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My question is in regards to the variational principle in approximating the wavefunction of Helium.

Some Background:

$$\hat{H}=-\frac{\hbar^2}{2m_{e}}\nabla_{1}^{2}-\frac{\hbar^2}{2m_{e}}\nabla_{2}^{2}-\frac{Ze^2}{4\pi\epsilon_{0}r_{1}}-\frac{Ze^2}{4\pi\epsilon_{0}r_{2}}+\frac{e^2}{4\pi\epsilon_{0}|r_{1}-r_{2}|}$$

$$\hat{H}\Psi(\vec{r_{1}},\vec{r_{2}}) =E\Psi(\vec{r_{1}},\vec{r_{2}})$$

For many electron atoms such as helium, the (nonrelativistic) electronic TISE cannot be solved analytically as a result of the repulsion term in the electronic Hamiltonian. The first approximation to the wavefunction of helium is done by completely neglecting the repulsive term. Neglection of the repulsive term in the electronic Hamiltonian is called the orbital approxiamtion and it allows for the Hamiltonian to separate into the sum of single electron Hamiltonians and the total wavefunction to take the form as the product of single-electron wavefunctions. $$\hat{H}=-\frac{\hbar^2}{2m_{e}}\nabla_{1}^{2}-\frac{\hbar^2}{2m_{e}}\nabla_{2}^{2}-\frac{Ze^2}{4\pi\epsilon_{0}r_{1}}-\frac{Ze^2}{4\pi\epsilon_{0}r_{2}}=\hat{H}_{1}+\hat{H}_{2}$$ $$\Psi(\vec{r_{1}},\vec{r_{2}})=\psi(\vec{r_{1}})\psi(\vec{r_{2}})$$ $$\psi(\vec{r_{1}})=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{\frac{3}{2}}e^{-\frac{Z}{a_{0}}r_{1}}$$$$\psi(\vec{r_{2}})=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{\frac{3}{2}}e^{-\frac{Z}{a_{0}}r_{2}}$$

This approximation is poor, however, one such way of optimizing the approximation is by introducing an effective nuclear charge ζ in place of the nuclear charge Z. The effective nuclear charge is a parameter that can be optimized using the variational principle to produce the lowest energy. $$E=\frac{\int\phi^*\hat{H}\phi d\tau }{\int\phi^*\phi d\tau}$$

So here is my question:

When the variational principal is applied, the trial wavefunction ϕ is the product of single-electron wavefunctions derived from the orbital approximation with ζ as a parameter, however the Hamiltonian that is used is the full electronic Hamiltonian (including the repulsive term). Why is the full Hamiltonian used in the variational principle instead of the approximate Hamiltonain?

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This is because the variational approach seeks an optimal wavefunction for a fixed Hamiltonian.

What is unusual about the helium atom is that the guess function is a solution to the problem with interactions but its form is very closely related to the form of the exact solution without inteaction. The interactions here are fixed; only the function is parametrized.

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$\newcommand{\bra}[1]{\langle #1 \rvert}$ $\newcommand{\ket}[1]{\lvert #1 \rangle}$ $\newcommand{\amat}[4]{\left(\begin{matrix}#1 & #2 \\ #3 & #4 \end{matrix}\right)}$

Why is the full Hamiltonian used in the variational principle instead of the approximate Hamiltonian?

Because we seek an approximate solution to the full Hamiltonian. Any approximate solution to the ground state will have a higher energy, so we can hope to effectively and practically use the variation principle to find a state that is "close" to the ground state by minimizing the expectation value of the Hamiltonian with respect to the parameters of the trial solution.

One typical formulation of the variation principle takes a parametrized function $\psi$ (subject to $\bra{\psi}\psi\rangle=1$) and constructs: $$ \bra{\psi}\hat H\ket{\psi}\tag{1} $$

The quantity in Eq. (1) is then varied. If the variation is completely arbitrary, the variation recovers the TISE and its solution. If not, the function is not an exact solution and the variational estimate in Eq. (1) is always an upper bound for the lowest energy eigenvalue. Thus the variation principle may be especially useful for determining the ground state. (Upper bounds on the higher energy states can also be determined by using trial wavefunctions that are orthogonal to the exact eigenfunction of lower states). (See Bethe and Jackiw "Intermediate Quantum Mechanics" at pages 8-10, and 48.)

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