research30 min studyLesson 756 in Physics

Tunneling determinants and zero modes

Separate Gaussian fluctuation factors from collective-coordinate measures, with explicit boundary conditions and a distinction between level splitting and decay.

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What you’ll understand

  • Identify the fluctuation operator for a Euclidean tunneling saddle.
  • Replace a translational zero mode with a collective-coordinate integral.
  • Interpret determinant normalization and phase before converting a saddle contribution into an observable.

The exponential of a tunneling action gives the dominant suppression, but fluctuations determine its prefactor and sometimes its phase. A naive determinant vanishes when a saddle can be translated freely, so the corresponding Gaussian formula appears divergent. The correct treatment integrates the saddle’s position as a collective coordinate. We will use a unit-mass quantum-mechanical double well to state that replacement, track the determinant ratio’s dimensions, and distinguish an energy splitting in a stable symmetric system from the imaginary energy associated with a metastable decay.

Expand the Euclidean action on a specified domain

Take a particle with unit mass and Euclidean action equal to the integral of one half velocity squared plus a smooth nonnegative double-well potential. The degenerate minima have zero potential. A finite-action instanton approaches one minimum at negative infinite Euclidean time and the other at positive infinite time. Its action is the integral of the square root of twice the potential between the minima. This relation uses the zero Euclidean first integral selected by those endpoints.

Write a nearby path as the saddle plus a fluctuation. The quadratic term contains a second-order differential operator with potential given by the second derivative of the mechanical potential along the saddle. Its spectrum depends on boundary conditions. On a long finite interval one may impose fixed endpoint fluctuations and compare with a vacuum operator using the same regulator. The exact whole-line translation mode can become a near-zero mode at finite interval, so a finite-size determinant must be matched carefully to the collective-coordinate limit.

SE[x]=∫dτ[12x˙2+V(x)],S0=∫x−x+dx2V(x),MI=−d2dτ2+V′′(xI(τ))S_E[x]=\int d\tau\left[\frac12\dot x^2+V(x)\right],\qquad S_0=\int_{x_-}^{x_+}dx\sqrt{2V(x)},\qquad M_I=-\frac{d^2}{d\tau^2}+V''(x_I(\tau))
Mass is set to one in a consistent mechanical unit system, while hbar is retained in semiclassical weights. The fluctuation operator’s eigenvalues have inverse-time-squared units. Its determinant requires a common boundary and regularization prescription with the reference operator.↔ Scroll wide equations horizontally.

A zero eigenvalue is replaced, not divided out without compensation

Translation of the instanton center tau zero gives a fluctuation proportional to its time derivative. Differentiating the saddle equation shows that this is a zero mode. The unnormalized translation tangent is the instanton velocity. Its squared L2 norm is the integral of the instanton velocity squared, equal to S zero in the unit-mass convention because kinetic and potential contributions are equal on the solution. Changing variables from that mode’s Gaussian coefficient to the center coordinate produces a Jacobian containing the square root of action divided by two pi hbar.

A prime on the instanton determinant removes the translation eigenvalue, while the collective-coordinate measure restores the integration along that direction. The reference vacuum determinant contains its full spectrum. Removing one inverse-time-squared eigenvalue gives the determinant ratio a net inverse-time-squared dimension, so its square root supplies an inverse time. Normalization constants must be compared with the same measure and regulator; a bare infinite product of eigenvalues is not a complete observable.

dc02πℏ⟶S02πℏ dτ0,K=S02πℏ(det⁡M0det⁡′MI)1/2,ΔE≃2ℏKe−S0/ℏ\frac{dc_0}{\sqrt{2\pi\hbar}}\longrightarrow\sqrt{\frac{S_0}{2\pi\hbar}}\,d\tau_0,\qquad K=\sqrt{\frac{S_0}{2\pi\hbar}}\left(\frac{\det M_0}{\det' M_I}\right)^{1/2},\qquad \Delta E\simeq2\hbar K e^{-S_0/\hbar}
The displayed replacement includes the normalized path-measure factor. K has inverse-time units. The splitting formula is leading dilute-instanton semiclassics for a symmetric stable double well with an appropriate positive fluctuation spectrum after removing translation, not a generic decay-rate formula.↔ Scroll wide equations horizontally.

