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Archived Piecewise Potential Function

  1. Feb 14, 2014 #1
    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. jcsd
  3. May 6, 2016 #2

    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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