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Homework Help: Quantum mechanics for wave equation solution

  1. Apr 5, 2010 #1
    1. The problem statement, all variables and given/known data

    consider the differential d²ψ(x)/dx²=k²ψ(x); for which values of a is the equation e^(a*x) is a solution to the above equation.

    2. Relevant equations

    3. The attempt at a solution
    I have been working on this problem but I do not know how relate the 2 equations, or if I should use the Schrodinger equation. Any help is greatly appreciated.
  2. jcsd
  3. Apr 5, 2010 #2


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    Welcome to PF!

    Hi bfed ! Welcome to PF! :smile:

    (try using the X2 tag just above the Reply box :wink:)
    Just put ψ(x) = eax into d²ψ(x)/dx²=k²ψ(x), and solve for a. :wink:
  4. Apr 5, 2010 #3
    thanks tiny-tim,
    so i should take the second derivative of ψ(x) = eax before I substitute it into d²ψ(x)/dx²=k²ψ(x) and solve for a?
  5. Apr 5, 2010 #4


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    uhhh? :confused:

    just put it in! :smile:
  6. Apr 5, 2010 #5
    I think your confusion may come from from the idea of solutions to differential equations.
    A solution is anything that we can put into this equation where after taking the derivatives and everything, the two sides equal each other. i.e. we are testing to see if our proposed solution of [tex]\Psi(x)[/tex] is valid. A good example to show would be one where the proposed solution was not valid. lets consider the proposed solution:

    [tex]\Psi(x) = x^2[/tex]

    Now if we plug this into our differential equation we obtain:

    [tex]2 = k^2 x^2[/tex]

    If k is a constant this is obviously not always true(example x=0). Therefore our proposed solution is not valid. In reality we want to find solutions that are valid. Lets propose a solution:

    [tex]\Psi(x) = e^{ax}[/tex]

    Now you go as I did in the invalid case, and plug [tex]\Psi(x)[/tex] into our differential equation. You will find that 'a' must equal something for the solution to be valid.
  7. Apr 7, 2010 #6
    thanks all, got'er done with your help!
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