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Consider the Schrodinger Equation$$\psi_{xx}-(u-\lambda)\psi=0$$ with the condition

1.when $x\to|\infty|,u\to0,u_x\to0$

2.$\psi|_{x\to \infty}=0$ How to prove that spectrums are real?

3.$u(x,0)=f(x)$,$\Sigma_{i=0}^4\int_{-\infty}^{+\infty}|\frac{\partial^if}{\partial x^i}(x)|^2 dx<\infty$,$\int_{-\infty}^{+\infty}(1+|x|)|f(x)|<\infty$

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closed as too localized by Will Jagy, Yemon Choi, Andreas Blass, Ryan Budney, Andy Putman Apr 11 '12 at 2:38

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posted yesterday at math.stackexchange.com/questions/128994/… with insufficient explanation –  Will Jagy Apr 9 '12 at 5:32

2 Answers 2

This is really more of a hint than a fully fledged answer, but the way to go is:

1) rewrite the equation as an eigenvalue problem $H\psi = \lambda \psi$

2) prove that $H$ is self adjoint (use integration by parts and the boundary conditions).

3) use the standard argument that says that selfadjoint operators in Hilbert space have real eigenvalues (see e.g http://planetmath.org/encyclopedia/EigenvaluesOfAHermitianMatrixAreReal.html)

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(It says "eigenvalues of a Hermitian matrix" but the proof should carry through to Hermitian operators as long as everything is well behaved.) –  AlexArvanitakis Apr 9 '12 at 15:18

Since $u$ tends to $0$ as $x$ goes to infinity and it is apparently supposed to be $C^1$, it is bounded. Since the Laplace operator is a negative operator, the operator $L = \Delta - u$ is bounded from above, meaning $(Lu, u) \leq C$ for some constant constant $C$. Here, one can choose $C:= \min u$. It is a classical theorem that such operators, when densely defined, have a self-adjoint extension (this can be found in many books, if necessary I can give a citation). In this case, a dense domain would be the set of Schwartz functions, for example, and the theorem states that $L$ is essentially self-adjoint here.

Now, the spectrum of a self-adjoint operator is real, and clearly also bounded from above by $C$.

Regarding eigenvalues, however, I am afraid that your operator is not very well conditioned in general. If $u$ is a positive function, then there will be no eigenvalues at all, except possibly zero. However, if $C = \min u < 0$, then there can be eigenvalues in the interval [0, C], but this is not necessary.

To give some explanation, there is a theorem that states in your case, if $u$ tends to $0$ when $x \rightarrow \infty$, then the essential spectrum is bounded from above by $0$.

So, there is in general no reason, why this solution should have any solution in $L^2$, but if it does, it automatically fulfills your condition ii, as does every function in $L^2$. Also, as far as I know, there is little to no hope to write down any solution analytically.

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@kofi I need a citation,thanks. –  89085731 Apr 10 '12 at 0:57
    
How about looking at the references in en.wikipedia.org/wiki/Self-adjoint_operator –  Per Alexandersson Apr 10 '12 at 7:54
    
In Kato's book "Perturbation theory of linear operators" is a chapter on semiboundedness (in 1995th edition, it starts p310). Also, if you know german, you find the exact theorem in "Dirk Werner: Funktionalanalysis". There it is Theorem VII.2.11. Kato's book, by the way, is a standard reference and you will find everything there what I claimed in my post. –  Kofi Apr 10 '12 at 21:30

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