Quantum Tunneling: Decay rates

We calculate tunneling rates through a cubic potential, and compare to the WKB

exponent (the classical Instanton action).

We’re estimating the decay rate from a cubic potential,

V(y) = (1/2)(m Ω^2) (y^2-y^3/Q). This form guarantees that the frequency in

the well is Ω, and the turning point is at Q. We set Q to make the barrier

crossing rate reasonable.

Clear@Q, m, Ω, Ñ, n0DV@y_D :=;

a0 =;

Vmax = V@ ...D . Q ® n0 a0

S0OverÑ = Simplify@Integrate@ ..., 8y, 0, Q<D . Q ® n0 a0, m > 0 && Ω > 0D

Alternatively, we can change variables and do a numerical integration for S0/Ñ.

We change variables from y to z = y/Q; hence dy = Q dz. We factor out a

sqrt(m Ω^2 Q^2) from sqrt(V(y)), and another factor of sqrt(m)/Ñ outside.

S0OverÑNumerical =

Simplify@H ... dimensionful prefactors ...L NIntegrate@, 8z, 0, 1<D . Q ® n0 a0,

Assumptions ® 8m > 0, Ñ > 0, Ω > 0, n0 > 0<D

Set constants

Ñ := 1.054571628251774`*^-27;

Ω := 10.^12;

m := 1.672621637`*^-24; H* Proton mass *LmElectron := 9.1094 ´ 10^-28; H* Electron mass *Lc := 2.998 ´ 10^10;

Α := 1 H137.036LBohrRadius := Ñ HΑ mElectron cLRydberg := mElectron c^2 Α^2 2

How tall and wide a barrier for various lifetimes?

LifeTimeGuess = ...;

n0Guess@Lifetime_D = ...;

8 ...< . n0 ® n0Guess@ ...D8n0, S0OverÑ, Vmax Rydberg, n0 a0 BohrRadius< . n0 ® n0Guess@1.D8 ...< . n0 ® n0Guess@ ...D

85.72679, 8.74563, 0.0000587599, 19.2153<

810.1792, 27.631, 0.000185647, 34.1546<

815.6864, 65.6171, 0.000440866, 52.6331<

91.77551 ´ 10-8

, 5.29163 ´ 10-9=

91.05457 ´ 10-15

, 2.17999 ´ 10-11=

We set Q = 5 a0 (n0=5) to make the barrier crossing rate reasonable.

n0 := 5;

Q := n0 a0

This makes the barrier only slightly higher than the ground state energy

Vmax H ...LVmax H ...L

0.925926

1.85185

Real-space lattice of Np points x, from -L/2 to L/2.

L = ...;

Np = ...;

dx = ...;

x = ...;

2 WKBDecayMathematica_Hints.nb

To see a decay rate, we need to absorb the outgoing probability before it reflects

back. We can implement this by adding a negative imaginary part to the potential

energy Ie-i HH -i Vdamp@xDL tÑ= e

-Vdamp@xD tÑ-i H tÑM. To avoid reflecting off of the

damping part of the Hamiltonian, we make it a linear ramp, starting it at +- L/10.

(Note |x| + x = 2x if x>0, 0 if x<0.)

yD = L 10;

Vdamp@y_D = ... HAbs@-yD - yD + H-yD - yL + Abs@y - yDD + y - yDL ...;

v = V@xD - I Vdamp@xD;

ListPlot@8Transpose@8x, Re@vD<D, Transpose@8x, Im@vD<D<,

Joined ® True, PlotRange ® 8-2 Ñ Ω, 2 Ñ Ω<, AxesLabel ® 8"x", "VHxL"<,

PlotLegends ® Placed@8"ReHVL", "ImHVL = -Vdamp"<, 80.85, 0.75<DDPeriod = ...;

dt = ...;

UpotDtOver2 = ...;

Kinetic energy unitary operator Ukin(dt):

dk = ...;

k = ...;

k2 = ...;

UkinDt = ...;

Initial wavefunction Ψ[0] is ground state wavefunction in harmonic well, Gaussian

centered at x=0.

Evolve in small time steps dt, using the Baker-Campbell-Hausdorff formula:

Upot(dt/2) (with damping) applied before and after the kinetic term Ukin(dt).

Ψ@0D = ...;

ΨMax = Max@Abs@Ψ@0DDD;

T = ...;

Nt = Round@T dtD;

times = Table@ ...D;

For@n = 1, n < Nt, n++, Ψ@nD = ...;

PlotΨ2@n_D := ListPlot@Transpose@8x, Abs@Ψ@nDD^2<D, Joined ® True,

PlotRange ® 80, ΨMax^2<, PlotLabel ® "Ω t = " + Ω times@@n + 1DDD

How long does it take for the particle to leak out of the well? First plot |Ψ^2| at t

= 0, 200P, and 800P.

Show@PlotΨ2@ ...D, PlotΨ2@ ...D, PlotΨ2@ ...DD

We then animate ... We measure time relative to Ω.

ListAnimate@Table@PlotΨ2@nD, 8n, 0, Nt - 1, 1000<DD

WKBDecayMathematica_Hints.nb 3

We can quantify this by calculating the decay of the total probability:

Decay = Table@ ..., 8n, 0, Nt - 1<D;

DecayPlot = ListPlot@Transpose@ ...D, Joined ® True, PlotRange ® 8All, 80, All<<D

We now fit the data to an exponential decay. Note that at short times there is a

small excess decay (the initial Gaussian wavepacket isn’t exactly the right shape for

the anharmonic well). For a better answer, drop the first few points and fit to A

e-G t.

model = A Exp@-G tD;

dropPoints = 2000;

8GFit, AFit< = 8G, A< . FindFit@ ...D;

fitPlot = Plot@ ..., PlotStyle ® RedD;

Show@DecayPlot, fitPlotD

Both WKB and the instanton method tell us for decay problems that G = G0 e-S0Ñ,

with S0 = 2Ù Sqrt[2 m V(y)] dy, integrated between the bottom of the well and the

turning point. (This is twice the exponential suppression of coherent tunneling that

we calculated for the symmetric well. I think about it as the difference between

amplitude Ψ tunneling in the symmetric well and probability |Ψ 2 tunneling in this

problem: the decay goes as the square of the tunneling rate.) The prefactor G0 is

again complicated to calculate, but it should be of order Ω. We can use our

numerical method to estimate G0 / Ω:

G0 = GFit ...;

G0 Ω

4 WKBDecayMathematica_Hints.nb