- #1

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## Homework Statement

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.

**2. The attempt at a solution**

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