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Bounce Solutions

A bounce is a finite-action Euclidean solution that leaves a metastable minimum, reaches a turning point beyond the barrier, and returns to the same minimum. It is the canonical saddle for the exponentially small decay of a metastable state.

For a background-subtracted potential U(x)U(x), the bounce satisfies

mxB′′=U′(xB),mx_B'' = U'(x_B),

with

xB(τ)→xF,τ→±∞.\begin{gathered} x_B(\tau)\to x_F, \\ \tau\to\pm\infty. \end{gathered}

Here xFx_F denotes the false minimum. Unlike an instanton between degenerate minima, a bounce makes a round trip and has one negative fluctuation mode in addition to its translation zero mode.

This page owns the classical bounce boundary-value problem, its action exponent, the one-negative-mode diagnostic, and a complete cubic-potential example. Euclidean Time and Imaginary-Time Action owns the continuation and background-subtraction rules. Fluctuation Determinants Preview owns determinant normalization and collective-coordinate prefactors. False Vacuum Decay in Quantum Mechanics owns the resonance, survival-law, WKB attempt-frequency, and lifetime interpretation.

Consider a smooth one-dimensional potential with a local minimum at xFx_F. Shift its value so that

U(xF)=0.U(x_F)=0.

For a nondegenerate false minimum,

U′(xF)=0,U′′(xF)=mωF2>0.U'(x_F)=0, \qquad U''(x_F)=m\omega_F^2\gt0.

Suppose an escape direction contains a barrier top xbx_b and a point xtx_t beyond it such that

U′(xb)=0,U′′(xb)<0,Ub≡U(xb)>0,U(xt)=0,U(x)>0forxF<x<xt.\begin{gathered} U'(x_b)=0, \qquad U''(x_b)\lt0, \\ U_b\equiv U(x_b)\gt0, \\ U(x_t)=0, \\ U(x)\gt0 \quad \text{for} \quad x_F\lt x\lt x_t. \end{gathered}

The point xtx_t is the zero-energy turning point on the far side of the barrier. The simple real bounce described below exists because the inverted potential −U(x)-U(x) has the same value at xFx_F and xtx_t.

A metastable cubic potential with a false minimum, barrier top, and escape turning point beside the corresponding localized Euclidean bounce profile.

The false minimum xFx_F is separated from the escape region by a barrier. In Euclidean time the bounce approaches xFx_F at both ends, reaches xtx_t at its center τ0\tau_0, and returns. The cubic example below has an exact sech⁡2\operatorname{sech}^2 profile.

What the false vacuum means in zero spatial dimensions

Section titled “What the false vacuum means in zero spatial dimensions”

In field theory, “vacuum” denotes a state associated with a homogeneous field configuration. In ordinary quantum mechanics, the corresponding object is a wavepacket or resonance localized near a metastable well. The terminology is useful, but the local minimum itself is not an exact stationary quantum state.

Two common realizations must be distinguished:

  1. An open escape problem: the potential falls toward an unbounded or asymptotic region, and outgoing boundary conditions define a resonance with a complex energy.
  2. A closed multiwell problem: the full Hamiltonian has a real discrete spectrum, but a state initially localized in one well can remain trapped for a long time before tunneling, dephasing, and eventually recurring.

A local minimum alone does not guarantee a useful exponential decay regime. The barrier must produce a parametrically long timescale, normally requiring a bounce action BB with

Bℏ≫1.\frac{B}{\hbar}\gg1.

Subtract the false-minimum background and define

B[x]=∫−∞∞dτ[m2(x′)2+U(x)].B[x] = \int_{-\infty}^{\infty} d\tau \left[ \frac{m}{2}(x')^2 + U(x) \right].

Stationarity gives

mx′′=U′(x).mx'' = U'(x).

The bounce boundary conditions are

xB(τ)→xF,xB′(τ)→0,τ→±∞,xB(τ0)=xt,xB′(τ0)=0.\begin{gathered} x_B(\tau)\to x_F, \qquad x_B'(\tau)\to0, \\ \tau\to\pm\infty, \\ x_B(\tau_0)=x_t, \qquad x_B'(\tau_0)=0. \end{gathered}

The solution is symmetric about its center:

xB(τ0+σ)=xB(τ0−σ).x_B(\tau_0+\sigma) = x_B(\tau_0-\sigma).

