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 $e^{-S_I/\hbar}$. 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:

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

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 $S_I/\hbar=\mathcal S_I/g=4/(3g)$. 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 $S_I=4a^3\sqrt{2m\lambda}/3$, exactly as on the boundary-condition page.

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

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

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. The full calculation, including the same determinant normalization, is given in Mariño 2015, §§ 1.8–1.9, pp. 38–53.

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

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

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

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

At one loop,

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

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.

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 action $4/(3g)$.

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

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

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

Solution

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

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.