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A question in solving schrodinger's equation

  1. Jul 23, 2004 #1
    Hi everyone;
    I have a question about the seperation method in solving the wave equation:
    In fact when we assume the answer as a seperable function we just loose a subset of answers, because it is a RESTRICTION.
    Just tell me why this assumption is correct.
    Thanks alot.
    somy :smile:
     
  2. jcsd
  3. Jul 23, 2004 #2

    reilly

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    If you pursue things a bit more you will find that separation does yield all possible solutions -- given appropriate boundary conditions. One approach notes that the resulting ordinary DE's are Sturm-Liouville equations which produce complete sets of solutions -- any solution can be "fourier expanded" by means of these sets. The other is more formal, and is a standard existance proof. All of this can be found in any book dealing with Partial Differential Equations. Good question.
    Regards,
    Reilly Atkinson
     
  4. Jul 23, 2004 #3

    Tom Mattson

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    That's a good question. I wondered the same thing when I first learned it.

    I'm going to shoot from the hip here and offer what I think the answer is. Someone else can correct me if I'm wrong. When we find the solutions to the Schrodinger equation, those are basis vectors. Any linear combination of those vectors is also a solution. It can be shown that the set of basis vectors is complete (that is, the linear combinations of the basis vectors exhausts the solution set of the equation).

    So, I am thinking that the solutions that are "lost" in separation of variables are recovered by superposition.
     
  5. Jul 24, 2004 #4
    Dear Tom; thanks for the answer, but what is your reason to tell that 'the superposition will cover the answers'.

    another question (I know it is a bit silly!!!):
    why the answer is complete??? I mean when we say a set of answer is complete???

    thanks a lot.
    somy :smile:
     
  6. Jul 24, 2004 #5
    Thanks reilly;
    I read your answer in griffiths' quantum book. :smile:
     
  7. Jul 24, 2004 #6

    Tom Mattson

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    When you solve a PDE, you get a set of basis solutions. Your worry is that these basis solutions are not the only solutions--that we lose some solutions by imposing the separation technique. What I am pointing out is that the basis solutions (call them {φi(x,t)}) found by separation of variables are not the only solutions to the differential equation. We can take a linear combination of those basis vectors to construct more solutions, as follows:

    ψ(x,t)=Σaiφi(x,t)

    More below.

    When I say the basis is "complete", I mean that it spans the space of the Hamiltonian. That is, any possible solution of Hψ=(i*hbar)∂ψ/∂t can be constructed from the basis vectors. So it is really as Reilly said: You don't actually lose anything with separation of variables.
     
  8. Jul 25, 2004 #7
    Dear Tom I think you made a mistake!!!
    In fact , by superposition you don't get any new answer, because they are not independent anymore.
    Another possibility is that: I didn't get your answer!!!

    waiting for your reply.
    somy
     
    Last edited: Jul 25, 2004
  9. Jul 25, 2004 #8
    Maybe I'm just tired, but I don't follow your statement.
     
  10. Jul 25, 2004 #9

    Galileo

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    As Tom said: You don't really lose anything by seperation of variables.
    The Schrödinger equation is linear and homogenous is [itex]\psi[/itex], so if [itex]\psi_1, \psi_2[/itex] are solutions, so is [itex]\psi_1+\psi_2[/itex].

    [tex]i\hbar \frac{\partial \psi_1+\psi_2}{\partial t}=\frac{-\hbar^2}{2m}\frac{\partial^2 \psi_1+\psi_2}{\partial x^2}+V(\psi_1+\psi_2)[/tex]
    (check it for yourself)

    So any linear combination of solutions (obtained by seperation of variables) is also a solution. Completeness said that every solution can be written as a linear combination of the solutions obtained by seperation of variables.
     
    Last edited: Jul 25, 2004
  11. Jul 25, 2004 #10
    Thanks Galileo;
    I think I need to know more about completeness and its properties.
    Can you help me???!!!
    somy
     
  12. Jul 25, 2004 #11

    Tom Mattson

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    Look at the standard basis for R3:

    i=<1,0,0>
    j=<0,1,0>
    k=<0,0,1>

    Let's form a vector by taking a superposition of these basis vectors:

    v=axi+ayj+azk

    Now, would you say that v is not different from the basis vectors simply because it is a linear combination of them? Of course not. But that is what you are saying regarding the basis {φi} and the superposition ψ=Σaiφi.
     
  13. Jul 25, 2004 #12
    I got it!!!
     
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