Both k and the functional form would change.Some of the sines would change to cosines, but the k's would stay the same.The value of k would get smaller, but the wave function would have the same form.How would the wave functions change if the box were to go from -a/2 to a/2? The wave functions for particles in a rigid box between 0 and a are Asin(kx) with k = nπ/a. What happens to the energies of electrons in a box as the box gets narrower? They must be moving in the same direction.What must be true in order for them to form a standing wave? What are the differences and similarities in the standing waves allowed? Using the evolution operator one can construct a formally exact solution, evolving from an exactly solvable case, which is quite a common procedure when developing all kinds of perturbation expansions.Compare standing waves in a circle to those on a 1d string. impossible using modern computational power (but possibly solvable with a quantum computer). Note that this is not true for many-particle problems, where the problems easily get NP-hard, i.e. but once we accept that computers can be used, almost any one particle Schrödinger equation is solvable. When we need to calculate a sine, an exponent, or a Bessel function, most of us turn to a computer. However, this notion of "exact" ultimately traces itself to our ability to calculate the numbers. Some people would generalize this to using special functions - Bessel functions, hypergeometric functions, etc. One may question the very notion of an exact solution: usually it means a solution in terms of simple functions that can be handled with pan and paper. (However, some particular cases, such as the case of one hard wall at $x=0$ are still doable.) For example, the harmonic potential case, $V(x) = \frac$ is exactly solvable with open boundary conditions, but solving it with hard wall conditions, e.g., at $x= \pm a$, already requires solving transcendental equations i.e. The solutions for the same type of potential may be easy to get for one set of boundary conditions and hard or even impossible for the other. Let me add that, from the mathematical standpoint, the equation alone is not even sufficient to have a well posed problem it needs to be supplemented by the boundary conditions. If we view the Schrödinger equation simply as a second order differential equation, then no, it doesn't have a general solution that we can find.
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