Canonical–Functional Crosswalk for Regulated Systems

Canonical and functional calculations agree in the example below because they are two exact representations of the same finitely regulated oscillator system. For finitely many massive scalar modes sampled at finitely many Euclidean times, the exact transfer kernel and the vacuum wave functions define a normalized finite Gaussian integral. Its covariance is precisely the Euclidean time-ordered vacuum two-point function computed with creation and annihilation operators. The equality requires the same regulator, Hamiltonian and ordering, operator domain, state, boundary data, normalization, source, observable, and prescription. It is not a universal equivalence theorem for canonical and functional formulations of QFT.

Required background. Time Slicing and Transition Amplitudes supplies the normalized kernel construction and its ordering data. Gaussian Fields and Sources supplies finite Gaussian source derivatives and inverse-kernel checks. Quantizing the Real Scalar Field supplies the finite-mode oscillator algebra and selected vacuum.

Helpful background. Boundaries and State Preparation distinguishes a fixed-endpoint kernel from the vacuum matrix element obtained by attaching and integrating boundary wave functions.

The comparison fixes one finite oscillator system

Let a spatial regulator $R$ retain $N_R<\infty$ real orthonormal scalar modes $v_a(\mathbf x)$ with strictly positive frequencies $\omega_a$. A finite box, boundary condition, and mode cutoff are included in $R$. We use natural units, set the oscillator masses to one by the choice of canonical coordinates, and assume $m>0$ so that no periodic zero-frequency mode occurs.

We take the standard self-adjoint oscillator Hamiltonian on the canonical Hilbert space, with the following chosen invariant core:

$$\mathcal H_R=L^2(\mathbb R^{N_R},\mathrm d^{N_R}q), \qquad \mathcal D_R=\mathcal S(\mathbb R^{N_R}),$$

Here $\langle q'|q\rangle=\delta^{(N_R)}(q'-q)$, $\widehat q_a$ acts by multiplication, and $\widehat p_a=-i\partial/\partial q_a$ on $\mathcal D_R$. The regulated Hamiltonian has the following differential expression on that core:

$$\widehat H_R = \frac12\sum_{a=1}^{N_R} \left( \widehat p_a^2+\omega_a^2\widehat q_a^2 \right).$$

Euclidean sample times are $\tau_n=n\epsilon$ for $n=0,\ldots,L$, with $L\geq1$, $\epsilon>0$, and $T=L\epsilon$. Instead of assuming a continuum product measure, the functional side will contain only the $N_R(L+1)$ real coordinates $q_{a,n}$. Each interval uses the exact matrix element $\langle q_{n+1}|e^{-\epsilon\widehat H_R}|q_n\rangle$; no Trotter error is hidden in the main match. Sources $s_{a,n}$ couple directly as $\sum_{a,n}s_{a,n}q_{a,n}$. If a continuum source is later sampled on the grid, its relation to $s_{a,n}$ requires a declared quadrature rule.

The choices that must coincide are:

Datum	Canonical choice	Functional choice and check
System	The same $N_R$ real modes and frequencies	The same coordinates and quadratic kernel; spectra agree mode by mode
Algebra and domain	$\widehat q_a$, $\widehat p_a$, and $\widehat H_R$ on the stated core	Coordinate integrations over $\mathbb R^{N_R(L+1)}$ with no omitted boundary term
Ordering	Euclidean time ordering generated by $e^{-\epsilon\widehat H_R}$	Exact transfer kernels, or a separately declared ordered short-step approximation
State and boundary	The normalized vacuum $\lvert0_R\rangle$	The same ground-state wave function at both endpoints, integrated once each
Normalization	Normalized vacuum matrix elements	Every kernel and endpoint factor retained, so the zero-source measure integrates to one
Source and observable	Ordered products of the retained $\widehat q_a$	Derivatives with respect to the direct source $s_{a,n}$
Prescription and limits	Euclidean decay in the chosen vacuum; $R$ fixed	The same decay and fixed $R$; no spatial cutoff removal inferred

This is a comparison record, not extra dynamics. A disagreement in any row means that the two calculations are not yet computing the same object.

