# What is the angular momentum spectrum of an sp${}^3$ electron?

So, one thing has been annoying me ever since I learned about orbital hybridization: you explain the shape of molecules by postulating that the orbitals of multi-electron atoms are linear combinations of the orbitals of the hydrogen atom. Fine.

Here's what gets me, and I've asked several chemistry professors about this, and I haven't gotten an answer. If I measure the orbital angular momentum of an electron in a hydrogen atom, I get the value $L = \hbar \sqrt{\ell(\ell +1)}$, so an s orbital has angular momentum $0$ and a p orbital has angular momentum $\hbar$.

So, what happens when I measure the orbital angular momentum of a sp${}^{3}$ electron? Do I get $\frac{3\hbar}{4}$ with probability 1? Or do I get $\hbar$ with probability 0.75, and $0$ with probability 0.25? Are the original orbitals really the fundamental thing, and the hybridized orbitals fundamentally just quantum superpositions of these things? Or does the multi-electron potential create a state that just looks like a linear combination of one s and 3 p orbitals?

Or is the actual answer more complicated and this a heuristic picture that I'm taking too seriously?

• The eigenvalue of $L^2$ can never be 3/4 of an allowed eigenvalue: there are no two eigenvalues that differ by a factor of 3/4 (except for two zeros). The allowed spectrum of $L^2$ is always $l(l+1)\hbar^2$ - it follows from basic SO(3) group theory. Mar 3, 2011 at 20:10

More like the latter than the former, although you've got one detail wrong. An $s$ orbital is $\ell=0$, so it corresponds to $L=0$, not $L=\hbar$. A $p$ orbital is $\ell=1$, corresponding to $L=\hbar\sqrt{2}$. If the electron is in a state that is a superposition of $s$ and $p$, then a measurement of angular momentum will yield either 0 or $\hbar\sqrt 2$, with probabilities given by the squares of the amplitudes of the different parts of the superposition. If it's an equally-weighted superposition (and I confess I don't know enough chemistry to know whether that's correct), then the probabilities are 25% and 75% as you say.

• Ugh, right. Correcting the original question. Mar 3, 2011 at 18:24

You're definitely taking this picture too seriously. Hybridization of atomic orbitals is just an approximation based on an independent particle model created by Pauling to rationalize some structural trends in chemistry with Quantum Mechanics. We can only assign orbitals unambiguously for one-electron systems, though of course independent particle models were widely employed and still are to explain reactivity trends, etc.

Anyway, if you prepare an electron in an sp3 state, which is an equally weighted linear combination of the px, py pz and s orbitals, the probability to find and electron at a certain angular momentum might be calculated by taking the square of the projection of this angular momentum on the wave function, as QM tells us to do.

Don't worry too much if you find weak spots in general theories developed to explain molecules, reactivity, etc, since they're highly approximated in most cases (unless you do an expensive computation for a system, but then it is not general anymore), and the guys that developed those such as Pauling already knew that. As you know these highly approximated theories (e.g., Molecular Orbital and Valence Bond theory) were and still are very successful, even if they're not very rigorous.

It gets really hard to be rigorous in chemistry beyond a certain point...

@TedBunn

Of course the basic formalism of QM would predict properties of this hypothetical electronic state, I did not say anything that would imply the opposite.

Still, to think about bonding in terms of sp3 hybrid orbitals such as one does in general chemistry is really a very crude picture, unless you're dealing with molecules that have very special properties such as Td symmetry (methane, for example), and still in those cases if you perform a calculation by using VB or MO (Hartree-Fock) theory (and it does not even need to be with a computer, since symmetry is going to make life very easy here) with only the SP3 orbitals of carbon and S of hydrogen in your set of basis functions, you will see a very big quantitative error in predicted ionization energies when you compare to photoelectron spectroscopy measurements of ionization energies. Another good test would be to perform high level quantum chemistry calculation of methane using a software and employing two different basis sets, one containing only s and p functions centered on carbon and s functions centered on hydrogen atoms, and another in which the basis set has functions of several angular momenta centered on each atom. If you compare your results with experiments you will see that the first method will have a much lower accuracy compared to the second one. S and P functions might still be the ones that contribute mostly to the bonding molecular orbitals in the different orbital configurations (set of occupied molecular orbitals) that contribute to a good description of methane, but we can only say that for sure for very special cases. When you go to molecules with very distinct geometries, it is vital to have d functions centered on carbon, for example, in order to determine even qualitatively right chemistry trends.

Quantum chemistry semi-empirical methods widely used in the past often had this problem of only employing valence basis functions centered on an atom in molecule calculations, not giving enough flexibility for the description of molecular geometries, and therefore providing bad results whenever the geometry deviated too much from what the qualitative models (such as the hybridization model) predicted.

In fact, I'd really be surprised (but not too much) if any modern paper that discusses chemical bonding, or that qualitatively discusses bonding orbitals to explain computed trends, do that by employing hybridization theory arguments. Those that I have seen almost always do that by looking at molecular orbitals or some refined valence bond treatments.

Therefore, although hybridization is a deep and important topic that every chemist should dominate, in the way it was developed by Pauling it has severe limitations that people are, or should be, aware of.

• I'm not quite sure I agree that he's taking the picture too seriously. I think it's true, to an excellent approximation, that the electron state in question is a superposition of the given l=0 and l=1 states. There are probably higher-l states in the superposition too, but with low amplitude. The basic formalism of quantum mechanics surely applies to such a state, and it makes sense to talk about the probabilities of getting various results when performing a (hypothetical) orbital angular momentum measurement. (Not that I have a clue how one would perform such a measurement in practice ...) Mar 4, 2011 at 2:37
• @TedBunn I think electron paramagnetic resonance is the tool to study the actual electron charge distribution inside a molecule. From an EPR spectrum you get a justification for the sp3 hybridisation and get a hint on the molecule shape. So an expert for EPR/NMR is needed for the initial question Mar 1, 2012 at 23:04

In the Hydrogen atom $L^2$ commutes with the Hamiltonian so that an eigenstate of $H$ can be at the same time an eigenstate of $L^2$. Additionally, the energy does not depend on the quantum number $l$, so that a superposition of s and p orbitals is still an eigenstate of $H$ with not well-defined angular momentum. In a sp$^3$ state if you measure $L^2$ you will observe $l=0$ 25% of the times, and $l=\sqrt{2}\hbar$ on 75% of the times, as you suggested.

On the other hand, in a molecule, neither $l_i^2$ (the orbital angular momentum of a single electron) nor $L^2$ (the total orbital angular momentum) commute with $H$. (Molecules are not spherical). You typically write your energy eigenstate as an antisymmetrized product of orbitals, which are eigenstates of the Fock operator (the self-consistent one-electron Hamiltonian). Additionally, each orbital can be written as a linear combination of whatever basis (e.g. atomic orbitals, as is usually done in Chemistry). Obviously $l_i^2$ is not well defined for this orbital, nor it shouldn't since you want to obtain faithful representations of the energy eigenstates, nor of the angular momentum operator.

Now if you still want to measure the orbital angular momentum of a single electron in the molecule, you first need to ionize an electron. The angular momentum will depend on the dynamics that led to the ionization. You may still ask what are the angular momentum components of whatever orbital of the molecule, although this information is not physical (I believe there can not be any measurement associated to it as a matter of principle). Then you need to choose an axis first (e.g. whatever bond axis, etc.) There is nothing terribly difficult in doing this, but usually you don't get much interesting information, expect perhaps in diatomic molecules or in studying molecules that are breaking along a particular bond.