Calculating energy for particle in finite box.

In summary, the conversation discusses calculating the energy for a particle in a finite box potential, where the solution is a transcendental equation that needs to be solved graphically or numerically. The task defines various variables such as k and l, and the solution for the equation is given for odd and even cases. The conversation also mentions using MATLAB for the numerical calculations and the difficulties faced in plotting the results. Eventually, the person solves the problem by using the formula E=\frac{\hbar ^2 k^2}{2m} and expressing the energy in terms of electron volts.
  • #1
knut-o
17
0
[solved] Calculating energy for particle in finite box.

Homework Statement


I am going to look at a particle in a finite boxpotential where V(x)=-V0 for -a<x=<a
And they are bound, in other words -V0<E<0. This is a transcendtale equation that need to be solved graphicly/numericly.
I am aware of two solutions, one of odd, and one of even.

Homework Equations


To make calculation easier, the task defines [tex]k=\frac{\sqrt{2m|E|}}{\hbar}, l=\frac{\sqrt{2m(V_0+E)}}{\hbar}=\frac{\sqrt{2m(V-|E|)}}{\hbar}[/tex]
Also told that solution for the equation is [tex]ltan(la)=k[/tex] (even) and [tex]lcot(la)=-k (odd)[/tex]
Rewriting these to, for odd: [tex]zcot(z)=-\sqrt{z_0 ^2-z}[/tex] and for even: [tex]ztan(z)=\sqrt{z_0 ^2-z^2}[/tex] where z=l*a and [tex]z_0 ^2=\frac{2mV0a^2}{\hbar ^2}[/tex]

Now, I am tolk to solve these numericly for V0=10eV and a=1nm, and m as electronic mass:


The Attempt at a Solution


Since it's a numerical task, I am writing it in matlab, but getting some horrible answers.
Since I am told to express the energi to the end I use the relevant units for my program (this could be the issue, but my graph is just a straight line).
So here is my program (agian, it's written in MATlab, but the pseudocode should be easy to recognize for other languages):

figure(1) %For even functions:
V0=10; %eV
a=1; nm
E=-V0:0.1:0; %I am really unsure about this one
z=0:.0.1:5*pi/2; %This one aswell, but the tan/cot-functions make this a sensible choise?
m=511000; %eV/c^2
h=6.582*10^(-16); eVs
k=sqrt(2*m*abs(E))/h;
l=sqrt(2*m*(V0+E))/h;
z0=(a^2/h^2)*2*m*V0;

y11=z.*tan(z); %Gives the array for left side of the equation
y12=sqrt(z0^2-z.^2); % Should give array for right side of equation
plot(z,y11,'-r') %Plots it into figure 1 with a red line
hold 'on'
plot(z,y12) %Plots it into same figure with a blue (default) line.




The problem I end up with, is that it's in the size order of 1037, so plotting it on same graph as something with size order of about 1-10 is just silly. So I am wondering where the program fails. It should be quite easy to calcute, considering z is just an array running from 0 to (a number) and z0 is a constant, but I end up with a straight line for the right side, which according to my book (David J. Griffits, Introduction to Quantum Mechanics, figure 2.18: page 80) says it two quite defined graphs, which is also where I got my z-array from, but the plot he is using ([tex]tan(z)=\sqrt{\frac{z0}{z} ^2-1}[/tex]) gives distinct points where they cross, whereas mine deosn't look anything like it.

I am only suppoosed to find the first one (where they cross) and then calcutate E, but what formula do I use for that? Do I put it into [tex]z=l\cdot a=\frac{\sqrt{2m(V0+E)}}{\hbar}a\Rightarrow E=\frac{(\frac{z\hbar}{a})^2}{2m}-V0[/tex]?
And then how do I express it in eV without having to calculate it to Joule first, then divide by the value of charge of an electron?

Halp.

Edit: I've done some breakthrough when it comes to the task, turns out, the z0 constant is easy to calculate, so I am very curious, and stressing about what I should use for z, amd what about the E? Should I make a loop that is just slooow?

Edit2: No need to respond, I solved it all by myself now :) .
 
Last edited:
Physics news on Phys.org
  • #2
I found out that I was calculating the wrong thing, so I needed to use the formula E=\frac{\hbar ^2 k^2}{2m} to solve the equation, and then I just used the value of l that I needed, which was given by z=la. Then I just calculated the energy with the formula above, and I got the right answer.
 

Related to Calculating energy for particle in finite box.

1. What is a particle in a finite box?

A particle in a finite box is a theoretical model used in quantum mechanics to study the behavior of a particle confined within a finite space. This can be applied to various systems, such as electrons in a semiconductor or atoms in a laser cavity.

2. How is energy calculated for a particle in a finite box?

The energy of a particle in a finite box is calculated using the Schrödinger equation, which is a mathematical equation that describes the behavior of quantum systems. The solution to this equation gives the allowed energy levels for the particle in the finite box.

3. What are the boundary conditions for a particle in a finite box?

The boundary conditions for a particle in a finite box are that the wavefunction (a mathematical representation of the particle's state) must be continuous at the boundaries of the box, and it must also go to zero at the boundaries to ensure that the particle remains confined within the box.

4. How do the energy levels of a particle in a finite box change with box size?

The energy levels of a particle in a finite box are quantized, meaning they can only take on certain discrete values. As the box size increases, the energy levels also increase and become more closely spaced together. This is known as the quantization of energy.

5. What is the significance of calculating energy for a particle in a finite box?

Calculating the energy levels for a particle in a finite box allows us to understand the behavior of quantum systems and make predictions about their properties. This is important in fields such as materials science, where the behavior of electrons in a solid can be described using the particle in a finite box model.

Similar threads

  • Advanced Physics Homework Help
Replies
8
Views
1K
Replies
2
Views
1K
Replies
7
Views
2K
  • Advanced Physics Homework Help
Replies
24
Views
2K
  • Advanced Physics Homework Help
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
9
Views
1K
  • Advanced Physics Homework Help
Replies
4
Views
3K
  • Advanced Physics Homework Help
Replies
5
Views
872
  • Advanced Physics Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
13
Views
2K
Back
Top