scatter3 error i not understand

I want to draw the hydrogen orbitals with the 3D-scatter-diagram. The probability density P should determine the size of the points. But I get the error-message

Error using scatter3scatter3 (line 135)

Error setting property 'SizeData' of class 'Scatter':

Value must be a vector of positive numeric type or nan

Error in HydrogenEigen2 (line 885)

scatter3(X,Y,Z,P)

And my X, Y, Z and my P are double-type vectors. What I do wrong?

% quantum numbers
n = 3; 
l = 1;  %  0 <= l < n 
m = 0;  %  -l <= m <= l

a = 1;  % Bohr radius

%Grid
[x,y,z] = meshgrid(-15*a:a:15*a, -15*a:a:15*a, -15*a:a:15*a);
%Grid Spherical coordinates
r=sqrt(x.^2+y.^2+z.^2); 
theta=acot(z./(sqrt(x.^2+y.^2)));
phi=atan2(y,x);

%Spherical Harmonics
b1 = (2*l+1)*factorial(l-m);
b2 = 4*pi*factorial(l+m);
b3 = sqrt(b1/b2);
Pl = legendre(l,cos(theta));
%Plm =reshape(Pl(m+1,:,:,:),size(phi));
Plm=squeeze(Pl(m+1,:,:,:));
V= b3 .* Plm .* exp(1i*m*phi);

%Radial Part
R =sqrt((2 / (a * n))^3 * factorial(n-l-1) / (2 * n * factorial(n + l))).*exp(-r / (a * n)) .* (2 * r / (a * n)).^l * 1 / factorial(n-l-1 + 2 * l + 1).*laguerreL(n-l-1, 2 * l + 1, 2 * r / (a * n));

%Ges
Phi=R.*V;
p=Phi.*conj(Phi);

%Plot preparation
Nx=size(x);
Ny=size(y);
Nz=size(z);

X=[];
for k=1:(Nz(1));
for i=1:(Ny(1));    
X=[X;x(:,i,k)];
end
end
X=squeeze(X(:,1));

Y=[];
for k=1:(Nz(1));
for i=1:(Ny(1));    
Y=[Y;y(:,i,k)];
end
end
Y=squeeze(Y(:,1));

Z=[];
for k=1:(Nz(1));
for i=1:(Ny(1));    
Z=[Z;z(:,i,k)];
end
end
Z=squeeze(Z(:,1));

P=[];
for k=1:(Nz(1));
for i=1:(Ny(1));    
P=[P;p(:,i,k)];
end
end
P=squeeze(P(:,1));

X=(X.');
Y=(Y.');
Z=(Z.');
P=((P.')/1.85*1e8);

%Plot
scatter3(X,Y,Z,P)

Answer 1

[John D'Errico]

Hmm. It took me a minute to see what the answer was from the error that was generated.

scatter3(X,Y,Z,P)
Error using scatter3 (line 135)
Error setting property 'SizeData' of class 'Scatter':
Value must be a vector of positive numeric type or nan 

This line confused me at first.

Error setting property 'SizeData' of class 'Scatter':

But next, I checked to see if your data was perhaps complex. No. But then, I looked at your data more closely.

min(P)
ans =
     0

Now read the eror message.

Value must be a vector of positive numeric type or nan

so...

find(P == 0)
ans =
  Columns 1 through 12
        9130       10926       10930       10986       10994       11110       11118       11174       11178       12846       12854       13094
  Columns 13 through 24
       13102       14710       14890       14896       14902       15082       16690       16698       16938       16946       18614       18618
  Columns 25 through 31
       18674       18682       18798       18806       18862       18866       20662

There are 31 elements in P that are exactly zero.

Then I tried:

scatter3(X,Y,Z,max(P,1.e-15))

As is often the case, the answer is found in the error message, if we read the error message carefully. Not like the "good" old days, when an error message was written in code, that could only be deciphered by consulting a manual that listed all possible error codes, and even then the document often did not explain what the real problem was or how to fix it. ;-)

[Severin Handelshauser]

Ok, the size must be bigger than 0. = 0 does not work either. Thank you very much

[John D'Errico]

"0." typivally means in MATLAB that the number was only zero to within a tolerance.

format long 
X = [0 1.e-20 1.e-16 1.e-15 .1 1 10]
X =
                   0   0.000000000000000   0.000000000000000   0.000000000000001   0.100000000000000   1.000000000000000  10.000000000000000

As you can see only the first case do we have an explicit true zero. The others are just small numbers.

Feel free to accept this answer if it answered your question.