Potential step

We consider a potential step at $x_1=0$ with $N=1$, $V_1=0$ and $V_2=V>0$ in (2.42).

a) For $E>V$, we have $k_1=\sqrt{(2m/\hbar^2)E}$ and $k_2=\sqrt{(2m/\hbar^2)(E-V)}$ such that from the transfer matrix $T_1=M$, Eq. (2.46), we obtain

$$M_{11}=(T_1)_{11}=\frac{1}{2}\left(1+\frac{k_2}{k_1}\right)e^{i(k_2-k_1)x_1}= \frac{1}{2}\left(1+\frac{k_2}{k_1}\right).$$ (136)

This yields the transmission and reflection coefficients

$$T = \frac{k_{2}}{k_1}\frac{1}{\vert M_{11}\vert^2}= \frac{k_2}{k_1}\frac{4k_1^2}{(k_1+k_2)^2} = \frac{k_2}{k_1}\frac{4}{(1+k_2/k_1)^2} =$$

$$= \left[ \frac{k_2}{k_1}=\sqrt{1-V/E}\right]= \frac{4\sqrt{1-(V/E)}}{(1+\sqrt{1-(V/E)})^2}$$

$$R = \left\vert\frac{M_{21}}{M_{11}}\right\vert^2 = \left\vert\frac{k_1-k_2}{k_1+k_2}\right\vert^2 = \frac{(1-\sqrt{1-(V/E)})^2}{(1+\sqrt{1-(V/E)})^2},$$ (137)

and we recognize that

$$T+R=1.$$ (138)

Compare this result to the case of a classical particle running from the bottom to the top of a (soft) step: if its energy $E$ is sufficient $(E>V)$, it overcomes the barrier and continues to run on the top of the step, if its energy is too small, is rolls back and is reflected. In quantum mechanics, for $E>V$ there is a finite probability for the particle being reflected!

b) For $E<V$ we see that $k_2$ becomes imaginary and there are no longer running waves for $x>0$: the particle then is in the classically forbidden zone. With $k_2=i\kappa_2$, $\kappa_2= \sqrt{(2m/\hbar^2)\vert E-V\vert}$, the wave function on the right is $\psi(x>0)=a_2e^{-\kappa_2x} +b_2e^{\kappa_2x}=a_2e^{-\kappa_2x}$ because we had set $b_2$ anyway. We therefore can still apply our scattering formalism to obtain the reflection coefficient

$$R = \left\vert\frac{k_1-k_2}{k_1+k_2}\right\vert^2 = \left\vert\frac{k_1-i\kappa_2}{k_1+i\kappa_2}\right\vert^2 = 1.$$ (139)

On the right side $x>0$, we don't have running waves any longer for $E<V$ and therefore cannot apply (2.55) for the transmission coefficient. The particle current density (2.51) $j(x>0)=0$, however, which means $T=0$. Again, we have $T+R=1$.

Compare this case to total reflection of waves in optics!