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Chapter 9: Problem 12

Use Moller-Plesset perturbation theory to obtain an expression for the ground- state wavefunction corrected to first order in the perturbation.

Short Answer

Expert verified
The ground-state wavefunction corrected to first order in the perturbation can be expressed as \( \Psi = \Psi^0 + \lambda \frac{1}{E^0 - H^0} V \Psi^0 \).

Step by step solution

01

Understanding Moller-Plesset perturbation theory

First, it is critical to understand what Moller-Plesset perturbation theory is. In quantum mechanics, it involves 'perturbing' the Hamiltonian, i.e., changing it slightly, and then solving the Schrodinger equation with the adjusted Hamiltonian. Assuming the Hamiltonian \( H = H^0 + \lambda V \) of system where \( H^0 \) is the unperturbed Hamiltonian, \( V \) is the perturbation and \( \lambda \) is a small constant.
02

Writing out the zeroth-order wavefunction

The zeroth-order wavefunction is the solution to the Schrodinger equation with the unperturbed Hamiltonian. This wavefunction, denoted as \( \Psi^0 \), is the ground-state wavefunction in unperturbed system, it is solved from \( H^0 \Psi^0 = E^0 \Psi^0 \) where \( E^0 \) is zeroth order energy.
03

Formula for first-order corrected wavefunction

To get the first-order corrected wavefunction, called \( \Psi^1 \), we can use the following formula derived from the Moller-Plesset series. \( \Psi^1 = \frac{1}{E^0 - H^0} V \Psi^0 \), this indicates how the wavefunction changes due to the small perturbation.
04

Writing out the final corrected wavefunction

The final corrected wavefunction for the ground state at first order correction can be expressed as \( \Psi = \Psi^0 + \lambda \Psi^1 \).

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Most popular questions from this chapter

Prove Brillouin's theorem; that is, show that hamiltonian matrix elements between the HF wavefunction \(\Phi_{0}\) and singly excited determinants are identically zero. Hint. Use the Condon-Slater rules.

Consider the two-clectron integrals over the basis functions defined in eqn 9.17 . If the basis functions are taken to be real, a number of the integrals are equivalent; for example, \((a b | c d)=(a d | c b)\). Find the other integrals that are equal to \((a b | c d)\)

Consider two Slater determinants \(\Phi_{1}\) and \(\Phi_{2}\) that differ by only one spinorbital; that is, $$\begin{array}{l} \Phi_{1}=\left\|\cdots \varphi_{m} \varphi_{i} \cdots\right\| \\ \Phi_{2}=\left\|\cdots \varphi_{p} \varphi_{i} \cdots\right\| \end{array}$$ Derive the following Condon-Slater rule: $$\begin{aligned} \left\langle\Phi_{1}|H| \Phi_{2}\right\rangle=&\left\langle\varphi_{m}(1)\left|h_{1}\right| \varphi_{p}(1)\right\rangle+\sum_{i}\left\\{\left[\varphi_{m} \varphi_{i} | \varphi_{p} \varphi_{i}\right]\right.\\\ &\left.-\left[\varphi_{m} \varphi_{i} | \varphi, \varphi_{p}\right]\right\\} \end{aligned}$$ where we have used the notation $$\left[\varphi_{4} \varphi_{b} | \varphi_{c} \varphi_{d}\right]=j_{0} \int \varphi_{a}^{*}(1) \varphi_{b}^{*}(2)\left(\frac{1}{r_{12}}\right) \varphi_{c}(1) \varphi_{d}(2) \mathrm{d} r_{1} \mathrm{d} r_{2}$$

Use the AM1 and PM3 semiempirical methods to compute the equilibrium bond lengths and enthalpies of formation of (a) ethanol, (b) 1,4 -dichlorobenzene.

A single Slater determinant is not necessarily an cigenfunction of the total clectron spin operator. Therefore, even within the Hartree-Fock approximation, for the wavefunction \(\Phi_{0}\) to be an eigenfunction of \(S^{2},\) it might have to be expressed as a linear combination of Slater determinants. The linear combination is referred to as a spin-adapted configuration. As a simple example, consider a two-electron system with four possible Slater determinants: $$\begin{array}{l} \Phi_{1}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \alpha(1) \psi_{2}\left(r_{2}\right) \alpha(2)\right| \\ \Phi_{2}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \alpha(1) \psi_{2}\left(r_{2}\right) \beta(2)\right| \\ \Phi_{3}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \beta(1) \psi_{2}\left(r_{2}\right) \alpha(2)\right| \\ \Phi_{4}=\left(\frac{1}{2}\right)^{1 / 2} \operatorname{det}\left|\psi_{1}\left(r_{1}\right) \beta(1) \psi_{2}\left(r_{2}\right) \beta(2)\right| \end{array}$$ (a) Show that the Slater determinants \(\Phi_{1}\) and \(\phi_{4}\) are themselves cigcnfunctions of \(S^{2}\) with cigenvaluc \(2 \hbar^{2}\) (corresponding to \(S=1\) ). (b) From \(\Phi_{2}\) and \(\Phi_{3}\), determine two linear combinations, one of which corresponds to \(S=1, M_{s}=0\) and the other of which corresponds to \(S=0\) \(M_{s}=0\)
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