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In elementary chemistry textbooks you often have pictures like the following one:

Orbitals

Are there any conventions how to get them?

I am not sure, but I guess that it are contour plots with only one iso-surface of constant probability density $|\psi|^2$ such that the (integrated) probability for an electron to be inside of on the blue volumes, is 90 %.

If so, give those conditions a unique result and how to solve it numerically? Maybe you could add an example using python mayavi or something like that.

Note that my question is not of how to visualize the probability density via colormap or scatter plots, that's conceptually clear.

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Your intuition is right, for example in 3D Orbitals (German Wikipedia) the caption explicitly states that 90% iso-surfaces are used. I have however seen different percentages before where the results look similar.

Did you check the Mayavi Example Atomic Orbital? If you remove the phase-coloring and find the additional parameter to contour that sets the cutoff, it produces the textbook plots.

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I am not aware of a cut off parameter. But that's the problem; how dows such a cutoff parameter work? – student Mar 21 at 12:21
    
Thanks for the link to the german wikipedia. Do you have an idea why it makes sense to visualize it in this way? – student Mar 21 at 12:23
    
In the german wikipedia article there is also another condition: Choosing the smallest volume such that the probability is > 0.9 beeing in that volume. I am not sure why this is the same as choosing an isosurface in such a way that the probability of beeing inside it > 0.9. – student Mar 21 at 12:36
2  
@student: there's an infinite amount of volumes that have probability = 0.9. Consider that there's a volume VA enclosed within this volume whose probability is 0.01, and a another volume VB outside this volume whose probability is also 0.01 (since there's still 10% chance for the electron to be outside this volume). Now If I construct a new volume by excluding VA and including VB, the probability of the electron being in this new volume is 0.9 - 0.01 + 0.01 = 0.9. In general, ∫∫∫ p(v) dV = P has no unique solution for 0<P<1 – MSalters Mar 21 at 14:06
2  
@student: Well, for my example the volume of VA would be smaller than the volume of VB, because the probability density in VA is higher. And yes, there's a unique solution if the probability density is continuous. Every point inside the volume has a higher probability density then any point outside that volume. You can create a proof by contradiction by considering small volumes with a radius epsilon round both points. – MSalters Mar 21 at 16:03

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