Canonical operators give the vacuum covariance

For one retained mode, suppress the label $a$ and write

$$\widehat q = \frac{\widehat a+\widehat a^\dagger}{\sqrt{2\omega}}, \qquad \widehat H = \omega\left(\widehat a^\dagger\widehat a+\frac12\right), \qquad \widehat a|0\rangle=0.$$

Euclidean evolution gives

$$\widehat q(\tau) = e^{\tau\widehat H}\widehat q e^{-\tau\widehat H} = \frac{ e^{-\omega\tau}\widehat a +e^{\omega\tau}\widehat a^\dagger }{\sqrt{2\omega}}.$$

For $\tau_n\geq\tau_m$, only $\langle0|\widehat a\widehat a^\dagger|0\rangle=1$ contributes. Euclidean time ordering therefore yields

$$C^{\mathrm{can}}_{nm} = \langle0|\mathcal T_\tau \widehat q(\tau_n)\widehat q(\tau_m)|0\rangle = \frac{e^{-\omega|\tau_n-\tau_m|}}{2\omega}.$$

For all retained modes, define

$$\widehat\phi_R(\tau,\mathbf x) = \sum_{a=1}^{N_R}v_a(\mathbf x)\widehat q_a(\tau).$$

The canonical vacuum two-point function is then the finite sum

$$\begin{aligned} G^{\mathrm{can}}_{E,R} &(\tau,\mathbf x;\tau',\mathbf y) \\ &= \sum_{a=1}^{N_R} v_a(\mathbf x)v_a(\mathbf y) \frac{e^{-\omega_a|\tau-\tau'|}}{2\omega_a}. \end{aligned}$$

Every term is an ordinary oscillator matrix element. The spatial cutoff makes the sum finite, and $\omega_a>0$ makes the selected Gaussian vacuum normalizable.

Exact Euclidean kernels give a finite Gaussian chain

For the same one-mode Hamiltonian, the normalized ground-state wave function and energy are

$$\psi_0(q) = \left(\frac{\omega}{\pi}\right)^{1/4} e^{-\omega q^2/2}, \qquad E_0=\frac{\omega}{2}.$$

The exact Euclidean transfer kernel over one interval is

$$\begin{aligned} \mathcal K_\epsilon(q',q) ={}& \left[ \frac{\omega}{2\pi\sinh(\omega\epsilon)} \right]^{1/2} \\ &\times \exp\!\left\{ -\frac{\omega}{2\sinh(\omega\epsilon)} \left[ (q'^2+q^2)\cosh(\omega\epsilon)-2q'q \right] \right\}. \end{aligned}$$

It is the coordinate matrix element of $e^{-\epsilon\widehat H}$, with the endpoint values and normalization shown explicitly (Zinn-Justin 2021, § 2.3, pp. 24–25). Attach the same vacuum at both endpoints and integrate every sampled coordinate once:

$$\begin{aligned} \mathrm d\nu_{\epsilon,L}(q_0,\ldots,q_L) ={}& e^{E_0T}\psi_0(q_L)\psi_0(q_0) \\ &\times \prod_{n=0}^{L-1} \mathcal K_\epsilon(q_{n+1},q_n) \prod_{n=0}^{L}\mathrm dq_n. \end{aligned}$$

This is a finite positive measure. Its normalization is an operator identity:

$$\int\mathrm d\nu_{\epsilon,L} = e^{E_0T} \langle0|e^{-T\widehat H}|0\rangle =1.$$

Coordinate resolutions, fixed endpoints, boundary states, and the ordering encoded by the transfer operator are the ingredients that connect canonical matrix elements to functional representations (Weinberg 1995, §§ 9.1–9.4, pp. 378–398). Here the exact kernel makes the bridge finite-dimensional at every stage.

The structure becomes transparent after setting

$$r=e^{-\omega\epsilon}, \qquad 0<r<1.$$

The joint measure factorizes into a stationary initial density and normalized conditional densities:

$$\mathrm d\nu_{\epsilon,L} = \pi(q_0)\,\mathrm dq_0 \prod_{n=0}^{L-1} P_\epsilon(q_{n+1}|q_n)\,\mathrm dq_{n+1},$$

where

$$\pi(q) = \left(\frac{\omega}{\pi}\right)^{1/2}e^{-\omega q^2},$$

and

$$P_\epsilon(q'|q) = \left[ \frac{\omega}{\pi(1-r^2)} \right]^{1/2} \exp\!\left[ -\frac{\omega(q'-rq)^2}{1-r^2} \right].$$

The factors follow by completing the square in the exact kernel. They exhibit the boundary preparation, all normalization constants, and the finite-dimensional integration domain without introducing a symbol such as $\prod_\tau\mathrm dq(\tau)$.