Negative modes require a contour prescription

For positive Gaussian eigenvalues, the real Euclidean fluctuation integral converges and the determinant square root has the positive real branch. A zero eigenvalue signals a flat direction and requires separate treatment. A negative eigenvalue makes a real Gaussian divergent; taking the absolute determinant discards the phase information rather than solving that problem. The appropriate deformed contour or analytic continuation must follow from the observable and the definition of the metastable or oscillatory integral.

The Gel’fand–Yaglom method can obtain a one-dimensional determinant ratio from initial-value differential equations under specified boundary conditions, avoiding an explicit product over eigenvalues. It does not remove the need to handle zero and negative modes or renormalization in higher-dimensional field theories. In gauge theories gauge fixing and ghost determinants enter as well. Keeping these ingredients explicit prevents a real tunneling amplitude, a stable level splitting, and a false-vacuum imaginary contribution from being treated as interchangeable numbers.

Keep the prefactor’s time scale

In a synthetic symmetric double-well semiclassical calculation let S0/ℏ=8S_0/\hbar=8 and the regulated ratio det⁡M0/det⁡′MI=9/t02\det M_0/\det' M_I=9/t_0^2, with positive t zero. Find K and the leading splitting, leaving the exponential exact.

  1. The collective-coordinate factor is 8/(2π)=2/π\sqrt{8/(2\pi)}=2/\sqrt\pi. It is dimensionless because action is divided by hbar.

  2. The determinant-ratio square root is 3/t03/t_0, so K=6/(π t0)K=6/(\sqrt\pi\,t_0). The missing eigenvalue is what permits the inverse-time dimension.

  3. The leading stable-well splitting is ΔE=12ℏe−8/(π t0)\Delta E=12\hbar e^{-8}/(\sqrt\pi\,t_0). Its units are energy, and no imaginary decay rate has been inferred from this positive-spectrum problem.

The result

K is 6/(√π t zero), and the leading energy splitting is 12 hbar e⁻⁸/(√π t zero) under the stipulated determinant normalization.

Explore the model

Interactive lab

Interactive lab · function

Positive, flat, and negative quadratic directions

Reset, set Coefficient a = 1 and Position x = 1, and predict positive y = 1. For a fresh comparison Reset, set a = −1 and x = 1, and predict negative y = −1. For a third fresh run Reset, set a = 0 and x = 1, and predict zero cost in a flat quadratic direction.

Positive, flat, and negative quadratic directionsThis 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 chosen parabola illustrates the sign of one quadratic coefficient. It does not evaluate a functional determinant, identify a normalizable zero mode, choose an integration contour, or calculate a tunneling prefactor.

Check your understanding

What must accompany removing a translational zero eigenvalue from an instanton determinant?

Try it yourself

For an ordinary two-variable Gaussian with hbar = 1, evaluate ∫R2dx dy exp⁡[−(x2+4y2)/2]\int_{\mathbb R^2}dx\,dy\,\exp[-(x^2+4y^2)/2]. Then explain what changes if the coefficient 4 is replaced by zero or by −4 while the contour remains real.

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

Give me a hint

Multiply the two convergent one-dimensional Gaussian integrals before considering the altered signs.

Compare with the solution

The positive Gaussian is (2π)(2π/4)=π(\sqrt{2\pi})(\sqrt{2\pi/4})=\pi. With zero y coefficient the integral over y is infinite, reflecting an untreated flat direction. With coefficient −4 the integrand grows exponentially along real y and the integral diverges. A rotated contour can define a different continued problem with a phase, but that contour and branch must be specified; replacing the negative coefficient by its magnitude is not the original real integral.

Take it with you

  • A fluctuation determinant depends on domain, boundary conditions, measure, and regulator.
  • Collective coordinates replace symmetry zero modes with Jacobians.
  • Negative-mode phases and the observable’s contour distinguish stable splitting from metastable decay.

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