Quantum-Mechanical Instantons and Tunnel Splitting

In a symmetric double well, perturbation theory about either minimum produces the same formal energy series and cannot split the nearly degenerate pair. The missing scale is $e^{-S_I^{\rm phys}/\hbar}=e^{-\mathcal S_I/g}$. For a fixed normalization, a single instanton supplies the transition fugacity, and a dilute instanton–anti-instanton sum exponentiates it into the even–odd level splitting.

Required background. Euclidean tunneling saddles and boundary conditions supplies the heteroclinic boundary-value problem. Multi-saddle sums and dilute ensembles supplies the combinatorics and the independent diluteness test.

Helpful background. Renormalized saddle contributions and validity tests separates loop control from ensemble control. WKB and turning-point matching gives the canonical one-dimensional alternative.

The normalized double well

Use the same dimensionless normalization throughout the semiclassical chapters:

$$\frac{S_E^{\rm phys}[x]}{\hbar} =\frac{\mathcal S_E[x]}{g} =\frac{1}{g}\int d\tau \left[\frac12\dot x^2+\frac12(x^2-1)^2\right], \qquad g\ll1.$$

The minima are $x=\pm1$, and small oscillations have frequency $2$. The zero-energy first-order equation is

$$\dot x=1-x^2.$$

Its increasing solution is

$$x_I(\tau)=\tanh(\tau-\tau_0), \qquad \mathcal S_I = \int_{-1}^{1}dx\,(1-x^2) =\frac43,$$

so the physical exponent is $S_I^{\rm phys}/\hbar=\mathcal S_I/g=4/(3g)$. The anti-instanton has the opposite sign and reverses the endpoints.

This normalization follows from the dimensional potential $V(q)=\lambda(q^2-a^2)^2$ after setting $q=ax$ and $\tau=\omega t/2$, with

$$\omega^2=\frac{8\lambda a^2}{m}, \qquad \frac1g=\frac{a^3\sqrt{2m\lambda}}{\hbar}.$$

Thus $S_I^{\rm phys}=4a^3\sqrt{2m\lambda}/3$, exactly as on the boundary-condition page. In these variables the dimensionless Hamiltonian is

$$\widehat H \equiv \frac{2H_{\rm phys}}{\hbar\omega} =-\frac g2\frac{d^2}{dx^2} +\frac{(x^2-1)^2}{2g}.$$

All energies below are eigenvalues of $\widehat H$; physical energies are $E_{\rm phys}=(\hbar\omega/2)\widehat E$.

The translation mode and determinant

Write $x=x_I+\sqrt g\,\eta$. The quadratic operators about the instanton and either vacuum are

$$\mathcal M_I =-\frac{d^2}{d\tau^2}+4-6\operatorname{sech}^2(\tau-\tau_0), \qquad \mathcal M_0=-\frac{d^2}{d\tau^2}+4.$$

Differentiating the saddle equation shows

$$\mathcal M_I\dot x_I=0.$$

The zero eigenvalue cannot remain in the Gaussian determinant. Replacing its amplitude by the center $\tau_0$ yields the Jacobian

$$J_I=\left(\frac{\mathcal S_I}{2\pi g}\right)^{1/2}.$$

For this Pöschl–Teller operator, the regulated determinant ratio is

$$\frac{\det{}'\mathcal M_I}{\det\mathcal M_0}=\frac1{48}.$$

The normalization map matters. Mariño uses operators $\widetilde{\mathcal M}$ with unit vacuum frequency, whereas the operators here obey $\mathcal M=4\widetilde{\mathcal M}$. In a common finite-interval regulator, the primed numerator has one fewer eigenvalue than the denominator, so

$$\frac{\det{}'(4\widetilde{\mathcal M}_I)} {\det(4\widetilde{\mathcal M}_0)} =\frac14 \frac{\det{}'\widetilde{\mathcal M}_I} {\det\widetilde{\mathcal M}_0} =\frac14\cdot\frac1{12} =\frac1{48}.$$

This one factor of $1/4$, rather than a cancellation of all powers of four, is the diagnostic effect of removing the translation eigenvalue.

Combining the determinant, Jacobian, and normalization of the asymptotic harmonic kernel gives the one-event density per unit Euclidean time

$$\kappa_{\text{1-loop}} = 4\sqrt{\frac{2}{\pi g}}\, \exp\!\left(-\frac{4}{3g}\right).$$

The prime removes only the translation zero mode. No negative eigenvalue is present: $\dot x_I=\operatorname{sech}^2(\tau-\tau_0)$ is nodeless and therefore the lowest eigenfunction. Mariño 2015, §§ 1.8–1.9, pp. 38–53 gives the unit-frequency determinant and the rescaling rule used above.