The finite functional covariance matches mode by mode

The conditional Gaussian can be represented as

$$q_{n+1}=rq_n+\eta_n, \qquad \langle\eta_n\rangle=0, \qquad \langle\eta_n\eta_m\rangle = \frac{1-r^2}{2\omega}\delta_{nm},$$

with the $\eta_n$ independent of earlier coordinates. The initial density has $\langle q_0^2\rangle=1/(2\omega)$, so the variance remains stationary:

$$\langle q_n^2\rangle = r^2\langle q_{n-1}^2\rangle +\frac{1-r^2}{2\omega} = \frac{1}{2\omega}.$$

For $n\geq m$, repeated conditioning gives

$$C^{\mathrm{fun}}_{nm} = \int\mathrm d\nu_{\epsilon,L}\,q_nq_m = r^{n-m}\langle q_m^2\rangle = \frac{e^{-\omega|\tau_n-\tau_m|}}{2\omega}.$$

Thus

$$C^{\mathrm{fun}}_{nm}=C^{\mathrm{can}}_{nm}$$

for every $L$, every $\epsilon>0$, and every pair of sampled times. There is no time-slicing limit in this equality because each interval used the exact transfer kernel.

The normalized finite source integral is consequently

$$\begin{aligned} Z_{\epsilon,L}[s] &= \int\mathrm d\nu_{\epsilon,L} \exp\!\left(\sum_{n=0}^{L}s_nq_n\right) \\ &= \exp\!\left( \frac12\sum_{n,m=0}^{L}s_nC_{nm}s_m \right), \end{aligned}$$

and

$$\left. \frac{\partial^2\log Z_{\epsilon,L}} {\partial s_n\partial s_m} \right|_{s=0} =C_{nm}.$$

If a continuum source is sampled by $\int_0^T j(\tau)q(\tau)\,\mathrm d\tau\approx\sum_nw_nj_nq_n$, then $s_n=w_nj_n$. For example, trapezoidal sampling has $w_0=w_L=\epsilon/2$ and $w_n=\epsilon$ at interior nodes. Therefore

$$\left. \frac{1}{w_nw_m} \frac{\partial^2\log Z_{\epsilon,L}} {\partial j_n\partial j_m} \right|_{j=0} =C_{nm}.$$

The direct finite source $s_n$ needs no continuum quadrature; differentiating with respect to $j_n$ as though it were $s_n$ would compare different source conventions.

For the multimode figure below, we use the compact definitions

$$\mathrm d\nu_{R,\epsilon,L} := \prod_{a=1}^{N_R}\mathrm d\nu^{(a)}_{\epsilon,L}, \qquad Z_{R,\epsilon,L}[s] := \int\mathrm d\nu_{R,\epsilon,L} e^{\sum_{a,n}s_{a,n}q_{a,n}}.$$

Taking the product of these normalized chains over $a=1,\ldots,N_R$ gives the declared free-scalar result:

$$\boxed{ G^{\mathrm{fun}}_{E,R} (\tau_n,\mathbf x;\tau_m,\mathbf y) = G^{\mathrm{can}}_{E,R} (\tau_n,\mathbf x;\tau_m,\mathbf y) }$$

with both sides equal to the same finite mode sum displayed above. The usual Euclidean oscillator source functional and its vacuum covariance are obtained by the same Gaussian calculation (Zinn-Justin 2021, § 2.6, pp. 30–31).

The inverse-kernel contact check closes the derivation

The finite chain itself supplies an independent Gaussian check. Its exponent can be written as $-q^{\mathsf T}\mathsf Q_\epsilon q/2$, up to the normalized constant factors, with tridiagonal precision matrix

$$\begin{aligned} (\mathsf Q_\epsilon)_{00} =(\mathsf Q_\epsilon)_{LL} &= \frac{2\omega}{1-r^2}, \\ (\mathsf Q_\epsilon)_{nn} &= \frac{2\omega(1+r^2)}{1-r^2}, &&1\leq n\leq L-1, \\ (\mathsf Q_\epsilon)_{n,n+1} =(\mathsf Q_\epsilon)_{n+1,n} &= -\frac{2\omega r}{1-r^2}. \end{aligned}$$

All other entries vanish. Substituting $C_{nm}=r^{|n-m|}/(2\omega)$ gives

$$\mathsf Q_\epsilon C=I_{L+1}.$$

For an interior row, the equality follows from

$$\begin{aligned} (\mathsf Q_\epsilon C)_{nm} = \frac{2\omega}{1-r^2} \big[ &(1+r^2)C_{nm} \\ &-rC_{n-1,m}-rC_{n+1,m} \big], \end{aligned}$$

which vanishes for $n\neq m$ and equals one for $n=m$; the two endpoint rows work because the vacuum wave functions supplied the endpoint entries of $\mathsf Q_\epsilon$. Omitting those wave functions changes the inverse problem.

