I have the question:
Consider a beam of particles of mass $m$ and energy $E>0$ incident from the right on the potential well
$$
V(x) =
\begin{cases}
0 & \text{if } x \leq 0\\
-V_0 & \text{if } 0<x<1\\
0 & \text{if } x \geq 1
\end{cases}
$$
where $V_0 > 1$.
(a) Write down the general solution to the time-independent Schrodinger equation in the region I ($x<0$), region II ($0<x<1$) and region III ($x>1$) in terms of real wavenumbers $k_1, k_2, k_1$ respectively using constants $\alpha$ and $\beta$ as the amplitudes of the incident and transmitted beams of particles. State the continuity conditions that the wavefunction must satisfy at $x=0$ and $x=1$.
(b) Calculate the probability current of incident and transmitted beams of particles.
(c) When $E=4\pi^2\hbar^2/m$ and $V_0=17\pi^2\hbar^2/(2m)$ you are given that $$ \frac{\beta}{\alpha} = ie^{-2i\sqrt{2}\pi} .$$ Calculate the probability of transmission. Comment on the effect the potential well has on the beam of particles for this case.
This is what I have:
(a) The wavefunction in region I, $\psi_1$, satisfies $$\frac{-\hbar^2}{2m}\frac{\mathrm{d}^2\psi_1}{\mathrm{d}x^2} = E \psi_1,$$ and so we have $$\psi_1(x)= Ae^{ik_1x}+Be^{-ik_1x},$$ where $k_1=2mE/\hbar^2$. There is no incident beam from the left so $A=0$ and since the amplitude of the transmitted beam is $\beta$ we have $B=\beta$.
The wavefunction in region II, $\psi_2$, satisfies $$\frac{-\hbar^2}{2m}\frac{\mathrm{d}^2\psi_2}{\mathrm{d}x^2} -V_0\psi_2 = E \psi_2,$$ and so we have $$\psi_2(x) = Ce^{k_2x}+De^{-k_2x},$$ where $(k_2)^2=2m(V_0+E)/\hbar^2$.
The wavefunction in region III, $\psi_3$, satisfies $$\frac{-\hbar^2}{2m}\frac{\mathrm{d}^2\psi_3}{\mathrm{d}x^2} = E \psi_3,$$ and so we have $$\psi_3(x)= Fe^{ik_1x}+Ge^{-ik_1x}.$$ Since the amplitude of the incident beam is $\alpha$ we have $G=\alpha$.
We require that the wavefunction be continuous and smooth so we must impose the following conditions at the boundaries of the the regions: $$\psi_1(0) = \psi_2(0) \implies \beta = C+D$$ $$\psi_1'(0) = \psi_2'(0) \implies ik_1\beta = (C-D)k_2$$ $$\psi_1(1) = \psi_2(1) \implies Ce^{k_2}+De^{k_2} = Fe^{ik_1}+\alpha e^{-ik_1}$$ $$\psi_1'(1) = \psi_2'(1) \implies k_2(Ce^{k_2}+De^{k_2}) = ik_1(Fe^{ik_1}+\alpha e^{-ik_1})$$
(b) We have that the probability current is given by $$ j = \frac{\hbar}{2mi} (\psi^\star\nabla\psi - \psi\nabla\psi^\star), $$ so we find the probability current of the incident beam to be
$$\begin{equation} \begin{split} j_I & = \frac{\hbar}{2mi} \left( (\alpha e^{-ik_1x})^\star \frac{\mathrm{d}}{\mathrm{d}x} (\alpha e^{-ik_1x}) - (\alpha e^{-ik_1x})[\frac{\mathrm{d}}{\mathrm{d}x}(\alpha e^{-ik_1x})]^\star \right) \\ & = \frac{\hbar}{2mi} \left( (\alpha^\star e^{ik_1x})(-ik_1\alpha e^{-ik_1x}) - (\alpha e^{-ik_1x})(ik_1\alpha^\star e^{ik_1x}) \right) \\ & =0 \end{split} \end{equation}.$$ This doesn't seem correct. Shouldn't I get something of the form $\frac{\hbar k_1}{m} |\alpha|^2$?
Similarly for $j_T$ I get that it equals zero.
(c) Now $P_T = \frac{j_T}{j_I}$, but this gives $0/0$ so I've definitely gone wrong. Obviously I should get something involving $\frac{\beta}{\alpha}$ which leads me to believe $j_I$ should definitely involve $|\alpha|^2$ and $j_T$ should involve $|\beta|^2$. Then $P_T$ will involve $\big| \frac{\beta}{\alpha}\big|^2$ which is just $$\frac{\beta}{\alpha} \left(\frac{\beta}{\alpha}\right)^\star = ie^{-2i\sqrt{2}\pi} (-ie^{2i\sqrt{2}\pi}) = 1,$$ i.e. all of the incident beam is transmitted. This of course makes sense as the energy of the beam is greater than the potential at the top of the well, so the well shouldn't block any of the beam.
So any help on where I've gone wrong with the probability currents would be much appreciated!
Thanks.
