As we all know, the wave function of the stationary state a quantum particle trapped in a rigid box (with infinite potential outside the box) cannot have a non-zero value outside the box. So can we achieve the same thing (constraining the particle in a certain area) in a finite potential field?
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2$\begingroup$ From memory, no: exponential decay outside the box, but this doesn't mean it goes to zero at a finite distance. $\endgroup$– David Roberts ♦Commented May 30, 2021 at 7:06
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5$\begingroup$ no solution with compact support is possible. $\endgroup$– Carlo BeenakkerCommented May 30, 2021 at 7:07
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1 Answer
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Trapping means that an eigenfunction has compact support. For locally bounded potential this is impossible, because of uniqueness theorem. Suppose for simplicity that our system is 1-dimensional. Then our wave function is a solution of the 1-dimensional stationary Schrodinger equation $$y''=p(x)y.$$ If the potential $p$ is bounded, then the RHS satisfies Lipschitz condition, as a function of $y$, and we have uniqueness of solutions by the classical uniqueness theorem for ODE. (If a solution is zero on an open set then it coincides everywhere with the zero solution).
For higher dimension, the corresponding property is called the unique continuation property, see T. Carleman, Sur les systèmes linéaires aux dérivées partielles du premier ordre à deux variables, C. R. Acad. Sc. 197 (1933), p. 471-474, and
C. Muller, On the behavior of the solutions of the differential equation $\Delta u = F(x, u)$ in the neighborhood of a point, Comm. Pure Appl. Math. vol. 7 (1954) pp. 505-551.
Ph. Hartman, A. Wintner, On the local behavior of solutions of non-parabolic partial differential equations. III. Approximations by spherical harmonics. Amer. J. Math. 77 (1955), 453–474.
These results were very much generalized, look under ``unique continuation property''.
answered May 30, 2021 at 14:57
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$\begingroup$ I think something happened with "systemes liniaires aux derivles partielles". $\endgroup$– LSpiceCommented May 31, 2021 at 3:28
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$\begingroup$ If I'm not mistaken the sentence in brackets (If a solution is zero on an open set then it coincides everywhere with the zero solution) is essentially the definition of unique continuation property. How does uniqueness imply UCP? $\endgroup$– lcvCommented May 31, 2021 at 3:39
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1$\begingroup$ If a solution $y$ is zero on an open set, then $y(x_0)=0$ and $y'(z_0)=0$ for some $x_0$ in this open set. So by uniqueness it coincides with the solution $y_0(x)\equiv 0$, since they both satisfy the same ODE and same initial condition. $\endgroup$ Commented May 31, 2021 at 3:42
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