,5/4]$, the cheapest half-unit of processing consists of the last half-unit interval $[\tau -1/2,\tau ]$: every earlier point has a price at least as large as every selected point. Exchanging any earlier powered interval with a later unpowered cheaper one proves this rearrangement claim, also for fractional controls.

Write $h=3/2-\tau \in [1/4,1]$. This minimal cost minus reward iswithIt has its unique minimum at $h=1/2$, where $J^{\prime }=0$. The selected interval is then $[1/2,1]$ and its threshold is $h^2$. ConsequentlyThe terminal condition also checks: $1/4+R^{\prime }(1)=1/4-1/4=p(1)=0$.

For a time-independent reference Hamiltonian, define interaction-picture states and observables byThe transformation is unitary, so matrix elements and expectation values are unchanged; the two pictures make the same physical predictions. Differentiating the state and using the Schrödinger equation cancels the $H_0$ term and gives the interaction picture evolutionIntegrating once and replacing the evolving state in the integral by its initial state gives $|{\psi }_I(t)⟩=|n⟩-(i/\hslash ){\int }_0^t{V}_I(t^{\prime })|n⟩d{t}^{\prime }+O(V^2)$. For $m\ne n$, the first-order transition amplitude therefore yields

For the oscillator, $a|n⟩=\sqrt{n}|n-1⟩$ and $a^{†}|n⟩=\sqrt{n+1}|n+1⟩$. Only neighboring levels have a first-order amplitude. Write $\Delta =\nu -\omega$. The two phase integrals are ${\int }_0^t{e}^{i\Delta t^{\prime }}d{t}^{\prime }$ and its conjugate, whose squared modulus is $4{\sin }^2(\Delta t/2)/{\Delta }^2$. To order ${\lambda }^2$,At resonance take their continuous limits, ${\lambda }^{2}n{t}^2$ and ${\lambda }^2(n+1)t^2$. All other distinct-level probabilities vanish to this order. Normalization gives $P_{n\to n}=1-4{\lambda }^2(2n+1){\sin }^2(\Delta t/2)/{\Delta }^2$ to this order, and the downward probability is zero when $n=0$.

The allowed transition rate is proportional to $t{\mathrm{sinc}}^2(\Delta t/2)$, so its central peak has width of order $1/t$ about $\nu =\omega$. Distributionally, $4{\sin }^2(\Delta t/2)/(t{\Delta }^2)\to 2\pi \delta (\Delta )$. Thus both allowed transitions become sharply resonant at the oscillator frequency. Near the peak, perturbation theory still requires ${\lambda }^2(n+1)t^2\ll 1$; its coherent $t^2$ growth cannot be extrapolated indefinitely at fixed coupling.

A periodic lattice Hamiltonian commutes with lattice translations. The Bloch theorem therefore classifies its states by characters of the translation group:where $u$ is lattice periodic. For each crystal momentum $\mathbf{k}$, solving on a primitive cell with the corresponding quasiperiodic boundary condition gives discrete eigenvalues indexed by $n$. Their continuous dependence on $\mathbf{k}$ produces the energy bands $E_n(\mathbf{k})$ as $\mathbf{k}$ ranges through the Brillouin zone. Bragg mixing opens gaps at suitable degeneracies, although bands can overlap in three dimensions and gaps are not guaranteed between every pair.

The quantum numbers are band index, wavevector modulo reciprocal lattice vectors, and spin where spin is conserved. In a finite crystal the allowed wavevectors are discrete; in the thermodynamic limit they become dense and the band description emerges. Spin-orbit coupling can replace a separately conserved spin label by an appropriate internal band label.

Dilute randomly placed impurities break exact translation symmetry, so crystal momentum is no longer an exact quantum number. Most extended states retain approximately the host band structure but acquire impurity scattering and disorder broadening. An impurity potential can also produce impurity levels in a crystal, often in a band gap: levels associated with a donor impurity lie near the conduction-band edge and can supply electrons, while those associated with an acceptor impurity lie near the valence-band edge and can create holes. The impurity's species and potential determine whether such levels exist and how deep they are; not every impurity necessarily makes a gap state. At low density these levels are nearly isolated, while overlap at increasing density can broaden them into an impurity band. Added or removed carriers shift the chemical potential as well.

