3.3.2 WKB Approximation#

Worked solutions for the homework problems in the 3.3.2 WKB Approximation lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Hamilton-Jacobi from WKB. Substitute the WKB ansatz \(\psi(x) = A(x)\,\mathrm{e}^{\mathrm{i}W(x)/\hbar}\) into the time-independent Schrödinger equation \(-\frac{\hbar^{2}}{2m}\psi'' + V(x)\psi = E\psi\).

(a) Collect terms by powers of \(\hbar\) and show that at leading order (\(\hbar^{0}\)) the phase \(W(x)\) satisfies the Hamilton-Jacobi equation \(\frac{1}{2m}(W')^{2} + V(x) = E\).

(b) At next order (\(\hbar^{1}\)), show that \(2A'W' + AW'' = 0\). Integrate this to obtain \(A \propto 1/\sqrt{W'} = 1/\sqrt{p(x)}\).

Solution.

(a) Differentiate the ansatz \(\psi = A\,\mathrm{e}^{\mathrm{i}W/\hbar}\) twice. The first derivative is

and differentiating once more,

Insert this into \(-\frac{\hbar^{2}}{2m}\psi'' + (V-E)\psi = 0\) and cancel the common factor \(\mathrm{e}^{\mathrm{i}W/\hbar}\):

The three groups carry distinct powers of \(\hbar\): the \((W')^{2}\) and \((V-E)\) terms are \(\hbar^{0}\), the bracket multiplying \(\mathrm{i}\hbar\) is \(\hbar^{1}\), and the \(A''\) term is \(\hbar^{2}\). Treating \(\hbar\) as a small expansion parameter, each power must vanish separately. At order \(\hbar^{0}\),

so \(\dfrac{1}{2m}(W')^{2} + V(x) = E\),

the Hamilton-Jacobi equation. Solving for the phase gradient gives the local classical momentum, \(W' = p(x) = \pm\sqrt{2m(E-V)}\).

(b) At order \(\hbar^{1}\) the bracketed term must vanish:

Divide through by \(A\,W'\) (nonzero away from turning points):

i.e. \(\dfrac{\mathrm{d}}{\mathrm{d}x}(2\ln A + \ln W') = 0\).

Hence \(2\ln A + \ln W' = \mathrm{const}\), i.e. \(A^{2}W' = \mathrm{const}\). In the allowed region take \(p(x) = |W'(x)| = \sqrt{2m(E-V)} > 0\) (the sign of \(W'\) tracks propagation direction), so

Equivalently, multiply \(2A'W' + AW'' = 0\) by \(A\) to get \(2AA'W' + A^{2}W'' = \frac{\mathrm{d}}{\mathrm{d}x}(A^{2}W') = 0\) directly. The constant \(A^{2}p\) is the probability current \(j = A^{2}p/m\): the WKB amplitude law is nothing but current conservation.

2. Allowed and forbidden regions. A particle of energy \(E\) moves in a slowly varying potential \(V(x)\).

(a) In the classically allowed region (\(E > V\)), write the WKB wavefunction \(\psi \sim p^{-1/2}\,\exp(\pm\frac{\mathrm{i}}{\hbar}\int^{x}p\,\mathrm{d}x')\) and explain physically why the amplitude \(1/\sqrt{p}\) means the particle is more likely found where it moves slowly.

(b) In the classically forbidden region (\(E < V\)), set \(p \to \mathrm{i}\kappa\) with \(\kappa = \sqrt{2m(V-E)}\) and show that the wavefunction decays as \(\psi \sim \kappa^{-1/2}\,\exp(-\frac{1}{\hbar}\int^{x}\kappa\,\mathrm{d}x')\).

Solution.

(a) Where \(E > V\), the momentum \(p(x) = \sqrt{2m(E-V)}\) is real and the phase \(W = \int p\,\mathrm{d}x'\) accumulates as a genuine oscillation. Combining the phase with the amplitude \(A = C/\sqrt{p}\) from Problem 1,

so the probability density is \(\vert\psi\vert^{2} \propto 1/p(x)\).

