As a follow-up to my answer here, I'd like to ask what exactly does it mean for a wavefunction to "respect the symmetry" of the system.

The original context is: immediately after ionisation of $\ce{CH4}$, the wavefunction of $\ce{CH4+}(T_\mathrm{d})$ cannot be adequately described by removal of an electron from an $\mathrm{sp^3}$-type hybrid orbital, as the resulting Slater determinant $|\phi_1\phi_2\phi_3\phi_4\overline{\phi}_1\overline{\phi}_2\overline{\phi}_3|$ does not reflect the tetrahedral symmetry of the system immediately after ionisation ($\phi_i$ being the four hybrid orbitals).

Intuitively, it is clear that this is related to the fact that the four C–H bonds must be equivalent. Mathematically, how is this formalised, i.e. what is the criterion that the wavefunction must obey?

As a side question, how does the criterion above apply to a boron atom, for example? Is the wavefunction $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_x^1$ acceptable on symmetry grounds? My gut feeling is that it's not (because the electron density needs to be spherically symmetric?), but I might really be overthinking this.

  • $\begingroup$ Hmmm... Something seems weird about this. The $T_d$ symmetry which comes from the $\ce{CH4}$ will change to $C_{2v}$, which I know you are aware of. But the point is that the wavefunction is not initially in a bound state. You would need to use the time-dependent Schrodinger equation to follow the time-evolution of the wavefunction. I'm not sure what physical meaning any constraint on this initial state would mean because independent of the constraint, the wavefunction will relax to the $C_{2v}$ state. I don't know the answer, I'm just throwing out a thought which might be useful. $\endgroup$
    – jheindel
    Commented Sep 14, 2017 at 21:48
  • $\begingroup$ @jheindel I believe what he is getting at is what does it mean for a wavefunction to reflect the symmetry of the system. What types of wavefunctions are disqualified from describing tetrahedral, octahedral, etc molecules? $\endgroup$
    – Tyberius
    Commented Sep 14, 2017 at 22:12
  • $\begingroup$ I think you constrain the coefficients of the AO of symmetry-equivalent atoms. $\endgroup$ Commented Sep 15, 2017 at 2:33
  • $\begingroup$ @jheindel Actually, I sort of get what you mean. I was also thinking about the particle in a 1-D box (box lies between $-a$ and $+a$); since the potential is an even function of $x$, the Hamiltonian must commute with the parity operator, and all stationary states must be eigenfunctions of the parity operator, i.e. either even or odd. This is what Ivan was describing in his answer. But that of course only applies to stationary states, and it is obviously possible to construct a wavefunction that is neither even nor odd by taking an appropriate superposition of even and odd stationary states... $\endgroup$ Commented Sep 15, 2017 at 15:07
  • $\begingroup$ and surely the same thing applies here? The wavefunction immediately after ionisation is obviously not in a stationary state, so why does it have to obey the symmetry of the system? I don't actually know, and the more I think about it the more confused I get, but I guess it is a problem for another question. $\endgroup$ Commented Sep 15, 2017 at 15:08

2 Answers 2


I use quantum chemistry about as much as a lumberjack uses math (that is, just enough not to fell a tree on one's own head), so my answer will be short and hopefully useful as a tl;dr to someone else's more comprehensive take.

In quantum chemistry, if two operators commute, their values are simultaneously observable. In other words, there is a basis of functions which are eigenfunctions to both operators at once. For an opposite example, think of coordinate and momentum; they do not commute, hence the Heisenberg uncertainty and stuff.

When our problem has some symmetry (including, but not limited to, geometrical symmetry like $T_d$), this means that symmetry operators commute with its Hamiltonian. Therefore the eigenfunctions of the latter must also be eigenfunctions of the former. In other words, the application of symmetry operator to any of our states is equivalent to a multiplication by some number. That's the constraint you were after. Since a symmetry operator of a finite point group must have finite order (by applying it a few times, you get back where you started), the said number must be a root of unity.

That's about the size of it. The numbers for different symmetry operators can be chosen in a multitude of ways, each corresponding to some representation. The character tables tell you how many functions of each representation do you have. Since the functions of different representations are orthogonal to each other and don't interact, this helps us to split a huge problem into smaller chunks.