Hamilton's equations are ${\dot{q}}_j=\partial H/\partial p_j$ and ${\dot{p}}_j=-\partial H/\partial q_j$. The Arnold-Liouville theorem states that $n$ functionally independent first integrals in involution on a $2n$-dimensional symplectic manifold give local integrability; a compact connected regular common level is an $n$-torus. Near such a torus there are action-angle variables $(I,\vartheta )$ with $H=H(I)$, $\dot{I}=0$ and $\dot{\vartheta }=\partial H/\partial I$.

For the displayed diagonal oscillator Hamiltonian set $E_k=(p_k^2+{\omega }_k^2{q}_k^2)/2$. Each is constant, since ${\dot{q}}_k=p_k$ and ${\dot{p}}_k=-{\omega }_k^2{q}_k$. Different $E_k$ depend on disjoint canonical pairs, so their Poisson brackets vanish. Their differentials are independent wherever every $d{E}_k$ is nonzero, an open dense set; this proves complete integrability. For nonzero frequencies put ${\Omega }_k=|{\omega }_k|$. A positive-energy level in each pair is an ellipse with semiaxes $\sqrt{2E_k}$ and $\sqrt{2E_k}/{\Omega }_k$. Its enclosed area is $2\pi E_k/{\Omega }_k$, so the action variables areFor the usual positive oscillator frequencies this is $I_k=E_k/{\omega }_k$. If a printed “constant” frequency is allowed to be zero, that pair is a free particle: $p_k$ is a commuting integral, but its generic level is noncompact and has no closed-cycle oscillator action. Complete integrability persists, while the torus/action formula requires nonzero frequencies.

Assume the velocity-to-momentum Legendre map is locally invertible. Differentiate the Hamiltonian while regarding $q,p,t$ as independent:The velocity variations cancel because $p_i=\partial L/\partial {\dot{q}}_i$. Combining the remaining coefficients with the Euler-Lagrange equations gives Hamilton's equationsAn ignorable coordinate does not occur explicitly in the Lagrangian, so its conjugate momentum is conserved. For the spherical motion, $\varphi$ is ignorable and the Hamiltonian has no explicit time dependence. The two constants are $p_{\varphi }$ and $H$.

The polar Hamilton equations areA constant polar angle therefore requires $p_{\theta }=0$ and vanishing of the second right-hand side. For $0<\theta <\pi$ with $\cos \theta \ne 0$, this isThe expression must be nonnegative to represent a real momentum. Once it holds, $\varphi$ advances uniformly at $p_{\varphi }/(m{a}^2{\sin }^2\theta )$, so it is also sufficient with those initial data. At the equator the undivided equation instead requires $V^{\prime }(\pi /2)=0$, with any $p_{\varphi }$; dividing by cosine would wrongly discard this special case. The polar coordinate chart itself excludes the poles.

Put $\pi =p-eA/c=m\dot{q}$. The Hamilton equations areDifferentiate the mechanical momentum and substitute:With $B=\nabla \times A$ and $E=-c^{-1}{\partial }_{t}A$ in this zero-scalar-potential gauge, this is $\boxed{m\ddot{q}=eE+(e/c)\dot{q}\times B}$, the Lorentz force in the units of the Hamiltonian.

For $A=(-y{B}_0(z,t),0,0)$, $x$ is cyclic, so $p_x$ is conserved. Choose $p_x=0$, giving $\dot{x}=\Omega y$, where $\Omega =e{B}_0/(mc)$. Since $p_y=m\dot{y}$, its Hamilton equation gives ${\dot{p}}_y=-(e/c)B_0\dot{x}=-m{\Omega }^{2}y$, henceFor nonzero constant $B_0$, the $(y,p_y)$ orbit is an ellipse of semiaxes $\sqrt{2E^{\prime }/(m{\Omega }^2)}$ and $\sqrt{2m{E}^{\prime }}$. Its enclosed phase-space area is $2\pi E^{\prime }/|\Omega |$, so the action variable isThe absolute value keeps the action positive if the signed charge or magnetic field is negative.

When the field varies slowly along the trajectory, with $|d\Omega /dt|\ll {\Omega }^2$ and no zero-frequency crossing, adiabatic invariance of the action makes $I$ approximately constant. Thus $E^{\prime }\simeq I|\Omega (z,t)|$ and the slow motion has effective HamiltonianThe effective potential is the second term. Direct averaging checks its force: the exact equation is $m\ddot{z}=-m\Omega {\partial }_z\Omega y^2$, while the oscillator has $⟨y^2⟩=E^{\prime }/(m{\Omega }^2)$. Hence $m\ddot{z}=-I{\partial }_z|\Omega |$. Spatial field variation along the slow motion must satisfy the same adiabatic condition; slow explicit time variation alone does not control passage through a sharp spatial gradient. A time-dependent effective potential need not conserve total energy.

