advanced27 min studyLesson 662 in Physics

Reduce a regulated Gaussian to its eigenmodes

Separate normalization determinants, covariances, zero modes, and boundary conditions.

On this page

What you’ll understand

  • Evaluate a positive finite-dimensional Gaussian.
  • Identify the inverse quadratic form as covariance.
  • Explain why continuum determinants need regulation and boundary data.

Quadratic actions are the solvable core of functional methods. At a finite regulator they are simply quadratic forms in a finite vector of integration variables. Diagonalizing that form explains both the determinant in the normalization and the inverse matrix in correlations. It also reveals exactly when the construction fails: a zero eigenvalue has no restoring weight, while a negative direction prevents an ordinary real Euclidean Gaussian from converging. A formal determinant symbol does not repair either problem.

Diagonalize before taking any limit

Let x contain n real, dimensionless variables and let A be a real symmetric positive-definite matrix. Work in units where ℏ=1 and the Euclidean action is xᵀAx/2. An orthogonal change of variables diagonalizes A and has absolute Jacobian one. Each eigenmode contributes the familiar one-dimensional Gaussian integral, so the product is controlled by the product of eigenvalues. This is the origin of the inverse square root determinant for real bosonic variables.

A general nonorthogonal coordinate change has a nontrivial Jacobian, which must accompany the transformed quadratic form. Keeping only the changed determinant would make the answer coordinate dependent. Overall measure conventions can supply additional constants, so state whether each integration uses dx or dx/√(2π). Ratios may cancel shared constants, but that cancellation is justified only when numerator and denominator use compatible measures and the same regulated degrees of freedom.

ZA=∫Rndnx  e−xTAx/2=(2π)n/2det⁡A,A=AT>0Z_A=\int_{\mathbb R^n}d^nx\;e^{-x^{\mathsf T}Ax/2}=\frac{(2\pi)^{n/2}}{\sqrt{\det A}},\qquad A=A^{\mathsf T}>0
This is a finite-dimensional Euclidean integral with dimensionless x and action. The measure is ordinary Lebesgue measure, with no hidden factors of √(2π).↔ Scroll wide equations horizontally.

Correlations invert the quadratic kernel

Normalize the Gaussian by its partition integral. A small linear source can be removed by completing the square; differentiating its resulting exponential gives the moments. The mean at zero source vanishes and the two-point covariance is A inverse. Off-diagonal entries can be negative even when A is positive definite: positivity constrains the variance of every linear combination, not the sign of every cross correlation. Higher even moments follow by pairing, while odd centered moments vanish.

A time-sliced quadratic action supplies a particular matrix A whose entries depend on spacing and endpoint conditions. Its inverse is a regulated Green function. Dirichlet, periodic, or other boundary conditions generally change its spectrum and hence its determinant and inverse. Calling the Green function merely an inverse differential operator without specifying the domain leaves the problem incomplete. An operator with a constant zero mode, for example, has no inverse on that full domain.

⟨xixj⟩=(A−1)ij,⟨(v⋅x)2⟩=vTA−1v≥0\langle x_ix_j\rangle=(A^{-1})_{ij},\qquad \langle(\mathbf v\cdot\mathbf x)^2\rangle=\mathbf v^{\mathsf T}A^{-1}\mathbf v\ge0
Angle brackets denote the normalized zero-source Gaussian. A positive-definite A gives a positive covariance matrix, although individual off-diagonal entries need not be positive.↔ Scroll wide equations horizontally.

Functional determinants need a definition

With more lattice sites or retained modes, the finite determinant becomes a growing product. Its continuum limit commonly diverges, even when suitably normalized correlations have sensible limits. A cutoff, determinant ratio, or another stated regularization must define the expression, followed by any required counterterms. Zero modes require separate treatment through constraints, collective coordinates, or a reduced determinant. Gauge redundancies are an especially important source of such modes, not extra physical Gaussian fluctuations to integrate repeatedly.

