Categories: Physics, Quantum mechanics.

WKB approximation

In quantum mechanics, the Wentzel-Kramers-Brillouin or simply the WKB approximation is a technique to approximate the wave function ψ(x)\psi(x) of the one-dimensional time-independent Schrödinger equation. It is an example of a semiclassical approximation, because it tries to find a balance between classical and quantum physics.

In classical mechanics, a particle travelling in a potential V(x)V(x) along a path x(t)x(t) has a total energy EE as follows, which we rearrange:

E=12mx˙2+V(x)    m2(x)2=2m(EV(x))\begin{aligned} E = \frac{1}{2} m \dot{x}^2 + V(x) \quad \implies \quad m^2 (x')^2 = 2 m (E - V(x)) \end{aligned}

The left-hand side of the rearranged version is simply the momentum squared, so we define the magnitude of the momentum p(x)p(x) accordingly:

p(x)=2m(EV(x))\begin{aligned} p(x) = \sqrt{2 m (E - V(x))} \end{aligned}

Note that this is under the assumption that E>VE > V, which is always true in classical mechanics, but not necessarily in quantum mechanics. We rewrite the Schrödinger equation:

0=d2ψdx2+2m2(EV)ψ=d2ψdx2+p22ψ\begin{aligned} 0 = \dvn{2}{\psi}{x} + \frac{2 m}{\hbar^2} (E - V) \psi = \dvn{2}{\psi}{x} + \frac{p^2}{\hbar^2} \psi \end{aligned}

If V(x)V(x) were constant, and by extension p(x)p(x) too, then the solution is easy:

ψ(x)=ψ(0)exp(±ipx/)\begin{aligned} \psi(x) = \psi(0) \exp(\pm i p x / \hbar) \end{aligned}

This form is reminiscent of the generator of translations. In practice, V(x)V(x) and p(x)p(x) vary with xx, but we can still salvage this solution by assuming that V(x)V(x) varies slowly compared to the wavelength λ(x)=2π/k(x)\lambda(x) = 2 \pi / k(x), where k(x)=p(x)/k(x) = p(x) / \hbar is the wavenumber. The solution then takes the following form:

ψ(x)=ψ(0)exp ⁣( ⁣± ⁣i0xχ(ξ)dξ)\begin{aligned} \psi(x) = \psi(0) \exp\!\Big(\!\pm\! \frac{i}{\hbar} \int_0^x \chi(\xi) \dd{\xi} \Big) \end{aligned}

χ(ξ)\chi(\xi) is an unknown function, which intuitively should be related to p(x)p(x). The purpose of the integral is to accumulate the change of χ\chi from the initial point 00 to the current position xx. Let us write this as an indefinite integral for convenience:

ψ(x)=ψ(0)exp ⁣( ⁣± ⁣i(χ(x)dxC))\begin{aligned} \psi(x) = \psi(0) \exp\!\bigg( \!\pm\! \frac{i}{\hbar} \Big( \int \chi(x) \dd{x} - C \Big) \bigg) \end{aligned}

Where C=χ(x)dxx=0C = \int \chi(x) \dd{x} |_{x = 0} is the initial point of the definite integral. For simplicity, we absorb the constant CC into ψ(0)\psi(0). We can now clearly see that:

ψ(x)=±iχ(x)ψ(x)    χ(x)=±iψ(x)ψ(x)\begin{aligned} \psi'(x) = \pm \frac{i}{\hbar} \chi(x) \psi(x) \quad \implies \quad \chi(x) = \pm \frac{\hbar}{i} \frac{\psi'(x)}{\psi(x)} \end{aligned}

Next, we insert this ansatz for ψ(x)\psi(x) into the Schrödinger equation to get:

0=±id(χψ)dx+p22ψ=±iχψ±iχψ+p22ψ=±iχψ12χ2ψ+p22ψ\begin{aligned} 0 &= \pm \frac{i}{\hbar} \dv{(\chi \psi)}{x} + \frac{p^2}{\hbar^2} \psi = \pm \frac{i}{\hbar} \chi' \psi \pm \frac{i}{\hbar} \chi \psi' + \frac{p^2}{\hbar^2} \psi = \pm \frac{i}{\hbar} \chi' \psi - \frac{1}{\hbar^2} \chi^2 \psi + \frac{p^2}{\hbar^2} \psi \end{aligned}

Dividing out ψ\psi and rearranging gives us the following, which is still exact:

±iχ=p2χ2\begin{aligned} \pm \frac{\hbar}{i} \chi' = p^2 - \chi^2 \end{aligned}

Next, we expand this as a power series of \hbar. This is why it is called semiclassical: so far we have been using full quantum mechanics, but now we are treating \hbar as a parameter which controls the strength of quantum effects:

χ(x)=χ0(x)+iχ1(x)+2i2χ2(x)+...\begin{aligned} \chi(x) = \chi_0(x) + \frac{\hbar}{i} \chi_1(x) + \frac{\hbar^2}{i^2} \chi_2(x) + ... \end{aligned}

The heart of the WKB approximation is its assumption that quantum effects are sufficiently weak (i.e. \hbar is small enough) that we only need to consider the first two terms, or, more specifically, that we only go up to \hbar, not 2\hbar^2 or higher. Inserting the first two terms of this expansion into the equation:

±iχ0=p2χ022iχ0χ1\begin{aligned} \pm \frac{\hbar}{i} \chi_0' &= p^2 - \chi_0^2 - 2 \frac{\hbar}{i} \chi_0 \chi_1 \end{aligned}

Where we have discarded all terms containing 2\hbar^2. At order 0\hbar^0, we then get the expected classical result for χ0(x)\chi_0(x):

0=p2χ02    χ0(x)=p(x)\begin{aligned} 0 = p^2 - \chi_0^2 \quad \implies \quad \chi_0(x) = p(x) \end{aligned}

While at order \hbar, we get the following quantum-mechanical correction:

±iχ0=2iχ0χ1    χ1(x)=12χ0(x)χ0(x)\begin{aligned} \pm \frac{\hbar}{i} \chi_0' = - 2 \frac{\hbar}{i} \chi_0 \chi_1 \quad \implies \quad \chi_1(x) = \mp \frac{1}{2} \frac{\chi_0'(x)}{\chi_0(x)} \end{aligned}

Therefore, our approximated wave function ψ(x)\psi(x) currently looks like this:

ψ(x)ψ(0)exp ⁣( ⁣± ⁣iχ0(x)dx)exp ⁣( ⁣± ⁣χ1(x)dx)\begin{aligned} \psi(x) &\approx \psi(0) \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int \chi_0(x) \dd{x} \Big) \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big) \end{aligned}

We can reduce the latter exponential using integration by substitution:

exp ⁣( ⁣± ⁣χ1(x)dx)=exp ⁣( ⁣ ⁣12χ0(x)χ0(x)dx)=exp ⁣( ⁣ ⁣121χ0dχ0)=exp ⁣( ⁣ ⁣12ln ⁣(χ0(x)))=1χ0(x)=1p(x)\begin{aligned} \exp\!\Big( \!\pm\! \int \chi_1(x) \dd{x} \Big) &= \exp\!\Big( \!-\! \frac{1}{2} \int \frac{\chi_0'(x)}{\chi_0(x)} \dd{x} \Big) = \exp\!\Big( \!-\! \frac{1}{2} \int \frac{1}{\chi_0}\:d\chi_0 \Big) \\ &= \exp\!\Big( \!-\! \frac{1}{2} \ln\!\big(\chi_0(x)\big) \Big) = \frac{1}{\sqrt{\chi_0(x)}} = \frac{1}{\sqrt{p(x)}} \end{aligned}

In the WKB approximation for E>VE > V, the solution ψ(x)\psi(x) is thus given by:

ψ(x)Ap(x)exp ⁣( ⁣± ⁣ip(x)dx)\begin{aligned} \boxed{ \psi(x) \approx \frac{A}{\sqrt{p(x)}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big) } \end{aligned}

What if E<VE < V? In classical mechanics, this is just not allowed; a ball cannot simply go through a potential bump without the necessary energy. On the other hand, in quantum physics, particles can tunnel through barriers.

Luckily, the only thing we need to change for the WKB approximation is to let the momentum take imaginary values:

p(x)=2m(EV(x))=i2m(V(x)E)\begin{aligned} p(x) = \sqrt{2 m (E - V(x))} = i \sqrt{2 m (V(x) - E)} \end{aligned}

And then take the absolute value in the appropriate place in front of ψ(x)\psi(x):

ψ(x)Ap(x)exp ⁣( ⁣± ⁣ip(x)dx)\begin{aligned} \boxed{ \psi(x) \approx \frac{A}{\sqrt{|p(x)|}} \exp\!\Big( \!\pm\! \frac{i}{\hbar} \int p(x) \dd{x} \Big) } \end{aligned}

In the classical region (E>VE > V), the wave function oscillates, and in the quantum-physical region (E<VE < V) it is exponential. Note that for EVE \approx V the approximation breaks down, because of the appearance of p(x)p(x) in the denominator.


  1. D.J. Griffiths, D.F. Schroeter, Introduction to quantum mechanics, 3rd edition, Cambridge.
  2. R. Shankar, Principles of quantum mechanics, 2nd edition, Springer.