Translation invariance makes τ0\tau_0 arbitrary. This continuous family of equal-action solutions is the origin of the translation zero mode.

The Euclidean equation can be written

mxB′′=−ddxB[−U(xB)].mx_B'' = - \frac{d}{dx_B} \left[ -U(x_B) \right].

In the mechanical analogy, a particle starts asymptotically at the unstable equilibrium on top of −U-U at xFx_F, rolls to xtx_t, stops there, and retraces its path. It takes infinite Euclidean time to leave or return to the top because xFx_F is approached exponentially.

The analogy constructs a stationary path. It does not describe the real-time motion of the quantum particle during tunneling.

Because the Euclidean Lagrangian has no explicit τ\tau dependence,

EE=m2(xB′)2−U(xB)\mathcal E_E = \frac{m}{2}(x_B')^2 - U(x_B)

is conserved. The asymptotic boundary conditions set EE=0\mathcal E_E=0, so

m2(xB′)2=U(xB).\frac{m}{2}(x_B')^2 = U(x_B).

On either half of the bounce,

∣dxBdτ∣=2U(xB)m.\left| \frac{dx_B}{d\tau} \right| = \sqrt{ \frac{2U(x_B)}{m} }.

The implicit profile is therefore

∣τ−τ0∣=m2∫xB(τ)xtdyU(y).\left| \tau-\tau_0 \right| = \sqrt{\frac{m}{2}} \int_{x_B(\tau)}^{x_t} \frac{dy}{\sqrt{U(y)}}.

Near xFx_F,

U(x)=mωF22(x−xF)2+O((x−xF)3),U(x) = \frac{m\omega_F^2}{2} (x-x_F)^2 + O\left( (x-x_F)^3 \right),

so the integral diverges logarithmically and

xB(τ)−xF∝e−ωF∣τ−τ0∣.x_B(\tau)-x_F \propto e^{-\omega_F|\tau-\tau_0|}.

The zero-energy first integral reduces the action:

B=∫−∞∞dτ[m2(xB′)2+U(xB)]=∫−∞∞dτ 2U(xB).\begin{aligned} B &= \int_{-\infty}^{\infty} d\tau \left[ \frac{m}{2}(x_B')^2 + U(x_B) \right] \\ &= \int_{-\infty}^{\infty} d\tau\, 2U(x_B). \end{aligned}

The outward and return branches contribute equally. Hence

B=2∫xFxtdx 2mU(x).B = 2 \int_{x_F}^{x_t} dx\, \sqrt{2mU(x)}.

The factor of two is structural. A monotone instanton crosses once between degenerate minima; a bounce leaves and returns to the same minimum.

For a one-dimensional barrier, define the fixed-energy WKB action

W(E)=∫x1(E)x2(E)dx 2m[U(x)−E].W(E) = \int_{x_1(E)}^{x_2(E)} dx\, \sqrt{2m[U(x)-E]}.

At the classical false-minimum energy E=0E=0,

B=2W(0).B = 2W(0).

Therefore the bounce factor

e−B/ℏe^{-B/\hbar}

matches the leading WKB probability suppression e−2W/ℏe^{-2W/\hbar}. A WKB wavefunction amplitude contains only e−W/ℏe^{-W/\hbar}.

The false-well ground energy is actually of order ℏωF/2\hbar\omega_F/2, not exactly zero. Using W(EF)W(E_F) shifts some subleading terms between exponent and prefactor. Agreement claims must state the approximation order and energy convention.

Worked Example: Cubic Metastable Potential

Section titled “Worked Example: Cubic Metastable Potential”

Consider

U(x)=mω22x2−g3x3,g>0.U(x) = \frac{m\omega^2}{2}x^2 - \frac{g}{3}x^3, \qquad g\gt0.

The false minimum is at xF=0x_F=0. The barrier top satisfies

xb=mω2g,x_b = \frac{m\omega^2}{g},

with height

Ub=m3ω66g2.U_b = \frac{m^3\omega^6}{6g^2}.

The nonzero root of U(x)=0U(x)=0 is

xt=3mω22g.x_t = \frac{3m\omega^2}{2g}.