This is exactly the classical dwell time. Picture a classical particle of energy \(E\) sweeping through \(V(x)\): the time it spends in an interval \(\mathrm{d}x\) is \(\mathrm{d}t = \mathrm{d}x/v = m\,\mathrm{d}x/p\). The fraction of a long observation during which the particle is found in \(\mathrm{d}x\) is therefore \(\propto 1/p\). Where the potential rises and the particle slows down, \(p\) is small, \(\mathrm{d}t\) is large, and the particle lingers; where it speeds up, it passes through quickly. The WKB density \(\vert\psi\vert^{2}\propto 1/p\) reproduces this dwell-time distribution — the quantum particle is most likely found precisely where its classical counterpart moves slowest.

(b) Where \(E < V\), the quantity under the square root flips sign: \(2m(E-V) < 0\). Write \(p = \mathrm{i}\kappa\) with \(\kappa = \sqrt{2m(V-E)}\) real and positive (momentum scale; the dimensionless tunneling exponent is \(\gamma = \int\kappa\,\mathrm{d}x/\hbar\)). Substituting into the allowed-region form, the phase integral becomes

a real exponent — the oscillation has turned into exponential growth or decay. The prefactor likewise becomes \(p^{-1/2} = (\mathrm{i}\kappa)^{-1/2} = \kappa^{-1/2}\,\mathrm{e}^{-\mathrm{i}\pi/4}\), a constant phase that can be absorbed into the overall normalization. The two signs give two independent real solutions; the physical (normalizable, decaying) branch takes the lower sign,

The wavefunction does not vanish abruptly at the classical boundary — it leaks into the forbidden region and dies off exponentially on the scale \(\hbar/\kappa\), which is the origin of quantum tunneling.

3. Validity criterion. The WKB approximation requires \(\vert\mathrm{d}\lambda/\mathrm{d}x\vert \ll 2\pi\), where \(\lambda(x) = 2\pi\hbar/p(x)\) is the local de Broglie wavelength.

(a) Show that this is equivalent to the dimensionful condition \(\hbar\,\vert V'(x)\vert \ll 2\sqrt{2m}\,(E - V(x))^{3/2}\) in the allowed region.

(b) For a linear potential \(V(x) = Fx\) (constant force), identify where the criterion fails. Relate this to the turning point \(x_{0} = E/F\).

(c) For a Gaussian barrier \(V(x) = V_{0}\,\mathrm{e}^{-x^{2}/a^{2}}\) with energy \(E = V_{0}/4\), identify the classical turning points and discuss whether WKB is reliable away from them.

Solution.

(a) Differentiate \(\lambda = 2\pi\hbar/p\):

The validity condition \(\vert\mathrm{d}\lambda/\mathrm{d}x\vert \ll 2\pi\) becomes \(2\pi\hbar\,\vert p'\vert/p^{2} \ll 2\pi\), i.e.

Now express \(p'\) through the potential. With \(p = \sqrt{2m(E-V)}\),

so \(\vert p'\vert = m\,\vert V'\vert/p\). Substituting,

so \(\hbar\,m\,\vert V'\vert \ll p^{3} = (2m(E-V))^{3/2}\).

Divide both sides by \(m\) and use \((2m)^{3/2}/m = 2^{3/2}m^{1/2} = 2\sqrt{2m}\):

The condition says the potential energy must change by little over one wavelength compared with the kinetic energy scale — the semiclassical “slowly varying potential” assumption.

