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First Order Homogeneous ODE

  1. Aug 30, 2010 #1
    1. The problem statement, all variables and given/known data

    Solving the following differential equation with the given boundary conditions:

    [tex]\hbar^2 \frac{d^2}{dx^2}\psi (x) = 2mE\psi (x), \ \ \ \ \ \forall \ \hbar^2,\ m,\ E > 0 [/tex]

    [tex]\psi(a) = \psi(-a) = 0 [/tex]

    2. Relevant equations



    3. The attempt at a solution

    [tex]\hbar^2 \frac{d^2}{dx^2}\psi (x) = 2mE\psi (x) [/tex]

    [tex]\frac{d^2}{dx^2}\psi (x) = \frac{2mE}{\hbar^2}\psi (x) [/tex]

    [tex]\frac{d^2}{dx^2}\psi (x) - \frac{2mE}{\hbar^2}\psi (x) = 0 [/tex]

    Let

    [tex]\frac{2mE}{\hbar^2} = \rho[/tex]

    Then

    [tex]\frac{d^2}{dx^2}\psi (x) - \rho \psi (x) = 0 [/tex]

    The characteristic polynomial for this ODE shall be

    [tex]r^2 - \rho = 0 [/tex]

    [tex]r^2 = \rho ,\ r_1 = -\sqrt{\rho},\ r_2 = \sqrt{\rho} [/tex]

    Therefore

    [tex]\psi (x) = c_1e^{r_1 x} + c_2e^{r_2 x} [/tex]

    [tex]\psi (x) = c_1^{-\sqrt{\rho} x} + c_2e^{\sqrt{\rho} x} [/tex]

    By using the given conditions the only possible solution I can get is

    [tex]c_1 = c_2 = 0,\ \psi (x) = 0 [/tex]

    As I get to something like

    [tex]e^{4 \sqrt{\rho} a} = 1 [/tex]

    And since

    [tex]\rho,\ a > 0[/tex]

    So

    [tex]\psi(x) = 0 [/tex]

    seems to be the only solution.

    Is this it?
     
    Last edited: Aug 30, 2010
  2. jcsd
  3. Aug 30, 2010 #2

    Dick

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    The characteristic equation isn't r^2-rho*r=0. It's r^2-rho=0. Where did the second r come from?
     
  4. Aug 30, 2010 #3

    CompuChip

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    I think that the characteristic polynomial will be
    [tex]r^2 - \rho = 0[/tex]
    rather than
    [tex]r^2 - \rho r = 0[/tex]
    which kinda makes some difference.
     
  5. Aug 30, 2010 #4
    Yeah I kinda saw that too, I've already corrected it and did all the work again and I still got nowhere.
     
  6. Aug 30, 2010 #5

    Dick

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    Yes, I think you are correct that the only solution is psi(x)=0. That looks like Schroedinger's equation with a minus sign missing. If there were an extra minus then you would get sin and cos type solutions and wouldn't be having these problems. Is it a typo?
     
  7. Aug 30, 2010 #6
    The equation posted in the op is not a first order homogeneous ODE as claimed in the title.
     
  8. Aug 30, 2010 #7
    The differential equation and boundary conditions are typed here exactly as in my assignment, I couldn't tell if there is a 'typo', but if there is it isn't mine. But is [tex]\psi (x) = 0[/tex] acceptable as a solution in terms of a quantum mechanical interpretation for a single particle in a box of length '2a'?

    Hah, you're right it's a second order, my bad, is there any way to correct that?
     
  9. Aug 30, 2010 #8

    epenguin

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    How I would look at it. By the physical meaning of the terms as I understand your constant rho is by nature positive. Often it is preferred to keep this evident by defining your parameter as rho2. If it is wanted to indicate a negative coefficient define a constant as -rho2.

    If instead your rho were negative, the maths of your equation would be formally almost identical to that for simple harmonic motion. (Only x instead of t, psi instead of x. Meaning changed, maths identical. The usual 1-D QM particle-in-box equation is like that too.) We know pretty well how that behaves.

    If instead your rho is positive, the equation is that of bacterial growth. Which is usually given as a 1st order eq. but your 2nd order equation will be true of it too. And we know how that behaves too. Exponential growth. And in exponential growth the population is never zero at any time unless it is zero at all times. Or think of a repulsive force that increases proportionately to the distance to a particle.

    So the conclusions mentioned are not so puzzling.
     
    Last edited: Aug 31, 2010
  10. Aug 31, 2010 #9

    CompuChip

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    Or, if you look at it as the Schrödinger equation with a minus sign missing, you could also interpret it as the Schrödinger equation for a particle with negative energy E. This would mean that the particle is in a bound state everywhere. Then imposing that the wave function be zero at two different points, means that it must be zero everywhere.

    In answer to your question: in QM wavefunction zero is not allowed. Mathematically, there is nothing wrong with the constantly zero function being a solution to the differential equation, but the function does not contain any physical content (e.g. you cannot use it to calculate sensible expectation values) and basically describes the "no particle at all" situation
     
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