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Archived Piecewise Potential Function

  • Thread starter stefan10
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1. Homework Statement
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. Homework 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.
 

vela

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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##.
 

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