The fixed terminal height makes $-y(T)+T=T-1$, so this is a minimum-time problem. An optimal path can be taken entirely in $y\ge 1/2$. To justify the positive-$y$ calculation despite the absolute value in the dynamics, the time of a spatial path is its length in the metric $ds/\sqrt{|y|}$. Reflection to $|y|$ preserves this time; clipping the reflected height below $1/2$ to $1/2$ leaves the endpoints unchanged and cannot increase time, since it decreases vertical length and increases the speed available. Any excursion below $1/2$ strictly wastes time. Thus an optimum does not meet $y=0$.

Use $H=\sqrt{y}(pu+qv)-1$. Its maximum is $\sqrt{y}\sqrt{p^2+q^2}-1$, attained atThe costate equations and free-terminal-time transversality condition giveThe last identity holds throughout because the maximized autonomous Hamiltonian is constant. An abnormal extremal would instead have maximized Hamiltonian $\sqrt{y}\sqrt{p^2+q^2}=0$, forcing both costates to vanish and contradicting their nontriviality. Thus the normal choice is valid. Also $p\ne 0$, since otherwise $x$ cannot reach one; in fact $p>0$ because $\dot{x}$ has its constant sign.

Now express both requested derivatives using the costates:Consequently the optimal trajectory satisfiesThe maximizing controls above and the fixed endpoints determine the relevant extremal; the displayed equation is the requested necessary trajectory equation.

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

Take the floor at $z=0$, gas-particle mass $m$, $N$ atoms and box side $L$. The one-particle Hamiltonian is $p^2/(2m)+mgz$. With $\beta =1/(k_{B}T)$ and horizontal area $A=L^2$, the partition function isThis follows by independent Gaussian momentum integrals and the elementary height integral. Taking $E=-{\partial }_{\beta }\log Z_N$ givesFor an arbitrarily tall box compared with the gravitational scale height, $L\gg k_{B}T/(mg)$, the upper-wall correction is negligible andThe energy includes gravitational potential relative to the floor; the kinetic part alone is $3N{k}_{B}T/2$. Retaining the finite-height correction also gives the ordinary ideal-gas limit as $g\to 0$.

Write $H=F(P,Q)$ with $P=\sum p_k^2$ and $Q=\sum q_k^2$. The angular momentum component $L_i={ϵ}_{ijk}{q}_j{p}_k$ has ${\partial }_{p_{\ell }}{L}_i={ϵ}_{ij\ell }{q}_j$ and ${\partial }_{q_{\ell }}{L}_i={ϵ}_{i\ell k}{p}_k$. Thereforesince a symmetric product contracts to zero with the antisymmetric Levi-Civita symbol. All three components of angular momentum are conserved by this rotationally invariant Hamiltonian.

Use ${\eta }_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =c=1$. In the interaction picture, fields evolve with the free Hamiltonian, while interactions enter the state evolution through the Dyson series. Write the Klein-Gordon field as $\varphi ={\varphi }^{(+)}+{\varphi }^{(-)}$, with the positive-frequency part of a quantum field containing annihilation operators and ${\varphi }^{(-)}=({\varphi }^{(+)}{)}^{†}$ containing creation operators. The mode normalization in the PDF corresponds to $[a_{\mathbf{p}},a_{\mathbf{q}}^{†}]=(2\pi {)}^3{\delta }^3(\mathbf{p}-\mathbf{q})$ and $E_{\mathbf{p}}=\sqrt{|\mathbf{p}{|}^2+m^2}$.

The time-ordered product places fields with the latest time on the left. For two bosonic insertions,The normal-ordered product $:\varphi (x_1)\cdots \varphi (x_n):$ instead moves every creation operator to the left of every annihilation operator, without adding the commutators generated by that rearrangement. Its vacuum expectation value is zero whenever a nonempty set of free-field factors remains.

The mode commutator gives the positive-frequency Wightman functionDefine the Feynman propagator here to be the two-point function itself, rather than that function divided by $i$. ThenEquivalently,The positive-energy pole is below and the negative-energy pole above the real axis. Closing the energy contour integral below for positive time separation and above for negative time separation gives the two terms above. Thus this is the scalar Feynman propagator pole prescription, and $(\square +m^2){\Delta }_F=-i{\delta }^4$. In particular, $T\{\varphi (x)\varphi (y)\}=:\varphi (x)\varphi (y):+{\Delta }_F(x-y)$: the difference between time ordering and normal ordering is the Wick contraction.

