What is the WKB approximation?

In quantum mechanics, Schrödinger’s equation for certain systems with varying parameters cannot be solved directly. In such a case, an approximation method is used to obtain the approximate solution. WKB method is a special technique used to obtain the approximate solution for Schrödinger’s time-independent equation. In general, the WKB method is used to solve linear differential equations with varying coefficients. This method is called semiclassical approximation in quantum mechanics.

Semiclassical region

In quantum mechanics, we know that the particles have a wave nature. The wavelength associated with a particle is known as de Broglie wavelength. The de Broglie wavelength is given as,

$\lambda =\frac{h}{p}$

Where h is the Plank’s constant.

p is the momentum of the particle.

For classical bodies, the wavelength is very small. As the wavelength tends to zero, the particle obeys classical mechanics well.

In some cases, the de Broglie wavelength varies with respect to position. If the de Broglie wavelength of a particle becomes very small the particle behaves as a classical particle in such region. This region is called the semiclassical region.

Consider a particle of mass m, let E be the total energy of the particle and V(x) be the potential energy of the particle. The potential energy depends upon the position of the particle.

The kinetic energy of the particle is,

$U=E-V\left(x\right)$

The kinetic energy in terms of momentum p is,

$U=\frac{p^2}{2m}$

Hence, the expression for momentum can be obtained by equating and solving for p.

$p\left(x\right)=\sqrt{2m\left[E-V\left(x\right)\right]}$

The momentum depends upon the position of the particle. This equation gives the classical momentum of the particle.

When E>V, the quantity [E-V(x)] is positive. Hence, the momentum is real. This region is called the semiclassical region.

If E<V, then [E-V(x)] is negative. Hence, the momentum is imaginary or a complex number. Hence the momentum is not defined. Hence, this region is not semiclassical.

If E=V, then the classical momentum of the particle is zero. This point act as the boundary of the classical region and is called the turning point. At the turning point the de Broglie wavelength is infinite.

The de Broglie wavelength of the particle is,

$\lambda \left(x\right)=\frac{h}{\sqrt{2m\left[E-V\left(x\right)\right]}}$

Solving equations using the WKB approximation method

The time-independent one-dimensional Schrödinger equation for the particle is given as follows,

$-\frac{{\hslash }^2}{2m}\frac{d^2\psi }{d{x}^2}+V\left(x\right)\psi =E\psi$

The equation is written in terms of momentum as shown below,

$\frac{d^2\psi }{d{x}^2}=-\frac{p^2}{{\hslash }^2}\psi$    ...(1)

where $p^2=\sqrt{2m\left[E-V\left(x\right)\right]}$

The wave function $\psi$ can be written as follows,

$\psi \left(x\right)=A\left(x\right)e^{i\phi \left(x\right)}$

Differentiating this equation with respect to x we get,

$\frac{d\psi }{dx}=Ai{e}^{\phi \left(x\right)}dx+A\text{'}{e}^{i\phi \left(x\right)}$

Again, differentiating with respect to x we get,

$\frac{d^2\psi }{d{x}^2}=A\text{'}\text{'}+2iA\text{'}\phi \text{'}+iA\phi \text{'}\text{'}-A\phi {\text{'}}^2$     ...(2)

$A\text{'}\text{'}+2iA\text{'}\phi +iA\phi \text{'}\text{'}-A\phi {\text{'}}^2=-\frac{p^2}{{\hslash }^2}A$

By separating the real and imaginary parts and equating we get,

$A\text{'}\text{'}=\left[{(\phi \text{'})}^2-\frac{p^2}{{\hslash }^2}\right]A$²

Since A’’ is very small compared to ${(\phi \text{'})}^2$, the LHS of the equation is equated to zero.

${\left(\phi \text{'}\right)}^2=\frac{p^2}{{\hslash }^2}$     ...(3)

and

$(A^2\phi \text{'})\text{'}=0$

solving for A we get,

$A=\frac{C}{\sqrt{\phi \text{'}}}$     ...(4)

here C is the constant of integration.

