# Quantum tunnelling decay time

1. Jun 29, 2011

### FunkyDwarf

Hi All,

I have a question regarding the WKB method for computing tunnelling through barriers.

I understand the method and the ability to arrive at a solution as given in the first part (summary) of the first page here:

http://www.physics.udel.edu/~msafrono/425/Lecture 18.pdf

Is it correct to say that the tunneling probability is given by
$$\left| \frac{\psi(b)}{\psi(a)}\right|^2$$ where a is the classical turning point and b is the end of the barrier?

If so, and one had some arbitrary potential barrier, does one not need to take into account the $$\frac{1}{\sqrt{p}}$$ factors evaluated at these end points (i.e. higher order terms in the WKB approx)? Does this give the accurate prefactor to the transmission probability T?

Cheers!

2. Jun 29, 2011

### SpectraCat

The tunneling probability is given by the square modulus of the ratio of the *coefficients* of the outgoing to incoming waves, not the ratio of the wavefunction evaluated at the classical turning points. Does that answer the rest of your question? Also, that equation for T is an approximation that is only valid when the tunneling probability is small. The full expression is:

$$T=\frac{e^{-2\gamma}}{|1+\frac{1}{4}e^{-2\gamma}|^2}$$

So it reduces to $T=e^{-2\gamma}$ when T << 1.

3. Jun 29, 2011

### FunkyDwarf

Well in my case I have a standing wave so I have waves with equal magnitudes in both directions, but anyway isn't it so that the sqrt momentum pre-factor would contribute to the amplitudes in both regions? Or is it instead folded into the phase by moving it to the exponent (in which case the phase is more complicated)? In the real standing wave case the reflected wave has the same amplitude as the incoming, and so one cannot deduce the transmission coefficient from that, so all you can do is consider relative amplitudes in Region I compared to Region III right? (region II being the barrier part I guess)

Cheers for your help!

EDIT: Nm figured it out! Taking psi* psi gets rid of the radial dependence and leaves you with the amplitude, hurr! =) Cheers dude!

Last edited: Jun 29, 2011