Simple Tunneling

Electron wave

An electron travels toward negative values of $x$ having a kinetic energy $E=2 eV$. Please write the equation of this electron wave and describe its characteristic values.

The general equation of the wave travelling in the negative x direction (towards left) is

$\psi(x,t)=A\exp[i(kx+\omega t)]$  ($A$ is the wave amplitude)

For a fixed point of the wave that has a constant value of $\psi$, the exponential has to maintain its value, so that for time t increasing, $x$ has to decrease, for the wave to travel towards left).

For wave number and frequency we write:

$k=2\pi/\lambda$, $\omega=2\pi/T$, $\nu=\lambda/T=\omega/k$

From the de Broglie relation

$\lambda =h/p=h/\sqrt{2mE}=\frac{6.62*10^{-34}}{\sqrt{2*9.1*10^{-31}*2*1.6*10^{-19}}}=8.67*10^{-10} m$   $k=7.24*10^9 (1/m)$



$\omega=kv=6.07*10^{15} (1/s)$

Tunneling Proton

A proton that is tunneling through a potential $V=10 MeV$. It has a kinetic energy $E_k= 2MeV (<V)$. Please find the tunneling probability if the width of the potential $V$ is $10^{-14} m$. What would be the new width of the potential barrier if the tunneling probability is to increase 10 times?

Proton Tunneling

The real transmission coefficient for tunneling through a square barrier can be found elsewhere. A simpler reasoning (with a simpler result, but with sufficient precision) is the following:

Consider that you have a proton situatied to the left of the barrier. The proton is traveling towards right. Its associated wave form is complex $\psi_1(x)=A\exp(ik_1*x)$

Now the proton enters the barrier. Its wave function has to decrease exponentially inside the barrier.


Then the proton exits the barrier and it is again a wave


(To the left and to the right of the barrier since the proton is situated in the free space the wave number is the same $k_1$)

The probability of tunneling is thus simply the ratio of its wavefunction amplitudes


Inside the barrier the wavenumber is:


Thus the probability for $L=10^{-14} m$ is


If the tunneling probability is increased 10 times we obtain:


$l-a=\frac{ln(10)}{2k_2}=1.85*10^{-15} m$

$a=(1.4-0.185)*10^{-15}=1.215*10^{-15} m$