WKB Approximation

Also called the Semiclassical Approximation for Quantum Mechanics.

\(S(\vec{x},t)=\pm\int^x dx' \sqrt{2m(E-V(x'))}-Et = W(x)-Et\).

Continuity of One-Dimension

\(\frac{\partial}{\partial t}\mathcal{P} + \frac{\partial}{\partial x}j = 0\).

\(j=\frac{\mathcal{P}}{m}\frac{\partial S}{\partial x}\).

Stationary Case

\(\frac{\partial \mathcal{P}}{\partial t} = 0\).

So, \(\frac{\partial j}{\partial x} = 0\). Hence, \(\frac{\partial}{\partial x}\left[\mathcal{P}\frac{\partial S}{\partial x}\right] = 0\). So, \(\rho\frac{d W}{dx}\) is a constant, \(\mathcal{C}\). Then, \(\sqrt{\mathcal{P}} = \frac{\mathcal{C}}{\sqrt[4]{E-V(x)}}\), so \(\mathcal{P}=\frac{\mathcal{C}^2}{\sqrt{E-V(x)}}\).

Thus, we are able to get a wavefunction without any solving: \(\Psi = \frac{\mathcal{C}}{\sqrt[4]{E-V(x)}}\exp(\pm\frac{i}{\hbar}\int^xdx' \sqrt{2m(E-V(x'))}-\frac{i}{\hbar}Et)\). Wentzel-Kramers-Brillouin solution. Thus, \(\mathcal{P}\sim\frac{1}{v_{classical}}\).

WKB Criteria

\(\hbar\left|\frac{d^2}{dx^2}W\right|\ll\left|\frac{dW}{dx}\right|^2\). So, \(\hbar\frac{2m\left|\frac{dV}{dx}\right|}{2\sqrt{2m(E-V(x))}}\ll|2m(E-V(x))|\). Hence, \(\frac{\hbar}{\sqrt{2m(E-V(x))}}=\frac{1}{k}=\frac{\lambda_B}{2\pi}\ll \frac{2(E-V(x))}{\left|\frac{dV}{dx}\right|}\), where \(k\) is the wavenumber, related to the momentum by \(\hbar\). This gives us a deBrolie wavelength to compare.

So, WKB is valid in the low wavelength cases.

Problem Areas

Turning points pose problem areas for WKB, but they can be solved as linear wavefunction pieces to splice the region solutions together.

Small Example

Finite Well but with rounded edges, energy such that it is a bound state.

For regions I, III: \(\Psi=\frac{\mathcal{C}}{\sqrt[4]{2m(V(x)-E)}}\exp(\pm \frac{1}{\hbar}\int^x \sqrt{2m(V(x')-E)})dx'\)

For regions II: \(\Psi=\frac{\mathcal{C}}{\sqrt[4]{2m(E-V(x))}}\exp(\pm \frac{i}{\hbar}\int^x \sqrt{2m(E-V(x'))})dx'\)

For turning point solutions to splice together the regions: \(\Psi = Ax+B\) \(-\frac{\hbar^2}{2m}\Psi''=(E-V(x))\Psi(x) \Rightarrow\) Linear solutions.

Alternatively, \(-\frac{\hbar^2}{2m}\Psi''+\frac{dV}{dx}|_{x_{1,2}}(x-x_{1,2})\Psi=0\). Gives a Bessel equation - Airy function.


Rigid \(\Rightarrow\) \(V\to\infty\).

\(\int_{x_1}^{x_2}dx\sqrt{2m(E_n-V(x))} = \pi\hbar g_n\).

Case 1 - No Rigid Walls


Case 2 - 1 Rigid Wall


Case 3 - 2 Rigid Walls



If \(n\) is large, each \(g_n\) is approximately equal.

Also, as \(n\) increases, we get closer to classical typically.


Energies of QMHO

\(E=\frac{m\omega^2 x_{1}^2}{2}\). \(\int_{-x_1}^{x_1}dx\sqrt{2m(\frac{m\omega^2 x_1^2}{2}-\frac{m\omega^2x^2}{2})}=\frac{\pi E}{\omega}=\pi\hbar(n+1/2)\). \(x_1=\pm\sqrt{\frac{2E}{m\omega^2}}\). WKB: \(E=\hbar\omega(n+1/2)\)

Bouncing Neutrons

Hard floor. So, the potential is infinite in the negative region and a linear potential at \(x\gt 0\).


Use WKB to find the wavefunction for a more complicated potential barrier, use typical solutions (if possible) for outside propogators.

\(T\approx \exp(-2\gamma), \gamma=\frac{1}{\hbar}\int_{x_1}^{x_2}\sqrt{2m(V(x)-E)}dx\)

Author: Christian Cunningham

Created: 2023-06-25 Sun 02:28