Infinite-Finite Potential Well

[AKG]

I need help with this question. I'm not sure exactly what it wants (what does it mean by bound state) and how should I start the problem? Here it is:

Consider a particle of mass m moving in the following potential:

• [itex]\infty[/itex] for [itex]x \leq 0[/itex]
• [itex]-V_0[/itex] for [itex]0 < x \leq a \ (V_0 > 0)[/itex]
• [itex]0[/itex] for [itex]x > a[/itex]

Calculate the minimum value for [itex]V_0[/itex] (in terms of a, m, and the Planck constant) so that the particle will have one bound state.

I guess what they're asking for is the smallest value for [itex]V_0[/itex] such that some particle will have energy E such that [itex]-V_0 < E < 0[/itex]. So, if I can find the energy of the particle that is negative but closest to zero, that value will be [itex]-V_0[/itex]. Is this right so far? If so, how do I go about finding E?

 

[Galileo]

The way I would go about this problem is by solving the Schrödinger equation and finding the energies. Assuming [itex]-V_0<E<0[/itex].

Then find the value of [itex]V_0[/itex] for which there is only one state with an energy<0.

I solved the finite potential well in the past, but don't remember it well enough to know if this is doable.

 

[wais]

The problem is asking for the minimum value of the potential energy, V_0, in order for the particle to have one bound state. A bound state is a state in which the particle is confined to a finite region, in this case, between x = 0 and x = a. This means that the particle's energy must be negative, since it is confined within the potential well.

To start the problem, we can use the Schrödinger equation for a one-dimensional infinite potential well:

-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V(x)\psi=E\psi

Where \hbar is the reduced Planck constant, m is the mass of the particle, V(x) is the potential function, E is the energy of the particle, and \psi is the wave function.

In our case, the potential function is given by:

V(x)=\begin{cases} \infty & x \leq 0 \\ -V_0 & 0 < x \leq a \\ 0 & x > a \end{cases}

Since the potential is infinite for x \leq 0, the wave function must be zero in this region. Similarly, since the potential is zero for x > a, the wave function must also be zero in this region. This means that the wave function will only be non-zero in the region 0 < x \leq a.

To solve for the energy, we need to solve the Schrödinger equation in this region. We can do this by separating the equation into two parts: one for the region 0 < x \leq a and one for the region x > a. For the first part, we can use the general solution for a particle in a potential well:

\psi(x)=A\sin(kx)+B\cos(kx)

Where k=\sqrt{\frac{2mE}{\hbar^2}}. We can then apply the boundary conditions \psi(0)=0 and \psi(a)=0 to solve for the constants A and B. This will give us a quantized energy level for the particle, given by:

E_n=\frac{n^2\pi^2\hbar^2}{2ma^2}

Where n is a positive integer. This means that the particle can only have certain discrete energy levels, and the lowest energy level