4.1 The Potential Step

Supposed we have a potential step

$$ U(x)=\begin{cases} 0 & x<0,\\ U_0 & x>0. \end{cases} $$

If there is a wave traveling from left to right, we need to consider following two scenarios (i) $E>U_0$, and (ii) $E

(i) $E>U_0$

Similar to the finite well case, let the wavefunction at where $U(x)=0$ be $\psi_I(x)$, and the wavefunction at where $U(x)=U_0$ be $\psi_{II}(x)$. To solve the wavefunction, we start with time-independent Schrödinger Equation as usual. For $\psi_I(x)$ we have

$$ \frac{d^2}{dx^2}\psi_I(x)=-\frac{2mE}{2\hbar}\psi_I(x). $$

Let $k^2={2mE}/{2\hbar}$, and then we have a general solution

$$ \psi_I(x)=A\mathrm{e}^{ikx}+B\mathrm{e}^{-ikx}. $$

For $\psi_{II}(x)$, we have

$$ \frac{d^2}{dx^2}\psi_{II}(x)=-\frac{2m(E-U_0)}{2\hbar}\psi_{II}(x). $$

Let $k'^2={2m(E-U_0)}/{2\hbar}$, and a solution can be

$$ \psi_{II}(x)=C\mathrm{e}^{ik'x}. $$

We can treat terms in the form $\exp(+ikx)$ as waves propagating to the right, and terms in the form $\exp(-ikx)$ as waves propagating to the left. The reason there are two terms in $\psi_I(x)$ but only one term in $\psi_{II}(x)$ is as follow:

Force and potential are related by $F=-\nabla U$. Thus the abrupt change of potential at $x=0$, cause a force that behaves like a Dirac-delta function.
This force can change the propagation direction of the wave which is initially traveling to the right, or let it pass through but changes its other properties.
Thus in the region $x<0$, there is a "reflected" wave traveling to the left.
In the region $x>0$, $U(x)$ again is a constant, so there is no additional force that can reflect the "transmitted" wave.

With reasoning above, we can denote terms of wavefunctions as incident wave, reflected wave, and transmitted wave

$$ \psi_{inc}=A\mathrm{e}^{ikx},\ \psi_{ref}=B\mathrm{e}^{-ikx},\psi_{trans}=C\mathrm{e}^{ik'x}. $$

Applying boundary conditions

$$ \begin{cases} \psi_I(0)=\psi_{II}(0),\\ \psi'_I|_{x=0}=\psi'_{II}|_{x=0}, \end{cases} $$

gives following relationship between amplitudes

$$ \begin{cases} A+B=C,\\ k(A-B)=k'C. \end{cases} $$

We can define two physical quantities: transmission probability $T$ and reflection probability $R$ as

$$ T=\frac{\text{transmition per time}}{\text{reflection per time}}, $$

$$ R=\frac{\text{reflection per time}}{\text{reflection per time}}. $$

Here we can multiply velocities to numerator and denominator to ensure those terms are "per time". Then, as $v=p/m=\hbar k/m$, we are able to use wavevector $k$ to force those terms are "per time" as desired. Thus, $T$ and $R$ can be expressed as

$$ T=\frac{|\psi_{trans}|^2k'}{|\psi_{inc}|^2k}=\frac{C^*Ck'}{A^*Ak}, $$

$$ R=\frac{|\psi_{ref}|^2k}{|\psi_{inc}|^2k}=\frac{B^*B}{A^*A}. $$

With the relationship between amplitudes, above relation becomes

$$ T=\frac{4kk'}{(k+k')^2}, $$

$$ R=\frac{(k-k')^2}{(k+k')^2}. $$

In terms of $U_0$ and $E$,

$$ \begin{gather} \bbox[silver]{ T=4\frac{\sqrt{E(E-U_0)}}{(\sqrt{E}+\sqrt{E-U_0})^2},\\ R=4\frac{(\sqrt{E}-\sqrt{E-U_0})^2}{(\sqrt{E}+\sqrt{E-U_0})^2}. }\end{gather} $$

(ii) $E<U_0$

When $E<U_0$, the wave equation $\psi_{II}$ becomes

$$ \psi_{II}(x)=C\mathrm{e}^{-\alpha x}, $$

where

$$ \alpha=\sqrt{\frac{2m(U_0-E)}{\hbar^2}}, $$

and the term with positive exponent ignored. This is because as a well behaved function, the function should converge as $x\rightarrow+\infty$. Again applying boundary conditions gives

$$ \begin{cases} A+B=C,\\ k(A-B)=-\alpha C. \end{cases} $$

It turned out that in this case $R=1$ and $T=0$. However this does not imply that there is no possibility to find a particle at $x>0$, as $|\psi_{II}|$ still has value greater than 0.