advanced27 min studyLesson 668 in Physics

Derive quantum equations with their contact terms

Use finite integration by parts to separate exact identities from closure assumptions.

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What you’ll understand

  • Derive the finite Euclidean integration-by-parts identity.
  • Identify contact terms in field correlation equations.
  • Audit boundary contributions and approximate moment closures.

Schwinger–Dyson identities express the fact that a change of integration variable cannot alter a well-defined integral. Their finite-dimensional version is integration by parts. The derivative can act on an inserted observable as well as on the weight, producing the contact terms that distinguish quantum correlation equations from classical equations of motion. An infinite hierarchy of such relations is exact under its assumptions. Closing that hierarchy by replacing higher correlations with guessed products is an additional approximation, however accurately the resulting finite algebraic equation is solved.

Keep the total derivative until boundaries are checked

Use a finite Euclidean integral with weight exp(−S(x)+Jx), normalized by its convergent partition function and with ℏ=1. Differentiate an insertion f times that weight and integrate over the declared domain. If the boundary term vanishes, the average derivative of f equals the average of f times S′−J. For the full real line, decay of the product at both ends is sufficient. A restricted interval or a rapidly growing insertion can invalidate that step.

In several dimensions there is one identity for each coordinate. The field-theory notation follows by replacing a regulated coordinate derivative with a functional derivative and retaining the regulator’s measure convention. A nontrivial measure can contribute its own logarithmic derivative; it should not be treated as a flat product silently. Gauge fixing, constraints, and possible anomalous Jacobians require particular care before a formal shift of every field variable is assumed harmless.

0=∫dx ddx[f(x)e−S(x)+Jx]⟹⟨f′⟩J=⟨f(S′−J)⟩J0=\int dx\,\frac d{dx}\left[f(x)e^{-S(x)+Jx}\right]\quad\Longrightarrow\quad \langle f'\rangle_J=\langle f(S'-J)\rangle_J
This dimensionless finite identity assumes the boundary term is zero and all indicated integrals exist. It does not assert that S′ vanishes at every configuration or inside every inserted average.↔ Scroll wide equations horizontally.

Insertions create contact contributions

Choosing f to be another coordinate gives a constant derivative, so the two-point identity has a nonzero right-hand side. In continuum language, differentiating an inserted field produces a delta function when the varied field point coincides with that insertion. These contact terms ensure that a Gaussian two-point kernel is the inverse of the quadratic action operator. Dropping them would incorrectly turn the propagator equation into a homogeneous classical equation everywhere.

At real time with weight exp(iS) the same integration-by-parts calculation introduces a factor i. Its sign must be derived from that exponent rather than copied from the Euclidean identity. Time ordering also makes coincident-point distributions important. Renormalized composite operators can require local counterterms, so products of continuum fields at the same point are not automatically finite quantities merely because their regulated identity was exact.

⟨δSEδϕ(x)F[ϕ]⟩=⟨δFδϕ(x)⟩,Aik⟨xkxj⟩=δij\left\langle\frac{\delta S_E}{\delta\phi(x)}F[\phi]\right\rangle=\left\langle\frac{\delta F}{\delta\phi(x)}\right\rangle,\qquad A_{ik}\langle x_kx_j\rangle=\delta_{ij}
Here J=0 and ℏ=1. The functional equation abbreviates a regulated identity with a specified measure; for F=φ(y), its right side is the corresponding delta distribution. The matrix example is finite and exactly Gaussian.↔ Scroll wide equations horizontally.

A closure is a model for missing information

A quartic action couples an equation for the second moment to the fourth moment, whose equation involves the sixth, and so on. A Gaussian closure replaces the fourth moment by three times the variance squared in a centered state. This can provide a useful estimate, but it discards non-Gaussian cumulants and is not the exact Schwinger–Dyson hierarchy. Solving the closed equation to many digits tests only that approximation’s internal algebra; it does not bound its error against the original integral.

The root-finding lab provides a narrow comparison. Reset and set target c to 1 and upper bound U to 2, then Step once; predict midpoint 1 and zero bracket bound because the midpoint is an exact root. Reset and set c to 2 and U to 2, then Step once; predict midpoint 1.5 and bound 0.5. These certified brackets apply to x²−c. They do not certify a chosen hierarchy truncation or solve the quartic moment closure discussed here.

S(x)=x22+gx44⟹⟨x2⟩+g⟨x4⟩=1,⟨x4⟩≈3⟨x2⟩2S(x)=\frac{x^2}{2}+\frac{g x^4}{4}\quad\Longrightarrow\quad \langle x^2\rangle+g\langle x^4\rangle=1,\qquad \langle x^4\rangle\approx3\langle x^2\rangle^2
The exact relation uses f=x, zero source, and g≥0 for convergence. The final approximate equality is a centered Gaussian closure, not an identity for the interacting quartic measure.↔ Scroll wide equations horizontally.

Solve an explicitly approximate closure

For S=x²/2+x⁴/2, derive the exact second-moment identity and estimate its variance using the centered Gaussian fourth-moment closure. State what the estimate proves.

  1. Here g=2, so integration by parts with f=x gives v+2⟨x⁴⟩=1, where symmetry makes the mean zero and v=⟨x²⟩ is the variance.

  2. Impose the additional approximation ⟨x⁴⟩≈3v². The closed equation is 6v²+v−1=0, with roots (−1±5)/12. Positivity selects v=1/3 and rejects −1/2.

  3. The associated fourth-moment estimate is 3(1/3)²=1/3, making the closed identity sum to one. This consistency check does not establish that either moment equals the exact quartic integral; at this coupling no small-error claim has been supplied.

The result

The exact identity is v+2⟨x⁴⟩=1. Gaussian closure gives v≈1/3, an approximate positive solution without a certified truncation error.

Explore the model

Interactive lab

Interactive lab · root-finding

Compare a bracket with a closure claim

Reset and set c=1,U=2, then Step once; predict midpoint 1 and bound 0. Reset and set c=2,U=2, then Step once; predict midpoint 1.5 and bound 0.5.

Compare a bracket with a closure claimBisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Current readings: Midpoint after 0 refinements: 1.5; Bracket error bound: 1.5; Computed residual · x² − c: 0.25. All parameters are adjustable with the labeled controls below.Bisection on x squared minus 2: bracket from 0 to 3 after 0 refinements.00.751.52.253−2.360.192.745.297.84x · bracket stays nonnegativef(x) = x² − cn = 0 · [ 0, 3 ]
Quadratic / midpointCurrent bracketPositive root
Midpoint after 0 refinements
1.5
Bracket error bound
1.5
Computed residual · x² − c
0.25
t = 0 s

Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forBisection certifies only the bracketed quadratic root. It does not bound the physical error made by discarding higher connected moments.

Check your understanding

Why does a propagator’s Schwinger–Dyson equation have a delta-function contact term?

Try it yourself

Normalize exp(−x²/2) on the half-line x≥0. Evaluate ⟨x⟩ and explain why the full-line f=1 identity ⟨S′⟩=0 fails.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Keep the boundary term [exp(−x²/2)] from 0 to infinity and use the half-Gaussian normalization.

Compare with the solution

The normalization is √(π/2), and the numerator integral of x exp(−x²/2) is 1. Thus ⟨x⟩=√(2/π)≈0.7979. The integrated total derivative equals −1, not zero, so 0=−⟨S′⟩ cannot be inferred. The missing boundary term precisely supplies the nonzero result.

Take it with you

  • Schwinger–Dyson equations follow from a regulated change of variables.
  • Inserted observables generate contact terms.
  • A solved closure remains an approximation until its discarded correlations are controlled.

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