At continuous Euclidean time, the same invariant check is

$$\left(-\partial_\tau^2+\omega^2\right) \frac{e^{-\omega|\tau|}}{2\omega} = \delta(\tau).$$

For $0<\tau'<T$, its restriction $C(\tau,\tau')=e^{-\omega|\tau-\tau'|}/(2\omega)$ also satisfies the vacuum endpoint conditions

$$(\partial_\tau-\omega)C(0,\tau')=0, \qquad (\partial_\tau+\omega)C(T,\tau')=0.$$

These are the continuous counterparts of the two special endpoint rows of $\mathsf Q_\epsilon$; the differential equation alone does not select this inverse.

The derivative jump is

$$\partial_\tau C(0^+)-\partial_\tau C(0^-)=-1.$$

Mode by mode, this tests the oscillator normalization and equal-time commutator. For the field, the spatial sum reconstructs the regulated identity kernel

$$\delta_R(\mathbf x,\mathbf y) = \sum_{a=1}^{N_R}v_a(\mathbf x)v_a(\mathbf y),$$

not an unregulated spatial delta distribution.

The matched data make the equality conditional

Read the diagram as a data-matching test: trace the operator and Gaussian routes to their common regulated covariance, then inspect every equality gate and the dashed list of excluded claims.

For a fixed spatial regulator and a finite set of Euclidean samples, the solid routes are exact because every interval uses the same transfer operator. The shared endpoint is the covariance $C_a(\tau)=e^{-\omega_a|\tau|}/(2\omega_a)$ and its contact equation. The dashed boundary marks claims not established by this comparison: spatial regulator removal, interacting equivalence, reconstruction, or uniqueness of infinite-system representations. The diagram is schematic and not to scale.

Read the two routes from top to bottom. Route A ends at $\langle0_R|\mathcal T_\tau\widehat q_a(\tau_n) \widehat q_b(\tau_m)|0_R\rangle$. Route B uses the exact Gaussian kernels, integrates both endpoint wave functions once over $\mathbb R^{N_R(L+1)}$, and ends at $\left.\partial_{s_{a,n}}\partial_{s_{b,m}} \log Z_{R,\epsilon,L}[s]\right|_{s=0}$. Its normalization is $\int\mathrm d\nu_{R,\epsilon,L}=1$. The gates require equality of the regulator and spectrum; Hamiltonian, ordering, and operator domain; vacuum, endpoint data, and integration domain; normalization and source convention; observable and prescription; and fixed-$R$ claim and limit order. Only then do the routes yield the common covariance, whose independent checks are $\mathsf Q_\epsilon C=I$ and $(-\partial_\tau^2+\omega_a^2)C_a=\delta$. The dashed box lists the excluded continuum, interacting, reconstruction, and representation claims. Open the SVG alone for scalable zoom and pan.

A mismatch in any defining datum stops the comparison

The exact calculation has a sharp domain of validity:

Mismatch	What changes
Remove the endpoint vacuum wave functions	The finite Gaussian has different endpoint rows and no longer computes the vacuum matrix element
Identify the two endpoints and integrate once	The object becomes a trace; at finite Euclidean extent it gives a thermal covariance
Replace the exact kernel by a short-step action	Equality acquires the product-formula error and ordering convention of that approximation
Change the spatial cutoff or boundary condition	The frequency list and regulated identity kernel change, so the systems differ
Set some $\omega_a=0$	The ground-state Gaussian and $1/(2\omega_a)$ covariance cease to exist in that direction
Couple $w_nj_nq_n$ but differentiate as though the source were $s_nq_n$	The quadrature weights spoil the comparison
Use a Lorentzian bulk action without a contour or $i0$	The oscillatory Gaussian and its inverse are not yet specified
Use a constrained or gauge-redundant system	The unconstrained coordinate measure and oscillator domain no longer apply

For the matched Feynman vacuum prescription, define $D_F(t):=\langle0|\mathcal T\{\widehat q(t)\widehat q(0)\}|0\rangle$. The one-mode Lorentzian result is

$$D_F(t) = \frac{e^{-i\omega|t|}}{2\omega}, \qquad \left(\partial_t^2+\omega^2\right)D_F(t) = -i\delta(t).$$

Its contact term differs from the Euclidean one. A retarded, advanced, thermal, or fixed-boundary inverse has another boundary condition even when the denominator looks the same. The time-slice spacing $\epsilon$ controls a product approximation when one is used; $i0$ selects a Lorentzian boundary value. They are not the same limit.

For nonquadratic momentum dependence, coordinate-dependent kinetic terms, momentum sources, or composite insertions, the finite-slice ordering and momentum integrations must be carried explicitly. Only suitable nonsingular quadratic momentum dependence reduces those integrations to Gaussian determinants. For an interacting or infinite system, agreement of one regulated two-point function does not construct the limit, identify representations, or prove reconstruction.