The zero-energy equation is

(xB′)2=ω2xB2(1−xBxt).(x_B')^2 = \omega^2x_B^2 \left( 1-\frac{x_B}{x_t} \right).

The solution centered at τ0\tau_0 is

xB(τ)=xtsech⁡2[ω2(τ−τ0)].x_B(\tau) = x_t \operatorname{sech}^2 \left[ \frac{\omega}{2} (\tau-\tau_0) \right].

Indeed,

xB′=−ωxBtanh⁡[ω2(τ−τ0)],1−xBxt=tanh⁡2[ω2(τ−τ0)],\begin{aligned} x_B' &= - \omega x_B \tanh \left[ \frac{\omega}{2} (\tau-\tau_0) \right], \\ 1-\frac{x_B}{x_t} &= \tanh^2 \left[ \frac{\omega}{2} (\tau-\tau_0) \right], \end{aligned}

so the first integral is satisfied on both branches.

Using the round-trip formula,

B=2mω∫0xtdx x1−xxt=815mωxt2=6m3ω55g2.\begin{aligned} B &= 2m\omega \int_0^{x_t} dx\, x \sqrt{ 1-\frac{x}{x_t} } \\ &= \frac{8}{15} m\omega x_t^2 \\ &= \frac{6m^3\omega^5}{5g^2}. \end{aligned}

In terms of the barrier height,

B=365Ubω.B = \frac{36}{5} \frac{U_b}{\omega}.

The natural dimensionless coupling is

λ≡ℏg2m3ω5,\lambda \equiv \frac{\hbar g^2} {m^3\omega^5},

for which

Bℏ=65λ.\frac{B}{\hbar} = \frac{6}{5\lambda}.

The bounce expansion is controlled when λ≪1\lambda\ll1.

Write

x(τ)=xB(τ)+η(τ).x(\tau) = x_B(\tau) + \eta(\tau).

The quadratic fluctuation operator is

MB=−md2dτ2+U′′(xB).\mathcal M_B = - m\frac{d^2}{d\tau^2} + U''(x_B).

Differentiate the bounce equation with respect to τ\tau:

mxB′′′=U′′(xB)xB′.m x_B''' = U''(x_B)x_B'.

Therefore

MBxB′=0.\mathcal M_B x_B' = 0.

The zero mode shifts the center τ0\tau_0. It must be removed from the ordinary determinant and replaced by an integral over τ0\tau_0.

The zero mode xB′x_B' changes sign once, at the center of the bounce. For a one-dimensional Sturm–Liouville operator on the Euclidean line, eigenfunctions are ordered by their number of nodes. A normalizable zero eigenfunction with one node has one lower eigenfunction with no nodes. Its eigenvalue is negative.

Thus a simple one-dimensional bounce has exactly one negative mode. This statement assumes the usual isolated bounce and self-adjoint fluctuation problem. Additional negative modes can signal a different saddle, an excited or oscillating solution, or an incorrectly chosen decay configuration.

For the cubic example,

MBm=−d2dτ2+ω2×[1−3sech⁡2(ω2(τ−τ0))].\begin{aligned} \frac{\mathcal M_B}{m} &= - \frac{d^2}{d\tau^2} + \omega^2 \\ &\quad\times \left[ 1 - 3\operatorname{sech}^2 \left( \frac{\omega}{2} (\tau-\tau_0) \right) \right]. \end{aligned}

With

z=ω2(τ−τ0),z = \frac{\omega}{2} (\tau-\tau_0),

this becomes the Pöschl–Teller operator

MB=mω24[−d2dz2+4−12sech⁡2z].\mathcal M_B = \frac{m\omega^2}{4} \left[ - \frac{d^2}{dz^2} + 4 - 12\operatorname{sech}^2z \right].

Its negative and zero modes can be displayed explicitly:

modeeigenvalueη−(z)∝sech⁡3z−5mω2/4η0(z)∝sech⁡2z tanh⁡z0\begin{array}{c|c} \text{mode} & \text{eigenvalue} \\ \hline \eta_-(z)\propto\operatorname{sech}^3z & -5m\omega^2/4 \\ \eta_0(z)\propto \operatorname{sech}^2z\,\tanh z & 0 \end{array}

The zero mode is proportional to xB′x_B'. The remaining discrete positive mode and continuum affect the determinant prefactor but not the classical exponent.

