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While reviewing some quantum mechanics, I cam across a very interesting situation. For a potential barrier, if a particle has an energy $E$ less than the potential barrier $V_0$, it is possible to measure it inside the potential barrier or the classically forbidden region quantum mechanically.

But, if we calculate reflection and transmission coefficients for the wave, we find out $R=1$ and $T=0$ which means the wave will be fully reflected and no transmission will take place. But still there is probability of finding it inside the potential barrier. How?


marked as duplicate by Brandon Enright, Qmechanic Dec 16 '13 at 10:34

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  • $\begingroup$ there should be some transmission for a potential barrier. If you are thinking of just a single step, then there is of course no transmission, but you can still find particles in the high potential region. $\endgroup$ – Brian Moths Sep 17 '13 at 2:02
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/11188/2451 and links therein. $\endgroup$ – Qmechanic Sep 17 '13 at 2:19

This is wholly analogous to the evanescent optical field that arises in the classically (i.e. computed by raytracing) forbidden region beyond a totally internally reflecting interface between two optical mediums. I analyse this situation in my answer here and there is also a great plot of the situation in Ruslan's answer here.

Let's think of a 1D barrier and nonrelativistically (the relativistic Dirac equation leads to the Klein paradox, so we'll let that sleeping hound lie). What's happenning is that on the low energy side (say $U=0$) of the barrier, the 1D Schrödinger equation has the form (setting all the constants to unity):

$$({\rm d}_x^2 + E)\psi(x) = 0$$

whence solutions of the form $\psi = A e^{\pm i \sqrt{E} x}$. These have real well-defined momentum (they are momentum eigenstates of the observable -i {\rm d_x}$ so they represent travelling waves.

But on the high energy side we've got:

$$({\rm d}_x^2 + E-U)\psi(x) = 0$$

with $U>E$ so that we get evanescent waves i.e. plane waves $\psi = A e^{-\sqrt{U-E} x}$ with imaginary wavenumber and imaginary momentum.

What this means is that the particle is not propagating in these classically forbidden regions: it is "standing still" and its field of influence decays swiftly with increasing depth into the forbidden region. In the optical, total internal reflexion case, an evanescent wave is one comprising pulsating electric and magnetic field energy shuttling periodically to and fro between neighbouring regions. Here the situation is mathematically precisely analogous, but we haven't got shuttling energy, we've got shuttling probability density. The regions of highest provbability quivver, there are instantanous probability current fluxes back and forth between neighbouring regions such that averaged over a period the nett probability flux is nought. So if there is an incident wave, all its probabilty flux will ultimately get reflected, owing to the lack of nett average flux in the classically forbidden region, just as a totally internally reflected wave is theoretically 100% energy efficient (there is no loss) notwitstanding the penetration of the field of influence. Therefore, one expects the reflexion coefficient to be unity for an infinitely thick classical region. But as in my and Ruslan's answers, if there is only a thin forbidden region, the evanescent waves reach the other side and become propagating waves with real momentum again, again precisely analogously to the optical situation.


Reflection and transmission coefficients always refer to plane waves, i.e. the components of the wave function that go straight to infinity. Which means, if you measure sufficiently deep inside of the barrier, the probability will indeed approach zero.

However, such waves are not the only solutions to the Schrödinger equation inside a constant potential. With energy below $V_0$, you rather get exponentially decaying solutions. Which is obviously unphysical in an unbounded space range: the exponential is unbounded, the wavefunction would thus not be renormalisable.

But that need not be an issue close to a boundary: if you "cut off" the unbounded part, the remaining "tail", which approaches zero quickly, is normalisable all right. So at a boundary between $0$ and $V_0>E$, you get a transition between a plane-waves solution and an exponential one, called evanescent wave. It is this evanescent wave that shows the amplitude of tunneling into the barrier. It also shows how unlikely tunnelling gets if you measure a little farther from the boundary, due to the exponential decay. In that sense, the transmission is indeed zero: the particle doesn't properly enter the barrier, it's rather "squeezed in but bounces back out immediately", for a classical analogy.

  • $\begingroup$ Have a look at hyperphysics.phy-astr.gsu.edu/hbase/quantum/barr.html . The exponential decay is within the barrier. on the other side it is again a propagating wave with imposed continuity conditions on the boundary.Note it is a probability wave. $\endgroup$ – anna v Dec 16 '13 at 8:19

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