Common pitfalls

Matching only the differential operator. Boundary conditions, state, and prescription select its inverse. Equal denominators do not identify Feynman, retarded, thermal, Dirichlet, and vacuum-projected kernels.

Comparing a kernel with a vacuum expectation value. A fixed-endpoint kernel leaves $q_0$ and $q_L$ unintegrated. The vacuum matrix element attaches $\psi_0(q_0)$ and $\psi_0(q_L)$ and integrates both endpoints exactly once.

Cancelling normalization before the objects match. Determinants and kernel prefactors cancel in a normalized source ratio only when the kernel, boundary data, and reference measure are unchanged. The finite chain above keeps every factor until its integral is demonstrably one.

Forgetting which source is differentiated. A direct discrete source $s_n$ and a quadrature-sampled continuum source $j_n$ differ by $s_n=w_nj_n$. Source derivatives must be translated before their correlators are compared.

Promoting fixed-regulator agreement to universal equivalence. The proof covers a finite free oscillator system and a selected observable. Spatial cutoff removal, interacting existence, reconstruction, and representation comparison require separate hypotheses and theorems.

Check your understanding

1. Starting from the conditional density $P_\epsilon(q'|q)$, verify its mean, variance, and stationary density.

   Solution

   Completing the square shows that $q'|q$ is Gaussian with mean $rq$ and variance $(1-r^2)/(2\omega)$. If $q$ has variance $1/(2\omega)$, then $q'$ has variance $r^2/(2\omega)+(1-r^2)/(2\omega)=1/(2\omega)$. Its mean remains zero, so $\pi(q')=(\omega/\pi)^{1/2}e^{-\omega q'^2}$ is stationary.

2. Multiply one interior row of $\mathsf Q_\epsilon$ by $C_{nm}=r^{|n-m|}/(2\omega)$.

   Solution

   For $n\neq m$, set $k=|n-m|\geq1$. The three terms are proportional to $(1+r^2)r^k-r\,r^{k-1}-r\,r^{k+1}=0$. For $n=m$, the bracket is $(1+r^2)-2r^2=1-r^2$ times $1/(2\omega)$; the prefactor $2\omega/(1-r^2)$ makes the result one.

3. Apply $-\partial_\tau^2+\omega^2$ to $e^{-\omega|\tau|}/(2\omega)$, including the origin.

   Solution

   Away from the origin the two terms cancel. The first derivative jumps from $+1/2$ on the left to $-1/2$ on the right, so its jump is $-1$. Distributionally, $-\partial_\tau^2 C$ therefore contributes $+\delta(\tau)$.

4. Replace the two endpoint vacuum wave functions by periodic identification $q_L=q_0$ and one integration over that common coordinate. What object is computed?

   Solution

   Removing the compensating factor $e^{E_0T}$, the integrations form $\operatorname{Tr}e^{-T\widehat H}$ rather than $\langle0|e^{-T\widehat H}|0\rangle$. If that factor is retained from the displayed vacuum measure, the result is $e^{E_0T}\operatorname{Tr}e^{-T\widehat H}$; it cancels in a normalized source ratio. In either convention, normalized insertions give a thermal correlator at inverse temperature $T$ (with units restored), not the exact vacuum correlator at finite $T$.

5. Set one retained frequency to zero. Which steps fail first?

   Solution

   The proposed ground-state wave function becomes nonnormalizable, the stationary density disappears, and the covariance $1/(2\omega)$ diverges. The zero mode needs a separate infrared definition before either side of the comparison is meaningful.

Where this crosswalk leads

• Scalar Propagators, Ordered Correlators, and Sources distinguishes the Feynman, retarded, advanced, and Wightman kernels that a common denominator does not distinguish.
• The Generating Functional develops normalized source derivatives beyond this finite free match.
• Wick Rotation and Analytic Continuation states the analytic and contour hypotheses needed to relate Euclidean and Lorentzian objects.
• Lattice Regulators and Target Continuum Theories develops spatial cutoff removal and continuum-target evidence.
• Equivalence, Uniqueness, and Comparison Notions distinguishes regulated agreement from unitary equivalence, quasiequivalence, reconstruction, and other framework-level claims.

References

• Weinberg, Steven. The Quantum Theory of Fields, Volume I: Foundations. Cambridge University Press, 1995. doi:10.1017/CBO9781139644167.
• Zinn-Justin, Jean. Quantum Field Theory and Critical Phenomena. 5th ed. Oxford University Press, 2021. doi:10.1093/oso/9780198834625.001.0001.