core34 min studyLesson 373 in Physics

Double-well splitting and localized states

Separate stationary parity eigenstates from localized superpositions and derive tunnelling in an isolated two-state approximation.

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What you’ll understand

  • Diagonalize a symmetric two-well effective Hamiltonian.
  • Compute a first transfer time from the doublet splitting.
  • Explain how detuning and higher states limit a localized two-state picture.

Two equivalent potential minima do not usually produce two exactly degenerate quantum ground states for a particle on a connected line with a finite intervening barrier. Instead, the lowest wavefunctions extend across both wells and have slightly different energies. A particle initially localized on one side is a superposition of those states. The relative phase between them then changes, converting a stationary spectral splitting into time-dependent motion between the wells.

Parity states survive a finite barrier

For a real symmetric potential on the full line with ordinary decaying bound-state conditions, bound eigenvalues are nondegenerate and the eigenfunctions have definite parity. The ground state has no interior nodes and can be taken positive. In a symmetric double well, its two lobes therefore have the same sign. The next state's opposite-sign lobes give odd parity and a central node. These facts depend on the connected one-dimensional problem; an infinite barrier splitting the domain changes the assumptions.

If the lowest pair is well separated in energy from higher states, construct orthonormal localized basis states ∣L⟩|L\rangle and ∣R⟩|R\rangle within that doublet. Their effective Hamiltonian has equal diagonal energy E0E_0 and an off-diagonal coupling chosen as −J-J with J>0J>0. This sign convention makes the symmetric state lower. The matrix is exact as a specified two-level model; treating it as an approximation to a particular spatial well requires verifying isolation of the doublet and computing its coupling.

Heff=(E0−J−JE0),∣±⟩=∣L⟩±∣R⟩2,E+=E0−J,E−=E0+JH_{\rm eff}=\begin{pmatrix}E_0&-J\\-J&E_0\end{pmatrix},\qquad |\pm\rangle=\frac{|L\rangle\pm|R\rangle}{\sqrt2},\qquad E_+=E_0-J,\quad E_-=E_0+J
E₀ and J are energies in the same units. The plus label denotes the symmetric vector and the minus label the antisymmetric vector, not the algebraic sign of the energy shift. The splitting is ΔE = 2J; no decoherence or coupling to additional states is included.↔ Scroll wide equations horizontally.

Localization is a relative-phase condition

An initial left state contains equal symmetric and antisymmetric amplitudes. Each amplitude acquires its own stationary phase under time evolution. Factoring out their common phase leaves a cosine left amplitude and an imaginary sine right amplitude. Squaring gives oscillating populations whose sum remains one. A stationary parity eigenstate itself never transfers its probability from left to right; time dependence requires the superposition. The apparent paradox disappears once an eigenstate is distinguished from a localized initial preparation.

The first complete transfer occurs when the relative phase becomes π\pi. A full return needs another equal interval. Increasing barrier separation or height usually reduces low-energy wavefunction overlap and hence the splitting in a weak-overlap regime, making this time long. That qualitative connection does not make the effective coupling universal. A detailed spatial potential, or an independently justified approximation to it, is needed before a barrier height can be turned into a numerical transfer time.

PR(t)=sin⁡2 ⁣(Jtℏ)=sin⁡2 ⁣(ΔE t2ℏ),tL→R=πℏΔEP_R(t)=\sin^2\!\left(\frac{Jt}{\hbar}\right)=\sin^2\!\left(\frac{\Delta E\,t}{2\hbar}\right),\qquad t_{L\to R}=\frac{\pi\hbar}{\Delta E}
This exact result for the symmetric two-level model assumes the initial state is |L⟩. The first transfer time is in seconds when ℏ and energy use compatible units. The population recurrence time is twice this value, and the discarded common phase has no effect on populations.↔ Scroll wide equations horizontally.

Detuning and environment change the conclusion

If one well is offset by detuning δ\delta, diagonal energies become E0±δ/2E_0\pm\delta/2. The gap is then δ2+4J2\sqrt{\delta^2+4J^2}, but the largest possible transfer probability from a localized initial state drops to 4J2/(δ2+4J2)4J^2/(\delta^2+4J^2). A larger gap therefore need not mean more complete tunnelling. Part of that gap can come from bias rather than coupling. At zero coupling the states remain localized even if their energies differ substantially.

