I am going through Hartree-Fock calculations in detail, but the textbooks I am referring all considered homonuclear molecules as examples. So I am trying to come up with my own the Slater determinant for Methane. Kindly let me know if my application is correct!

For Methane, we have in total 10 electrons (Carbon 6e, and four Hydrogen with 1e each). We also know that valence shell electrons of C and H engage in bonding. The $1s$ shell of Carbon does not engage in bonding. I am also going to denote $+$ for spin-up and nothing for spin-down. Please note that $1s$ for first two columns in the following matrix describes the $1s$ orbital of carbon which does not involve in bonding.

$$\begin{pmatrix}1s(1)^+ & 1s(1) & S_1(1) & S_1^+(1) & S_2^+(1) & S_2(1) & S_3^+(1) & S_3(1) & S_4(1) & S_4^+(1) &\\\ 1s(2)^+ & 1s(2) & S_1(2) & S_1^+(2) & S_2^+(2) & S_2(2) & S_3^+(2) & S_3(2) & S_4(2) & S_4^+(2) \end{pmatrix}$$

For brevity, I am not writing out the rest of the eight rows in matrix. Now $S_n$ (spatial part of the wavefunction $S_n$), above in the matrix, is defined as the following:

$S_n=c_{n,1}1s + c_{n,2}2s + c_{n,3}2p_x + c_{n,4}2p_y + c_{n,5}2p_z$, where $n$ runs from $1$ to $5$. Why? because we know that these electrons engage in bonding, hence we have to take a superposition of orbitals (mixing). So in total we have 5 equations e.g. $S_1, S_2,...,S_5$.

Now each such superimposed MO (the $S_n$) can hold 2e each for a total of 10 electrons. Now the carbon $1s$ took away 2 electrons, so 8 electrons remained. And I have hit redundancy, because theoretically each $S_n$ can take 2e, so I have 10 electrons which can be accommodated, but I have only 8. Moreover in the Slater matrix above, I can't find a way to fit in $S_5$.

I am missing something. I would be glad if someone can throw some light on this.

  • $\begingroup$ +1 If you don't get an answer here within a few days, you might consider migrating the question to matter modelling SE. (This is not to say in any way that the question is off-topic here) $\endgroup$
    – S R Maiti
    Commented May 24, 2021 at 12:49

1 Answer 1


Whenever some says they are just starting out with electronic structure theory, I always recommend that they read Szabo and Ostlund's Modern Quantum Chemistry. The book is great introduction to HF and post-HF methods (some of the discussion of coupled cluster is a bit outdated) and its very affordable. It is nice to have around as a reference.

Getting to your actual question, I think you are confusing yourself by trying to hybridize the orbitals. During the HF procedure, you will solve for a set of MOs that is are linear combinations of the AOs, so there isn't a need to form these $S_n$ functions beforehand.

There is also the issue of actually forming the Slater determinant; while the solution to HF will be a Slater determinant, you generally don't explicitly write it out this way. Generally, you assume that each MO in the Slater determinant is a linear combination of the AOs and you solve for the coefficients of each AO, such that the energy is minimized. So the Slater determinant for methane will have 10 (initially unknown) MOs in it to account for the 10 electrons and you will then need to build these MOs using the Roothaan-Hall form of the Hartree-Fock equations.


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