User:Gareth Owen/WKB approximation

In physics, the WKB (Wentzel-Kramers-Brillouin) approximation, also known as WKBJ approximation, is the most familiar example of a semiclassical calculation in quantum mechanics in which the wavefunction is recast as an exponential function, semiclassically expanded, and then either the amplitude or the phase is taken to be slowly changing.

This method is named after physicists Wentzel, Kramers, and Brillouin, who all developed it in 1926. In 1923, mathematician Harold Jeffreys had developed a general method of approximating linear, second-order differential equations, which includes the Schrödinger equation. But since the Schrödinger equation was developed two years later, and Wentzel, Kramers, and Brillouin were apparently unaware of this earlier work, Jeffreys is often neglected credit. Early texts in quantum mechanics contain any number of combinations of their initials, including WBK, BWK, WKBJ and BWKJ.

Derivation

We beginning with a one dimensional, time-independent wave equation in which the local wavenumber varies. Such an equation can typically be written as

$\epsilon ^{2}{\frac {d^{2}}{dx^{2}}}\Psi (x)+K^{2}(x)\Psi (x)=0$,

where K is O(1) and $\epsilon$ is small.

We recast the wavefunction as the exponential of another function Φ (which is closely related to the action):

$\Psi (x)=\displaystyle {e^{\Phi (x)}}$

The function Φ must then satisfy

$\epsilon ^{2}\left[\Phi ''(x)+\left[\Phi '(x)\right]^{2}\right]+K^{2}(x)=0$

where Φ' indicates the derivative of Φ with respect to x. Now let us separate $\Phi '(x)$ into real and imaginary parts by introducing the real functions A and B:

$\Phi '(x)=A(x)+iB(x)$

The amplitude of the wavefunction is then $e^{A(x)}$ while its phase is $B(x)$. The governing equation implies that these functions must satisfy:

$A'(x)+A(x)^{2}-B(x)^{2}+K^{2}(x)=0$

and since the right hand side of the differential equation for Φ is real,

$B'(x)+2A(x)B(x)=0.$

Next we want to find an asymptotic approximation to solve this. That means we expand each function as a power series in $\epsilon$. From the equations we can already see that the power series must start with at least an order of $\epsilon ^{-1}$ to satisfy the real part of the equation.

$A(x)=\sum _{n=-1}^{\infty }\epsilon ^{n}A_{n}(x)$

$B(x)=\sum _{n=-1}^{\infty }\epsilon ^{n}B_{n}(x)$

To first order in this expansion, the conditions on A and B can be written.

$A_{-1}(x)^{2}-B_{-1}(x)^{2}+K^{2}(x)=0,$

$A_{-1}(x)B_{-1}(x)=0$

Clearly then, the second of these equations tells us that either $A_{-1}(x)$ or $B_{-1}(x)$ must be identically zero. Since both functions are real, examination of the first equation tells us that if $K^{2}>0$ then $A_{-1}\equiv 0$ and if $K^{2}(x)<0$, then $B_{-1}\equiv 0$

Oscillatory Regime

In the former case, $\displaystyle K^{2}>0$ the leading order term

$\epsilon ^{-1}B_{-1}(x)=\pm {\frac {1}{K(x)}},$

constitutes a rapid variation in phase. Thus, the solutions to the equation are predominately oscillatory in nature. For this case, we can calculate the next order correction. Taking the real and imaginary parts of term of order $\epsilon$, and noting that $A_{-1}=A'_{-1}=0$ we have

$B_{0}(x)B_{-1}(x)=0and:<math>B'_{-1}(x)+2A_{0}(x)B_{-1}(x)=0,$

giving us

$B_{0}(x)=0\quad \mathrm {and} \quad A_{0}(x)=-{\frac {1}{2}}{\frac {B'_{-1}(x)}{B_{-1}(x)}}-{\frac {1}{2}}{\frac {K'(x)}{K(x)}}$

Thus, to the first two orders

$\Phi '(x)={\frac {i}{\epsilon }}B_{-1}(x)+A_{0}(x)=\pm {\frac {i}{\epsilon }}{\frac {1}{K(x)}}-{\frac {1}{2}}{\frac {K'(x)}{K(x)}}$

