Some (potentially) really simple algebra I can't figure out Okay so, I couldn't be more specific in the title because honestly I can't make it fit in a way that makes sense.
We've been told that:
$$A + B = C + D \tag{1}$$
and
$$ik_1A - ik_1B = ik_2C-ik_2D  \tag{2}$$
I'm trying to show that:
$$\frac{A+B}{A-B} = \frac{k_1}{k_2}\frac{C+D}{C-D} = \frac{k_1^2}{k_2^2} $$
So basically I've rearranged equation $(2)$ to show that
$$ A-B=\frac{k_2}{k_1}(C-D) $$
and so we can take equation $(1)$, divide both sides by $A-B$ and then substitute in the expression for $A-B$ we just found, at which point we get
$$ \frac{A+B}{A-B} = \frac{C+D}{A-B} = \frac{k_1}{k_2}\frac{C+D}{C-D} \\$$
This is where I hit a dead end. I can show that
$\dfrac{k_1}{k_2} = \dfrac{C-D}{A-B},\quad$ but I can't  show
$\quad\dfrac{C+D}{C-D}
= \dfrac{k_1}{k_2}$
(which would give me the last part) and honestly I've been banging my head against this all morning and just making more of a mess.
Can someone nudge me and put me out of my misery? This isn't even a real part of the question it's like the preamble bit  (The question overall is to do with tunnelling and scattering in Quantum Mechanics)
 A: The main reason you're having such a hard time proving it is that
it's false.
To keep the notation simple, write
$$
\lambda = \frac{k_1}{k_2}.
$$
(This is valid, because the question requires us to assume
$k_2 \ne 0$ in any case.)
Thus (assuming also that $i \ne 0$ - which
is true in particular if $i$ denotes a square root of $-1$) we are given:
\begin{align*}
A + B & = C + D, \\
\lambda(A - B) & = C - D.
\end{align*}
One hint that
something is wrong is that there is nothing here to imply that
$A \ne B$ or $C \ne D,$ both of which conditions are required in
order for the conclusion to make sense.
An even stronger hint that something is wrong is that if
$(A, B, C, D)$ is any solution of (1) and (2), then so is
$$
(A', B', C', D') = (A + h, B + h, C + h, D + h),
$$
for any number $h,$ but if $h \ne 0,$ then
$$
\frac{A' + B'}{A' - B'} = \frac{A + B}{A - B} + \frac{2h}{A - B}
\ne \frac{A + B}{A - B},
$$
so it cannot be true that
$$
\frac{A + B}{A - B} = \lambda^2
$$
for every solution $(A, B, C, D)$ of (1) and (2).
A: Start with solutions to Schrodingar Equation for Finite step potential with E>V:
$x<0:\psi_1=Ae^{ikx}+Be^{-ikx}$
$x>0: \psi_2=Ce^{ipx}+De^{-ipx}$
The wave function and first derivative are continuous for bounded potentials, so:
$i) A+B=C+D$
$ii) ik(A-B)=ip(C-D)$
We have two equations in two unknowns, so we should be able to express two of the unknowns in terms of the other two unknowns. Let's find C and D in terms of A and B.
$p(A+B)=p(C+D)$
$k(A-B)=p(C-D)$
So: $p(A+B)+k(A-B)=2pC$
And: $p(A+B)-k(A-B)=2pD$
$\frac{A+B}{A-B}=\frac{k}{p}\frac{C+D}{C-D}$
This is  as far as you can go without other principles of quantum mechanics. The stated ratio of $C+D$ to $C-D$ implies a ratio between the sum and difference of A and B. That in turn allows expressing B in terms of A and then C and D exclusively in terms of A. This is allowed only if you have 3 equations for the 4 unknowns.
These wave functions can't be normalized. For a scattering problem though, there is no incoming wave from the opposite side of the Barrier from the incoming wave. Because of symmetry, we can assume the wave is inbound from left to right.
So we can allow $D=0$.
This combined with your target equation implies $k_1/k_2$ is 0 or +/-1 which is not possible for a non-zero, finite potential.
We can still find the transmission and reflection coefficients.
$T=\frac{pCC^*}{kAA^*} ; R=\frac{BB^*}{AA^*}$
Here, we only  care about the ratios, so WLOG, $A=1$.
$ (p-k)A+(p+k)B=0$.
So $B=A\frac{k-p}{k+p}$
$p$ and $k$ are real, so $B^*=A^*\frac{k-p}{k+p}$
So $R=\frac{(k-p)^2}{(k+p)^2}$
Since $1+B=C$, $C=\frac{2k}{k+p}$
So the Transmission Coefficient is $T=\frac{p}{k}\frac{4k^2}{(p+k)^2}=\frac{4pk}{(p+k)^2}$
As expected by the conservation of probability, we have $1=R+T$.
Your target equation looks like a way to calculate the Transmission and Reflection coefficients but interchanging the wave numbers with the coefficients. I vaguely recall multiplying ratios of wave numbers playing a role in the barrier case, as opposed to the step potential.
What text are you using to study QM? I've found Griffith's very approachable a good supplement for Bransden and Jochain. It's worth the extra cost if you can swing it.
