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Re: Exact Solution to the double well potential problem in mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg132451] Re: Exact Solution to the double well potential problem in mathematica
• From: Bob Hanlon <hanlonr357 at gmail.com>
• Date: Wed, 19 Mar 2014 04:23:48 -0400 (EDT)
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• References: <20140315074705.D13256A16@smc.vnet.net>

hbar = 105*10^-36;
m = 96*911*10^-36;
a = 5*10^-9;
b = 5*10^-10;
V = 3*16*10^-21;
k = Sqrt[(2*m*e)/((hbar)^2)];
l = Sqrt[(2*m*(V - e))/(hbar)^2];


f[e_] = k*Cot[(-k)*a] - l*Tanh[(-l)*b/2] // Simplify;


Plot[f[e], {e, 0, 10^-18}]


FindRoot[f[e], {e, #}, WorkingPrecision -> 25] & /@
 Table[est, {est, {6, 15, 35, 57, 85}*10^-18}]


{{e -> 5.795052681985343157761426*10^-18}, {e ->
   1.522389834765808661449499*10^-17}, {e ->
   3.474550534236380053861731*10^-17}, {e ->
   5.707872905176405519130226*10^-17}, {e ->
   8.492682781188889164140891*10^-17}}



Bob Hanlon




On Sat, Mar 15, 2014 at 3:47 AM, Ragavendran Nagarajan <
ragavendran.nagarajan at gmail.com> wrote:

> A little background. This is a Schrodinger equation problem to find the
> energy stages e in the double well where there is a step potential inside a
> infinite rectangular well. After solving the boundary conditions
> analytically, I tried to solve the following program in mathematica
>
> hbar := 1.05*10^-34;
> m := 0.096*9.11*10^-31;
> a := 5*10^-9;
> b := 0.5*10^-9;
> V := 0.3*1.6*10^-19;
> k = Sqrt[(2*m*e)/((hbar)^2)];
> l = Sqrt[(2*m*(V - e))/(hbar)^2];
> Block[{e}, e /. First@Solve[k*Cot[(-k)*a] - l*Tanh[(-l)*b/2] == 0, e]]
>
> As you can see since the values of the constants are very small, I am
> getting a lot of errors and I am not sure how to get the solution.
>
> Please let me know if there is a way to solve this equation using
> Mathematica. I have to submit this assignment in a couple of days, so I
> wold be really really grateful if you guys can show me the way to solve
> this.
>
> Thanks for your time
>
> Raga
>
>

• References:
  • Exact Solution to the double well potential problem in mathematica
    • From: Ragavendran Nagarajan <ragavendran.nagarajan@gmail.com>