For a long Euclidean interval of duration TT, the arbitrary center produces a factor proportional to TT. Widely separated bounces can then be summed in a dilute approximation. The negative mode specifies how the integration contour passes through the saddle and produces an imaginary contribution to the analytically continued false-well energy.

The resulting semiclassical structure is

Eres=EF−iℏΓ2,E_{\mathrm{res}} = \mathcal E_F - \frac{i\hbar\Gamma}{2},

with

Γ∼Ae−B/ℏ[1+O(ℏB)].\Gamma \sim A e^{-B/\hbar} \left[ 1 + O\left( \frac{\hbar}{B} \right) \right].

The exponent is fixed by the classical bounce. The prefactor AA must have units of inverse time and depends on fluctuation determinants, zero-mode normalization, the negative-mode contour, and the definition of the metastable state.

This formula does not assert exact exponential decay at all times. Unitary quantum mechanics gives nonexponential behavior at sufficiently short and long times. The bounce result describes the resonance-dominated regime in which a decay rate is a useful emergent quantity.

A one-bounce calculation is parametrically credible when:

Bℏ≫1,ΓωF≪1.\frac{B}{\hbar}\gg1, \qquad \frac{\Gamma}{\omega_F}\ll1.

The first condition suppresses each event; the second separates the bounce width, of order 1/ωF1/\omega_F, from the mean time between events. If bounces overlap strongly, their interactions and quasi-zero modes invalidate a naive Poisson sum.

The symbolic form Ae−B/ℏA e^{-B/\hbar} hides several logically distinct operations.

IngredientWhy it matters
Reference determinantNormalizes fluctuations relative to the false-well saddle
Translation zero modeReplaces a vanishing determinant eigenvalue by an integral over τ0\tau_0
One negative modeSelects a contour and produces the imaginary part associated with decay
Boundary conditionsDistinguish a resonance, fixed-endpoint kernel, thermal trace, and closed system
State normalizationFixes whether the answer is a kernel coefficient, energy width, or rate
Multi-bounce sumConverts the one-event contribution into long-time extensive behavior

The familiar schematic zero-mode replacement,

(det⁡MB)−1/2⟶J0∫dτ0×(det⁡′MB)−1/2.\begin{aligned} (\det\mathcal M_B)^{-1/2} &\longrightarrow J_0 \int d\tau_0 \\ &\quad\times (\det{}'\mathcal M_B)^{-1/2}. \end{aligned}

contains a Jacobian J0J_0 whose precise form depends on the path-integral measure and normalization convention. The negative Gaussian direction is not convergent on the original real fluctuation contour; its continuation supplies an imaginary factor and a prescription-dependent half factor.

These details cannot be reconstructed from BB alone. Fluctuation Determinants Preview is the canonical home for the determinant ratio and collective-coordinate framework.

For a general one-dimensional potential, the first integral is usually more stable than direct shooting.

  1. Find xFx_F, the barrier top, and the first root xtx_t beyond the barrier with U(xt)=U(xF)U(x_t)=U(x_F).
  2. Shift U(xF)U(x_F) to zero.
  3. Evaluate
B=2∫xFxtdx 2mU(x).B = 2 \int_{x_F}^{x_t} dx\, \sqrt{2mU(x)}.
  1. Reconstruct the profile from
∣τ−τ0∣=m2∫xB(τ)xtdyU(y).\left| \tau-\tau_0 \right| = \sqrt{\frac{m}{2}} \int_{x_B(\tau)}^{x_t} \frac{dy}{\sqrt{U(y)}}.

The integrable square-root behavior at xtx_t and logarithmic divergence at xFx_F should be handled with endpoint-aware quadrature or a change of variables.

On a finite symmetric interval [−L,L][-L,L], impose

xB′(0)=0,xB(L)≈xF,x_B'(0)=0, \qquad x_B(L)\approx x_F,

and solve only on [0,L][0,L]. The center value approaches xtx_t as L→∞L\to\infty. A boundary-value solver is usually more stable than launching exactly from xtx_t with zero velocity, because the latter is sensitive to roundoff.

