Finite potential well transcendental graph

Hello,

I am trying to find the energies solution to the transcendental equation by plotting both graphs and finding their points of intersection. I don't why but my graphs are very weirdly shaped and don't follow the expected solution that is shown in textbooks. The tan graph itself is coming out to be a weird shape.

My train of though was to just vary E, plug that numbers in the equations and then plot the solutions against E (in eV).

The code is pasted below:

%%%%%%%%%%%%%%%%%%%%%

clear all;

e=1.6*10^-19; % eV to Joules

V=20*e;

L=6e-10;

m=9.11e-31;

h=(6.63e-34)/(2*pi);

a=(sqrt(2*m)*L)/(2*h);

E=[0:0.0001*e:V];

for i=1:length(E)

y1(i)=sqrt(E(i))*tan(a*E(i)^0.5);

y2(i)=(V-E(i))^0.5;

end

%Back to eV

E=E/e;

y3=y1/e;

y4=y2/e;

plot(E,y4)

xlim([0 20]); grid on;

%%%%%%%%%%%%%%%%%%%%%

Any help will be greatly appreciated.

[Star Strider]

Please enlighten us.

What is ‘the expected solution that is shown in textbooks’?

[Salman Azam]

I used the following the resource to help me with ithttps://www.macmillanlearning.com/studentresources/college/physics/tiplermodernphysics6e/more_sections/more_chapter_6_1-graphical_solution_of_the_finite_square_well.pdf

Also, Wikipedia has solved the problem although the approach there is slightly.

https://en.wikipedia.org/wiki/Finite_potential_well

Thanks

[Taharat]

Hi,

Could you please let me know what would happen if an static electric field of 10V/um is applied on to this well?

How the energies are to be calculated?

Thanks

Answer 1

[Ashish Kumar]

clear all;
e=1.6*10^-19; % eV to Joules
V=10*e;       
L=1.8*10^(-9)/2;
m=9.11e-31;
h_bar=(6.63e-34)/(2*pi);
a=(sqrt(2*m)*L)/(h_bar);
E=[0:0.01*e:V];
for i=1:length(E)
    alpha_by_k(i)=sqrt((V-E(i))/E(i));
    y1(i)=tan(a*sqrt(E(i)));
    y2(i)=-cot(a*sqrt(E(i)));
end
%Back to eV
E=E/e;

y3=y1;
y4=y2;
alphaK=alpha_by_k;
%plot(E,y4)
figure('Name','V=10eV');
plot(E,alphaK,E,y3,E,y4);
%plot(E,y3);
xlim([0 10.2]);
ylim([-10 10]);
legend('alpha/k','tan','-cot');

This might help others.

[David Goodmanson]

Hi Ashish,

If you are going for posterity I think it could be made clearer that e has the value 1 eV. Also it might help if you pointed out that E and V are negative, meaning that what you denote as E and V are actually abs(E) and abs(V); that what you denote by h is usually denoted by hbar; and that L is the half width of the well, not the full width.

As verification, if V is in eV and L is in angstroms (both V and L considered to be dimensionless numbers at this point), then the expected number of bound states is approximately

n = sqrt(V)*L/pi

(except there is always at least one bound state) which in this case gives n= 18, close to what the plot shows.

[Ashish Kumar]

thanks for pointing out :)