# The spectrum of Schrodinger Equation [closed]

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

-

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

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

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

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)

-
(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.

-
@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. – Matthias Ludewig Apr 10 '12 at 21:30