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advanced27 min studyLesson 663 in Physics

Generate moments with a declared source convention

Derive source differentiation on a finite regulator before using functional notation.

Before you begin
On this page

The question for today

What you’ll understand

  • Obtain moments from finite source derivatives.
  • Track normalization and factors of i between Euclidean and real time.
  • Relate source response to fluctuations without inventing a physical measurement.

A generating functional packages many correlation functions into one source-dependent object. Its usefulness comes from a simple finite-dimensional fact: differentiating an exponential with a linear source inserts the corresponding integration variable. Every factor of i or ℏ follows from the source convention in that exponential. Declaring the convention before differentiating avoids ambiguities that otherwise spread through propagators, connected functions, and effective actions. The source can be a formal bookkeeping device even when no laboratory field with that value is applied.

Start with finitely many independent sources

Choose a convergent Euclidean integral in n real variables, a source vector J, and dimensionless units with ℏ=1. Normalize its generating function by the zero-source partition integral, so its value at J=0 is exactly one. Differentiate before setting J to zero. Repeated derivatives generate full moments, including disconnected contributions; no logarithm has yet been taken. Differentiating only after setting J to zero would discard the very source dependence being used.

At nonzero source, divide derivatives by the source-dependent generating function to obtain normalized tilted expectations. The normalization depends on J and must be differentiated too when calculating response. For example, differentiating the mean gives a covariance rather than the uncentered second moment. Interchanging differentiation and integration requires suitable convergence, satisfied for positive Gaussians and for many stable finite polynomial actions in a neighborhood of zero source, but not guaranteed for every formal expression.

ZE(J)=∫dnx e−SE(x)+J⋅x∫dnx e−SE(x),∂Ji1⋯∂JirZE∣J=0=⟨xi1⋯xir⟩0\mathcal Z_E(J)=\frac{\int d^nx\,e^{-S_E(x)+J\cdot x}}{\int d^nx\,e^{-S_E(x)}},\qquad \left.\partial_{J_{i_1}}\cdots\partial_{J_{i_r}}\mathcal Z_E\right|_{J=0}=\langle x_{i_1}\cdots x_{i_r}\rangle_0
All variables, sources, and actions are dimensionless here, with ℏ=1. The source has a plus sign in the Euclidean exponent, and the expectation on the right is normalized at zero source.↔ Scroll wide equations horizontally.

A Gaussian is an exact source test

For a positive quadratic action, complete the square by shifting x by A inverse times J. The resulting source dependence is an exponential quadratic form. Its first derivative gives the shifted mean, while the covariance remains A inverse. Linear response is therefore exact for the mean of a Gaussian. A nonlinear action generally changes both the mean and fluctuation pattern, so the same constant susceptibility cannot simply be carried into an interacting problem.

In a lattice field regulator, attach a source to each field coordinate. A continuum source integral contains volume weights; the functional derivative is defined so that varying the source term inserts the field with the corresponding delta-function convention. These weights must be absorbed consistently into the source coordinates or derivative definition. Writing a delta function while silently dropping lattice cell volumes can introduce incorrect dimensions even when the finite-vector algebra was correct.

SE=12xTAx⟹ZE(J)=eJTA−1J/2,⟨x⟩J=A−1JS_E=\tfrac12x^{\mathsf T}Ax\quad\Longrightarrow\quad \mathcal Z_E(J)=e^{J^{\mathsf T}A^{-1}J/2},\qquad \langle x\rangle_J=A^{-1}J
A is real symmetric positive definite. This normalized source function contains no determinant because the same zero-source Gaussian normalization has been divided out.↔ Scroll wide equations horizontally.

Real time changes the differentiation factors

For a Minkowski vacuum generating functional, adopt a source term plus Jφ inside i times the action, and retain the vacuum boundary prescription. Each source derivative then inserts iφ when ℏ=1, so recovering a time-ordered correlator requires a factor 1/i for every insertion. Restoring ℏ replaces that factor by ℏ/i. A Euclidean derivative rule copied unchanged into this convention would therefore have incorrect phases. Vacuum persistence amplitudes are not positive probability distributions.

The function lab illustrates local source sensitivity only. Reset and set coefficient to 0.5 and position to 1; predict value 0.5 and slope 1. Reset and set coefficient to 1 and position to 1; predict value 1 and slope 2. A quadratic log generating function can have this shape, but the instrument itself displays y=ax², its tangent, and a signed integral. It does not insert fields, choose vacuum boundary conditions, or generate quantum correlators.

ZM[J]=∫Dϕ eiSM+i∫Jϕ∫Dϕ eiSM,⟨Tϕ(x1)⋯ϕ(xr)⟩=(1i)rδrZMδJ(x1)⋯δJ(xr)∣J=0\mathcal Z_M[J]=\frac{\int\mathcal D\phi\,e^{iS_M+i\int J\phi}}{\int\mathcal D\phi\,e^{iS_M}},\qquad \langle T\phi(x_1)\cdots\phi(x_r)\rangle=\left.\left(\frac1i\right)^r\frac{\delta^r\mathcal Z_M}{\delta J(x_1)\cdots\delta J(x_r)}\right|_{J=0}
This is formal continuum notation after regulation, with ℏ=1 and a specified vacuum iε prescription. The finite-regulator identity precedes the continuum limit; derivatives create time-ordered vacuum correlations.↔ Scroll wide equations horizontally.

Work it through

Tilt one Gaussian

Let SE(x)=x² in dimensionless units. At source J=1, find the normalized source function, mean, variance, and second moment.

  1. The quadratic coefficient is A=2, since SE=Ax²/2. Completing the square gives −x²+Jx=−(x−J/2)²+J²/4.

  2. The normalized generating function is exp(J²/4), so at J=1 it equals exp(1/4). The tilted mean is J/2=1/2, obtained from its logarithmic derivative.

  3. The variance stays A⁻¹=1/2. The full second moment is variance plus mean squared, namely 1/2+1/4=3/4; it must not be confused with the derivative of the mean alone.

The result

At J=1, the source function is exp(1/4), mean 1/2, variance 1/2, and second moment 3/4.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

Inspect a quadratic response curve

Reset and set a=0.5,x=1; predict y=0.5 and slope 1. Reset and set a=1,x=1; predict y=1 and slope 2.

Inspect a quadratic response curveThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. 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 forThe fixed parabola illustrates a value and derivative. It does not evaluate a source integral or supply Euclidean-to-Minkowski conversion factors.

Pause & reason

Check your understanding

With a real-time source appearing as exp(i∫Jφ) and ℏ=1, which factor removes the insertion phase from each source derivative?

Your turn

Try it yourself

A finite variable x takes values ±1 with equal zero-source weights. Tilt them by exp(Jx), and choose J=(ln3)/2. Find the two probabilities, mean, and variance.

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

Give me a hint

The ratio of positive to negative weights is exp(2J); normalize that ratio before taking moments.

Compare with the solution

The weight ratio is 3, giving probabilities 3/4 and 1/4. The mean is 1/2 and the second moment is exactly 1, so variance is 3/4. This discrete source problem has the same differentiation logic as a finite integral but a non-Gaussian response; its mean is tanh J rather than a globally linear function.

Keep the idea

Take it with you

  • Source derivatives insert fields with factors fixed by the exponent.
  • Normalized response requires differentiating the normalization too.
  • Finite-regulator identities need consistent volume and delta-function conventions in a continuum limit.

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