(b) For \(V(x) = Fx\) the force is constant, \(V'(x) = F\), so the left side \(\hbar F\) is fixed while the right side is \(2\sqrt{2m}\,(E - Fx)^{3/2}\). The criterion reads

It holds comfortably deep in the allowed region where \(E - Fx\) is large, but fails as \(x \to x_{0} = E/F\), the classical turning point, because \((E-Fx)^{3/2} \to 0\) there and the right side cannot dominate \(\hbar F\). Setting the two sides comparable locates the edge of the WKB-trustworthy zone: with \(E - Fx = F(x_{0} - x)\),

so \(x_{0} - x \sim (\hbar^{2}/2mF)^{1/3} \equiv \ell\).

So WKB breaks down within a layer of thickness of order the Airy length \(\ell = (\hbar^{2}/2mF)^{1/3}\) around the turning point. Outside that thin layer the linear potential is varying slowly enough and WKB is fine; inside it one must use the Airy-function matching of Problem 6.

(c) The turning points are where \(V(x) = E\):

giving \(\mathrm{e}^{-x^{2}/a^{2}} = 1/4\), so \(x^{2} = a^{2}\ln 4\),

giving two symmetric turning points

Since the barrier peaks at \(V(0) = V_{0} > E\), the region \(\vert x\vert < x_{+}\) is classically forbidden and \(\vert x\vert > x_{+}\) is allowed. WKB is unreliable in the Airy layers around \(x_{\pm}\), where \(E - V \to 0\). Away from those points, the test of reliability is the criterion of part (a). The barrier varies on the length scale \(a\), so \(\vert V'\vert \sim V_{0}/a\), and away from the turning points \(\vert E - V\vert\) is of order \(V_{0}\). The criterion then reads, in order of magnitude,

i.e. \(\hbar/(a\sqrt{2mV_{0}}) \ll 1\),

i.e. the barrier width \(a\) must greatly exceed the penetration/decay length \(\hbar/\sqrt{2mV_{0}}\). If the barrier is wide and tall (\(a\sqrt{2mV_{0}}/\hbar \gg 1\)), WKB is accurate everywhere except in the two narrow Airy zones, and the tunneling exponent \(\gamma\) can be trusted. If the barrier is thin (width comparable to a de Broglie wavelength), the turning-point zones overlap, there is no genuine semiclassical interior, and WKB fails throughout.

4. Square-barrier tunneling (heuristic). A particle of mass \(m\) and energy \(E < V_{0}\) encounters a rectangular barrier \(V(x) = V_{0}\) for \(0 < x < a\) and \(V = 0\) elsewhere. The discontinuous walls violate the WKB validity criterion \(\hbar|V'(x)| \ll 2\sqrt{2m}\,(E-V(x))^{3/2}\), so the formula below captures only the dominant exponential factor — the true transmission coefficient has an additional reflection-coefficient prefactor of order unity.

(a) Apply \(T \approx \mathrm{e}^{-2\gamma}\) with \(\gamma = \frac{1}{\hbar}\int_{0}^{a}\kappa\,\mathrm{d}x\) to obtain \(T \approx \exp\!\left(-\frac{2a}{\hbar}\sqrt{2m(V_{0} - E)}\right)\).

(b) Evaluate \(T\) for an electron (\(m \approx 9.1\times 10^{-31}\,\text{kg}\)) with \(V_{0} - E = 2\,\text{eV}\) and \(a = 0.2\,\text{nm}\). Is tunneling significant?

(c) Show that doubling the barrier width \(a\to 2a\) replaces \(T\) by \(T^{2}\) (much smaller). Explain why tunneling is exponentially sensitive to barrier width.

Solution.

(a) Inside the barrier the potential is constant, \(V = V_{0}\), so the decay rate \(\kappa = \sqrt{2m(V_{0}-E)}\) is constant. The tunneling exponent is then a trivial integral,

The transmission probability is

(b) With \(V_{0} - E = 2\,\text{eV} = 3.20\times 10^{-19}\,\text{J}\), \(m = 9.1\times 10^{-31}\,\text{kg}\), \(a = 0.2\,\text{nm} = 2.0\times 10^{-10}\,\text{m}\), and \(\hbar = 1.055\times 10^{-34}\,\mathrm{J\cdot s}\):

Therefore

About 5% of incident electrons tunnel through — small but absolutely significant. A sub-nanometre barrier of a few eV is no obstacle to electrons, which is exactly why electron tunneling is observed routinely (field emission, tunnel diodes, STM). The order-unity prefactor omitted by the heuristic formula does not change this conclusion.

