plot the probability distribution from Schrodinger Equation

Hi everyone,

I have been trying t plot this probability of a function in 3D space with the following coding, but it always gives a blank plot. I'll appriciate your help.

The function is:

f(x,y,z) = 8*sqrt(sin(k*x)*sin(k*y)*sin(k*z)) on the interval: 0<x<1, 0<y<1, and 0<z<1. Note that k = n*pi/a, in this case a =1 and n = 1.

>> n=1; a=1;
>> x = linspace(0,0.01,1);
>> y = linspace(0,0.01,1);
>> z = linspace(0,0.01,1);
>> [X,Y,Z] = meshgrid(x,y,z);
>> f = 8.*sqrt(sin(n*pi*x/a).*sin(n*pi*y/a).*sin(n*pi*z/a));
>> figure
>> scatter3(f,X,Y,Z)

Answer 1

[Star Strider]

It gives a blank plot because the linspace calls are only producing one element. The third argument to linspace is the number of elements desired in the vector it produces. An argument of 1 returns the beginning point (first argument) and stops.

I am not certain what the desired result is, however here are two possibilities —

n=1; a=1; 
x = linspace(0,0.01,25);
y = linspace(0,0.01,25);
z = linspace(0,0.01,25);
[X,Y,Z] = meshgrid(x,y,z);
f = @(x,y,z) 8.*sqrt(sin(n.*pi.*x/a).*sin(n.*pi.*y/a).*sin(n.*pi.*z/a));
figure
scatter3(X(:),Y(:),f(X(:),Y(:),Z(:)), '.')

figure
isosurface(X,Y,Z,f(X,Y,Z), 5E-13)
grid on
view(60,30)

I leave the rest to you.

.

Answer 2

[Paul]

scatter3() is used to plot points in 3D space. But f is really a function of three variables, so you'll need an alternative to visualize f. Check the doc page for visualizing volume data.