Similarly, solving of $\phi \text{'}$ we get the following equation,

$\phi \text{'}=\pm \frac{p}{\hslash }$

$\phi =\pm \frac{1}{\hslash }\int p\left(x\right)dx$

Hence, the wavefunction can be written using equation (3) and (4), in terms of path integral as follows,

$\psi \left(x\right)=\frac{C}{\sqrt{p\left(x\right)}}{e}^{\pm \frac{1}{\hslash }\int p\left(x\right)dx}$

This is the approixmate solution of the one-dimensional Schrödinger equation using WKB approximation.

The connection formulae

The WKB approximation method is used to study the behavior of the particle in both classical and nonclassical regions. But at the turning point i.e. when E=V the WKB approximation method fails. It is impossible to define the state of the particle at the turning point using the WKB method.

Let the point x=0 be the turning point. The positive x-axis is the non-classical region and the negative x-axis is the classical region.

If x<0, then

$\psi \left(x\right)=\frac{1}{\sqrt{p\left(x\right)}}\left[B{e}^{\frac{1}{\hslash }\int p\left(x\right)dx}+C{e}^{-\frac{1}{\hslash }\int p\left(x\right)dx}\right]$

If x>0, then

$\psi \left(x\right)=\frac{D}{\sqrt{p\left(x\right)}}{e}^{\pm \frac{1}{\hslash }\int p\left(x\right)dx}$

A function called patching wave function is defined to obtain the state of the turning point. This patching wave function is obtained by combining the wave function of the classical and nonclassical regions.

Let ${\psi }_p$ be the patching function that is defined near a turning point.

Let us consider that at some regions near the turning point where the potential changes linearly. This region is called the patching region. The linear potential is given as follows,

$V\left(x\right)=E+V\text{'}\left(0\right)x$

Substituting the linear potential in the Schrödinger equation we get,

$-\frac{{\hslash }^2}{2m}\frac{d^2{\psi }_p}{d{x}^2}+\left[E+V\text{'}\left(0\right)x\right]{\psi }_p=E{\psi }_p$

By simplifying and rearranging we get,

$\frac{d^2{\psi }_p}{d{x}^2}=V\text{'}\left(0\right)\frac{2m}{{\hslash }^2}x{\psi }_p$

$\frac{d^2{\psi }_p}{d{x}^2}={\alpha }^{3}x{\psi }_p$

where ${\alpha }^3=V\text{'}\left(0\right)\frac{2m}{{\hslash }^2}$

or

$\frac{d^2{\psi }_p}{d{x}^2}=z{\psi }_p$       where $z={\alpha }^{3}x$

This equation is called Airy’s equation. The solutions to Airy’s equation are called Airy function.

The patching wave function is obtained using Airy function as,

${\psi }_p\left(x\right)=\alpha Ai\left(\alpha x\right)+bBi\left(\alpha x\right)$

Where Ai and Bi are the special Airy function, and a and b are arbitrary constants.

The final solution is the linear combination of Airy function. This wave function describes the state at the linear region. There exists a region i.e. at the neighborhood of turning point called overlap region. At overlap region the patching function and the WKB method hold good.

Now, using the linear potential equation and the semiclassical equation for the momentum we get,

$p\left(x\right)=\sqrt{2m\left[E-E-V\text{'}\left(0\right)x\right]}=\hslash {\alpha }^{\frac{3}{2}}\sqrt{-x}$

For the nonclassical region, the path integral of momentum is,

$\int \left|p\left(x\right)\right|dx=\int \hslash {\alpha }^{\frac{2}{3}}\sqrt{x}=\frac{2}{3}\hslash {\left(\alpha x\right)}^{\frac{3}{2}}$

Combining this result with the equation obtained from the WKB approximation we get,

