Quick intro:

A 1D quantum particle is subject to the potential

$$ V(x) = \begin{cases} 0 \;\;\;\;\; x\leq 0\\ V_0 \;\;\; x > 0 \end{cases} $$

I am trying to understand the definition of transmission and reflection coefficients depending on the energy of the particle.

We know that if $E>V_0$, then the Schrödinger equation is solved by a combination of plane waves (scattering solutions), namely

$$ \psi(x) = Be^{ik'x}+Ce^{-ikx} $$

where the coefficient $B$ is associated to the plane wave moving to the left and $C$ to the one moving to the right. Say I am sending a free particle from the left side, having wave function

$$ \psi(x) = Ae^{ikx} $$

then we write the reflection and transmission coefficients as

$$ \begin{cases} T = \frac{|B|^2}{|A|^2}\\ R = \frac{|C|^2}{|A|^2} \end{cases} $$

and this makes sense to me, since these ratios give us the probabilities of finding the particle on the left and right side of the barrier after the interaction has happened.

Now let's say that $0<E<V_0$ and we are still sending a free particle moving to the right, now the solution for $x>0$ will be a decaying exponential

$$ \psi(x) = Ae^{-kx} $$

while we still have a free particle reflected back to the left.

My question:

Now there is a finite probability of finding the particle at $x>0$, although this is decaying very quickly to zero, so do we just say that the transmission coefficient is zero because the probability of finding the particle on the right side far from $x=0$ is zero? In other words, how are the transmission and reflection defined in general?

  • $\begingroup$ Two things to think about: 1.) conservation requires $1-A=T+R$ where $A$ is absorption. 2.) how would you measure transmission? $\endgroup$
    – garyp
    Apr 17, 2022 at 14:00
  • $\begingroup$ I would put a detector to a distance which is far bigger than the characteristic length of the exponential, so this means that for the transmission we should take into account only the wave function at infinity? $\endgroup$
    – Andrea
    Apr 17, 2022 at 14:22
  • $\begingroup$ Well, yes, but haven't you concluded that already in your last paragraph? Perhaps I'm not really understanding what you are getting at. $\endgroup$
    – garyp
    Apr 18, 2022 at 11:12

1 Answer 1


In general, $R$ and $T$ can be defined with probability current.

In 1D,

$$j = \frac{\hbar}{ 2mi} ( \psi^* \frac{\partial \psi}{\partial x} - \psi \frac{\partial \psi^*}{\partial x} )$$

In 3D, partial derivative will be replaced to gradient, and it satisfies contonuity equation.

$$ \frac{\partial \psi\psi^*}{\partial t} + \nabla \cdot \vec{j}=0 $$

$R$ and $T$ are ratio of these probability current.

$$R = \frac{j_{\text{reflect}}}{j_{\text{incident}}}$$ $$T = \frac{j_{\text{transmit}}}{j_{\text{incident}}}$$

These definition include your expressions with $|A|^2, |B|^2$, and $|C|^2$. Also, for decaying wave function case, probability current is zero because it is real-valued function.


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