A converged solution should satisfy:

  • EE=m(xB′)2/2−U(xB)≈0\mathcal E_E=m(x_B')^2/2-U(x_B)\approx0 along the path;
  • direct τ\tau integration and configuration-space quadrature agree for BB;
  • the endpoint error falls as e−ωFLe^{-\omega_F L};
  • the smallest odd fluctuation eigenvalue approaches zero as the grid is refined;
  • exactly one lower eigenvalue remains negative;
  • BB is stable under independent changes of LL, mesh density, and quadrature rule.

The translation mode is an especially sensitive end-to-end check because it tests the profile, derivatives, fluctuation operator, and boundary truncation at once.

Several real bounces can exist when a potential has multiple escape directions or intermediate wells. At leading exponential order, the smallest admissible action dominates:

Bdom=min⁡αBα,B_{\mathrm{dom}} = \min_\alpha B_\alpha,

provided no symmetry, contour, or boundary condition excludes that saddle. Nearly equal actions require prefactors and interference information.

Other possibilities include:

  • periodic bounces at finite temperature;
  • energy-dependent periodic instantons;
  • multidimensional bounces with more than one path through configuration space;
  • complex saddles when no relevant real Euclidean solution exists;
  • oscillating solutions with extra negative modes.

Not every stationary Euclidean solution controls a decay rate. Boundary data, contour accessibility, and fluctuation signature remain part of the selection rule.

For one scalar field in dd spatial dimensions,

SE[ϕ]=∫dτ ddx×[12(∂τϕ)2+12(∇ϕ)2+V(ϕ)].\begin{aligned} S_E[\phi] &= \int d\tau\,d^d x \\ &\quad\times \left[ \frac12(\partial_\tau\phi)^2 + \frac12(\boldsymbol\nabla\phi)^2 \right. \\ &\qquad\left. + V(\phi) \right]. \end{aligned}

Under the standard assumptions for a single scalar, the least-action bounce can be taken to be rotationally symmetric in D=d+1D=d+1 Euclidean dimensions. With

ρ=τ2+∣x∣2,\rho = \sqrt{ \tau^2+|\mathbf x|^2 },

the radial equation is

d2ϕBdρ2+dρdϕBdρ=V′(ϕB),\frac{d^2\phi_B}{d\rho^2} + \frac{d}{\rho} \frac{d\phi_B}{d\rho} = V'(\phi_B),

with

ϕB′(0)=0,ϕB(ρ)→ϕFasρ→∞.\begin{gathered} \phi_B'(0)=0, \\ \phi_B(\rho)\to\phi_F \quad \text{as} \quad \rho\to\infty. \end{gathered}

The term dϕB′/ρd\phi_B'/\rho acts like friction in the inverted-potential analogy. The mechanical energy obeys

ddρ[12(ϕB′)2−V(ϕB)]=−dρ(ϕB′)2≤0.\frac{d}{d\rho} \left[ \frac12(\phi_B')^2 - V(\phi_B) \right] = - \frac{d}{\rho} (\phi_B')^2 \leq0.

Quantum mechanics is the case d=0d=0, where the friction term vanishes and the exact first integral is recovered. In field theory the center value is not generally the equal-potential turning point; it must be tuned so that friction leaves the solution at ϕF\phi_F as ρ→∞\rho\to\infty. This is the basis of the overshoot–undershoot construction.

The field-theory exponent is

B=SE[ϕB]−SE[ϕF].B = S_E[\phi_B] - S_E[\phi_F].

Gauge fields, fermions, renormalization, multiple scalar fields, gravity, and finite temperature add substantial structure. Bridge to QFT Instantons maps the saddle dictionary, while From Euclidean Time to Euclidean QFT owns the broader Euclidean-field and reconstruction framework.