The matrix lab checks a change-of-variables illustration. Reset and set diagonal scale to 1 and shear to 0; predict final-A determinant 1 and singular values 1,1. Reset and set scale to 2 and shear to 0; predict determinant 2 and singular values 2,1. Readouts describe the selected final A. Geometry starts at identity at Reset, and Play or Step interpolates it toward A and back. This triangular transformation family does not compute a quadratic-action spectrum or functional determinant.

log⁡det⁡Adet⁡A0=∑r=1nlog⁡λrλ0r,det⁡′A=∏λr≠0λr\log\frac{\det A}{\det A_0}=\sum_{r=1}^{n}\log\frac{\lambda_r}{\lambda_{0r}},\qquad \det{}'A=\prod_{\lambda_r\ne0}\lambda_r
The first equality concerns matched finite positive spectra; a continuum ratio still needs a regulator and convergence analysis. The prime denotes an explicitly excluded zero-mode sector, not an automatic prescription for its missing integral.↔ Scroll wide equations horizontally.

Two independent widths

For A=diag(1,4), evaluate the two-dimensional Gaussian normalization with measure dxdy and find both variances. Check the determinant factor directly.

  1. The quadratic action is x²/2+2y². Its two eigenvalues are 1 and 4, both positive, so the real integration converges independently in each direction.

  2. The x integral is √(2π) and the y integral is √(2π/4). Their product is π, equal to 2π/√det A because det A=4.

  3. The inverse matrix is diag(1,1/4). Thus x has variance 1, y variance 1/4, and their covariance vanishes. Narrowing the y distribution changes the normalization as well as the corresponding variance.

The result

The normalization is π and the variances are 1 and 1/4, with zero cross covariance.

Explore the model

Interactive lab

Interactive lab · matrix

Check a finite Jacobian

Reset and set a=1,b=0; predict final determinant 1 and singular values 1,1. Reset and set a=2,b=0; predict determinant 2 and singular values 2,1.

Check a finite JacobianThe selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. Current readings: Area scale · det A: 1.4; Singular values · σmax, σmin: 1.597, 0.876; Condition number · κ₂(A): 1.823. All parameters are adjustable with the labeled controls below.Matrix A with first row 1.4, 0.6 and second row zero, one; display interpolates between identity and A while final determinant remains 1.4.−4−2024−2−1012x · equal geometric scaleyA = [ 1.4 0.6 ; 0 1 ]Display blend: 0% · final values below
Interpolated geometryOriginal unit squareSelected final transform
Area scale · det A
1.4
Singular values · σmax, σmin
1.597, 0.876
Condition number · κ₂(A)
1.823
t = 0 s

The selected matrix is A = [[a,b],[0,1]]. Playback interpolates geometry between the identity and A; readouts always describe A. Its determinant is a and its singular values come from AᵀA. Positive a makes every supported matrix invertible. This two-dimensional family is not a general eigensolver or a learned network. 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 forReadouts describe final A. Reset geometry is identity; Play/Step interpolates identity↔A. This transformation family is a Jacobian illustration, not a functional-determinant solver.

Check your understanding

What prevents a normalized real Gaussian on the full space when its quadratic form has an exact zero eigenvalue?

Try it yourself

Use A=[[2,1],[1,2]]. Find the normalization, both variances, and cross covariance. Would replacing it by [[1,1],[1,1]] still define a normalized Gaussian on the plane?

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

Give me a hint

Invert the first matrix using its determinant, then inspect the eigenvalues of the replacement.

Compare with the solution

The first determinant is 3, so Z=2π/√3. Its inverse is [[2,−1],[−1,2]]/3: both variances are 2/3 and cross covariance is −1/3. The replacement has eigenvalues 2 and 0, leaving an undamped direction; its full-plane integral diverges. Removing that direction changes the integration problem and must be stated.

Take it with you

  • Finite positive quadratic forms define Gaussian normalization and covariance together.
  • Boundary conditions and zero modes belong to the operator definition.
  • A continuum determinant is a regulated quantity rather than an unqualified infinite product.

Read further

Original sources and deeper explanations to follow at your own pace.

Optional progress, saved in this browser when available. Revisit any lesson at any time.

Lesson not marked complete.