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I Linear equations and superposition of wavefunctions

  1. Aug 15, 2016 #1

    dyn

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    Hi. I have read many times that the Schrodinger equation is a linear equation and so if Ψ1 and Ψ2 are both solutions to the equation then so is Ψ1 + Ψ2. Is this use of the word linear the same as generally used for differential equations ? As the Schrodinger equation is also an eigenvalue equation for the Hamiltonian.

    My main confusion is why a superposition of wavefunctions such as eikx + e-ikx is not a solution to the momentum eigenvalue equation as this also looks like a linear equation ?
    Thanks
     
  2. jcsd
  3. Aug 15, 2016 #2

    Krylov

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    Yes.
    Because the summands correspond to different eigenvalues.
     
  4. Aug 15, 2016 #3

    dyn

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    Thanks. So the condition that the superposition of wavefunctions also solves the eigenvalue equation is that it is both linear and the wavefunctions also have the same eigenvalues ?
     
  5. Aug 15, 2016 #4

    Krylov

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    Yes. (Eigenvalue equations are usually linear, by the way. There does exist something called a "nonlinear eigenvalue problem", but that is not relevant here.)

    The SE is a partial differential equation, but for your understanding perhaps consider the setting of a linear ordinary differential equation. In the simplest case (constant coefficients, homogeneous) a linear ODE can be written as
    $$
    \frac{dx}{dt}(t) = Ax(t) \qquad (\ast)
    $$
    (Think, for example, about the simple harmonic oscillator.) If ##x_1## and ##x_2## are both solutions, then their sum is also a solution and hence satisfies ##(\ast)##. This is superposition.

    Regarding the corresponding eigenvalue problem, if ##\lambda## and ##\mu## are distinct eigenvalues of ##A## with eigenvectors ##u## and ##v##, then ##u + v## is not an eigenvector of ##A##. (Try it out.) In contrast, if ##\lambda = \mu## then it is true that
    $$
    A(u + v) = \lambda(u + v)
    $$
    In case of the SE similar remarks hold, but the matrices are replaced by (unbounded) operators that act on function spaces instead of ##\mathbb{R}^n## or ##\mathbb{C}^n##. Does this clarify matters a bit?
     
  6. Aug 15, 2016 #5

    dyn

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    Yes. Thanks for your help.
     
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