With methane, the qualitative result could have been envisioned in a simpler way, almost without any calculations. Think of a carbon atom with four $sp^3$ orbitals. They are all equivalent and orthogonal to each other. Now bring in the four H atoms. This will produce four $\sigma$ orbitals (and $\sigma^*$ too, we just don't care about those at the moment). But wait, those $\sigma$ orbitals are no longer orthogonal, so they must interact, if only a little. And because of symmetry, each two of them overlap by exactly the same amount. So it is as simple as Hückel can be: find the eigenvalues of the matrix $$\begin{pmatrix} 0&-1&-1&-1\\ -1& 0&-1&-1\\ -1&-1& 0&-1\\ -1&-1&-1& 0 \end{pmatrix}$$ The symmetry approach I outlined above still applies; it is just that the system became so small that one can manage without it. The thing can be solved by hand. Or it can be noted that we're dealing with a complete graph, and their spectra are well known (just remember that math people have it upside down, for they use $\mathbf1$s instead of $\mathbf{-1}$s). Either way, the eigenvalues are $\{-3,1,1,1\}$, which kinda explains the 1:3 thing.

So it goes.

  • $\begingroup$ Obligatory reference to a lumberjack. $\endgroup$
    – Jan
    Commented Sep 15, 2017 at 9:52
  • $\begingroup$ Yes... thank you! I thought something along these lines too (cf. my comments on the question, in response to jheindel). I think I am still a bit unsure on how to deal with degeneracy, which is what I was aiming for with the boron example; eg. the wavefunction $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_x^1$ is not an eigenfunction under, say, a 90° rotation about the y-axis (which transforms it into $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_z^1$). How can we deal with this? (I realise spherical symmetry isn't the easiest example to use, but maybe we just restrict it to 90° rotations.) $\endgroup$ Commented Sep 15, 2017 at 18:33

If you have degenerate orbitals, they need to have same occupation numbers. Mathematically, the wave function still musst obey the symmetry of the molecule. Sometimes you can only achieve this by mixing some determinants together.

Lets stick with the Boron example (I happend to have a calculation about that lying around). The trick here is to do a CASSCF calculation (in $D_{2h}$), with active orbitals 2px 2py 2pz. Thus the orbitals can be state-averaged and describe all required states in a balanced way. This will results in some fractional occupation numbers for the orbitals.

Due to symmetry we need the 3 p orbitals to be degenerate, and we also expect 3 degenerate electronic states (because we have a $^2P$ term).

The resulting state-averaged orbital occupations and energies are:

1s  2.00000    -7.697613
2s  2.00000    -0.495495
px  0.33333     0.025103 
py  0.33333     0.025103 
pz  0.33333     0.025103 

And the CI vectors are (first column is orbial occupation, a for $\alpha$ spin, 0 for empty)

 CI vector for state symmetry 1

 a 0 0           1.0000000

 TOTAL ENERGIES                       -24.52694952

 CI vector for state symmetry 2

 0 a 0           1.0000000

 TOTAL ENERGIES                       -24.52694952

 CI vector for state symmetry 3

 0 0 a           1.0000000

 TOTAL ENERGIES                       -24.52694952

In principle, one should be able to do this within HF as well, but this requires a calculation that can deal with the full point group of the spherical atom.

So for $\ce{CH4+} (T_d)$ we can construct the WF from four configurations, one for each $sp^3$ hybrid missing one electron.

  • 1
    $\begingroup$ I've been wondering about the boron case for a long time, so thanks! Now, if I interpret your data correctly (not a computational chemist), the CI wavefunction is then an equal admixture of the three determinants corresponding to $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_x^1$, $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_y^1$, and $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_z^1$. I guess I am still slightly confused, because putting in three determinants should give three linear combinations. What happens to the other two - or am I on an entirely wrong track? $\endgroup$ Commented Sep 15, 2017 at 18:23
  • $\begingroup$ In this case we have 3 states with only one configuration each, due to symmetry. But the orbitals are optimized for all 3 states simultaneously (state-averaged), it's what MCSCF (CASSCF) does. $\endgroup$
    – Feodoran
    Commented Sep 15, 2017 at 20:14

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