I would like to study time-dependence (TD) in linear combinations of atomic orbitals (LCAO).

The Hückel method enables quick and dirty determination of MOs for suitable systems (view link for assumptions). The basis function $\phi$ is the 2pz orbital.

Any molecular orbital (MO) is described as

$$\psi_n(r) = \sum_k \phi_k(r)\ c_{k,n}$$

where $c_{k,n}$ is the coefficient on the $k$th atom at energy level (MO number) $n$. For benzene, the first MO is (calculation omitted):

$$\psi_1(r) = \frac{1}{\sqrt{6}}\sum_{k=1}^6 \phi_k(r)$$

Pretty straightforward.

Notice that the functions are all time-independent. I want to construct the TD wavefunction describing benzene in its ground state.

$$\psi_{tot}(r,t) = \sum_j \psi_j \mathbf{C}_j\ e^{-itE_j}$$

Benzene has 6 $\pi$ electrons, thus the first three MOs will be occupied in its ground state, each containing 2 electrons. From the previous formula, the total wavefunction should be:

$$\psi_{tot}(r,t) = 2\sum_j^3 \psi_j \mathbf{C}_j\ e^{-itE_j}$$

How do I determine the second set of coefficients $\mathbf{C}_j$?

  • $\begingroup$ @Qmechanic I'm still unsure if this belongs on Physics.SE or Chemistry.SE :/ $\endgroup$
    – CHM
    Commented Aug 6, 2012 at 18:56
  • $\begingroup$ I think it is fine on either site, but please don't cross-post. If you after a couple of days haven't received any good answers here, flag it for migration. $\endgroup$
    – Qmechanic
    Commented Aug 6, 2012 at 19:04

2 Answers 2


It does not work like that. The wave function of 6 electrons is the product, not the sum, of orbitals (wave functions of single electrons). However, since electrons are fermions the overall wave function must be antisymmetric. Thus, the simplest wave function that you can write for the 6 electron-benzene approximation (in the spirit of the H\"uckel method) is the Slater determinant: $$\psi({\bf{r}_1,\ldots,\bf{r}_6}) = \left| \begin{array}{cccccc} \phi_1(r_1) \alpha(1) & \phi_1(r_1)\beta(1) & \phi_2(r_1)\alpha(1) & \phi_2(r_1) \beta(1) & \phi_3(r_1)\alpha(1) & \phi_3(r_1)\beta(1) \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots \\ \phi_1(r_6)\alpha(6) & \phi_1(r_6)\beta(6) & \phi_2(r_6)\alpha(6) & \phi_2(r_6)\beta(6) & \phi_3(r_6)\alpha(6) & \phi_3(r_6)\beta(6) \end{array} \right|$$ where $\phi_k$ are the H\"uckel single electron wavefunctions (the MO-LCAO in H\"uckel's approach), $\bf{r}_i$ is the position and spin variable of electron $i$, and $\alpha(i)$ and $\beta(i)$ refer to the spin function (up and down, respectively). The Slater determinant has 6 rows: you change the electron index in each row.

Following the simple H\"uckel theory the energy of the system is $E = 2(E_1 + E_2 + E_3)$ where $E_k$ is the energy of the orbital. (Of course this is a rough approximation that doesn't take into account interelectron repulsion or the double counting of these repulsions if they are included in a self-consistent way in $E_k$.)

Therefore the time-dependent wave function is $\Psi({\bf{r}_1,\ldots,\bf{r}_6},t) = \psi({\bf{r}_1,\ldots,\bf{r}_6}) e^{-iEt/\hbar}$


The ground state of benzene is an eigenstate of the hamiltonian and as such its time dependence is trivial - it only acquires a phase with time at a rate equal to the ground state energy (modulo $\hbar$), and if it's only the ground state you're interested in then you can ignore this phase.

  • $\begingroup$ The Hamiltonian I'm really interested into is time-dependent. $\endgroup$
    – CHM
    Commented Aug 7, 2012 at 4:07
  • $\begingroup$ Then you've got a completely different problem on your hands, and the ground state is not a particularly relevant quantity. If the hamiltonian changes adiabatically then the state will follow the instantaneous eigenstates, but the time dependence will either be trivial or quite hard to visualize. If the hamiltonian changes quickly then you need to formulate a TDSE problem, including initial conditions, more precisely. $\endgroup$ Commented Aug 7, 2012 at 20:52

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