The spectrum of Schrodinger Equation - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-25T16:04:17Z http://mathoverflow.net/feeds/question/93543 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/93543/the-spectrum-of-schrodinger-equation The spectrum of Schrodinger Equation 89085731 2012-04-09T05:15:49Z 2012-05-09T04:32:38Z <p>Consider the Schrodinger Equation$$\psi_{xx}-(u-\lambda)\psi=0$$ with the condition</p> <p>1.when $x\to|\infty|,u\to0,u_x\to0$</p> <p>2.$\psi|_{x\to \infty}=0$ How to prove that spectrums are real? </p> <p>3.$u(x,0)=f(x)$,$\Sigma_{i=0}^4\int_{-\infty}^{+\infty}|\frac{\partial^if}{\partial x^i}(x)|^2 dx&lt;\infty$,$\int_{-\infty}^{+\infty}(1+|x|)|f(x)|&lt;\infty$</p> http://mathoverflow.net/questions/93543/the-spectrum-of-schrodinger-equation/93575#93575 Answer by AlexArvanitakis for The spectrum of Schrodinger Equation AlexArvanitakis 2012-04-09T15:16:33Z 2012-04-09T15:16:33Z <p>This is really more of a hint than a fully fledged answer, but the way to go is:</p> <p>1) rewrite the equation as an eigenvalue problem $H\psi = \lambda \psi$</p> <p>2) prove that $H$ is self adjoint (use integration by parts and the boundary conditions).</p> <p>3) use the standard argument that says that selfadjoint operators in Hilbert space have real eigenvalues (see e.g <a href="http://planetmath.org/encyclopedia/EigenvaluesOfAHermitianMatrixAreReal.html" rel="nofollow">http://planetmath.org/encyclopedia/EigenvaluesOfAHermitianMatrixAreReal.html</a>)</p> http://mathoverflow.net/questions/93543/the-spectrum-of-schrodinger-equation/93580#93580 Answer by Kofi for The spectrum of Schrodinger Equation Kofi 2012-04-09T16:01:30Z 2012-04-09T16:01:30Z <p>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.</p> <p>Now, the spectrum of a self-adjoint operator is real, and clearly also bounded from above by $C$. </p> <p>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 $C = \min u &lt; 0$, then there can be eigenvalues in the interval [0, C], but this is not necessary.</p> <p>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 $0$.</p> <p>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.</p>