## to find the root of quantum transcendental equation

### physics learner (view profile)

on 11 Apr 2019
Latest activity Edited by David Goodmanson

on 12 Apr 2019

### David Goodmanson (view profile)

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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physics learner

### physics learner (view profile)

on 11 Apr 2019
but how to find Ee
David Goodmanson

### David Goodmanson (view profile)

on 11 Apr 2019
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)
Walter Roberson

### Walter Roberson (view profile)

on 11 Apr 2019
Do you have the symbolic toolbox?

### David Goodmanson (view profile)

on 12 Apr 2019
Edited by David Goodmanson

### David Goodmanson (view profile)

on 12 Apr 2019

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])

### David Wilson (view profile)

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);

physics learner

### physics learner (view profile)

on 12 Apr 2019
your code is wrong for each value of R yield same result
Torsten

### Torsten (view profile)

on 12 Apr 2019
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