2.4.1 WKB approximation for bound states

In a bound state problem with potential $V\left(x\right)$, for a given energy $E$, we can divide space into classically allowed regions, for which $E>V\left(x\right)$, and classically forbidden regions for which $E<V\left(x\right)$. For simplicity we will assume that there are only three regions in total, classically forbidden for $x<a$ and $x>b$, and classically allowed for $a<x<b$.

In the classically allowed region $a<x<b$ the wave function will be oscillating and we can write it either as a superposition of right- and left-moving complex exponentials or as

For the particular case of a well with infinite sides the solution must vanish at the boundaries, so (taking the lower limit of integration as $a$ for definiteness; any other choice just shifts $\varphi$) $\varphi =\left(n^{\prime }+\frac{1}{2}\right)\pi$ and ${\int \; <!--nolimits-->}_a^{b}k\left(x^{\prime }\right)\text{d}{x}^{\prime }+\varphi =\left(n^{″}+\frac{1}{2}\right)\pi$; in other words ${\int \; <!--nolimits-->}_a^{b}k\left(x^{\prime }\right)\text{d}{x}^{\prime }=\left(n+1\right)\pi$, with integer $n\ge 0$. Of course for constant $k$ this gives $k=n\pi ∕\left(b-a\right)$, which is exact.

For a more general potential, outside the classically allowed region we will have decaying exponentials. In the vicinity of the turning points these solutions will not be valid, but if we approximate the potential as linear we can solve the Schrödinger equation exactly (in terms of Airy functions). Matching these to our WKB solutions in the vicinity of $x=a$ and $x=b$ gives the surprisingly simple result that inside the well

which can only be satisfied if $A^{\prime }=\pm A$ and ${\int \; <!--nolimits-->}_a^{b}k\left(x^{\prime }\right)\text{d}{x}^{\prime }=\left(n+\frac{1}{2}\right)\pi$. This latter is the quantisation condition for a finite well; it is different from the infinite well because the solution can leak into the forbidden region. (For a semi-infinite well, the condition is that the integral equal $\left(n+\frac{3}{4}\right)\pi$. This is the appropriate form for the $l=0$ solutions of a spherically symmetric well.) Unfortunately we can’t check this against the finite square well though, because there the potential is definitely not slowly varying at the edges, nor can it be approximated as linear. But we can try the harmonic oscillator, for which the integral gives $E\pi ∕\hslash \omega$ and hence the quantisation condition gives $E=\left(n+\frac{1}{2}\right)\hslash \omega$! The approximation was only valid for large $n$ (small wavelength) but in fact we’ve obtained the exact answer for all levels.

Details of the matching process are given in section 2.4.1.1, since I’ve not found them in full detail in any textbook. They are not examinable.

• Gasiorowicz Supplement 4 A
• Shankar ch 16.2

2.4.1.1 Matching with Airy Functions

This section is not examinable. More about Airy functions can be found in section A.7.

If we can treat the potential as linear over a wide-enough region around the turning points that, at the edges of the region, the WKB approximation is valid, then we can match the WKB and exact solutions.

Consider a linear potential $V=\beta x$ as an approximation to the potential near the right-hand turning point $b$. We will scale $x={\left({\hslash }^2∕\left(2m\beta \right)\right)}^{1∕3}z$ and $E={\left({\hslash }^2{\beta }^2∕2m\right)}^{1∕3}\mu$, so the turning point is at $z=\mu$. Then the differential equation is $y^{″}\left(z\right)-zy\left(z\right)+\mu y\left(z\right)=0$ and the solution which decays as $z\to \infty$ is $y\left(z\right)=A\text{Ai}\left(z-\mu \right)$. This has to be matched, for $z$ not too close to $\mu$, to the WKB solution. In these units, $k\left(x\right)=\left(\mu -z\right)$ and ${\int \; <!--nolimits-->}_b^{x}k\left(x^{\prime }\right)\text{d}{x}^{\prime }={\int <!--nolimits-->}_{\mu }^z\left(\mu -z^{\prime }\right)\text{d}{z}^{\prime }=-\left(2∕3\right){\left(\mu -z\right)}^{3∕2}$, so

(We chose the lower limit of integration to be $\mu$ in order that the constant of integration vanished; any other choice would just shift $\varphi$.) Now the asymptotic forms of the Airy function are known:

so

and these will match the WKB expressions exactly provided $C=2B$ and $\varphi =\pi ∕4$.

At the left-hand turning point $a$, the potential is $V=-\beta x$ (with a different $\beta$ in general) and the solution $y\left(z\right)=A\text{Ai}\left(-z-\mu \right)$. On the other hand the WKB integral is ${\int \; <!--nolimits-->}_a^{x}k\left(x^{\prime }\right)\text{d}{x}^{\prime }={\int <!--nolimits-->}_{\mu }^z\left(\mu +z^{\prime }\right)\text{d}{z}^{\prime }=2∕3{\left(\mu +z\right)}^{3∕2}$. So in the classically allowed region we are comparing

which requires $\varphi =-\pi ∕4$. (Note that $\varphi$ is different in each case because we have taken the integral from a different place.)

It is worth stressing that though the exact (Airy function) and WKB solutions match “far away” from the turning point, they do not do so close in. The ${\left(z-\mu \right)}^{-1∕4}$ terms in the latter mean that they blow up, but the former are perfectly smooth. They are shown (for $\mu =0$) below, in red for the WKB and black for the exact functions. We can see they match very well so long as $\left|z-\mu \right|>1$; in fact $z\to \infty$ is overkill!

So now we can be more precise about the conditions under which the matching is possible: we need the potential to be linear over the region $\Delta x\sim {\left({\hslash }^2∕\left(2m\beta \right)\right)}^{1∕3}$ where $\beta =\text{d}V∕\text{d}x$. Linearity means that $\Delta V∕\Delta x\approx \text{d}V∕\text{d}x$ at the turning point, or $\left|{\text{d}}^{2}V∕\text{d}{x}^2\right|\Delta x\ll \text{d}V∕\text{d}x$ (assuming the curvature is the dominant non-linearity, as is likely if $V$ is smooth). For the harmonic oscillator, $\left|{\text{d}}^{2}V∕\text{d}{x}^2\right|\Delta x∕\left(\text{d}V∕\text{d}x\right)=2{\left(\hslash \omega ∕E\right)}^{2∕3}$ which is only much less than 1 for very large $n$, making the exact result even more surprising!

[next] [up] [top]