3.4.3 Instantons#

Prompts

What is an instanton? Why does the Euclidean equation of motion \(m\ddot{x} = +\mathrm{d}V/\mathrm{d}x\) admit a real solution crossing the barrier, while the real-time equation does not?
Walk me through the three-step derivation of the energy splitting: how does the transition amplitude \(\langle{-a}\vert\mathrm{e}^{-\hat{H}T/\hbar}\vert{a}\rangle\) give a \(\sinh\) from both the eigenstate expansion and the instanton gas?
Why is the energy splitting \(\Delta \propto \mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\) called non-perturbative? What does it mean that perturbation theory cannot produce this result?
How does the instanton result compare to WKB tunneling — same exponent, different pre-factor. What physical information is contained in the pre-factor \(K\)?

Lecture Notes#

Overview#

Instantons encode quantum tunneling non-perturbatively. By studying the Euclidean equations of motion for a double-well potential, one discovers a finite-action classical solution — the instanton — that interpolates between the two minima. Summing over all instanton contributions (the dilute instanton gas) yields the ground-state energy splitting \(\Delta \propto \mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\). This is the paradigm calculation of non-perturbative quantum mechanics.

The Double-Well Potential#

The symmetric double-well potential is

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

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.

Classical Equation of Motion in Imaginary Time#

In real time, no classical path crosses the barrier at \(E < V_{\max}\). However, the Wick rotation \(t \to -\mathrm{i}\tau\) flips the sign of the potential in the equation of motion. The Euclidean equation of motion is

(67)#\[ m\ddot{x} = +\frac{\mathrm{d}V}{\mathrm{d}x} \]

This corresponds to a particle moving in the inverted potential \(-V(x)\): the barrier becomes a hill the particle can roll over, and a smooth finite-action trajectory exists.

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 (66), the unique (up to center position \(\tau_0\)) finite-action kink is

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

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

Derivation: Instanton from Separation of Variables

Starting from \(\dot{x} = \sqrt{2\lambda/m}\,(a^2 - x^2)\), separate variables:

\[ \frac{\mathrm{d}x}{a^2 - x^2} = \sqrt{\frac{2\lambda}{m}}\,\mathrm{d}\tau \]

Integrating both sides (with partial fractions on the left):

\[ \frac{1}{2a}\ln\frac{a + x}{a - x} = \sqrt{\frac{2\lambda}{m}}\,(\tau - \tau_0) \]

Inverting gives \(x(\tau) = a\tanh(\omega(\tau - \tau_0)/2)\) where \(\omega = 2a\sqrt{2\lambda/m}\). \(\checkmark\)

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

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

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

Derivation: Computing \(S_{\mathrm{inst}}\)

Using the first-order equation \(\dot{x} = \sqrt{2V/m}\), the kinetic and potential terms are equal along the instanton trajectory: \(\frac{m}{2}\dot{x}^2 = V(x)\). Therefore

\[ S_{\mathrm{inst}} = 2\int_{-\infty}^{+\infty} V(x(\tau))\,\mathrm{d}\tau = \int_{-a}^{a}\sqrt{2mV(x)}\,\mathrm{d}x \]

where we changed variables from \(\tau\) to \(x\) using \(\mathrm{d}x = \dot{x}\,\mathrm{d}\tau = \sqrt{2V/m}\,\mathrm{d}\tau\). Evaluating:

\[ S_{\mathrm{inst}} = 2\int_0^a \sqrt{2m\lambda}\,(a^2 - x^2)\,\mathrm{d}x = 2\sqrt{2m\lambda}\left(a^2 \cdot a - \frac{a^3}{3}\right) = \frac{4\sqrt{2\lambda m}\,a^3}{3} \]

\(\checkmark\)

Discussion

The first-order equation \(\dot{x} = \sqrt{2V/m}\) is the Euclidean analog of energy conservation with \(E = 0\). In the inverted potential \(-V\), this means the “particle” starts at rest at the top of one hill (\(x = -a\)) and arrives at rest at the other (\(x = +a\)). Why must the energy be exactly zero for a finite-action solution? What happens if the energy is slightly positive — does the solution still connect the two minima?