For $n$ free bosonic fields, the Wick theorem stateswhere $P$ runs over all sets of disjoint pairs of indices, including the empty set, and $V(P)$ is the set of paired indices. The empty normal-ordered product equals one. This is an operator identity for free fields, understood with smearing or a regulator when products of operator-valued distributions require it.

Here is an inductive proof of the scalar Wick theorem. Sort the arguments chronologically so that $x_1^0$ is the latest. For any remaining set $I$, commuting the annihilation part of ${\varphi }_1$ through the creation parts in a normal-ordered product givesThe creation part of ${\varphi }_1$ is already in the correct position. Every commutator on the right is a scalar, so no further operator rearrangement is needed for it. Because $x_1$ is latest, that commutator is ${\Delta }_F(x_1-x_j)$. Assume the Wick theorem for the other $n-1$ fields and multiply its expansion by ${\varphi }_1$. The first term of the displayed identity gives every pairing in which index 1 is unpaired; the sum gives every pairing in which 1 is paired with exactly one $j$. Each final partial pairing has exactly one such construction. The one-field identity supplies the base case, proving the theorem by mathematical induction. Applying time ordering restores arbitrary argument order.

Taking the vacuum expectation value retains only complete pairings. Odd free correlators vanish; for example the free scalar four-point function is ${\Delta }_{12}{\Delta }_{34}+{\Delta }_{13}{\Delta }_{24}+{\Delta }_{14}{\Delta }_{23}$. These facts explain why propagators are the edges of Feynman diagrams.

For the interacting quartic scalar field theory, the interaction Hamiltonian density is $\lambda {\varphi }^4/4!$. With a single time ordering acting on external insertions and interaction vertices, the unnormalized numerator expands asApply the proved Wick theorem term by term. Each complete pairing connects external insertions and four-legged vertices by Feynman propagators. There are $4!$ assignments of four distinct incident legs at a vertex, canceling its factorial. Permuting identical integrated vertices cancels the corresponding part of $v!$. Permutations still leaving the labeled graph unchanged give its Feynman-diagram symmetry factor. For example,whereas the two-point tadpole diagram has coefficient $-i\lambda /2$, because choosing the two vertex legs contracted with the labeled external fields gives $4\cdot 3$ possibilities and the other two contract with each other.

Thus the position-space rules attach ${\Delta }_F$ to every edge, $-i\lambda \int d^{4}z$ to every four-legged interaction vertex, and divide by the graph's Feynman-diagram symmetry factor. Fourier transformation gives the momentum-space Feynman rules:Enforce momentum conservation at each vertex, retaining the overall four-momentum delta function. External lines of these correlation functions remain unamputated. The unnormalized expression includes Vacuum Feynman diagrams; dividing by $N_0=⟨0|S|0⟩$ cancels those disconnected vacuum factors. They should not be discarded without that normalization.

There is a time-ordering convention in the printed correlation-function notation worth making explicit. The usual interacting correlator uses the single global time ordering in $N_n$ above. If the displayed $T\{{\varphi }_1\cdots {\varphi }_n\}S$ is read literally with $S$ outside that ordering, the external block and the vertex block are separately time ordered: their cross-contractions are $D^{+}(x_i-z)$ because each external field stands to the left of every vertex field. The same pairing construction then uses ${\Delta }_F$ for lines within either block and $D^{+}$ for lines between blocks; in momentum space the latter is $2\pi \theta (p^0)\delta (p^2-m^2)$. This literal product is different from the conventional globally time-ordered interacting correlator, and does not have the ordinary external Feynman propagators derived above.

Use the free Maxwell Lagrangian $\mathcal{L}=-F_{\mu \nu }{F}^{\mu \nu }/4$ in mostly-minus signature. In terms of the spatial electromagnetic four-potential, define $\mathbf{E}=-\dot{\mathbf{A}}-\nabla A_0$ and $\mathbf{B}=\nabla \times \mathbf{A}$. Thus $\mathcal{L}=({\mathbf{E}}^2-{\mathbf{B}}^2)/2$. Taking the Cartesian components of $\mathbf{A}$ as coordinates gives the canonical momentaThis sign convention treats $A_i$ here as Cartesian spatial-vector labels; it is not a claim that covariant and contravariant four-vector components coincide.

