Plotting the Probability Density of the Coulomb Wave Function

[tomdodd4598]

Hey there - I think I have an issue with my 3D density plots of the probability density of the Coulomb wave function. The reason I think something is going wrong is because my plots of |ψ(n=2, l=1, m=-1)|² and |ψ(2, 1, 1)|² are identical, while I would expect them to have the same shape but be rotationally symmetric along different orthogonal axes.

The above images are Mathematica's plots of |ψ(2, 1, -1)|², |ψ(2, 1, 0)|², |ψ(2, 1, 1)|², respectively. As you can see, the first and third are identical, and not the shape of 2p orbitals, while the second plot actually looks like what I would expect - one of the three 2p orbitals.

Here is my wave function - it's possible that the conversion from spherical to Cartesian coordinates is a problem, but I'm not sure:

If the above needs clarifying, do ask. Thanks for any help in advance ;)

 

[fzero]

If you look up the explicit form of the spherical harmonics, you'll find that $Y_{\ell,\pm m}$ differ by a relative phase that is $\pm e^{\pm 2 i \phi}$. So their complex modulus is the same. See also the normalization conventions at https://en.wikipedia.org/wiki/Spherical_harmonics#Orthogonality_and_normalization, which would give a general proof of this.

 

[jtbell]

If you take suitable linear combinations of $\psi(2,1,-1)$ and $\psi(2,1,+1)$, and then find the probability densities, you do get the $p_x$ and $p_y$ orbitals.

 

[tomdodd4598]

If you take suitable linear combinations of $\psi(2,1,-1)$ and $\psi(2,1,+1)$, and then find the probability densities, you do get the $p_x$ and $p_y$ orbitals.

Ah, I've managed to make linear combinations which give the x and y-direction p orbitals, but I am still confused - does that mean some orbital wave functions aren't energy energy eigenfunctions? How/why would that be true?

 

[DrClaude]

Ah, I've managed to make linear combinations which give the x and y-direction p orbitals, but I am still confused - does that mean some orbital wave functions aren't energy energy eigenfunctions? How/why would that be true?

They are still energy eigenfunctions. However, $p_x$ and $p_y$ are not eigenfunctions of $\hat{L}_z$.

 

[jtbell]

However, ##p_x## and ##p_y## are not eigenfunctions of ##\hat{L}_z##.

...which is of course related to the fact that $\hat L_x$, $\hat L_y$ and $\hat L_z$ do not commute with each other:

http://farside.ph.utexas.edu/teaching/qmech/Quantum/node71.html

 

[tomdodd4598]

They are still energy eigenfunctions. However, $p_x$ and $p_y$ are not eigenfunctions of $\hat{L}_z$.

Right, I understand that, but I've now got more questions - what are those donut-shaped probability densities? Why aren't they valid orbitals? Also, is there a systematic way for me to combine these 'base' wave functions to create the correct electron orbitals? I managed to make an intuitive guess with the other two p orbitals, but is there a general set of linear combinations which give all of the orbitals?

 

[DrClaude]

Right, I understand that, but I've now got more questions - what are those donut-shaped probability densities? Why aren't they valid orbitals?

They are completely valid orbitals. But they are complex functions, which can make them more difficult to work with in certain situations. The $p_x$ and $p_y$ orbitals are completely real functions, and their orientation along the Cartesian coordinates makes them useful to understand things like chemical bonding.

Also, is there a systematic way for me to combine these 'base' wave functions to create the correct electron orbitals? I managed to make an intuitive guess with the other two p orbitals, but is there a general set of linear combinations which give all of the orbitals?

See https://en.wikipedia.org/wiki/Atomic_orbital#Real_orbitals