3.4.3 Instantons#

Prompts

  • What is an instanton, and why does the Euclidean equation of motion admit a real, finite-action solution crossing the barrier while the real-time equation does not?

  • How does the transition amplitude \(\langle -a\vert\mathrm{e}^{-\hat{H}T/\hbar}\vert a\rangle\) give the energy splitting by matching its eigenstate expansion against the dilute instanton gas?

  • Why is the splitting \(\Delta\propto\mathrm{e}^{-S_{\text{inst}}/\hbar}\) called non-perturbative, and why does no finite order of perturbation theory in \(\hbar\) produce it?

  • How does the instanton exponent compare to the WKB tunneling exponent? Why can we leave the prefactor \(K\) symbolic without losing the key physics?

Lecture Notes#

Overview#

The Wick rotation of §3.4.1 revealed that imaginary-time motion obeys Newton’s law in the inverted potential \(-V(x)\) — barriers become hills. §3.4.2 put this geometric fact to dramatic use: demanding that the Euclidean black-hole geometry close smoothly into a thermal circle delivered the Hawking temperature without quantizing a single field. Here we exploit the inverted-potential trick for a third miracle, this time inside ordinary quantum mechanics: classically forbidden tunneling through a barrier becomes a classically allowed trajectory in imaginary time. The trajectory that crosses the barrier — the instanton — has finite Euclidean action, and summing over instanton paths (the dilute instanton gas) produces the ground-state energy splitting \(\Delta \propto \mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\). This is the paradigm calculation of non-perturbative quantum mechanics.

The Inverted Potential#

Two ingredients from §3.4.1 set up everything that follows. First, the Wick rotation \(t \to -\mathrm{i}\tau\) replaces the real-time action \(S = \int(\tfrac{1}{2}m\dot{x}^2 - V)\,\mathrm{d}t\) with the Euclidean action

(138)#\[ S_E[x] \;=\; \int\!\Bigl(\tfrac{1}{2}m\,x'(\tau)^2 + V(x)\Bigr)\,\mathrm{d}\tau, \]

in which kinetic and potential terms add. Second, varying \(S_E\) gives a second-order equation of motion with the opposite sign from real-time Newton’s law:

(139)#\[ m\,x''(\tau) \;=\; +V'(x). \]

In imaginary time the particle moves as if it were rolling in the inverted potential \(-V(x)\). Barriers in \(V\) become hills in \(-V\) that the particle can climb over with finite Euclidean action; trajectories that are classically forbidden in real time become classically allowed in imaginary time. The rest of this lesson exploits this single sign flip in the case of a double-well potential.

The Double-Well Potential#

The symmetric double-well potential is

(140)#\[ V(x) = \lambda(x^2 - a^2)^2 \]

with two degenerate minima at \(x = \pm a\) (where \(V(\pm a) = 0\)) and a barrier of height \(\lambda a^4\) at the origin.

Classical vs. quantum. Classically, a particle with energy \(E < \lambda a^4\) is trapped in one well. Quantum mechanically, tunneling splits the ground state into a symmetric combination \(\vert +\rangle\) (energy \(E_0 - \Delta\)) and an antisymmetric combination \(\vert -\rangle\) (energy \(E_0 + \Delta\)). The splitting \(2\Delta\) is non-perturbative in \(\hbar\): it is \(O(\mathrm{e}^{-1/\hbar})\) and invisible at any finite order in perturbation theory.

The Instanton Solution.

A first-order reduction (Bogomolny trick) simplifies the second-order equation. Any path satisfying

\[ \dot{x} = \sqrt{\frac{2V(x)}{m}} \]

automatically solves the Euclidean equation of motion (verify by differentiating). For the double-well \(V = \lambda(x^2 - a^2)^2\):

\[ \dot{x} = \sqrt{\frac{2\lambda}{m}}\,(a^2 - x^2) \]

Separating variables and integrating gives the instanton (kink) solution:

Instanton

An instanton is a classical solution in Euclidean time that interpolates between two distinct minima. For the double-well potential (140), the unique (up to center position \(\tau_0\)) finite-action kink is

(141)#\[ x_{\mathrm{inst}}(\tau) = a\tanh\!\left(\frac{\omega(\tau - \tau_0)}{2}\right) \]

where \(\omega = 2a\sqrt{2\lambda/m}\) is the oscillation frequency at either minimum. It satisfies \(x_{\mathrm{inst}}(-\infty) = -a\) and \(x_{\mathrm{inst}}(+\infty) = +a\).