Integrating directly, we have

$\Psi (x)=e^{\Phi (x)}=\exp \left({\frac {i}{\epsilon }}\int ^{x}ds/K(s)-{\frac {1}{2}}\log(K(x))\right)={\frac {\exp \left(\pm {\frac {i}{\epsilon }}\int ^{x}{\frac {ds}{K(s)}}\right)}{\sqrt {K(s)}}}$

Exponential Regime

On the other hand, if the phase varies slowly as compared to the amplitude, we set $B_{0}(x)=0$ and get

$A_{0}(x)=\pm {\sqrt {2m\left(V(x)-E\right)}}$

Which is obviously only valid when the potential energy is greater than the total energy (the regime in which quantum tunneling occurs). Grinding out the next order of the expansion yields

$\Psi (x)\approx {\frac {C_{+}e^{+\int dx{\sqrt {{\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)}}}+C_{-}e^{-\int dx{\sqrt {{\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)}}}}{\sqrt[{4}]{{\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)}}}$

It is apparent from the denominator, that both of these approximate solutions 'blow up' near the classical turning point where $E=V(x)$ and cannot be valid. What we have are the approximate solutions away from the potential hill and beneath the potential hill. Away from the potential hill, the particle acts similarly to a free wave - the phase is oscillating. Beneath the potential hill, the particle undergoes exponential changes in amplitude.

To be complete we must find the approximate solutions everywhere and match coefficients to make a global approximate solution. We have yet to approximate the solution near the classical turning points $E=V(x)$.

Let us label a classical turning point $x_{1}$. Now because we are near $E=V(x_{1})$, we can expand ${\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)$ in a power series.

${\frac {2m}{\hbar ^{2}}}\left(V(x)-E\right)=U_{1}(x-x_{1})+U_{2}(x-x_{1})^{2}+\cdots$

To first order, one finds

${\frac {d^{2}}{dx^{2}}}\Psi (x)=U_{1}(x-x_{1})\Psi (x)$

This differential equation is Airy equation, and the solution may be written in terms of Airy functions. Alternatively, with some trickery, it may be transformed into a Bessel equation of fractional order, leading to the solution:

$\Psi (x)={\sqrt {x-x_{1}}}\left(C_{+{\frac {1}{3}}}J_{+{\frac {1}{3}}}\left({\frac {2}{3}}{\sqrt {U_{1}}}(x-x_{1})^{\frac {1}{3}}\right)+C_{-{\frac {1}{3}}}J_{-{\frac {1}{3}}}\left({\frac {2}{3}}{\sqrt {U_{1}}}(x-x_{1})^{\frac {1}{3}}\right)\right)$

Hopefully this solution should connect the far away and beneath solutions. Given the 2 coefficients on one side of the classical turning point, we should be able to determine the 2 coefficients on the other side of the classical turning point by using this local solution to connect them. We should be able to find a relationship between $C,\theta$ and $C_{+},C_{-}$.

Fortunately the Airy/Bessel function solutions will asymptote into sine, cosine and exponential functions in the proper limits. The relationship can be found to be as follows (often referred to as "connection formulas"):

$C_{+}={\frac {1}{2}}C\cos {\left(\theta -{\frac {\pi }{4}}\right)}$

$C_{-}=-C\sin {\left(\theta -{\frac {\pi }{4}}\right)}$

Now we can easily construct global (approximate) solutions.

See also

Perturbation methods, Quantum tunnelling, Airy Function

References

• Razavy, Moshen (2003). Quantum Theory of Tunneling. World Scientific. ISBN 9812380191.
• Griffiths, David J. (2004). Introduction to Quantum Mechanics (2nd ed.). Prentice Hall. ISBN 0131118927.
• Liboff, Richard L. (2003). Introductory Quantum Mechanics (4th ed.). Addison-Wesley. ISBN 0805387145.
• Sakurai, J. J. (1993). Modern Quantum Mechanics. Addison-Wesley. ISBN 0201539292.

External links

• The W.K.B. Approximation (Note that in this webpage, ${\mbox{Eq.}}(4.x)\equiv (x+639)$: there are two sets of labels for the equations.)