Homework Help: Archived Piecewise Potential Function

1. Feb 14, 2014

stefan10

1. The problem statement, all variables and given/known data
Consider reflection from a step potential of height v-knot with E> v-knot, but now with an infinitely high wall added at a distance a from the step.

infinity < x < 0 => v(x) = 0
0≤ x≤ a => v = vknot
x=a => v= infinity

Solve the Schroedinger equation to find ψ(x) for x< 0 and 0 ≤ x ≤ a, solution should contain only one unknown constant.

2. Relevant equations

T-Independent Schrodinger EQ

General forms of a wave function.

3. The attempt at a solution

Is it correct to first assume that all constants are physically possible in both equations? You'll have a reflection and transmission at the first finite barrier, and a reflection (always) at the infinite barrier. That means there are four constants in both equations. If not, can you explain why?

I should have

ψ_1 = A1 cos (k1*x) + B1 sin(k1*x) (or the respective complex exponentials)
ψ_2= A2 cos(k2*x) + B2 sin(k2*x) (this is for the region 0≤x≤a

When I look at the three boundary conditions,

1. ψ_1(0_ = ψ_2(0)

2. dψ_1/dx (0) = dψ_2/dx (0)

3. ψ_2(a) = 0

I get a complicated algebraic relation between the constants that does not simplify.

So I assume I must get rid of one of the constants, but I'm unsure which one.

2. May 6, 2016

vela

Staff Emeritus
Solving the SE equation in the two regions, we get
\begin{align*}
\psi_1(x) &= Ae^{ik_1x}+B^{-ik_1x} \\
\psi_2(x) &= C\sin[k_2(x-a)] + D\cos[k_2(x-a)]
\end{align*} where $k_1^2 = \frac{2mE}{\hbar^2}$ and $k_2^2=\frac{2m(E-V_0)}{\hbar^2}$.

From $\psi(a)=0$, it follows that $D=0$. (Writing $\psi_2$ in terms of $x-a$ instead of just $x$ simplifies the algebra.) Continuity of the wave function and its derivative at $x=0$ yields
\begin{align*}
A + B &= -C \sin k_2 a \\
ik_1 A - ik_1 B &= C k_2 \cos k_2 a.
\end{align*} Since $e^{ik_1x}$ represents the incident wave, let's choose to solve for the other two constants in terms of $A$. Applying Cramer's rule to the system
\begin{align*}
B + C \sin k_2 a &= -A \\
B + C \left(\frac{k_2}{ik_1}\right) \cos k_2 a &= A,
\end{align*}we get
$$B = \frac {\begin{vmatrix} -A & \sin k_2 a \\ A & \left(\frac{k_2}{ik_1}\right) \cos k_2 a \end{vmatrix}} {\begin{vmatrix} 1 & \sin k_2 a \\ 1 & \left(\frac{k_2}{ik_1}\right) \cos k_2 a \end{vmatrix}} = A \frac{ \sin k_2 a - i\left(\frac{k_2}{k_1}\right) \cos k_2 a}{\sin k_2 a + i\left(\frac{k_2}{k_1}\right) \cos k_2 a} \\ C = \frac {\begin{vmatrix} 1 & -A \\ 1 & A \end{vmatrix}} {\begin{vmatrix} 1 & \sin k_2 a \\ 1 & \left(\frac{k_2}{ik_1}\right) \cos k_2 a \end{vmatrix}} = -A \frac{2}{\sin k_2 a + i\left(\frac{k_2}{k_1}\right) \cos k_2 a}.$$
After doing a problem like this, it's good to sanity-check your answers. For example, you can show that $\lvert B/A \rvert^2 = 1$, which indicates that particle is always reflected, as you'd expect because of the infinite potential. A good idea would also be to check what happens when $V_0=0$. There should be no reflection at $x=0$.