Before trusting a bounce exponent, verify:

  1. Reference configuration: Is the false-minimum action subtracted?
  2. Escape geometry: Does a turning point with the correct equal-potential value exist?
  3. Boundary data: Does the path return to the same metastable configuration?
  4. Finite action: Are the Euclidean tails normalizable and exponentially decaying?
  5. Round trip: Has the factor of two in the one-dimensional action been included?
  6. Semiclassical control: Is B/ℏ≫1B/\hbar\gg1?
  7. Mode count: Is there one translation zero mode and exactly one negative mode?
  8. Prefactor units: Does the completed result have units of a decay rate?
  9. Competing saddles: Are lower-action exits, complex saddles, or thermal saddles absent?
  10. Independent check: Does WKB or direct resonance numerics reproduce the exponent?
  • Calling any finite-action Euclidean solution a bounce without checking its endpoints.
  • Confusing an instanton between degenerate minima with a bounce that returns to one false minimum.
  • Omitting the return branch and losing a factor of two in BB.
  • Using the barrier top xbx_b as the turning point; the bounce turns at U(xt)=U(xF)U(x_t)=U(x_F).
  • Leaving the false-vacuum background in an infinite action.
  • Treating the inverted-potential path as a real-time tunneling trajectory.
  • Assuming the bounce is a minimum of SES_E even though it has one negative mode.
  • Including the translation zero eigenvalue in an ordinary determinant.
  • Quoting e−B/ℏe^{-B/\hbar} as a normalized rate without a prefactor and state convention.
  • Assuming exponential decay is exact at arbitrarily short or long times.
  • Copying the frictionless quantum-mechanical first integral into field theory.
  • Choosing the lowest numerical action without checking contour accessibility and boundary conditions.

Starting from

mx′′=U′(x),mx'' = U'(x),

derive the zero-energy first integral, the implicit bounce profile, and the round-trip action.

Solution

Multiply the equation by x′x':

ddτ[m2(x′)2−U(x)]=0.\frac{d}{d\tau} \left[ \frac{m}{2}(x')^2 - U(x) \right] = 0.

At τ→±∞\tau\to\pm\infty, both x′x' and U(x)U(x) vanish, so

m2(x′)2=U(x).\frac{m}{2}(x')^2 = U(x).

Separating variables on either branch gives

∣τ−τ0∣=m2∫x(τ)xtdyU(y).\left| \tau-\tau_0 \right| = \sqrt{\frac{m}{2}} \int_{x(\tau)}^{x_t} \frac{dy}{\sqrt{U(y)}}.

On shell, the kinetic and potential terms are equal. The two branches give

B=2∫xFxtdx∣x′∣ 2U(x)=2∫xFxtdx 2mU(x).\begin{aligned} B &= 2 \int_{x_F}^{x_t} \frac{dx}{|x'|} \,2U(x) \\ &= 2 \int_{x_F}^{x_t} dx\, \sqrt{2mU(x)}. \end{aligned}

Let

W(0)=∫xFxtdx 2mU(x).W(0) = \int_{x_F}^{x_t} dx\, \sqrt{2mU(x)}.

Explain why the bounce exponent is B=2W(0)B=2W(0) and why this matches a tunneling probability rather than a wavefunction amplitude.

Solution

The bounce traverses the forbidden interval twice: once from xFx_F to xtx_t and once on the return branch. Each branch contributes W(0)W(0), so

B=2W(0).B = 2W(0).

In WKB, crossing the barrier suppresses a wavefunction amplitude by e−W/ℏe^{-W/\hbar}. Squaring an amplitude produces a probability suppression

∣e−W/ℏ∣2=e−2W/ℏ=e−B/ℏ.\left| e^{-W/\hbar} \right|^2 = e^{-2W/\hbar} = e^{-B/\hbar}.

Prefactors and the false-well oscillation scale are still needed to turn this probability per attempt into a rate.

For

U(x)=mω22x2−g3x3,U(x) = \frac{m\omega^2}{2}x^2 - \frac{g}{3}x^3,

find xbx_b, UbU_b, and xtx_t. Verify the sech⁡2\operatorname{sech}^2 bounce and compute BB.

Solution

The stationary points obey

U′(x)=x(mω2−gx)=0,U'(x) = x(m\omega^2-gx) = 0,

so the barrier top is

xb=mω2g,Ub=m3ω66g2.x_b = \frac{m\omega^2}{g}, \qquad U_b = \frac{m^3\omega^6}{6g^2}.

The nonzero root of U(x)=0U(x)=0 is

xt=3mω22g.x_t = \frac{3m\omega^2}{2g}.

