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Homework Help: Griffiths Quantum 2.51

  1. Sep 6, 2007 #1
    [SOLVED] Griffiths Quantum 2.51

    This is problem 2.51 from Griffiths Introduction to Quantum Mechanics, 2nd ed. p89.

    The problem statement, all variables and given/known data
    Consider the potential

    [tex]V(x) = -\frac{\hbar^2 a^2}{m}sech^2(a x)[/tex]

    where [itex]a[/itex] is a positive constant, and "sech" stands for the hyperbolic secant.

    a) Graph this potential.

    b) Check that this potential has the ground state

    [tex]\psi_0(x) = A sech(a x)[/tex]

    and find its energy. Normalize [itex]\psi_0[/itex], and sketch its graph.

    c) Show that the function

    [tex] \psi_k(x) = A\left(\frac{i k - a tanh(a x)}{i k + a}\right)e^{i k x} [/tex]

    (where [itex] k = \sqrt{2 m E}/\hbar [/itex] as usual) solves the Schr\ddot{o}dinger equation for any (positive) energy [itex] E [/itex]. Since [itex] tanh z \rightarrow -1 [/itex] as [itex] z \rightarrow -\infty[/itex],

    [tex] \psi_k(x) \approx A e^{i k x} [/tex], for large negative x.

    This represents, then, a wave coming in from the left with no accompanying reflected wave (i.e., no term exp(-ikx)). What is the asymptotic form of [itex]\psi_k(x)[/itex] at large positive [itex]x[/itex]? What are R and T, for this potential? Comment: This is a famous example of a reflectionless potential - every incident particle, regardless of its energy, passes right through.

    The attempt at a solution

    Part a is easy, just draw the graph, intersecting the y axis at -\frac{\hbar^2 a^2}{m}.

    Part b is a little more difficult. I can show that it is a solution, but I'm not sure how to guarantee that it's the ground state. With that potential, I get that

    [tex] \hat H\psi_0 = -\frac{\hbar^2 a^2}{2 m} \psi_0[/tex]

    If this result was the same as the minimum potential, I could say for certain that it's the ground state.

    And for the normalization constant I get

    [tex] A = \sqrt{\frac{a}{2}} [/tex]

    Part c I don't have much clue for. I tried just putting the hamiltonian into mathematica with that wave function, to see what it gives me, but it didn't give me anything that I can see looks like the RHS of the equation, any suggestions on how to go about part c would be appreciated.
     
    Last edited: Sep 6, 2007
  2. jcsd
  3. Sep 6, 2007 #2

    Avodyne

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    Part b: do you recall any paricular property that the ground state wave function always has for any potential (in one dimension), and that no excited-state wave function has?

    Part c: plug in and grind should work to show that it is a solution.
     
  4. Sep 6, 2007 #3
    The lowest energy?

    Ok, I'll try it again and write back here later/tomorrow.
     
  5. Sep 6, 2007 #4
    The fact that it has no nodes guarantees that it's the ground state?
     
  6. Sep 6, 2007 #5

    Avodyne

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    Bingo!
     
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