The anti-instanton \(\bar{x}_{\mathrm{inst}}(\tau) = -a\tanh(\omega(\tau - \tau_0)/2)\) traverses the barrier in the opposite direction, from \(+a\) to \(-a\).

The Instanton Action.

The Euclidean action of the instanton is

Instanton Action

(142)#\[ S_{\mathrm{inst}} = \int_{-\infty}^{+\infty}\!\left(\frac{m}{2}\dot{x}^2 + V(x)\right)\mathrm{d}\tau = \frac{4\sqrt{2\lambda m}\,a^3}{3} \]

This equals the WKB tunneling integral \(\int_{-a}^{a}\sqrt{2mV(x)}\,\mathrm{d}x\), confirming that the instanton exponent matches the WKB result.

Energy Level Splitting#

The key is to compute the transition amplitude \(\langle{-a}\vert\mathrm{e}^{-\hat{H}T/\hbar}\vert{a}\rangle\) in two independent ways, then match.

Step 1 — Amplitude from Eigenstate Expansion.

Insert a complete set of energy eigenstates \(\{\vert n\rangle\}\):

\[ \langle -a \vert\,\mathrm{e}^{-\hat{H}T/\hbar}\,\vert a\rangle = \sum_n \psi_n(-a)\,\psi_n^*(a)\,\mathrm{e}^{-E_n T/\hbar} \]

For large \(T\), the sum is dominated by the two lowest states. Since \(V\) is symmetric, \(\psi_+\) is even and \(\psi_-\) is odd, so \(\psi_+(-a) = \psi_+(a)\) and \(\psi_-(-a) = -\psi_-(a)\). With \(E_\pm = E_0 \mp \Delta\):

\[ \langle -a \vert\,\mathrm{e}^{-\hat{H}T/\hbar}\,\vert a\rangle \xrightarrow{T\to\infty} \mathrm{e}^{-E_0 T/\hbar}\sinh\!\left(\frac{\Delta T}{\hbar}\right) \]

The \(\sinh\) arises from the sign difference between \(\psi_+\) and \(\psi_-\) at \(x = \pm a\).

Step 2 — Amplitude from the Dilute Instanton Gas.

The same amplitude equals the Euclidean path integral \(\int_{x(0)=a}^{x(T)=-a}\mathcal{D}[x]\,\mathrm{e}^{-S_E/\hbar}\).

Single instanton centered at \(\tau_0 \in [0,T]\): the dominant saddle. Including Gaussian fluctuations around the saddle (functional determinant), one instanton contributes

\[ \int_0^T \mathrm{d}\tau_0\; K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar} = K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\,T \]

where \(K\) is the pre-exponential factor from the functional determinant (it involves the ratio of the fluctuation operators with and without the instanton).

Many instantons. Over a large interval \(T\), the particle can tunnel back and forth \(n\) times. The \(n\)-instanton path has action \(\approx nS_{\mathrm{inst}}\) (dilute gas: widely separated instantons don’t interact), and contributes \((K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\,T)^n/n!\).

Going from \(x = +a\) to \(x = -a\) requires an odd number of crossings (\(n = 1, 3, 5, \ldots\)):

\[ \sum_{\substack{n=1,3,5,\ldots}} \frac{(K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\,T)^n}{n!} = \sinh\!\left(K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\,T\right) \]

Including the zero-point factor \(\mathrm{e}^{-E_0 T/\hbar}\):

\[ \langle -a\vert\,\mathrm{e}^{-\hat{H}T/\hbar}\,\vert a\rangle = \mathrm{e}^{-E_0 T/\hbar}\sinh\!\left(K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\,T\right) \]

Step 3 — Matching.