For

xB=xtsech⁡2z,z=ω2(τ−τ0),x_B = x_t\operatorname{sech}^2z, \qquad z = \frac{\omega}{2} (\tau-\tau_0),

one has

xB′=−ωxBtanh⁡zx_B' = -\omega x_B\tanh z

and

1−xBxt=tanh⁡2z.1-\frac{x_B}{x_t} = \tanh^2z.

Hence

(xB′)2=ω2xB2(1−xBxt),(x_B')^2 = \omega^2x_B^2 \left( 1-\frac{x_B}{x_t} \right),

which is the first integral. Finally,

B=2mωxt2∫01dy y(1−y)1/2=2mωxt2(415)=6m3ω55g2.\begin{aligned} B &= 2m\omega x_t^2 \int_0^1 dy\, y(1-y)^{1/2} \\ &= 2m\omega x_t^2 \left( \frac{4}{15} \right) \\ &= \frac{6m^3\omega^5}{5g^2}. \end{aligned}

Show that xB′x_B' is a zero mode of MB\mathcal M_B. Use its node count to argue that a simple one-dimensional bounce has one negative mode.

Solution

Differentiating

mxB′′=U′(xB)mx_B'' = U'(x_B)

gives

mxB′′′=U′′(xB)xB′.mx_B''' = U''(x_B)x_B'.

Therefore

[−md2dτ2+U′′(xB)]xB′=0.\left[ - m\frac{d^2}{d\tau^2} + U''(x_B) \right] x_B' = 0.

The bounce rises before τ0\tau_0 and falls after it, so xB′x_B' changes sign once. The Sturm–Liouville node theorem orders normalizable eigenfunctions by node number. A zero mode with one node is the first excited fluctuation state; one nodeless eigenfunction lies below it and has negative eigenvalue. No second negative eigenvalue can lie below without violating the node ordering.

Suppose the tail is

xB(τ)−xF≃Ce−ωF∣τ∣.x_B(\tau)-x_F \simeq C e^{-\omega_F|\tau|}.

Estimate the action omitted by truncating the Euclidean line to [−L,L][-L,L].

Solution

In the quadratic tail,

U≃mωF22(xB−xF)2,U \simeq \frac{m\omega_F^2}{2} (x_B-x_F)^2,

and the zero-energy relation makes the kinetic term equal to UU. On the positive tail,

δB+≃∫L∞dτ mωF2C2e−2ωFτ=mωFC22e−2ωFL.\begin{aligned} \delta B_+ &\simeq \int_L^\infty d\tau\, m\omega_F^2C^2 e^{-2\omega_F\tau} \\ &= \frac{m\omega_F C^2}{2} e^{-2\omega_F L}. \end{aligned}

The negative tail contributes equally, so

δB≃mωFC2e−2ωFL.\delta B \simeq m\omega_F C^2 e^{-2\omega_F L}.

The endpoint displacement is of order e−ωFLe^{-\omega_F L}, while the action error is of order e−2ωFLe^{-2\omega_F L}.

6. Derive Euclidean friction in field theory

Section titled “6. Derive Euclidean friction in field theory”

For a radial field-theory bounce satisfying

ϕB′′+dρϕB′=V′(ϕB),\phi_B'' + \frac{d}{\rho}\phi_B' = V'(\phi_B),

show that the inverted-potential mechanical energy decreases with ρ\rho. Explain why the quantum-mechanical equal-potential turning-point rule no longer holds for d>0d\gt0.

Solution

Define

E(ρ)=12(ϕB′)2−V(ϕB).\mathcal E(\rho) = \frac12(\phi_B')^2 - V(\phi_B).

Then

dEdρ=ϕB′[ϕB′′−V′(ϕB)]=−dρ(ϕB′)2≤0.\begin{aligned} \frac{d\mathcal E}{d\rho} &= \phi_B' \left[ \phi_B''-V'(\phi_B) \right] \\ &= - \frac{d}{\rho} (\phi_B')^2 \leq0. \end{aligned}

For d=0d=0, the right-hand side vanishes and energy is conserved, so the center and false-vacuum endpoints lie at equal potential values. For d>0d\gt0, friction removes mechanical energy as the trajectory moves outward in ρ\rho. The center value must start farther up the inverted potential so that the damped trajectory approaches ϕF\phi_F at infinity.

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