User:Thunderbird2/WKB approximation

In physics, the WKB (Wentzel-Kramers-Brillouin) approximation, also known as WKBJ (Wentzel-Kramers-Brillouin-Jeffreys) 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
Beginning with the one dimensional, time-independent Schrödinger equation:
 * $$-\frac{\hbar^2}{2m} \frac{\mathrm{d}^2}{\mathrm{d}x^2} \Psi(x) + V(x) \Psi(x) = E \Psi(x)$$

which can be rewritten as


 * $$\frac{\mathrm{d}^2}{\mathrm{d}x^2} \Psi(x) = \frac{2m}{\hbar^2} \left( V(x) - E \right) \Psi(x)$$

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


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

&Phi; must then satisfy


 * $$\Phi''(x) + \left[\Phi'(x)\right]^2 = \frac{2m}{\hbar^2} \left( V(x) - E \right)$$

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


 * $$\Phi'(x) = A(x) + i B(x) \;$$

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


 * $$A'(x) + A(x)^2 - B(x)^2 = \frac{2m}{\hbar^2} \left( V(x) - E \right)$$

and since the right hand side of the differential equation for &Phi; is real,


 * $$B'(x) + 2 A(x) B(x) = 0 \;$$

Next we want to take the semiclassical approximation to solve this. That means we expand each function as a power series in $$\hbar$$. From the equations we can already see that the power series must start with at least an order of $$\hbar^{-1}$$ to satisfy the real part of the equation. But as we want a good classical limit, we also want to start with as high a power of Planck's constant as possible.


 * $$A(x) = \frac{1}{\hbar} \sum_{n=0}^\infty \hbar^n A_n(x)$$


 * $$B(x) = \frac{1}{\hbar} \sum_{n=0}^\infty \hbar^n B_n(x)$$

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


 * $$A_0(x)^2 - B_0(x)^2 = 2m \left( V(x) - E \right)$$


 * $$A_0(x) B_0(x) = 0 \;$$

If the amplitude varies slowly as compared to the phase, we set $$A_0(x) = 0$$ and get


 * $$B_0(x) = \pm \sqrt{ 2m \left( E - V(x) \right) }$$

which is only valid when the total energy is greater than the potential energy, as is always the case in classical motion. After the same procedure on the next order of the expansion we get


 * $$\Psi(x) \approx C \frac{ e^{i \int \mathrm{d}x \sqrt{\frac{2m}{\hbar^2} \left( E - V(x) \right)} + \theta} }{\sqrt[4]{\frac{2m}{\hbar^2} \left( E - V(x) \right)}}$$

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 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


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

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


 * $$\frac{\mathrm{d}^2}{\mathrm{d}x^2} \Psi(x) = U_1 (x - x_1) \Psi(x)$$

This differential equation is known as the 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"):


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


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

Now we can construct global (approximate) solutions.