Understanding the Role of Gamma in the Semi-Infinite Potential Well Problem

In summary, we have a potential function V(x) with different values in different regions. The solutions in each region are represented by different mathematical expressions, and there are continuity conditions that need to be satisfied at the boundaries between regions. When the potential in one region tends to zero, the momentum in that region also tends to zero, and the phase shift constant also tends to zero. This means that the wave function in that region will tend to the wave function in the next region, and the Hamiltonian expression for that region will have the same form as the one for the previous region.
  • #1
CAF123
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Homework Statement


Let V(x) = +∞ for x ≤ 0, -V1 for 0 < x < b and 0 for x > b. V1 and b are positive. The solutions in each of the physical regions are ##\psi_1 = P \exp(ik_1 x) + Q \exp (-ik_1 x)## and ##\psi_2 = R \sin (k_2x + \gamma)##.

Show that ##\lim_{V_1 \rightarrow 0} \gamma = 0##

The Attempt at a Solution


I am trying first to understand what ##\gamma## actually represents. I know to obtain the solution in region 2, it is a phase shift constant and so that solution is equivalent to a LC of sin and cos solutions. I also derived a relation between ##k_1, k_2 ##and ##\gamma## using the continuity conditions, but I am not sure how (or if) to apply it here.
Any hints in the right direction would be great.
Thanks.
 
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  • #2
yeah, it sounds like you are on the right track. exactly what is the relation have you found so far? remember there is more than one continuity condition that needs to be satisfied.
 
  • #3
Hi BruceW,
BruceW said:
yeah, it sounds like you are on the right track. exactly what is the relation have you found so far? remember there is more than one continuity condition that needs to be satisfied.
The continuity condition I obtained by imposing the continuity of the first derivative of the wave function and the fact it must be single valued, was ##k_1 \tan (k_2 b + \gamma) = k_2 \tan (k_1 b)## The other condition I got was that simply P = -Q. But what should I do with these?

Thanks.
 
  • #4
right. that's good so far. You have used both of the continuity equations. Now, when ##V_1## tends to zero, what does ##k_1## tend to? hint: it's almost intuitive really, but you can use the Hamiltonian to find the exact relationship.
 
  • #5
BruceW said:
right. that's good so far. You have used both of the continuity equations. Now, when ##V_1## tends to zero, what does ##k_1## tend to? hint: it's almost intuitive really, but you can use the Hamiltonian to find the exact relationship.
It tends to ##k_2##. I agree that this is intuitive, but I think to prove it, we should take the limit as V1 tends to zero in the Hamiltonian expression for the second region. So $$-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} \lim_{V_1 \rightarrow 0} \psi_{1} - \lim_{V_1 \rightarrow 0} V_1 = E \lim_{V_1 \rightarrow 0} \psi_{1}$$ The second term on the LHS goes to zero since the potential is continuous in this region. The wave function in the second region will tend to the wave function in the third region. This means we end up with precisely the form of the Hamiltonian that would have previously described region 3.

The continuity condition then becomes as ##V_1## tends to zero: $$k_2 \tan (k_2 b + \gamma) = k_2 \tan(k_2 b)$$ For this to hold for all ##k_2##, ##\gamma \rightarrow 0## too. Is this okay?
 
  • #6
yep. looks good to me. although, technically, since tan is a periodic function, you can have gamma tend to other values, but it doesn't change the physics of the system, right? so maybe you should say a few words about that.
 
  • #7
Ok thanks. What do these quantities ##\gamma, k_1, k_2## actually represent? I think ##\gamma## is a phase shift and ##k_1, k_2## are a measure of the momentum of the particles in each region.
 
  • #8
yep. ##\hbar k_1,\hbar k_2## are definitely the absolute value of the momentum in each region. And just by looking at how ##\gamma## enters into the wavefunction, its role is to give a phase shift, as you say.
 

1. What is a semi-infinite potential well?

A semi-infinite potential well is a theoretical concept in quantum mechanics that represents a particle confined to a finite region between two infinitely high potential energy barriers. This means that the particle is free to move within the well, but cannot escape beyond the barriers.

2. How is a semi-infinite potential well different from a regular potential well?

In a regular potential well, the particle is confined between two finite potential barriers. However, in a semi-infinite potential well, one of the barriers is infinitely high, making it impossible for the particle to escape beyond that point.

3. What are the implications of a semi-infinite potential well in quantum mechanics?

The semi-infinite potential well is a simplified model that helps explain the behavior of particles in quantum systems. It demonstrates the concept of energy quantization, where the particle can only have certain discrete energy levels within the well.

4. How does the width of the potential well affect the particle's energy levels?

The width of the potential well directly affects the spacing between the particle's energy levels. A wider well will have larger energy level spacings, while a narrower well will have smaller energy level spacings. This is because the particle has a higher probability of being found in a smaller region, leading to a higher energy state.

5. Can a particle in a semi-infinite potential well have infinite energy?

No, a particle in a semi-infinite potential well cannot have infinite energy. While the potential barrier on one side is infinitely high, the particle still has a finite amount of energy within the well. Additionally, the particle's energy levels are quantized, meaning they can only exist at certain discrete levels within the well.

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