(c) Since \(\gamma = \kappa a/\hbar\) is linear in \(a\), doubling the width sends \(\gamma \to 2\gamma\), and

Because \(T < 1\), squaring makes it dramatically smaller — for the electron above, \(T \approx 0.055\) would drop to \(T^{2} \approx 0.003\). The deep reason is that the transmission depends on width through an exponential, \(T \sim \mathrm{e}^{-2\kappa a/\hbar}\): the wavefunction decays by a fixed factor per unit length inside the barrier, so equal increments of width multiply \(T\) by equal factors. A linear change in \(a\) produces a geometric (exponential) change in \(T\). This exponential sensitivity is what makes scanning tunneling microscopy possible — a tip-height change of a fraction of an ångström produces a measurable change in tunneling current.

5. Penetration depth. The characteristic decay length of the wavefunction in a forbidden region is \(\lambda_\text{decay} = \hbar/\kappa\) with \(\kappa = \sqrt{2m(V - E)}\).

(a) Compute \(\lambda_\text{decay}\) for an electron at a metal surface with work function \(V - E = 5\,\text{eV}\). Express the answer in angstroms.

(b) Explain why this length scale makes scanning tunneling microscopy (STM) work: the tunneling current depends exponentially on the tip-surface distance.

Solution.

(a) With \(V - E = 5\,\text{eV} = 8.01\times 10^{-19}\,\text{J}\) and \(m = 9.1\times 10^{-31}\,\text{kg}\), the decay rate is

so the penetration depth is

The electron wavefunction outside the metal decays on a scale of roughly one ångström — an atomic length.

(b) The tunneling current in STM is \(I \propto T \approx \mathrm{e}^{-2\gamma}\), and for a gap \(d\) between tip and surface \(\gamma = \kappa d/\hbar = d/\lambda_\text{decay}\), so

With \(\lambda_\text{decay} \approx 0.87\,\text{Å}\), increasing the gap by a single ångström multiplies the current by \(\mathrm{e}^{-2/0.87} \approx \mathrm{e}^{-2.3} \approx 0.1\) — a tenfold drop. The current is so steeply dependent on \(d\) that essentially all of it flows through the single closest atom of the tip, and height variations of a few hundredths of an ångström are detectable. That extreme sensitivity is what gives the STM its atomic-scale spatial resolution: a small geometric corrugation of the surface translates into a large, measurable current modulation.

6. Airy connection formula. At a soft turning point \(x_{0}\) with \(V'(x_{0}) > 0\), linearize the potential as \(V(x) \approx E + V'(x_{0})(x - x_{0})\).

(a) Show that the Schrödinger equation reduces to the Airy equation \(\psi'' = z\psi\) with \(z = (x - x_{0})/\ell\), and identify the length scale \(\ell = \bigl(\hbar^{2}/(2m\,V'(x_{0}))\bigr)^{1/3}\).

(b) The Airy function \(\mathrm{Ai}(z)\) decays exponentially for \(z\to+\infty\) (forbidden side) and oscillates for \(z\to-\infty\) (allowed side). Use its asymptotic forms to recover the connection formula \(\frac{1}{\sqrt{\kappa}}\mathrm{e}^{-\int_x^{x_0}\kappa\,\mathrm{d}x'/\hbar} \longleftrightarrow \frac{2}{\sqrt{p}}\sin\!\bigl[\int_{x_0}^{x}p\,\mathrm{d}x'/\hbar + \pi/4\bigr]\) with the \(\pi/4\) phase shift.