Both expressions have the same \(\mathrm{e}^{-E_0 T/\hbar}\sinh(\cdots T)\) structure. Identifying the arguments:

\[ \frac{\Delta T}{\hbar} = K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\,T \]

Energy Splitting

The half-splitting \(\Delta\) (where the full energy gap is \(2\Delta = E_- - E_+\)) is

(143)#\[ \Delta = \hbar K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar} \]

The exponent \(\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\) matches the WKB tunneling exponent. The pre-factor \(K\) (from the Gaussian fluctuation determinant around the instanton) goes beyond WKB and gives the correct coefficient — this is the main advantage of the instanton method.

For this course we leave \(K\) symbolic. Computing it requires evaluating an infinite-dimensional Gaussian determinant, including a careful treatment of the zero mode associated with the instanton’s center \(\tau_{0}\) — material standardly covered in graduate field theory but beyond our scope. The key qualitative physics — that the splitting is exponentially small in \(1/\hbar\), with the same exponent as WKB — is captured already by \(\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\).

Summary#

  • Double-well potential and tunneling: \(V(x) = \lambda(x^2-a^2)^2\) has degenerate minima; quantum tunneling splits them by \(2\Delta\), a non-perturbative effect invisible at any finite order in \(\hbar\).

  • Euclidean equation of motion: \(m\ddot{x} = +\mathrm{d}V/\mathrm{d}x\) inverts the potential; the kink solution \(x_{\mathrm{inst}}(\tau) = a\tanh(\omega\tau/2)\) tunnels the barrier classically in imaginary time.

  • Instanton calculation: The instanton action \(S_{\mathrm{inst}} = 4\sqrt{2\lambda m}a^3/3\) enters the splitting via instanton gas: matching eigenstate expansion (\(\sinh(\Delta T/\hbar)\)) to instanton gas (\(\sinh(K\mathrm{e}^{-S/\hbar}T)\)) yields \(\Delta = \hbar K\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\).

  • Non-perturbative suppression: The splitting is \(O(\mathrm{e}^{-1/\hbar})\)—exponentially small and inaccessible to any finite perturbation series, exemplifying tunneling as a purely quantum phenomenon.

  • 3.4.1 Wick Rotation — time becomes temperature; Euclidean path integral and the inverted-potential equation of motion.

  • 3.4.2 Black Hole Temperature — Wick rotation applied to spacetime geometry; another payoff of “imaginary time is periodic”.

  • 3.3.2 WKB Approximation — real-time semi-classical tunneling, for comparison with the instanton exponent.

See Also

  • 3.4.1 Wick Rotation — time becomes temperature; Euclidean path integral and the inverted-potential equation of motion.

  • 3.4.2 Black Hole Temperature — Wick rotation applied to spacetime geometry; another payoff of “imaginary time is periodic”.

  • 3.3.2 WKB Approximation — real-time semi-classical tunneling, for comparison with the instanton exponent.

Homework#

1. Instanton verification. For the double-well potential \(V(x) = \lambda(x^{2} - a^{2})^{2}\), verify that \(x_{\text{inst}}(\tau) = a\tanh(\omega\tau/2)\) satisfies the Euclidean equation of motion \(m\,x''(\tau) = +V'(x)\). (Hint: compute \(x'(\tau)\) and \(x''(\tau)\), then check both sides.) Also verify that \(x_{\text{inst}}\to\pm a\) as \(\tau\to\pm\infty\), so the kink interpolates between the two minima.

2. Instanton action. For the same double-well potential, compute the instanton action \(S_{\text{inst}} = \int_{-\infty}^{+\infty} \bigl(\tfrac{m}{2}\dot{x}^{2} + V(x)\bigr)\,\mathrm{d}\tau\) explicitly using the kink solution. Express your answer in terms of \(m\), \(\lambda\), and \(a\).

3. Inverted potential geometry. The Euclidean equation of motion \(m\,x''(\tau) = +V'(x)\) is the real-time motion in the inverted potential \(-V\).

(a) Sketch \(-V(x) = -\lambda(x^{2} - a^{2})^{2}\) and identify its two maxima.

(b) Use energy conservation in the inverted system, \(\tfrac{1}{2}m\,(x')^{2} - V(x) = -V(\pm a) = 0\), to derive the first-order equation \(m\,x' = \sqrt{2m\,V(x)}\) (BPS-type equation) for the kink trajectory.

