Plotting a 3D surface in Julia, using either Plots or PyPlot

Question

I would like to plot a two variable function(s) (e_pos and e_neg in the code). Here, t and a are constants which I have given the value of 1.

My code to plot this function is the following:

t = 1
a = 1

kx = ky = range(3.14/a, step=0.1, 3.14/a)

# Doing a meshgrid for values of k
KX, KY = kx'.*ones(size(kx)[1]), ky'.*ones(size(ky)[1])


e_pos = +t.*sqrt.((3 .+ (4).*cos.((3)*KX*a/2).*cos.(sqrt(3).*KY.*a/2) .+ (2).*cos.(sqrt(3).*KY.*a)));
e_neg = -t.*sqrt.((3 .+ (4).*cos.((3)*KX*a/2).*cos.(sqrt(3).*KY.*a/2) .+ (2).*cos.(sqrt(3).*KY.*a)));

using Plots


plot(KX,KY,e_pos, st=:surface,cmap="inferno")

If I use Plots this way, sometimes I get an empty 3D plane without the surface. What am I doing wrong? I think it may have to do with the meshgrids I did for kx and ky, but I am unsure.

Edit: I also get the following error:

Answer 1

I changed some few things in my code.

First, I left the variables as ranges. Second, I simply computed the functions I needed without mapping the variables onto them. Here's the code:

t = 2.8
a = 1

kx = range(-pi/a,stop = pi/a, length=100)
ky = range(-pi/a,stop = pi/a, length=100)

#e_pos = +t*np.sqrt(3 + 4*np.cos(3*KX*a/2)*np.cos(np.sqrt(3)*KY*a/2) + 2*np.cos(np.sqrt(3)*KY*a))

e_pos(kx,ky) = t*sqrt(3+4cos(3*kx*a/2)*cos(sqrt(3)*ky*a/2) + 2*cos(sqrt(3)*ky*a))
e_neg(kx,ky) = -t*sqrt(3+4cos(3*kx*a/2)*cos(sqrt(3)*ky*a/2) + 2*cos(sqrt(3)*ky*a))

# Sort of broadcasting?
e_posfunc = e_pos.(kx,ky);
e_negfunc = e_neg.(kx,ky);

For the plotting I simply used the GR backend:

using Plots
gr()

plot(kx,ky,e_pos,st=:surface)
plot!(kx,ky,e_neg,st=:surface, xlabel="kx", ylabel="ky",zlabel="E(k)")

I got what I wanted!