Finite potential well- well's depth?

[Someone88]

Homework Statement

The problem is to find a well's depth Vo that the electron which is trapped inside has two stable states. Well starts at x=0 and ends at x=L.

Homework Equations

The Attempt at a Solution

I tried to solve Schrödinger equation for each area (x<0 0<x<L x>L) but I still can't deduce potential Vo from given statement that electron has to have 2 stable states.

Please people, help =) thanks in advance

 

[fzero]

What is the equation that you obtain from the boundary conditions? How many solutions does it have?

 

[Someone88]

I'll start from the beginning.

Equation for the area:
x<0 Ψ(x)= Cexp(αx)
0<x<L Ψ(x)= Asinkx+Bcoskx
x>L Ψ(x)= Fexp(-αx)

For boundary values (0, L), when I equalize equations and their derivations I get:
C=B
Cα=Ak
AsinkL+BcoskL=Fexp(-αL)
A*k*coskL-B*k*sinkL= -α*Fexp(-αL)

So I have 4 equations with 4 unknown values (A,B,C,F). When I solve that system, and insert α=sqrt(2m(V-E)/h-bar^2) and k=sqrt(2mE/h-bar^2) I get:
V= E*sec^2(kL/2) for (n-1)pi<kL<npi (n=1,3,5,...)
V=E*csc^2(kL/2) for (n-1)pi<kL<npi (n=2,4,6...)

I have to use second equation I presume (because of the two stable states), but I don't know how, I don't have all needed values.

 

[fzero]

I'll start from the beginning.

Equation for the area:
x<0 Ψ(x)= Cexp(αx)
0<x<L Ψ(x)= Asinkx+Bcoskx
x>L Ψ(x)= Fexp(-αx)

For boundary values (0, L), when I equalize equations and their derivations I get:
C=B
Cα=Ak
AsinkL+BcoskL=Fexp(-αL)
A*k*coskL-B*k*sinkL= -α*Fexp(-αL)

So I have 4 equations with 4 unknown values (A,B,C,F). When I solve that system, and insert α=sqrt(2m(V-E)/h-bar^2) and k=sqrt(2mE/h-bar^2) I get:
V= E*sec^2(kL/2) for (n-1)pi<kL<npi (n=1,3,5,...)
V=E*csc^2(kL/2) for (n-1)pi<kL<npi (n=2,4,6...)

I have to use second equation I presume (because of the two stable states), but I don't know how, I don't have all needed values.

You need to use both equations because one solution comes from the odd series and the other from the even (these are the first symmetric/antisymmetric solutions). You will need to look at how you graphically solve these equations and look for a condition on V_0 such that n=3 is not a solution.

 

[Someone88]

Can you please explain this part in detail, because I don't understand it very well. How am I suppossed to solve these equations? I can draw them, but what then? If you have another way of solving this problem, without these equations I've got (which seem very complicated to solve), I would appreciate if you'd explain it to me.

 

[fzero]

Can you please explain this part in detail, because I don't understand it very well. How am I suppossed to solve these equations? I can draw them, but what then? If you have another way of solving this problem, without these equations I've got (which seem very complicated to solve), I would appreciate if you'd explain it to me.

It's really hard to try to explain this type of graphical solution in a forum post. So please reread your notes and text to get the details. The solution is also discussed in a wikipedia article, so I can refer to the figure at http://en.wikipedia.org/wiki/Finite_potential_well#Finding_wavefunctions_for_the_bound_state In the solution there, they first write the equations down in terms of a dimensionless number v, which is related to the energy of a solution, and a dimensionless number u_0 which is related to the value of the potential. The graphical solution looks for points of intersection between \sqrt{u_0^2 - v^2} and either v\tan v or -v \cot v.

In your case, while the numerical value of u_0 is not specified, it's clear that you can find the values of u_0 where the 2nd and 3rd solutions appear. This involves finding numerical roots and you can probably use wolframalpha.com or Newton's method if you want to do it by hand. You can then turn the special values of u_0 into expressions for the range of V_0 where there will be exactly 2 solutions.

 

[Someone88]

I know I probably ask too much, but can you solve this part for me, because I really don't understand it. It is very important for me...

 

[Someone88]

Can I shift potential well so that the center is in x=0? Does this make any difference? I've found some examples of finite potential well, but in every of them well goes from -L/2 to L/2. Mine, however, goes from 0 to L, and when I insert boundary values in equations and derivations, I get equations which differ from those in examples.

 

[fzero]

Can I shift potential well so that the center is in x=0? Does this make any difference? I've found some examples of finite potential well, but in every of them well goes from -L/2 to L/2. Mine, however, goes from 0 to L, and when I insert boundary values in equations and derivations, I get equations which differ from those in examples.

There's no need to shift. The only difference for you is that your dimensionless variables will be defined with a factor of 2 difference.

 

[Someone88]

I started all over again with a different approach and in the end I've got:

\frac{2\sqrt{E(V-E)}}{2E-V}=tan(\sqrt{\frac{2m*E*L^2}{hbar^2}})

Now, I have to plot left and right side of the equation, and look for the intersections, right? I'll try to guess value of V, so that there are only two intersections. But when I plot right side of the equation (in Matlab), I get something that doesn't look like tan function at all. I am obviously doing something wrong. What's the value of m and hbar? There are many values for m (mass of electron), I don't know which one to take.

 

[fzero]

I started all over again with a different approach and in the end I've got:

\frac{2\sqrt{E(V-E)}}{2E-V}=tan(\sqrt{\frac{2m*E*L^2}{hbar^2}})

Now, I have to plot left and right side of the equation, and look for the intersections, right? I'll try to guess value of V, so that there are only two intersections. But when I plot right side of the equation (in Matlab), I get something that doesn't look like tan function at all. I am obviously doing something wrong. What's the value of m and hbar? There are many values for m (mass of electron), I don't know which one to take.

You can avoid using the numerical constants if you define the dimensionless variables

v= \sqrt{\frac{2m E L^2}{\hbar^2}}, ~~u_0 = \sqrt{\frac{2m V_0L^2}{\hbar^2}}.

This will let you find solutions in terms of u_0,v and you can add the dimensionful constants into your solution later.

 

[Someone88]

I've made some progress, but I have another question. In one example I've seen that Hbar= 197,3 eV nm. Is this correct? Is there a connection between well's length L and hbar?

 

[fzero]

I've made some progress, but I have another question. In one example I've seen that Hbar= 197,3 eV nm. Is this correct? Is there a connection between well's length L and hbar?

Your solution will involve a particular value of u_0. The potential can be written by inverting the relationship above as

V_0 = \frac{\hbar^2u_0^2}{2m L^2}.