An environment can further destroy the coherence between the doublet components. A statistical half-left, half-right mixture and a coherent equal superposition can share basis populations but differ in interference and later evolution. Additional excited states can create several beating frequencies after an energetic preparation. The clean sine-squared result is consequently a useful controlled model for an isolated near-resonant pair, not a claim that every double-well experiment performs perfect periodic transfer indefinitely.

From an energy doublet to a transfer time

An isolated symmetric doublet is described by E0=0.50 eVE_0=0.50\,\mathrm{eV} and J=0.020 eVJ=0.020\,\mathrm{eV}. A particle begins in the left localized state. Use ℏ=0.658 eV fs\hbar=0.658\,\mathrm{eV\,fs} to find the eigenenergies, first complete-transfer time, and first return time.

  1. Diagonalizing the symmetric matrix gives the even energy E+=0.480 eVE_+=0.480\,\mathrm{eV} and odd energy E−=0.520 eVE_-=0.520\,\mathrm{eV}. Their gap is ΔE=0.040 eV\Delta E=0.040\,\mathrm{eV}, twice the coupling rather than equal to it.

  2. Initial localization means equal parity components. Complete right localization first occurs at relative phase ΔE t/ℏ=π\Delta E\,t/\hbar=\pi, so tL→R=π(0.658)/0.040 fs=51.7 fst_{L\to R}=\pi(0.658)/0.040\,\mathrm{fs}=51.7\,\mathrm{fs}. The average energy supplies only a common phase.

  3. The first full population return occurs at twice the transfer time, approximately 103 fs. Direct substitution gives Jt/ℏ=πJt/\hbar=\pi there, hence PR=0P_R=0 and PL=1P_L=1. This checks the distinction between a transfer and a recurrence.

The result

The doublet energies are 0.480 and 0.520 eV. The first left-to-right transfer takes about 51.7 fs, and the first population return about 103 fs within the isolated symmetric model.

Explore the model

Interactive lab

Interactive lab · quantum

Parity populations can hide localization phase

Interpret the lab basis as the two parity states. Reset and set both controls to mixing θ = 90° and phase φ = 0°; predict equal basis probabilities and Bloch x = +1. Reset and set both controls to θ = 90° and φ = 180°; predict the same probabilities and Bloch x = −1. These superpositions can label opposite localized combinations; the lab is a pure two-state geometry, without a spatial well or calibrated tunnelling clock.

Parity populations can hide localization phaseA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forRelative phase can distinguish opposite localized superpositions even with unchanged parity populations; the lab does not calculate a double-well splitting.

Check your understanding

Why can a particle initially localized in a symmetric double well move between its sides while a parity eigenstate has stationary density?

Try it yourself

In the detuned two-state model take J = 4 meV and diagonal-energy difference δ = 6 meV. Find the doublet gap and maximum probability of transfer from an initially left-localized state. Explain why the symmetric full-transfer conclusion no longer follows.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use the 6–8–10 triangle in the squared gap, since the off-diagonal contribution is 2J.

Compare with the solution

The gap is δ2+4J2=62+82=10 meV\sqrt{\delta^2+4J^2}=\sqrt{6^2+8^2}=10\,\mathrm{meV}. The maximum transfer is 4J2/(δ2+4J2)=64/100=0.644J^2/(\delta^2+4J^2)=64/100=0.64. Thus at least 36% remains on the initially occupied side even at the best transfer instant in this ideal model. Detuning changes the eigenvectors as well as the gap, so the initial localized state no longer contains the particular equal energy components needed for complete transfer. A larger frequency does not compensate for the reduced amplitude.

Take it with you

  • A finite symmetric barrier gives parity eigenstates separated by a doublet gap.
  • Localized states transfer through coherent relative-phase evolution.
  • Bias, higher states, and decoherence limit the isolated symmetric two-state result.

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