1
$\begingroup$

Consider a stream of particles (a plane wave) with energy $E$ incident on a potential barrier with height $V_0 > E$.

Mathematically this leads to an exponential decay of the probability of finding the particle in the classically forbidden region, i.e. beyond the barrier.

Is there a physical interpretation of this?

Classically the analogue is an evanescent wave in the case of total internal reflection. The physical meaning that I've attributed to this is that the evanescent field is the field generated by the material to cancel the incoming radiation and reach equilibrium. So the energy after the barrier is just determined by the spatial rearrangement of charge of the material, and it's not taken from the incident wave which is why the energy flux of the reflected wave is identical to the incident one.

Is there a similar reasoning for the quantum case?

$\endgroup$
  • $\begingroup$ Are you asking 'what happens if the particle appears in a forbidden region instead of tunnelling through the region to the inaccessible but otherwise fine region beyond it?' $\endgroup$ – gautampk Apr 22 '17 at 16:35
  • $\begingroup$ So, just a step potential. So finding the particle inside the step. $\endgroup$ – SuperCiocia Apr 22 '17 at 17:19
  • $\begingroup$ The physical interpretation is just the same as with an evanescent wave. In quantum mechanics, there isn't really a particle at all, just the wavefunction. So if you happen to do a measurement and find the particle there, it'll will, for that instant, just look like the particle was there. The wavefunction collapses to a position eigenstate, which then proceeds to evolve under the action of $e^{-iHt}$ just like normal. If $[H,x] = 0$ then $e^{-iHt} = e^{-iE_{x}t}$ where $E_{x}$ is the associated energy eigenvalue, otherwise you have more complex time evolution. $\endgroup$ – gautampk Apr 22 '17 at 17:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.