How Does a Particle Behave in a 2-Level Finite Potential Well?

In summary, the conversation discusses the derivation of a transcendental equation for energy eigenstates of a particle with a given potential function. The equations for the wavefunction in different regions are found, and boundary conditions are applied to determine the coefficients. However, the resulting equations are complicated and it is unclear how to eliminate the coefficients and obtain the desired transcendental energy solution.
  • #1
IainH
1
0
Consider a particle of mass m subject to the following potential function (taking Vo and L
to be positive):
V (x) =
40 Vo if x < 0;
0 if 0 < x < L/2;
2 Vo if L/2 < x < L;
40 Vo if x > L.

(a) Derive the transcendental equation for energy eigenstates having an energy
2 Vo < E < 40 Vo.
To simplify the mathematics and numerics, for the rest of this question take h = 1, L = 1,
m = 1, and choose (in this system of units) Vo = 1.

Homework Equations

The Attempt at a Solution


I've found the solution for the wavefunction in all areas.
For x<0,

## \psi_1(x) = A exp(k_1 x)+B exp(-k_1 x) ##

Using the condition that ##psi(x)##=0 when x approaches infinity,

##\psi_1(x) for x<0 = A exp(k_1 x)##

For region 2, (0<x<L/2), solution is oscillatory as E>V(x), so

## \psi_2(x) = \alpha sin(k_2*x) + \beta cos(k_2*x).##

Applying boundary conditions: at x=0, ## \psi_1(x)=\psi_2(x)##, and the derivatives of## (\psi_1)/\psi_2 ## must also be equal at x=0.

##A exp(0) = \alpha sin(0)+\beta cos(0)
k_1 A exp(0) = \alpha cos(0) -k_2 \beta sin(0)##
##\Rightarrow \beta = A,\alpha = {\frac{k_1}{k_2}}A##

## \Rightarrow \psi_2(x) = {\frac{k_1}{k_2}}A sin(k_2 x) + A cos(k_2 x).##

For region 3, (L/2<x<L), psi is also oscillatory as E>V(x), so

##\psi_3(x) = C sin(k_3 x)+D cos(k_3 x)##

Again, I applied the boundary conditions, where ##\psi_2(x=L/2) = psi_3(x=L/2)##, and the respective derivatives at x=L/2, which produced two equations. Dividing one equation by the other allows the 'A' coefficients to be cancelled.

## \psi_2(L/2) = \psi_3(L/2)\Rightarrow {\frac{k_1}{k_2}}A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}}) = C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})##(1)

Taking derivatives and letting x = L/2;

##{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2}) = {\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) +{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}## (2)

Dividing ##{\frac{(1)}{(2)}} \Rightarrow {\frac{{\frac{k_1}{k_2}} A sin({\frac{k_2 L}{2}}) + A cos({\frac{k_2 L}{2}})}{{\frac{k_1}{2}} A cos({\frac{k_2 L}{2}}) - {\frac{k_2}{2}} A sin({\frac{k_2 L}{2})}}} = {\frac{C sin({\frac{k_3 L}{2}}) + D cos({\frac{k_3 L}{2}})}{{\frac{k_3}{2}} C cos({\frac{k_3 L}{2}}) -{\frac{k_3}{2}} D sin({\frac{k_3 L}{2}})}}(3)##

For region 4, x>L, psi_4(x) is decaying, and applying the boundary condition as in region 1,

##\psi_4(x) = G exp(-k_1 x).##

I let ##psi_3(x=L) = psi_4(x=L)##, and the respective derivatives equal each other when x=L. This produced another two equations.

##C sin(k_3 L) + D cos(k_3 L) = G exp(-k_1 L)## (4)

##k_3 C cos (k_3 L) - k_3 D sin(k_3 L) = -k_1 G exp(-k_1 L)## (5)

Dividing (5) by (4) gives:

##{\frac{k_3 C cos (k_3 L) - k_3 D sin(k_3 L)}{C sin(k_3 L) + D cos(k_3 L)}} = -k_1## (6)

Dividing one by the other allows the 'G' coefficients to be removed as above.

I'm then left with a mess of sines and cosines with the factors C and D. I can't seem to get rid of the C and D coefficients to get the required transcendental energy solution. Most of my efforts has been trying to plug in (6) into (3), to cancel C and D, but I've had no luck so far.

Am I missing something here? Any help would be much appreciated. I was thinking that perhaps as ## 2 V_0 < E< 40 V_0,## the wave function won't be found below ## E = 2 V_0##, but I don't think this is correct.

Thanks very much.
 
Last edited:
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  • #2
Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

Related to How Does a Particle Behave in a 2-Level Finite Potential Well?

What is a 2 Level Finite Potential Well?

A 2 Level Finite Potential Well is a concept in quantum mechanics that describes a potential energy barrier with two distinct energy levels. It is often used to model the behavior of particles confined to a certain space, such as an electron in a semiconductor.

How is the energy of a particle in a 2 Level Finite Potential Well calculated?

The energy of a particle in a 2 Level Finite Potential Well is calculated using the Schrödinger equation. This equation takes into account the size and shape of the potential well, as well as the mass and energy of the particle.

What is the significance of a 2 Level Finite Potential Well in physics?

The concept of a 2 Level Finite Potential Well is important in understanding the behavior of particles at the quantum level. It helps explain phenomena such as tunneling and the quantization of energy levels. It is also used in various applications, such as in the design of electronic devices.

How does the depth and width of a 2 Level Finite Potential Well affect the behavior of a particle?

The depth and width of a 2 Level Finite Potential Well determine the energy levels that a particle can occupy. A deeper and narrower well will have more discrete energy levels, while a shallower and wider well will have fewer energy levels. This can affect the probability of a particle tunneling through the barrier.

Can a particle in a 2 Level Finite Potential Well have a negative energy?

Yes, it is possible for a particle in a 2 Level Finite Potential Well to have a negative energy. This means that the particle is confined to the well and cannot escape. It is commonly seen in semiconductor devices, where electrons are confined to a well of negative energy.

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