0
$\begingroup$

I have been confused several times by one case when considering particles in wells/with barriers in quantum mechanics, and I keep sweeping it under the carpet because I spend a very long time trying to find a satisfactory answer but fail. Hopefully someone can shed some light on this.

My difficulty is with a particle beam incident on a step barrier where $E<V_0$, so that there is a decaying exponential within the barrier.

What I am unable to figure out is why the transmission coefficient is said to be zero. The image below shows the final transmission and reflection amplitude coefficients:

enter image description here

(Unfortunately they use $j$ rather than $i$ for the unit imaginary number)!

So I can motivate this outcome (that the transmission coefficient is zero) from the following observations:

  1. For the conservation of flux

$|\frac{B}{A}|^2+\frac{i\kappa}{k_1}|\frac{C}{A}|^2=1$, hence if $|\frac{B}{A}|^2=1$ (which it is because the numerator and denomenator are complex conjugates) then it must be that $|\frac{C}{A}|^2=0$

  1. If I rationalise the denomenator of $|\frac{C}{A}|^2$, I get $\frac{2k_1^2-2i\kappa k_1}{k_1^2+\kappa ^2}$ which I see could be zero; well, it would be zero if $\kappa = k_1$. However I do not see why this would be the case...

before the step barrier, we have a wavevector of $k_1=\frac{\sqrt{2mE}}{\hbar}$ whereas afterwards $\kappa = \frac{\sqrt{2m(V-E)}}{\hbar}$. These would only be equal if $V=2E$, which is clearly not always the case.

Could someone please explain how the transmission coefficient is equal to 0?

$\endgroup$
  • $\begingroup$ $C$ is not zero. It represents a decaying exponential, not a plane wave propagating to the right, so there is no transmission. $\endgroup$ – knzhou Mar 31 '17 at 21:25
  • $\begingroup$ Think about light bouncing off a mirror; there's a small transient-like field near the mirror surface. But none of the light goes inside. $\endgroup$ – knzhou Mar 31 '17 at 21:27
  • $\begingroup$ @knzhou Thank you for your reply! The intuition helps- I think I get the situation that it penetrates a bit and reflects back (from what I do not know), but it always returns from within the barrier and no component is propagated to infinity. However I do not understand where this comes from, mathematically. The probability transmission coefficient T is defined as $T=(C/A)^2$, and in the image above it clearly states that T=0, which implies that $c/A=0$. I do not see how the spression above for $C/A$ (i.e. $C/A=\frac{2}{1-i\kappa/k_1}$ guanrantees this to be the case for all well values E<V... $\endgroup$ – 21joanna12 Mar 31 '17 at 21:36
  • $\begingroup$ Who told you that definition of $T$? I think you're using a related definition out of context. $\endgroup$ – knzhou Mar 31 '17 at 21:45
  • $\begingroup$ @knzhou This is what we were taught in our lectures. Stated in the lecture notes in front of me "...we can define probability reflection coefficient R=|B/A|^2, and probability transmission coefficient T=|C/A|^2. Using the conservation for the flux of waves then gives |B/A|^2+(k2/k1)|C/A|^2=1"... Although I admit this was stated in the section for the case of a barrier where E>V, but from the way it was stated it seemed to me that this was a formal definition of T and R for all cases. $\endgroup$ – 21joanna12 Mar 31 '17 at 21:50

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.