Infinite-Finite Potential Well

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SUMMARY

The discussion focuses on determining the minimum potential energy \( V_0 \) required for a particle of mass \( m \) to have one bound state in a finite potential well defined by \( V(x) \). The potential is infinite for \( x \leq 0 \), negative \( -V_0 \) for \( 0 < x \leq a \), and zero for \( x > a \). The solution involves applying the Schrödinger equation, leading to the quantized energy levels \( E_n = \frac{n^2\pi^2\hbar^2}{2ma^2} \). The goal is to find \( V_0 \) such that the particle's energy \( E \) is negative and closest to zero, ensuring a single bound state exists.

PREREQUISITES
  • Understanding of quantum mechanics concepts, particularly bound states.
  • Familiarity with the Schrödinger equation and its applications in potential wells.
  • Knowledge of the reduced Planck constant \( \hbar \) and its significance in quantum mechanics.
  • Basic grasp of wave functions and boundary conditions in quantum systems.
NEXT STEPS
  • Study the derivation of the Schrödinger equation for one-dimensional potential wells.
  • Learn about quantization of energy levels in quantum mechanics.
  • Explore the implications of boundary conditions on wave functions in confined systems.
  • Investigate the relationship between potential energy and bound states in quantum mechanics.
USEFUL FOR

Students and professionals in physics, particularly those specializing in quantum mechanics, as well as educators seeking to explain the concept of bound states in potential wells.

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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?
 
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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.
 


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
 

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