# Modelling the shape of an atom

1. Dec 8, 2009

### lavster

hi,

when modelling the shape of a hydrogen atom orbital, is it the real part of the spherical harmonics that i plot?

thanks

Last edited: Dec 8, 2009
2. Dec 8, 2009

### alxm

Well the wave function (being time-independent and without an external field) is real, and so are the spherical harmonics (m being an integer).

3. Dec 8, 2009

### lavster

but this only works if $$\phi$$ is pi o 2pi? but $$\phi$$ can take any value? or am i misunderstanding it?

4. Dec 8, 2009

### alxm

Ah, I was a bit confused. Right, the harmonics have a complex phase, but it's physically insignficant here. So $$e^{i\phi m}$$ can be taken to be $$sin(\phi m)$$ if m > 0, $$cos(\phi m)$$ if m < 0 (or vice versa) and $$\frac{1}{\sqrt{2}}$$ if m=0

5. Dec 8, 2009

### StElsewhere

You basically plot the probability-function, lYl2 = Y.Y*

6. Dec 8, 2009

### Feldoh

You want to plot the probability distribution for the probabilities of the electron.

This includes both the radial part and the spherical harmonics.

7. Dec 8, 2009

### alxm

Why are you guys telling the OP what he wants?
There's nothing wrong with plotting the wave function itself, it's done all the time (and there's a pedagogical point in doing so, for understanding MO theory).

The question was how to handle the complex-valued phase of the spherical harmonics when doing so. If you're plotting the density then the question is moot; they disappear in the integration. The less-trivial point is that the phase is irrelevant, and not only for plotting; you can stick that real-valued wave function back into any equation and you'll still get all the same observables.

Also, in numerical QM calculations, atomic wave functions are expressed as simple real-valued functions (namely gaussians) all the time.

8. Dec 8, 2009

### blkqi

This is what the OP asked for: how to model the shape of a hydrogen atom. And you do so by plotting the square modulus (≡probability/electron density) of the wavefunction, which is real valued.

9. Dec 8, 2009

### weejee

When we plot wavefunctions(not its absolute square), we have two options. We can either plot real and imaginary parts separately, or choose a basis so that all basis functions are real-valued (We can do this whenever there is a time reversal symmetry). The latter is what they often do, especially in chemistry books.

For l=1 case, Y_10 is already real, whereas Y_11 and Y_1,-1 are complex.
Y_10 ~ cos(theta) = z/r
Y_11 ~ sin(theta)exp(i*phi) = (x+iy)/r
Y_1,-1 ~ sin(theta)exp(-i*phi) = (x-iy)/r

Yet, if we take the linear combinations of Y_11 and Y_1,-1 we have real valued functions

Y_10 ~ cos(theta) = z/r
Y_11 + Y_1,-1 ~ sin(theta)cos(phi) = x/r
Y_11 - Y_1,-1 ~ sin(theta)sin(phi) = y/r

Now, we have all real orbitals! These orbitals are so called p_z, p_x and p_y orbitals.

The rule of thumb is that you leave m=0 state alone, and take the sum of and difference between all m and -m states for m!=0.
I guess you might want to do the same thing for l=2 and verify that we get x^2-y^2, 3z^2-r^2, xy, yz, zx states.

Last edited: Dec 8, 2009
10. Dec 11, 2009

### intervoxel

I would try something like this:

Inicialization:
- define the number of points in cloud
- define a maximum radial distance (since you cannot go as far as infinity)
- define the radial discretization
- define the azimuthal discretization
- define the polar discretization
- visit each point in the sphere with these discretization values
- calculate the volume element
- calculate psi modulus squared
- multiply the two numbers
- accumulate this value
- end each
- the accumulated value is the normalization factor

Plotting:
- for each point in cloud do
- calculate a random value between 0 and 1 (collapse)
- visit each point in the sphere with the discretization values
- calculate the volume element
- calculate psi modulus squared
- multiply the two numbers
- accumulate the value
- divide this value by the normalization factor. If it is greater than collapse value, plot point and break each
- end each
- end each

Obs.: the phase angle must be mapped to a specifc color

Last edited: Dec 11, 2009
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