The temporal momentum vanishes because no ${\dot{A}}_0$ occurs. This is the primary momentum constraint of the electromagnetic potential, so one cannot treat all four components as independent unconstrained oscillators. The Legendre transformation for the spatial components gives ${\Pi }_i{\dot{A}}_i-\mathcal{L}=({\mathbf{\Pi }}^2+{\mathbf{B}}^2)/2-\mathbf{\Pi }\cdot \nabla A_0$. After integration by parts, the Hamiltonian isSurface terms vanish for decaying fields or the corresponding regulated boundary conditions. The total constrained Hamiltonian also contains an arbitrary multiplier of ${\Pi }^0$. Preserving ${\Pi }^0=0$ under time evolution requireswhich is the Gauss law constraint in gauge theory. Both constraints are first-class constraints; they express the gauge invariance $\mathbf{A}\mapsto \mathbf{A}+\nabla \alpha$, $A_0\mapsto A_0-\dot{\alpha }$. The electric and magnetic fields are unchanged, so different gauge potentials need not describe different physical states.

One consistent canonical quantization of the electromagnetic field first reduces this redundancy. Choose Coulomb gauge $\nabla \cdot \mathbf{A}=0$. In the absence of charges, the Gauss law constraint in gauge theory gives ${\nabla }^2{A}_0=0$; decay at infinity sets $A_0=0$. Residual spatially constant gauge parameters do not change $\mathbf{A}$. The remaining variables are the transverse pairs ${\mathbf{A}}^T,{\mathbf{\Pi }}^T={\dot{\mathbf{A}}}^T$. Their gauge reduction of the Maxwell Hamiltonian isThe two transverse coordinates for each nonzero momentum are the physical degrees of freedom. Treating the constraints as operator identities while retaining unconstrained componentwise commutators would be inconsistent: differentiating $[A_i^T,{\Pi }_j^T]=i{\delta }_{ij}{\delta }^3$ would give a nonzero right side even though ${\partial }_i{A}_i^T=0$.

The reduced Dirac bracket, promoted to a canonical commutation relation, instead gives the transverse equal-time commutator:whereIndeed the spatial transverse projector of a vector field $P_{ij}^T={\delta }_{ij}-k_i{k}_j/|\mathbf{k}{|}^2$ is symmetric, has $P^T\mathbf{k}=0$ and $(P^T{)}^2=P^T$. Projecting both entries of the unconstrained canonical bracket onto the transverse subspace therefore produces $P^T{\delta }^3$, equivalently the constrained Dirac bracket. The spatial zero mode is excluded by the decay condition; in a finite-volume regulator it must be treated separately.

Choose two orthonormal transverse photon polarization vectors ${\mathbf{ϵ}}_r(\mathbf{k})$, so thatThe canonical transverse photon field has mode expansionDifferentiate in time to obtain ${\Pi }_i^T$: the annihilation coefficient acquires $-i{\omega }_{\mathbf{k}}$ and the creation coefficient $+i{\omega }_{\mathbf{k}}$. The oscillator canonical commutation relations areSubstitution into $[A_i^T,{\Pi }_j^T]$ gives $i/2$ times the sum of the two Fourier kernels with phases $\pm i\mathbf{k}\cdot (\mathbf{x}-\mathbf{y})$. The polarization completeness relation and the evenness of $P^T(\mathbf{k})$ make that sum exactly $i{\delta }_{ij}^T$. This derives, rather than presumes, the correspondence between the oscillator and field commutators.

Substitute the mode expansions into $H_T$. Spatial integration sets paired momenta equal or opposite. The $aa$ and $a^{†}{a}^{†}$ terms from the electric and magnetic energies cancel because ${\omega }_{\mathbf{k}}^2=|\mathbf{k}{|}^2$ and $\mathbf{k}\cdot {\mathbf{ϵ}}_r=0$. The remaining terms are harmonic-oscillator energies. Removing the state-independent zero-point energy by normal ordering givesThe corresponding momentum operator is $\mathbf{P}={\sum }_r\int d^{3}k\mathbf{k}{a}_r^{†}{a}_r/(2\pi {)}^3$. These expressions imply $[H_T,a_r^{†}(\mathbf{k})]={\omega }_{\mathbf{k}}{a}_r^{†}(\mathbf{k})$ and $[\mathbf{P},a_r^{†}(\mathbf{k})]=\mathbf{k}{a}_r^{†}(\mathbf{k})$.