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I'm a student majoring in Mathematics.But now I'm studying the KDV equation which uses Schrodinger Equation. My question is that in time-independent Schrodinger Equation$$\psi_{xx}-(u-\lambda)\psi=0$$,and when $x\to|\infty|,u\to0,u_x\to0$,there are two questions that I have:

  • Why are all the eigenvalues real?

  • Why are there discrete eigenvalues for $\lambda<0$ and continuous eigenvalues for $\lambda>0$?

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1 Answer 1

The eigenvalues are real because the Schroedinger operator $$u-\partial_{xx}$$ is self-adjoint.

For the second question, $\lambda<0$ corresponds to bound states for which $\psi \to 0$ as $x \to 0$ which are usually discrete, whereas $\lambda>0$ corresponds to scattering states. This is not rigorous. Perhaps someone can fill in a better answer.

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Can you give me a proof for Question 1? –  Gingerjin Apr 7 '12 at 13:12
    
@Gingerjin: unfortunately not, because it's not quite true. Physicists are sloppy because we use the equations for physics, not for maths (i.e. inverse scattering theory). A proper proof would need to be very careful about the domain of that operator and boundary conditions. –  genneth Apr 7 '12 at 13:52
    
@genneth the condition $x\to|\infty|,u\to0,u_x\to0$ may help –  Gingerjin Apr 7 '12 at 14:11
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@Gingerjin: not really. The problem is that the Schrodinger operator is unbounded, and the theory of unbounded operators is complicated. Vijay meant to say that it is Hermitian, which is not the same --- and the usual proof that we trot out to physics students about finite dimensional operators does not apply. If you are a current student, I suggest asking your local instructor about the level of detail necessary for you to grasp. –  genneth Apr 7 '12 at 19:39
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For a counterexample that illustrates @genneth's point, see e.g. Example 3 in this preprint arxiv.org/abs/quant-ph/9907069 –  Qmechanic Apr 8 '12 at 16:42

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