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to find the root of quantum transcendental equation

Asked by physics learner on 11 Apr 2019
Latest activity Edited by David Goodmanson on 12 Apr 2019
Vc = 0.5;
c = 3e8;
m0 = 0.511e6/c^2;
me1 = 0.067*m0;
me2=0.039*m0;
hr = 6.58e-16;
k= sqrt(2*me1*Ee/hr^2);
K= sqrt(2*me2*(Vc-Ee)/hr^2);
R=2e-9;
equation K*cot(K*R)= -k
how to find the root of this eqution

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Hi pl,
Evidently Ee and Vc are in eV, and it appears that Vc>Ee. Then k and K are in the neighborhood of 5e8 m^-1. So R = 2 produces a kR of about 1e9. Do you have any thoughts on the units and the size of R? (R subsequently changed by the OP from 2 to 2e-9)
Do you have the symbolic toolbox?

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2 Answers

Answer by David Goodmanson on 12 Apr 2019
Edited by David Goodmanson on 12 Apr 2019
 Accepted Answer

Hi pl,
The boundary condition K*cot((KR) = -k (denoted by bc1 below) is for a wave function that is proportional to sin(KR) in the potential well, giving the odd parity states. However, it appears that your potential well is not deep enough to support any of those states. Plot 1 does not show a crossing for the would-be first excited state.
The even parity states are proportional to cos(KR) in the well; that boundary condition is denoted by bc0 below. Plot 2 shows a crossing and there is a solution for the ground state.
Vc = .5;
Ee = 0:.01:Vc-.01;
c = 3e8;
m0 = 0.511e6/c^2;
me1 = 0.067*m0;
me2=0.039*m0;
hr = 6.58e-16;
R=2e-9;
k= sqrt(2*me1*Ee/hr^2);
K= sqrt(2*me2*(Vc-Ee)/hr^2);
bc1 = K.*cot(K*R); % = -k; first excited state and odd parity states
bc0 = -K.*tan(K*R); % = -k; ground state and even parity states
figure(1)
plot(Ee,bc1,Ee,-k); grid on
figure(2)
plot(Ee,bc0,Ee,-k); grid on
% solution
ks = @(Ee) sqrt(2*me1*Ee/hr^2);
Ks = @(Ee) sqrt(2*me2*(Vc-Ee)/hr^2);
fun = @(Ee) -Ks(Ee).*tan(Ks(Ee)*R) + ks(Ee);
Ee_groundstate = fzero(@(Ee) fun(Ee),[0 .5])

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Answer by David Wilson on 11 Apr 2019

Well you have given us quite a puzzler; little background info and no context.
I'm assuming you are looking for the value Ee since you didn't give us a value for it. But are you sure your equation is correct? I couldn't find a (real) solution, only complex, so I guessed you made a sign error.
If the equation really is K*cot(K*R)= +k (note sign change), then we have a solution as shown below.
Vc = 0.5;
c = 3e8;
m0 = 0.511e6/c^2; me1 = 0.067*m0; me2=0.039*m0;
hr = 6.58e-16;
R=2e-9;
k= @(Ee) sqrt(2*me1*Ee/hr^2);
K= @(Ee) sqrt(2*me2*(Vc-Ee)/hr^2);
f = @(Ee) K(Ee).*cot(K(Ee)*R)-k(Ee); % should equal zero
Ee0 = 0.01; % who knows ??
Ee = fzero(@(Ee) f(Ee), Ee0)
Ee = linspace(0,0.02)';
plot(Ee, f(Ee)); yline(0);

  2 Comments

your code is wrong for each value of R yield same result
Not true:
function main
Vc = 0.5;
c = 3e8;
m0 = 0.511e6/c^2; me1 = 0.067*m0; me2=0.039*m0;
hr = 6.58e-16;
k= @(Ee) sqrt(2*me1*Ee/hr^2);
K= @(Ee) sqrt(2*me2*(Vc-Ee)/hr^2);
f = @(Ee,R) K(Ee).*cot(K(Ee)*R)-k(Ee);
Ee0 = 0.01;
for i=1:180
R(i) = 2e-9 - i*1e-11;
Ee(i) = fzero(@(Ee) f(Ee,R(i)), Ee0)
Ee0 = Ee(i);
end
plot(R,Ee)
end

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