# Is the correct zeroth-order wave function $\phi^{(0)}$ an eigenfunction of the perturbation Hamiltonian $H'$?

Consider the perturbation treatment of He $$1s2s$$ excited state configuration. The quantum mechanical Hamiltonian for the He-atom is:

\begin{aligned} \hat{H} = \sum_{i=1}^2 h_i + \frac 1 {r_{1 2}} \\ h_i = -\tfrac 12 \nabla_i^2 - \frac 1 r_i \end{aligned}

In the following perturbation treatment, the unperturbed Hamiltonian $$\hat{H}^0$$ is $$\sum_{i=1}^2 h_i$$, while the perturbation Hamiltonian $$\hat{H}'$$ includes the inter-electron repulsion $$r_{1 2}^{-1}$$.

The four unperturbed degenerate wave functions $$\psi^{(0)}$$ are shown below as Slater determinants:

\begin{aligned} &\psi_1^{(0)} = \left|\begin{array}{} 1s(1) \overline{2s}(1) \\ 1s(2) \overline{2s}(2) \end{array}\right| \equiv \left| 1 \overline{2} \right\rangle & \psi_2^{(0)} = \left| \overline{1} 2 \right\rangle \\ &\psi_3^{(0)} = \left| 1 2 \right\rangle &\psi_4^{(0)} = \left| \overline{1} \overline{2} \right\rangle \end{aligned}

The correct zeroth-order wave functions $$\phi^{(0)}$$ for the perturbation Hamiltonian $$\hat{H}'$$ are linear combinations of unperturbed eigenfunctions such that $$\phi^{(0)} = \sum_i c_i \psi_i^{(0)}$$. Solving for $$\phi^{(0)}$$ using the linear variational method is equivalent to diagonalizing the perturbation Hamiltonian $$\hat{H}'$$ in a basis of $$\psi^{(0)}$$. The eigenvectors and eigenvalues of the $$\mathbf{H}'$$ matrix, whose elements are $$\left\langle \psi_m^{(0)} \middle| \hat{H}' \middle| \psi_i^{(0)} \right\rangle$$, are the correct zeroth-order wave functions and the corresponding first-order corrections to the energy, respectively.

Because $$\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$$ (the Condon-Slater rules), $$\psi_3^{(0)} = \left| 1 2 \right\rangle$$ is already the eigenvector of the perturbation Hamiltonian $$\hat{H}'$$ with the eigenvalue $$E_3^{(1)} = \left\langle 1 2 \mid \mid 1 2 \right\rangle = \left\langle 1 2 \mid 1 2 \right\rangle - \left\langle 1 2 \mid 2 1 \right\rangle$$. Could you suggest whether it is correct to go on to state the eigen-equation of $$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$$ ?

No, writing $$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$$ will be incorrect. I will give you two explanations for the same, one intuitive and other mathematical.

## Intuitive Explanation:

Assuming that you can write $$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$$. But, $$\hat{H}_0\psi_3^{(0)} = E_3^{(0)} \psi_3^{(0)}$$, and in an overall sense, you can write

$$\hat{H}\psi_3^{(0)} = (E_3^{(0)}+E_3^{(1)}) \psi_3^{(0)}$$

This essentially means that electron-electron repulsion is keeping the wavefunction same as that of single electron Hamiltonian. Now, since in single electron Hamiltonian, each electron does not "see" the other electron, and they are allowed to be in the same spatial position. Now, on bringing the $$1/r_{12}$$ term inside the Hamiltonian, the energy goes to $$\infty$$ for such a case, and hence a singularity is created in the system.

So, it means that the $$\psi_3^{(0)}$$ is not an eigenvector of total Hamiltonian, $$\hat{H}$$. Since $$\psi_3^{(0)}$$ is defined to be eigenvector of $$\hat{H}_0$$, so the initial assumption that $$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$$ is incorrect.

## Mathematical Explanation:

You have given $$\left\langle \psi_{m \neq 3}^{(0)} \middle| \hat{H}' \middle| \psi_3^{(0)} \right\rangle = 0$$, which is correct. Along with that, we know $$\left\langle \psi_{m}^{(0)} \middle| \psi_n^{(0)} \right\rangle = \delta_{mn}$$. However, you could write the given statement $$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$$ provided $$\psi_n^{(0)}$$ would be a complete basis set (for explanation, see below). However, $$\psi_n^{(0)}$$ is not a complete basis set, since for a He atoms, there are infinite number of ways to put two electrons in an orbital, since there are infinite number of orbitals. Since you are treating the orbitals as hydrogenic, it is not guaranteed that the orbitals will be orthogonal to each other once the $$1/r_{12}$$ term is pulled in.

Even for a hypothetical case if we consider only 1s and 2s orbitals are present and are orthogonal to other orbitals even after the presence of the $$1/r_{12}$$ term, still they would not be a complete basis set. The reason is simple: You have considered only $$1s^12s^1$$ configurations, and not $$1s^2$$ and $$2s^2$$.

For complete basis, proof:

Assuming $$\psi_n$$ is an othronormal and complete basis set of dimension $$M$$, one can write: $$\hat{H}'\psi_3^{(0)}=\sum_{i=1}^M c_i \psi_i^{(0)}$$ Now, $$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)} \Rightarrow C_m=0 > \forall m\neq 3$$ $$\Rightarrow \hat{H}'\psi_3^{(0)}= c_3 > \psi_3^{(0)}$$

However, this would have been true if, along with other conditions, $$\psi_n^{(0)}$$ would have been a complete basis set.

So, from the given information, it is impossible to conclude that $$\hat{H}'\psi_3^{(0)} = E_3^{(1)} \psi_3^{(0)}$$ mathematically, and from a chemistry standpoint, $$\hat{H}'\psi_3^{(0)} \neq E_3^{(1)} \psi_3^{(0)}$$.

• Hmm... I don't agree with your "intuitive explanation". The energy doesn't just go to infinity simply because there is a nonzero probability density corresponding to $\vec{r} = \vec{r}_1 - \vec{r}_2 = \vec{0}$. According to that logic, a 1s orbital should not exist either, because it has nonzero probability density at the nucleus, which should cause infinite attraction. Basically, the integral in question is not $\int_0^\infty (1/r) \,\mathrm{d}r$ (which would explode); it's $\int_0^\infty \psi_i^* (1/r) \psi_j\,\mathrm{d}r$, and those $\psi$'s depend on $r$ as well. Dec 30, 2020 at 2:22
• @Mitradip Hi, I mostly agree with your mathematical explanation. I was wondering if it would be appropriate to argue that the number of atoms is a very large number, and the number of orbitals is very large and approaches infinity. Here's a link: en.m.wikipedia.org/wiki/Electronic_band_structure Thank you for the answer. Dec 30, 2020 at 5:28
• @orthocresol Thank you for the comment. Since $\psi_3^{(0)}$ is derived from one-electron hamiltonians, there is no such term in the wave function that depends on $|r_1-r_2|$. Now, based on your argument, the wavefunction should vanish at $|r_1-r_2|=0$, irrespective of the value of $r_1$ or $r_2$. This essentially means the one-electron wavefunction should go to zero always. Jan 1, 2021 at 9:46