The passage from a saddle through collective coordinates and determinants to a usable measure is summarized on Instanton Measures, Zero Modes, and Determinants. The distinctions from gauge instantons and false-vacuum bounces are tabulated in the instanton–bounce boundary and mode comparison.

From the dilute sum to two energies

Let $|L\rangle$ and $|R\rangle$ be states localized near $x=-1$ and $x=+1$. A path that begins and ends in the same well contains an even number of alternating events; a path that changes wells contains an odd number. Neglecting interactions between well-separated events, the ordered-center integral gives $T^n/n!$. Therefore

$$\begin{aligned} \langle L|e^{-\widehat H T}|L\rangle &\simeq A(T)e^{-\widehat E_{\mathrm{pert}}T} \sum_{k=0}^{\infty}\frac{(\kappa T)^{2k}}{(2k)!} =A(T)e^{-\widehat E_{\mathrm{pert}}T}\cosh(\kappa T),\\ \langle R|e^{-\widehat H T}|L\rangle &\simeq A(T)e^{-\widehat E_{\mathrm{pert}}T} \sum_{k=0}^{\infty}\frac{(\kappa T)^{2k+1}}{(2k+1)!} =A(T)e^{-\widehat E_{\mathrm{pert}}T}\sinh(\kappa T). \end{aligned}$$

Here $A(T)$ contains the common perturbative normalization. Forming parity eigenstates

$$|{\rm even}\rangle=\frac{|L\rangle+|R\rangle}{\sqrt2}, \qquad |{\rm odd}\rangle=\frac{|L\rangle-|R\rangle}{\sqrt2},$$

and comparing the large-$T$ exponentials gives

$$\widehat E_{\rm even}=\widehat E_{\mathrm{pert}}-\kappa, \qquad \widehat E_{\rm odd}=\widehat E_{\mathrm{pert}}+\kappa, \qquad \Delta\widehat E=2\kappa.$$

At one loop,

$$\Delta\widehat E_{\text{1-loop}} =8\sqrt{\frac{2}{\pi g}}\, e^{-4/(3g)}.$$

The exponential is nonanalytic at $g=0$, which is why no finite perturbative expansion about one minimum can generate it. Coleman’s transfer-matrix derivation makes the same parity projection explicit in Coleman 1985, chapter 7, § 2.2, pp. 270–277.

Numerical diagonalization benchmark

The prefactor can be tested without fitting the exponent. On the half-line $x\ge0$, the even ground state obeys $\psi'(0)=0$, the odd state obeys $\psi(0)=0$, and both decay at large $x$. A midpoint finite-difference implementation uses $x_j=(j+\tfrac12)h$ and

$$(\widehat H_h\psi)_j =-\frac{g}{2h^2} (\psi_{j+1}-2\psi_j+\psi_{j-1}) +\frac{(x_j^2-1)^2}{2g}\psi_j.$$

Parity is imposed by the ghost values $\psi_{-1}=\psi_0$ in the even sector and $\psi_{-1}=-\psi_0$ in the odd sector. A Dirichlet wall halfway between the last physical point and its ghost gives $\psi_N=-\psi_{N-1}$. Solving the lowest tridiagonal eigenvalue in each sector avoids mixing the exponentially close parity pair; their difference is $\Delta\widehat E_{\rm num}=\widehat E_{\rm odd}-\widehat E_{\rm even}$.

The table uses $L=5$, the three spacings $h=10^{-3},5\times10^{-4},2.5\times10^{-4}$, and second-order Richardson extrapolation. The last column tests both the instanton exponent and its one-loop normalization:

$$R(g)=\frac{\Delta\widehat E_{\rm num}} {\Delta\widehat E_{\rm 1-loop}}.$$

$g$	$\Delta\widehat E_{\rm num}$	$\Delta\widehat E_{\rm 1-loop}$	$R(g)$
$0.20$	$1.5087781324\times10^{-2}$	$1.8164293248\times10^{-2}$	$0.83062859$
$0.15$	$1.9973566014\times10^{-3}$	$2.2729455093\times10^{-3}$	$0.87875252$
$0.10$	$3.0140910192\times10^{-5}$	$3.2691658714\times10^{-5}$	$0.92197556$
$0.08$	$1.2234903102\times10^{-6}$	$1.3038982124\times10^{-6}$	$0.93833268$

Across these rows the observed grid order is $p=2.0000$, the largest relative Richardson correction is $2.9\times10^{-6}$, and changing the half-domain from $L=3$ to $L=5$ changes the reported gap by less than $2\times10^{-7}$ relatively. An independent 60-digit harmonic-oscillator-basis calculation agrees through the displayed digits. The approach $R(g)\to1$ as $g$ decreases is the predicted weak-coupling trend; $1-R$ is mainly a physical higher-loop and correlated-event correction, not a discretization error. At still smaller $g$, a double-precision calculation will eventually lose the gap when subtracting nearly equal energies, so arithmetic precision must then become part of the convergence test.

