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Deriving the solution for a free particle from the TISE

  1. Nov 26, 2012 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    In my physics course, we have been given 'plausibility' arguments for the solutions of the time dependent/time independent SE. However, have done a Calculus course alongside, I feel I should really derive these solutions, since I have learnt about the techniques of second order diff eqns.

    3. The attempt at a solution
    Considering the simple case first: ##V = 0##. In such a region, the TISE reduces to $$\frac{-\hbar^2}{2m} \frac{d^2}{dx^2} \psi(x) - E\psi(x) = 0.$$ Take the auxiliary eqn of this to get ##-\frac{\hbar^2}{2m}r^2 - E = 0 => r = ±\frac{\sqrt{2mE}}{\hbar}i## and so the general solution is $$\psi(x) = A\cos(\frac{\sqrt{2mE}}{\hbar}) + B\sin(\frac{\sqrt{2mE}}{\hbar})$$
    First question:I recognise my expression for ##r## as the expression for ##k##, the wavenumber. Can I simply let ##r = k## to recover this expression? If so, why?
    Second question:I know the general soln in this case is ##\psi(x) = Ae^{ikx} + Be^{-ikx}. ##How does this follow from what I have? I thought about using the method of reduction of order to perhaps find this other term but then I noticed that the expression that I got did not have a term ##i\sin##, so what I have does not conform with the answer anyway.
    Many thanks.
     
  2. jcsd
  3. Nov 26, 2012 #2

    dextercioby

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    You can set r=k on dimensionality arguments.
    Use Euler's formula linking sin/cos and the complex exponential.
     
  4. Nov 26, 2012 #3

    CAF123

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    Are you talking about $$e^{ikx} = \cos(kx) + i\sin(kx)?$$
    In my case, I have ##Ae^{ikx} + Be^{-ikx} = A[\cos(kx) + i\sin(kx)] + B[\cos(kx) -i\sin(kx)] = \cos(kx)[A+B] +i\sin(kx)[A-B]##. In my expression I have no i. The only thing I can see of doing is saying ##A+B = ##some constant, ##C## and ##(A-B)i = ##some constant, ##D##, right?
     
  5. Nov 27, 2012 #4

    haruspex

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    What if you allow that A and B could be complex? Aren't the two solution formats then equivalent?
     
  6. Nov 27, 2012 #5

    CAF123

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    Is that not what I tried? Instead of A and B, I used C and D, so that in the end I get $$\psi(x) = C\cos(\frac{\sqrt{2mE}}{\hbar}) + D\sin(\frac{\sqrt{2mE}}{\hbar}) = (A+B)\cos(\frac{\sqrt{2mE}}{\hbar}) + (A-B)i\sin(\frac{\sqrt{2mE}}{\hbar})$$ Simplifying gives ##\psi(x) = Ae^{ikx} + Be^{-ikx}##.
     
  7. Nov 27, 2012 #6

    haruspex

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    OK. It wasn't clear to me that you had come to understand that the source of your difficulty was the assumption that A and B were real.
     
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