(c) Integrate \(m\,\mathrm{d}x/\sqrt{2m\,V(x)} = \mathrm{d}\tau\) explicitly for the double-well potential and recover \(x(\tau) = a\tanh(\omega\tau/2)\) with \(\omega = \sqrt{8\lambda a^{2}/m}\).

4. Numerical splitting. Treat a quantum-mechanical “atom” trapped in the double well: \(V(x) = \lambda(x^{2} - a^{2})^{2}\) with \(\lambda = 1\,\mathrm{eV/nm^{4}}\), \(a = 0.1\,\mathrm{nm}\), \(m = m_{e}\) (electron mass).

(a) Compute the small-oscillation frequency \(\omega = \sqrt{8\lambda a^{2}/m}\) in units of \(\mathrm{rad/s}\), and the harmonic level spacing \(\hbar\omega\) in eV.

(b) Compute \(S_{\text{inst}}/\hbar\) using the result of Problem 2.

(c) Estimate the splitting \(\Delta = \hbar K\,\mathrm{e}^{-S_{\text{inst}}/\hbar}\) taking \(K\sim\omega\) on dimensional grounds. Compare \(\Delta\) to \(\hbar\omega\) and explain why the suppression by \(\mathrm{e}^{-S_{\text{inst}}/\hbar}\) makes \(\Delta\) “non-perturbative” — invisible to any finite-order expansion in \(\hbar\) around either single well.

5. Even vs odd traversals. The dilute instanton gas sums over paths with \(n = 0, 1, 2,\ldots\) instantons placed at distinct times along the imaginary-time interval \(T\).

(a) Argue that paths with an odd number of instantons end at the opposite minimum from where they started, while paths with an even number return to the original minimum. (Use the kink structure \(\pm a \to \mp a\).)

(b) Sum the contributions of odd \(n\) to obtain \(\langle -a\vert\mathrm{e}^{-\hat{H}T/\hbar}\vert a\rangle = \mathrm{e}^{-E_{0}T/\hbar}\sinh\!\bigl(K\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}T\bigr)\), and the contributions of even \(n\) to obtain \(\langle a\vert\mathrm{e}^{-\hat{H}T/\hbar}\vert a\rangle = \mathrm{e}^{-E_{0}T/\hbar}\cosh\!\bigl(K\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}T\bigr)\).

(c) Compare the off-diagonal expression with the eigenstate expansion \(\mathrm{e}^{-E_{0}T/\hbar}\sinh(\Delta T/\hbar)\) to recover the splitting formula \(\Delta = \hbar K\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\).

6. When do instantons exist. Consider three potentials:

(i) \(V(x) = \lambda(x^{2} - a^{2})^{2}\) — degenerate double well (the lecture’s example),

(ii) \(V(x) = \lambda x^{4}\) — single well at \(x = 0\),

(iii) \(V(x) = \lambda(x^{2} - a^{2})^{2} - \epsilon x\) — asymmetric double well with small \(\epsilon > 0\).

(a) For each, sketch \(-V(x)\) and identify the asymptotic behavior. Which inverted potentials admit a finite-action solution \(x(\tau)\) interpolating between two maxima of \(-V\) as \(\tau \to \pm\infty\)?

(b) For (ii), explain why no finite-action kink exists: what feature of the inverted potential is missing?

(c) For (iii), the minima of \(V\) are split by \(\Delta V \approx 2\epsilon a\). Argue that the inverted-potential trajectory starting from rest at the higher maximum cannot end at rest at the lower one — energy conservation forbids it. Conclude that exact instantons require degenerate vacua.

7. WKB vs instanton. The WKB tunneling exponent is \(T \propto \exp(-2\gamma)\) with \(\gamma = \frac{1}{\hbar}\int_{x_{1}}^{x_{2}}\kappa(x)\,\mathrm{d}x\) and \(\kappa(x) = \sqrt{2m(V(x) - E)}\) (matching the §3.3.2 convention). For the double-well at energy \(E = 0\) (the bottom of either well), compare \(2\gamma\) to \(S_{\text{inst}}/\hbar\) and confirm both predict the same exponential dependence on \(\lambda\), \(m\), and \(a\).