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This started when i was wondering why p-orbitals have an unsymmetrical shape (which isn't clear to me yet). But then i came across Halliday/Resnick/Walker's Fundamentals of Physics. In it, they have said that the three p-orbitals show their individual shapes in presence of a magnetic field. But normally, in an isolated atom, the three p-orbitals have the same energy and are indistinguishable, and that their combined probability density is spherically symmetrical.

I am having trouble in visualizing this. The three dumbells do not look like they could complete a sphere! For example, what about the region in the direction of $\hat i + \hat j + \hat k$? Wouldn't gaps be left behind? Wouldn't this destroy the symmetry of the distribution?

A visual graph of the three separately and then the three together (like different coloured dots for different p-orbitals) would help. I have no resources or skills (not even remotely close!) to do it myself, so i am requesting this to the skilled people out here. Thank you very much!

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    $\begingroup$ I have no graph but if you mathematically add up the squares of the wavefunctions (to be more precise $\psi^*\psi$), you get a probability density that is independent of $\theta$ and $\phi$. The dumbbell diagrams are only pictorial representations of a concept (probability density function) that we cannot adequately depict in 2D. It is not surprising that when you add these limited diagrams you don't get the full truth. $\endgroup$ – orthocresol Jun 11 '16 at 17:25
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    $\begingroup$ The dumbbell pictures show where the probability density is greatest for each of the orbitals, but there is still some probability density in the area outside the dumbbells. When you add the orbitals together, the area inside each dumbbell has a large contribution from one orbital and near zero from the other two, while the "gaps" have a small contribution from each of the three orbitals, totalling the same amount. $\endgroup$ – f'' Jun 11 '16 at 19:08
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    $\begingroup$ To make orthocresol's comment more concrete, you can try to calculate $|Y_{1,0}|^2+|Y_{1,-1}|^2 + |Y_{1,1}|^2$, see this quantity is independent with $\theta,\phi$ $\endgroup$ – Rodriguez Jun 11 '16 at 21:22
  • $\begingroup$ @f'' Thank you for the comment! So this means the probability plot for the p-orbitals together would look like the s-orbitals with value of $n$ one less than their own? (That is to say, $3p$ orbitals would look like $2s$ orbitals and so on) $\endgroup$ – FreezingFire Jun 14 '16 at 17:31
  • $\begingroup$ @orthocresol Hmm, that makes sense! Then what is the factor that separates them in the first place? Thank you for your comment! $\endgroup$ – FreezingFire Jun 14 '16 at 17:33

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