- #1

CAF123

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## Homework Statement

Consider a one-dimensional quantum system described by the potential: $$V(x) = -V_o + \frac{1}{2}mw^2x^2\,\,,V_o > 0\,\,\text{for}\,\, |x| < b\,\,\text{and}\,\,0\,\,\text{otherwise}$$

Show that the state described by: $$\psi_{-}(x) = R_{-} \exp\left[-\frac{x^2}{2x_o^2}\right]\,\,\,\text{for}\,\,|x| < b$$

can be an eigenstate of the Hamiltonian describing a bound state (i.e. a state with

negative energy) only for a specific value of ##x_o##.

## Homework Equations

##\hat{H}\psi_{-}(x) = E\psi_{-}(x)## if ##\psi_{-}## is an eigenstate of the Hamiltonian.

## The Attempt at a Solution

I have sketched the potential and it looks parabolic in the region ##x \in (-b,b)## A bound state is a state of neg energy => E < 0 and so the path of some incoming particle intersects the potential plot in two places. For ##x \in (-b,b)## it makes sense that the wave function is a decaying exponential. Because ##V_o## is fixed, it also makes sense that the decaying exp is not proportional to m or w, since I think this means there is only one parabola that satisfies the boundary conditions. If ##\psi_{-}## is to be an eigenstate, it must satisfy the eigenvalue problem ##\hat{H}\psi_{-} = E \psi_{-}##. I tried subbing ##\psi_{-}## in, and this gave me an expression for E in terms of x. I am not really sure what else to try, any hints would be great.

Many thanks.

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