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

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

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

Discussion

The energy splitting \(\Delta = \hbar K\,\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\) is non-perturbative: it vanishes faster than any power of \(\hbar\) as \(\hbar \to 0\), so no finite-order perturbation theory can produce it.

The dilute gas approximation assumes instantons are widely separated. When the typical inter-instanton separation \(\sim T/n\) becomes comparable to the instanton width \(\sim 1/\omega\), what breaks down? Is there a regime where instantons interact?
The pre-factor \(K\) involves a functional determinant: the ratio of eigenvalues of the fluctuation operator with and without the instanton. One eigenvalue is zero (the translation zero mode, from shifting \(\tau_0\)). How is this zero mode handled, and why does it produce the factor \(T\) in the amplitude?
In QCD, instantons connect topologically distinct vacuum states and contribute to the axial anomaly and the \(\theta\)-vacuum. What is the analog of the double-well’s two minima, and what observable corresponds to \(\Delta\)?

Summary#

Double-well potential: \(V(x) = \lambda(x^2 - a^2)^2\) has degenerate minima at \(x = \pm a\); tunneling splits them by \(2\Delta\)
Euclidean EOM: \(m\ddot{x} = +\mathrm{d}V/\mathrm{d}x\) inverts the potential, enabling barrier crossing
Instanton: kink solution \(x_{\mathrm{inst}}(\tau) = a\tanh(\omega\tau/2)\) with action \(S_{\mathrm{inst}} = \frac{4\sqrt{2\lambda m}\,a^3}{3}\)
Three-step argument: eigenstate expansion gives \(\sinh(\Delta T/\hbar)\); instanton gas gives \(\sinh(K\mathrm{e}^{-S/\hbar}T)\); matching yields \(\Delta = \hbar K\mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\)
Non-perturbative: splitting is \(O(\mathrm{e}^{-1/\hbar})\) — invisible at any finite order in \(\hbar\)

Homework#

1. Verify that \(x_{\text{inst}}(\tau) = a\tanh(\omega\tau/2)\) satisfies the Euclidean equation of motion \(m\ddot{x} = dV/dx\) for the double-well potential \(V(x) = \lambda(x^2 - a^2)^2\). (Hint: Compute \(\dot{x}\) and \(\ddot{x}\), then check both sides.)

2. Show that the kink solution \(x_{\mathrm{inst}}(\tau) = a\tanh(\omega\tau/2)\) approaches the minima at \(\tau \to \pm\infty\): \(\lim_{\tau \to +\infty} x_{\text{inst}}(\tau) = a\) and \(\lim_{\tau \to -\infty} x_{\text{inst}}(\tau) = -a\).

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

4. Explain why the Euclidean equation of motion admits a real, finite-action solution (instanton), whereas the real-time equation \(m\ddot{x} = -dV/dx\) does not. Why is the sign change crucial?

5. The energy splitting in the double-well is \(\Delta = \hbar K \mathrm{e}^{-S_{\mathrm{inst}}/\hbar}\), where \(K\) is a pre-exponential factor from the fluctuation determinant. Which contribution dominates: the exponential or the pre-exponential? Why is the exponential part called the “non-perturbative” part?

6. In the instanton gas approximation, the partition function includes contributions from paths with zero, one, two, or more instantons. Why do two-instanton (round-trip) contributions give the same energy splitting as one-instanton contributions? (Hint: Think about imaginary-time periodicity and the partition function at temperature \(T = 1/\beta\).)

7. A particle in a potential \(V(x) = \lambda x^4\) (single well) has no instanton solution. Why not? Contrast this with the double-well case.

8. The WKB tunneling amplitude is \(T \propto \exp(-2\int_{x_1}^{x_2} \kappa(x) dx / \hbar)\), where \(\kappa = \sqrt{2m(V-E)}/\hbar\). Compare this to the instanton result \(\propto \exp(-S_{\text{inst}}/\hbar)\). For the double-well, do they give the same exponential dependence on potential parameters? Sketch how you would verify this.