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Variational Method and Bound States
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[QUOTE="Einj, post: 4592502, member: 425276"] [h2]Homework Statement [/h2] Consider a potential function [itex] V(x)[/itex] such that: $$ \begin{cases} V(x)\leq 0\text{ for }x\in[-x_0,x_0] \\ V(x)=0 \text{ for }x\not\in[-x_0,x_0] \end{cases} $$ Show, using the variational method that: (a) In the 1-dimensional case [itex]\lambda^2V(x)[/itex] always possesses at least one bound state. (b) In the 3-dimensional spherically symmetric case, [itex]V(|\vec r|)[/itex], it possesses no bound states if [itex]\lambda^2[/itex] is made sufficiently small. [b]2. The attempt at a solution[/b] The idea is to use the variational method, i.e.: $$ \frac{\langle\psi|H|\psi\rangle}{\langle \psi|\psi\rangle}\geq E_{ground}, $$ to show that the average value of the energy is negative and hence the ground state energy is negative. In my first attempt I used a gaussian trial function: $$ \psi(x)=\left(\frac{2a}{\pi}\right)^{1/4}e^{-ax^2}. $$ However the problem is that it turns out that the average value of the kinetic energy is [itex]\hbar^2a/2m[/itex] and so, in order to determine whether the average energy is negative or not we need to know [itex]a[/itex] explicitly. However, this is not possible since we have no insight on the actual shape of the potential. It seems to me that this thing will turn out to be a problem for every trial function. How can I do that? Thank you [/QUOTE]
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