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Why 1D well must has this solution?

  1. Nov 27, 2008 #1

    KFC

    User Avatar

    For 1D well with size a, the potential is V and when 0<=x<=a, V=0; othersize V = infinity


    the Schrodinger equation becomes

    [tex]
    \frac{d^2 f}{dx^2} + k^2 f = 0
    [/tex]

    where f is the wavefunction.

    For solving the equation, many textbook choose

    [tex]
    f = A\sin(kx) + B\cos(kx)
    [/tex]

    as the solution and apply the boundary conditions to find A and B. It is quite trival. However, I also try the solution

    [tex]
    f = A\exp(ikx) + B\exp(-ikx)
    [/tex]

    where [tex]i=\sqrt{-1}[/tex] and this is also a solution. But if I apply the boundary condition, I get

    [tex]
    B=-A, f=i2A\sin(kx)=0
    [/tex]

    and from this it gives the same condition for k but somone said this is wrong. I know that k could be positive or negative but for either case why can't I chose the general solution as

    [tex]
    f = A\exp(ikx) + B\exp(-ikx)
    [/tex]
     
  2. jcsd
  3. Nov 28, 2008 #2
    Your solution seems perfectly reasonable to me, it only differs from the textbook one with a phase factor, it seems, and as we know, for real physical quantities such factors doesn't matter. So I'd say you can use your solution.
     
  4. Nov 28, 2008 #3

    malawi_glenn

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    Science Advisor
    Homework Helper

    1) recall that A,B in these two eq's is not nessicary equal:
    [tex] f = A\sin(kx) + B\cos(kx)[/tex]
    [tex] f = A\exp(ikx) + B\exp(-ikx)[/tex]

    2) Also recall that you can write [tex] f = A\exp(ikx) + B\exp(-ikx)[/tex]
    as something with cos and sin. So the two solutions are the same, except that you can't have the same A and B...
     
  5. Nov 28, 2008 #4

    jtbell

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    Staff: Mentor

    The equivalence of the two solutions is more apparent if you use different names for the arbitrary constants:

    [tex]\psi = A \cos {kx} + B \sin {kx}[/tex]

    [tex]\psi = C e^{ikx} + D e^{-ikx}[/tex]

    Using Euler's formula

    [tex]e^{i \theta} = \cos \theta + i \sin \theta[/tex]

    you can derive equations for C and D in terms of A and B, and vice versa. So it's only a matter of convenience which solution you use.
     
  6. Nov 29, 2008 #5

    KFC

    User Avatar

    Thanks for all your reply. Well, I know that the solution (bound state) should be

    [tex]A\sin(kx) + B\cos(kx)[/tex]

    that why I need to take the real part of my solution mentioned before. But my question is: why wave function for bound state must be real?

    Thanks. Have a nice thanksgiving.
     
  7. Nov 30, 2008 #6

    malawi_glenn

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    Science Advisor
    Homework Helper

    It does not have to!! It is you who restrict the vaules of A and B to be real numbers.
    When introducing constants, one has to specify which set they belong to.
     
  8. Dec 1, 2008 #7
    It shouldn't matter if you have the factor of i in your bound state solution. When finding <psi \ psi> or any expectation value, you multiply by the complex congagate of the wave function anyway. This eventually results in the same value as if the i was never there. It may seem wierd that the wavefunction itself is imaginary, but all that really matters are the physically measurable values.
     
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