# Using solutions of 1-D TISE to calculate constants A, B, C [closed]

I'm tentatively putting this here as it's a quantum/Schrodinger problem however I'm unable to solve two equations in the way it's done in a text. At the minute I think that I can't do it because I've not understood something so I'm trying the wrong mathematical techniques. Here goes, it's on page 16 of K.S. Krane's Introductory Nuclear Physics...

Given the potential

$$V(x) = 0, x < 0, region 1$$ $$V(x) = V_0, x > 0, region 2$$

For region 1 you get the "typical" result for 1-D time-independent equation $\psi_1 = Ae^{ik_1x} + Be^{ik_1x}$ and $k_1=\sqrt{(2mE/\hbar^2}$.

And in region 2 the corresponding solutions are $\psi_2 = Ce^{ik_2x} + De^{ik_2x}$ and $k_2=\sqrt{(2m(E-V_0)/\hbar^2}$.

Then at the boundary condition x = 0 gives $A + B = C + D$ and $k_1(A-B) = k_2(C-D)$ from some equations earlier in the book. Obviously at the boundary condition the wave form is required to be continuous for it to be a valid solution. The text goes on to say the D term must equal 0 which therefore, as I understand it, makes the previous two equations now equal $A + B = C$ and $k_1(A - B) = k_2C$.

The next two lines say that when these equations are solved you end up with the following:

$$B = A\frac{1-\frac{k_2}{k_1}}{1+\frac{k_2}{k_1}}$$ and $$C = A\frac{2}{1+\frac{k_2}{k_1}}$$ Which has me completely lost. I've been unable to algebraically rearrange the equations to reach these solutions, what am I missing? I can only assume that there's a mathematical method that I don't know about (or I can't remember - festive brain fog!) which you can plug these numbers into and the right hand side of the top equation is equal to $C - A$ as $B = C - A$ or that there's a physical reason that I'm missing so I don't understand the concept enough therefore I'm missing an obvious step. The values of A and B are then used to calculate the probability of the wave being reflected at the barrier and A and C used to calculate the probability of transmittance (tunnelling) so I think it best that I understand how to get these results myself.

Any help much appreciated, it's been eluding me since Christmas Eve and I've wasted far too much paper thus far.

## closed as off-topic by Kyle Kanos, Jon Custer, AccidentalFourierTransform, Bill N, JamalSDec 30 '16 at 12:07

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## 1 Answer

Your starting equations are: $$A + B = C\tag{1}$$ $$k_1(A - B) = k_2C\tag{2}$$ Substitute $(1)$ into $(2)$: $$k_1(A - B) =k_2(A+B)$$ Multiply out: $$k_1A-k_1B=k_2A+k_2B$$ Sort the $A$ from the $B$ terms: $$(k_2+k_1)B=(k_1-k_2)A$$ $$B=A\frac{k_1-k_2}{k_2+k_1}$$ Divide both numerator and denominator by $k_1$: $$\implies \boxed{B = A\frac{1-\frac{k_2}{k_1}}{1+\frac{k_2}{k_1}}}\tag{3}$$ To get the expression for $C$, add $A$ to $(3)$.