# $sp^3$ Hybridization wavefunctions and probability density

I have plotted a hydrogen-like sp3-hybrid orbital probability density and it looks like this:

I can plot 4 overlapping probability densities in a tetrahedral shape:

So far it looks OK. But when I'm trying to sum the orbitals (wavefunctions) and square the resulting function to get the probability density, I always come up with a sphere:

So I have the following questions:

1.) Is this an expected result? If not, what am I doing wrong?

2.) What is usually depicted on the sp3-hybridization schemes: overlapping wavefunctions, sums of the wavefunctons, probability density or something else?

The $$sp^3$$ orbitals are a mix of an $$s$$ orbital and all three $$p$$ orbitals and form four equivalent hybrids. Due to orthogonality, it can be proven that the orbitals have the following form:
$$|h_1\rangle=\frac{1}{2}\left(|s\rangle+|p_x\rangle+|p_y\rangle+|p_z\rangle\right)\\ |h_2\rangle=\frac{1}{2}\left(|s\rangle+|p_x\rangle-|p_y\rangle-|p_z\rangle\right)\\ |h_3\rangle=\frac{1}{2}\left(|s\rangle-|p_x\rangle+|p_y\rangle-|p_z\rangle\right)\\ |h_4\rangle=\frac{1}{2}\left(|s\rangle-|p_x\rangle-|p_y\rangle+|p_z\rangle\right)$$
The addition of all hybrid orbitals is, wait for it, an $$|s\rangle$$ orbital! So yes, it should be a spheric orbital.
• Regarding orbital plots, I think that the most common way to represent them is by plotting a surface where the probability $|\psi|^2$ is constant. You should plot the orbitals independently and put them togother once you have calculated all of them. From the plots you uploaded I would say that you should take a lower value of $|\psi|^2$ as a boundary. – TheAverageHijano Feb 21 at 7:39