${\psi }_p\left(x\right)=\frac{D}{\sqrt{\hslash }{\alpha }^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{x}^{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}{e}^{-i{\left(\alpha x\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$

Similarly, the solution can be obtained for the classical region using,

$\int p\left(x\right)dx=\frac{2}{3}\hslash {(-\alpha x)}^{3/2}$

Using the WKB approximation for the classical region we get,

${\psi }_p\left(x\right)=\frac{\alpha }{\sqrt{\mathrm{\pi }}{(-\alpha )}^{{\displaystyle \frac{1}{4}}}}\sin \left[\frac{2}{3}{(-\alpha x)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+\frac{\mathrm{\pi }}{4}\right]$

${\psi }_p\left(x\right)=\frac{\alpha }{\sqrt{\mathrm{\pi }}{(-\alpha x)}^{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}}\frac{1}{2i}\left[e^{i\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{e}^{i\frac{2}{3}{(-\alpha x)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}-e^{-i\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{e}^{-i\frac{2}{3}{(-\alpha x)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}\right]$

Comparing the WKB equation with the obtained pathing equation we get the following set of equations,

$B=i{e}^{i\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}D$

$C=i{e}^{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}D$

where D is a constant.

These equations are called WKB connection formulae. The WKB connection formulae are used to combine the classical and nonclassical region solution that is obtained using the WKB approximation.

Formulas

The de Broglie wavelength is,

$\lambda =\frac{h}{p}$

The classical momentum equation is,

$p\left(x\right)=\sqrt{2m\left[E-V\left(x\right)\right]}$

The de Broglie wavelength in terms of energy is,

$\lambda \left(x\right)=\frac{h}{\sqrt{2m\left[E-V\left(x\right)\right]}}$

The solution of the Schrödinger equation using the WKB method is,

$\psi \left(x\right)=\frac{C}{\sqrt{p\left(x\right)}}{e}^{\pm \frac{1}{\hslash }\int p\left(x\right)dx}$

The connection formulae for WKB approximation,

$B=i{e}^{i\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}D$

$C=i{e}^{-\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}D$

Where D is a constant.

Context and Applications

This topic is significant in physics for both undergraduate and graduate courses, especially for Masters in physics, and Bachelors in physics.

Practice Problems

Question 1: WKB approximation in quantum mechanics is called ____.

(a) Path integral equation

(b) Semiclassical approximation

(c) Classical approximation

(d) Perturbation

Answer: Option (b) is correct.

Explanation: The WKB approximation method is used when the momentum of a system is the same as the classical momentum. Hence, it is also called semiclassical approximation.

Question 2: For classical bodies, the de Broglie wavelength ___.

(a) Very large

(b) Always positive

(c) Close to zero

(d) None of the above

Answer:  Option (c) is correct.

Explanation: As the de Broglie wavelength tends to zero, the particle obeys classical mechanics well. Hence for classical bodies, the de Broglie wavelength is close to zero.

Question 3: The WKB method is used to solve a system with ___.

(a) Varying potential

(b) Varying kinetic energy

(c) Constant energy

(d) Zero kinetic energy

Answer: Option (b) is correct.

Explanation: The WKB method is used to solve a system with varying potential. The potential varies with respect to the position. It is convenient to solve by the WKB method rather than solving it directly.

Question 4: At the classical region, which of the following conditions holds good?

(a) E<V

(b) E>V

(c) E=V

(d) None of the above

Answer: Option (b) is correct.

Explanation: When the total energy is greater than the potential energy i.e., E>V. The potential of the is real and resembles the classical potential. Hence this region is called the classical region.

Question 5: The WKB method fails at ____.

(a) Turning point

(b) Classical region

(c) Nonclassical region

(d) Path integral region

Answer: Option (a) is correct.

Explanation: At the turning point, the total energy is the same as the potential energy i.e. E=V at this point the momentum is zero and the wavelength is infinite. Hence the equation for WKB collapses at the turning point. The state of the system cannot be defined at this point.