Modelling the shape of an atom

In summary, the conversation discusses the process of modeling the shape of a hydrogen atom orbital and how to handle the complex-valued phase of the spherical harmonics when plotting the wave function. It is suggested to plot the square modulus of the wave function, which is real-valued, and to use a specific basis to make all basis functions real-valued. The rule of thumb is to leave the m=0 state alone and take the sum and difference between all m and -m states for m!=0. The conversation also includes a suggested method for plotting the wave function.
  • #1
lavster
217
0
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:
Physics news on Phys.org
  • #2
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
but this only works if [tex]\phi[/tex] is pi o 2pi? but [tex]\phi[/tex] can take any value? or am i misunderstanding it?
 
  • #4
Ah, I was a bit confused. Right, the harmonics have a complex phase, but it's physically insignficant here. So [tex]e^{i\phi m}[/tex] can be taken to be [tex]sin(\phi m)[/tex] if m > 0, [tex]cos(\phi m)[/tex] if m < 0 (or vice versa) and [tex]\frac{1}{\sqrt{2}}[/tex] if m=0
 
  • #5
You basically plot the probability-function, lYl2 = Y.Y*
 
  • #6
You want to plot the probability distribution for the probabilities of the electron.

This includes both the radial part and the spherical harmonics.
 
  • #7
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
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
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:
  • #10
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:

What is an atom?

An atom is the smallest unit of matter that retains all the properties of an element. It is composed of a central nucleus, which contains protons and neutrons, and electrons that orbit around the nucleus.

How do scientists model the shape of an atom?

Scientists use a variety of models to represent the shape of an atom. The most commonly used model is the Bohr model, which depicts the electrons as orbiting the nucleus in specific energy levels or shells. Other models include the quantum mechanical model, which represents the electrons as existing in a cloud-like region around the nucleus.

What factors influence the shape of an atom?

The shape of an atom is primarily influenced by the number of protons, neutrons, and electrons it contains. The arrangement of these particles in the atom's nucleus and electron shells also plays a role in determining its shape.

Why is it important to understand the shape of an atom?

Understanding the shape of an atom is crucial in explaining its properties and behavior. It also helps scientists predict how atoms will interact with each other, which is essential in fields such as chemistry and materials science.

Can the shape of an atom change?

Yes, the shape of an atom can change under certain conditions. For example, atoms can gain or lose electrons, which can alter their electron configuration and therefore, their shape. Additionally, atoms can undergo nuclear reactions, which can change the number of protons and neutrons in the nucleus and ultimately affect the atom's shape.

Similar threads

  • Quantum Physics
Replies
2
Views
365
  • Quantum Physics
Replies
21
Views
1K
Replies
26
Views
1K
  • Quantum Physics
Replies
5
Views
1K
  • Quantum Physics
Replies
10
Views
889
  • Quantum Physics
Replies
28
Views
3K
Replies
17
Views
2K
Replies
21
Views
969
  • Quantum Physics
Replies
6
Views
727
  • Quantum Physics
Replies
2
Views
915
Back
Top