Non trivial solution to Schrödinger equation for 1-D infinite well

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The discussion centers on solving the Schrödinger equation using MATLAB, specifically for a one-dimensional infinite potential well. The user encounters an issue where applying boundary conditions results in a trivial solution with both coefficients equal to zero. The suggested approach is to first solve the differential equation with one boundary condition, which allows for one integration constant. The second boundary condition can then be applied to determine the quantized energy levels. Additionally, it is noted that MATLAB may not be the best tool for symbolic computations, with alternatives like Mathematica and Maple recommended for more effective symbolic calculations. Suggestions for modifying the code to achieve the desired solution are also sought.
MrMuscle
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Hello, I am trying to find the solution of Schrödinger equation on matlab. However, when I apply boundary conditions, MATLAB only gives me the solution with both coefficients 0. I want to find the solution : Asin(n*pi*x/L)
You can see my code below. Could you please tell me where is my mistake?
Thanks in advance!
Hello, I am trying to find the solution of Schrödinger equation on matlab. However, when I apply boundary conditions, MATLAB only gives me the solution with both coefficients 0. I want to find the solution : Asin(n*pi*x/L)
You can see my code below. Could you please tell me where is my mistake?
Thanks in advance!

clear,clc %Solution of Schrödinger eqn with 1-D infinite well
syms WF(x) V(x) %WF is the wavefunction, V(x) is the potential
syms hbar m L E positive %E energy, L length of the well, m mass, hbar reduced Planck
V(x)=0 %Potential
Sch=((-hbar^2)/(2*m))*diff(WF,x,2)+V*WF==E*WF %Schrödinger equation

Solution1=dsolve(Sch) %Solution without applying boundary conditions

%BOUNDARY CONDITIONS
cond1=WF(0)==0;
cond2=WF(L)==0;
conds=[cond1 cond2];
Solution2=dsolve(Sch,conds) %Solution with applying boundary conditions.
 
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Matlab is trying to solve your setup for an arbitrary energy ##E##. This will generally not have a solution but ##E## will be quantised. What you want to do is to solve the differential equation using one of the boundary conditions, leaving you one integration constant. Then you can solve the second boundary condition for the energy. This will give you the quantised energy levels.

Note that MATLAB is not the ideal for symbolic computations.
 
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Thanks for the answer! what would you suggest for symbolic calculations?
Also, do you have any suggestions about how to modify the code?
 
Personally, I would use Mathematica. Maple is also popular.
 
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