(c) Explain in one sentence why omitting the \(\pi/4\) shift in the bound-state quantization rule of §3.3.3 would mis-predict the harmonic-oscillator zero-point energy.

Solution.

(a) Insert the linearized potential \(V(x) \approx E + V'(x_{0})(x-x_{0})\) into the time-independent Schrödinger equation. The constant \(E\) cancels against the \(E\psi\) on the right:

so \(\psi'' = \dfrac{2m\,V'(x_{0})}{\hbar^{2}}\,(x-x_{0})\,\psi\).

Introduce the dimensionless coordinate \(z = (x-x_{0})/\ell\). Then \(\mathrm{d}/\mathrm{d}x = \ell^{-1}\mathrm{d}/\mathrm{d}z\), so \(\psi'' = \ell^{-2}\,\mathrm{d}^{2}\psi/\mathrm{d}z^{2}\), and \(x - x_{0} = \ell z\):

i.e. \(\dfrac{\mathrm{d}^{2}\psi}{\mathrm{d}z^{2}} = \dfrac{2m\,V'(x_{0})\,\ell^{3}}{\hbar^{2}}\,z\,\psi\).

This is the Airy equation \(\psi'' = z\psi\) provided the coefficient equals one,

so \(\ell = (\hbar^{2}/2m\,V'(x_{0}))^{1/3}\).

This \(\ell\) is the width of the turning-point zone where neither the oscillatory nor the exponential WKB form is valid — the region the Airy function bridges.

(b) The Airy function has the asymptotic forms (standard results)

Now match these to the WKB forms. On the forbidden side (\(x > x_{0}\), \(z > 0\)), \(\kappa = \sqrt{2m(V-E)} = \sqrt{2m\,V'(x_{0})(x-x_{0})}\). Using \(2m\,V'(x_{0}) = \hbar^{2}/\ell^{3}\) and \(x - x_{0} = \ell z\),

so \(\dfrac{1}{\hbar}\int_{x_{0}}^{x}\kappa\,\mathrm{d}x' = \int_{0}^{z}z'^{1/2}\,\mathrm{d}z' = \tfrac{2}{3}z^{3/2}\).

So the decaying WKB solution \(\kappa^{-1/2}\exp[-\frac1\hbar\int\kappa]\) behaves as \(z^{-1/4}\,\mathrm{e}^{-\frac23 z^{3/2}}\) — exactly the \(z\to+\infty\) form of \(\mathrm{Ai}\). On the allowed side (\(x < x_{0}\), \(z < 0\)), \(p = \sqrt{2m\,V'(x_{0})(x_{0}-x)} = (\hbar/\ell)(-z)^{1/2}\), and

so the oscillatory WKB solution \(p^{-1/2}\sin[\frac1\hbar\int_{x}^{x_{0}}p + \tfrac{\pi}{4}]\) matches the \(z\to-\infty\) form of \(\mathrm{Ai}\), including the \(+\pi/4\) that appears in the Airy asymptotics. Comparing the amplitudes of the two limits, the oscillatory side carries prefactor \(1/\sqrt{\pi}\) and the exponential side \(1/(2\sqrt{\pi})\) — a ratio of \(2\). Restoring the WKB amplitudes \(1/\sqrt{p}\), \(1/\sqrt{\kappa}\), this is the connection formula

The \(\pi/4\) is not an extra assumption: it is forced by the asymptotic phase of the exact Airy function that interpolates between the two regions.

(c) Each soft turning point contributes a \(\pi/4\) phase, so a well bounded by two soft turning points carries an extra \(\pi/2\); dropping it changes the Bohr-Sommerfeld rule from \(\oint p\,\mathrm{d}x = 2\pi\hbar\,(n + \tfrac12)\) to \(\oint p\,\mathrm{d}x = 2\pi\hbar\,n\), which for the harmonic oscillator gives \(E_{n} = n\hbar\omega\) — a vanishing ground-state energy instead of the correct zero-point energy \(\tfrac12\hbar\omega\).

