The spectrum of Schrodinger Equation - MathOverflow [closed]

The spectrum of Schrodinger Equation

2012-04-09T05:15:49Z

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

Answer by AlexArvanitakis for The spectrum of Schrodinger Equation

2012-04-09T15:16:33Z

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)

Answer by Kofi for The spectrum of Schrodinger Equation

2012-04-09T16:01:30Z

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.