Control and limitations

Three scales must be separated:

1. Local loop control: $g\ll1$, equivalently $\mathcal S_I/g\gg1$.
2. Diluteness: $\kappa\ell_I\ll1$, where the core width is $\ell_I\sim1$ in these units.
3. Spectral projection: $T\gg1$ to suppress higher oscillator states, while the sum retains all powers of $\kappa T$.

The third condition is compatible with many instantons in a long interval: diluteness constrains their mean separation, not their total number. Corrections arise from higher loops around each event, interactions in close instanton–anti-instanton pairs, and additional saddles. Exact WKB and resurgence organize some of those corrections, but they are not needed to establish the leading splitting.

Common pitfalls

Keeping the zero eigenvalue in the determinant. The translation mode is integrated as $d\tau_0$ with its Jacobian. Including it in $\det\mathcal M_I$ would make the prefactor vanish and double-count the same direction.

Equating a one-instanton amplitude with an energy shift. A single event changes wells. The energy eigenvalues emerge only after summing the allowed even and odd sequences and projecting onto parity.

Using $\kappa T\ll1$ as the dilute criterion. Large $T$ is required for spectroscopy and may make $\kappa T$ large. The relevant small quantity is the overlap of neighboring event cores, $\kappa\ell_I$.

Exercises

1. Verify directly that $x_I=\tanh(\tau-\tau_0)$ solves the second-order Euclidean equation and has $\mathcal S_I/g=4/(3g)$.

Solution

The equation is $\ddot x=2x(x^2-1)$. For $x_I=\tanh u$, $\dot x_I=\operatorname{sech}^2u=1-x_I^2$, hence $\ddot x_I=-2x_I(1-x_I^2)=2x_I(x_I^2-1)$. Along this solution, $\dot x_I^2=(1-x_I^2)^2$, so

$$\frac{\mathcal S_I}{g}=\frac1g\int d\tau\,\dot x_I^2 =\frac1g\int_{-1}^{1}dx\,(1-x^2) =\frac{4}{3g}.$$

2. Starting from the even- and odd-event sums, show that the lower state is even and that $\Delta\widehat E=2\kappa$.

Solution

Adding the diagonal and off-diagonal kernels selects $\cosh(\kappa T)+\sinh(\kappa T)=e^{\kappa T}$, so the even channel behaves as $e^{-(\widehat E_{\rm pert}-\kappa)T}$. Subtracting selects $e^{-\kappa T}$, so the odd channel behaves as $e^{-(\widehat E_{\rm pert}+\kappa)T}$. Their difference is $2\kappa$, and positivity of $\kappa$ places the nodeless even state lower.

3. Suppose both fluctuation operators are rescaled by a positive constant, $\mathcal M=c\widetilde{\mathcal M}$. In a common finite-dimensional regulator with $n$ denominator eigenvalues, show that a determinant ratio with one numerator zero mode removed acquires the factor $c^{-1}$. Apply this to $c=4$ and the unit-frequency ratio $1/12$.

Solution

After the zero mode is removed, the numerator contains $n-1$ eigenvalues while the denominator contains $n$. Therefore

$$\frac{\det{}'(c\widetilde{\mathcal M}_I)} {\det(c\widetilde{\mathcal M}_0)} =c^{n-1-n} \frac{\det{}'\widetilde{\mathcal M}_I} {\det\widetilde{\mathcal M}_0} =\frac1c\frac1{12}.$$

For $c=4$, the ratio is $1/48$. The missing power of $c$ is a useful check that the translation mode was removed exactly once.

4. At $g=0.10$ and $g=0.08$, the benchmark gives $1-R\simeq7.8\%$ and $6.2\%$, respectively. Explain why these discrepancies should not be identified with numerical error.

Solution

Grid refinement, Richardson extrapolation, domain enlargement, and an independent basis calculation place the numerical uncertainty far below one percent. The one-loop formula, however, omits relative corrections beginning at $O(g)$ as well as correlated multi-event effects. The shrinking discrepancy as $g$ decreases is therefore evidence for the asymptotic prediction, while the residual is predominantly omitted physics.

References

• Coleman, Sidney. Aspects of Symmetry: Selected Erice Lectures. Cambridge: Cambridge University Press, 1985. DOI.
• Mariño, Marcos. Instantons and Large N: An Introduction to Non-Perturbative Methods in Quantum Field Theory. Cambridge: Cambridge University Press, 2015. DOI.