★ 7. Double-well tunnel splitting. Consider a symmetric double-well potential with two minima separated by a barrier of width \(\sim a\). Let \(\omega\) be the small-oscillation frequency in either well and \(\gamma = \frac{1}{\hbar}\int_\text{barrier}\kappa\,\mathrm{d}x\).

(a) Estimate the tunneling amplitude \(\Delta \sim \tfrac{1}{2}\hbar\omega\,\mathrm{e}^{-\gamma}\) from the WKB tunneling formula (the same energy scale as the uncoupled ground state in part (b)).

(b) Explain why \(\Delta\) lifts the degeneracy of the two single-well ground states, splitting them by \(\Delta E \approx 2\Delta\).

(c) For the ammonia molecule \(\mathrm{NH}_{3}\), the nitrogen tunnels with \(\gamma \approx 5\). Estimate \(\Delta E/(\hbar\omega)\). Is the splitting large or small compared to a single-well excitation?

Solution.

(a) Confined to one well, the particle is a harmonic oscillator with ground-state energy \(\tfrac{1}{2}\hbar\omega\). It bounces against the barrier with period \(2\pi/\omega\), so each cycle makes of order one tunneling “attempt.” The leaked amplitude is suppressed by \(\mathrm{e}^{-\gamma}\), where \(\gamma = \frac1\hbar\int_\text{barrier}\kappa\,\mathrm{d}x\) (transmission probability would be \(\mathrm{e}^{-2\gamma}\), but the splitting is set by the tunneling amplitude, not its square). Multiplying the in-well energy scale \(\tfrac{1}{2}\hbar\omega\) by this factor,

This is the order of magnitude of the coherent coupling between the wells.

(b) With an impenetrable barrier the two wells are independent; each has a ground state, \(\vert L\rangle\) localized in the left well and \(\vert R\rangle\) in the right, degenerate at energy \(E_{0} = \tfrac12\hbar\omega\). Tunneling provides a small matrix element coupling them. In the two-state basis \(\{\vert L\rangle, \vert R\rangle\}\) the effective Hamiltonian is

Diagonalizing, the eigenstates are the symmetric and antisymmetric combinations \(\vert\pm\rangle = (\vert L\rangle \pm \vert R\rangle)/\sqrt{2}\) with eigenvalues

The degeneracy is lifted: the symmetric (nodeless) state is pushed down to \(E_{0}-\Delta\), the antisymmetric (one-node) state up to \(E_{0}+\Delta\). The energy gap between the true stationary states is

The localized states \(\vert L\rangle, \vert R\rangle\) are not energy eigenstates once tunneling is allowed; a particle started in one well oscillates to the other and back at the angular frequency \(\Delta E/\hbar = 2\Delta/\hbar\).

(c) With \(\gamma \approx 5\) and \(\Delta E \approx 2\Delta \approx \hbar\omega\,\mathrm{e}^{-\gamma}\),

The tunnel splitting is only about \(7\times 10^{-3}\) times \(\hbar\omega\) — far smaller than the energy of a single-well vibrational excitation, which costs a full \(\hbar\omega\). The two inversion-doublet levels of \(\mathrm{NH}_{3}\) therefore sit very close together at the bottom of the double well, well separated from the next vibrational levels. (Experimentally the ammonia inversion splitting is about \(24\,\text{GHz}\) against a vibrational quantum of order \(30\,\text{THz}\), a ratio \(\sim 10^{-3}\) — the same small order of magnitude the crude \(\gamma\approx 5\) estimate produces.) The exponential factor \(\mathrm{e}^{-\gamma}\) is what makes tunnel splittings tiny: even a modest barrier (\(\gamma\) of a few) buries the splitting orders of magnitude below the natural oscillator scale.