The variational principle in quantum mechanics, lecture 2

1. Lecture 2: The helium atom

In this lecture we are going to apply the variational method to a more complicated example, the helium atom. The simplified example we consider only involves two particles but still exhibits many of the complications and subtleties that arise in the more general {N}-body problem. Lecture notes can be found in pdf format here.

1.1. The hamiltonian

A helium atom consists of a nucleus of charge {2e} surrounded by two electrons.

Let the nucleus lie at the origin of our coordinate system, and let the position vectors of the two electrons be {\mathbf{r}_1} and {\mathbf{r}_2}, respectively. The Hamiltonian of the system thus takes the form

\displaystyle H = -\frac{\hbar}{2m}(\nabla_1^2 + \nabla_2^2) - e^2\left(\frac{2}{|\mathbf{r}_1|} + \frac{2}{|\mathbf{r}_2|} - \frac{1}{|\mathbf{r}_1-\mathbf{r}_2|}\right), \ \ \ \ \ (1)

where we are using units so that {4\pi \epsilon_0 = 1}. We write {H} as {H = H_0 + H_{\text{int}}}, where

\displaystyle H_0 = -\frac{\hbar}{2m}(\nabla_1^2 + \nabla_2^2) - e^2\left(\frac{2}{|\mathbf{r}_1|} + \frac{2}{|\mathbf{r}_2|}\right) \ \ \ \ \ (2)


\displaystyle H_{\text{int}} = -\frac{e^2}{|\mathbf{r}_1-\mathbf{r}_2|}. \ \ \ \ \ (3)

The only(!) term which causes problems here is the interparticle interaction term {H_{\text{int}}}, which is the same order as the single-particle potentials. So, unfortunately, we cannot neglect it. However, let’s do so nonetheless: our strategy is to simply solve {H_0} first to get an idea of the shape of ground-state wavefunctions. Then, inspired by the form of the solution to {H_0}, we design a trial wavefunction which we subsequently use with the variational method.

1.2. Solving {H_0}

Notice that that hamiltonian {H_0} is separable:

\displaystyle H_0 = H_1\otimes \mathbb{I}_2 + \mathbb{I}_1\otimes H_2, \ \ \ \ \ (4)


\displaystyle H_j = -\frac{\hbar}{2m}\nabla_j^2 - \frac{2e^2}{|\mathbf{r}_j|}, \quad j = 1, 2. \ \ \ \ \ (5)

This means that the eigenfunctions of {H_0}, in the case where the particles are distinguishable, are products {\psi(\mathbf{r}_1, \mathbf{r}_2) = \phi_k(\mathbf{r}_1)\phi_{l}(\mathbf{r}_2)} of the eigenvectors {\phi_k(\mathbf{r}_1)} and {\phi_l(\mathbf{r}_2)} of {H_1} and {H_2}.

We can obtain an eigenfunction {\phi(\mathbf{r}_j)} of {H_j}, {j = 1, 2}, by solving the hydrogenic Schrödinger equation

\displaystyle H_{j} \phi(\mathbf{r}_j) = E\phi(\mathbf{r}_j), \ \ \ \ \ (6)

which reads

\displaystyle \left(-\frac{\hbar}{2m}\nabla_j^2 - \frac{2e^2}{|\mathbf{r}_j|}\right)\phi(\mathbf{r}_j) = E\phi(\mathbf{r}_j), \ \ \ \ \ (7)

which is nothing other than the hamiltonian for a “hydrogen” atom with nuclear charge {2e}. The lowest-energy eigenstate of {H_j} is given by

\displaystyle \phi(\mathbf{r}_j) = \psi_0(\mathbf{r}_j) = \frac{Z^{\frac32}}{\sqrt{\pi}a^{\frac32}}e^{-\frac{Zr_j}{a}}, \ \ \ \ \ (8)

where {r_j = |\mathbf{r}_j|} and {Z = 2} is the nuclear charge. The proof of this is left as an exercise (find a reference which solves the Schrödinger equation for the hydrogen atom and replace the nuclear charge {Z = 1} with {Z = 2} throughout; spherical coordinates are helpful here). The corresponding ground-state eigenvalue is {E_0 = -Z^2 \times 13.6 \text{eV}}.

\subsubsection{A note on particle statistics} Electrons are indistinguishable fermions. This means that their wavefunction must be antisymmetric under exchange of particles, i.e.,

\displaystyle \Psi(\mathbf{r}_1, \mathbf{r}_2) = -\Psi(\mathbf{r}_2, \mathbf{r}_1). \ \ \ \ \ (9)

The problem is that if we say the wavefunction for two electrons is a product:

\displaystyle \Psi(\mathbf{r}_1, \mathbf{r}_2) ``='' \phi(\mathbf{r}_1)\psi(\mathbf{r}_2) \ \ \ \ \ (10)

then we run into problems: this wavefunction isn’t antisymmetric under interchange of the two particles!

In order to ensure that the overall wavefunction for the two electrons is antisymmetric we write

\displaystyle \Psi(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}}\left(\phi(\mathbf{r}_1)\psi(\mathbf{r}_2) - \phi(\mathbf{r}_2)\psi(\mathbf{r}_1)\right). \ \ \ \ \ (11)

(This is a Slater determinant.) This expression vanishes when {\phi = \psi}, and is the manifestation of the Pauli exclusion principle. But the trial wavefunction we are going to study has exactly the form of a product of identical wavefunctions, which is incompatible with the particle statistics. In our special case, however, we can largely avoid consideration of the particle statistics by realising that there are two species of electrons, namely, spin-up and spin-down electrons: when magnetic fields and spin-orbit couplings are disregarded we can obtain a perfectly legitimate antisymmetric wavefunction for two — now distinguishable by their spin — electrons in the same wavefunction by writing

\displaystyle \Psi(\mathbf{r}_1, \mathbf{r}_2) = \frac{1}{\sqrt{2}}\left(\phi_{\downarrow}(\mathbf{r}_1)\phi_{\uparrow}(\mathbf{r}_2) - \phi_{\uparrow}(\mathbf{r}_1)\phi_{\downarrow}(\mathbf{r}_2)\right), \ \ \ \ \ (12)

i.e., the antisymmetry is carried by the spin degree of freedom. The system is said to be in a singlet state with total overall spin. But in this case we can do all our calculations, i.e., calculations of energy expectation values, as though {\Psi(\mathbf{r}_1, \mathbf{r}_2)} was a product (exercise!); the spin never features in the subsequent derivations.

1.3. The variational method, take 1: no variation

Here we take for our “trial wavefunction” the product of the solutions (8) of the hydrogenic single-particle hydrogenic Schrödinger equations for {H_j}:

\displaystyle \Psi(\mathbf{r}_1, \mathbf{r}_2) = \frac{Z^{3}}{\pi a^{3}}e^{-\frac{Z(r_1+r_2)}{a}}. \ \ \ \ \ (13)

Of course, this isn’t a variational wavefunction: there are no variational parameters! The energy expectation of value of {\Psi} is given by

\displaystyle \langle H\rangle = E_0 + E_0 + \int d\mathbf{r}_1d\mathbf{r}_2 |\Psi(\mathbf{r}_1, \mathbf{r}_2)|^2 \frac{e^2}{|\mathbf{r}_1-\mathbf{r}_2|} \ \ \ \ \ (14)

\displaystyle = 2\times -54.4\text{eV} + 34\text{eV} = -74.8 \text{eV}. \ \ \ \ \ (15)

The observed value is {E_{\text{obs}} = -78.9\text{eV}}, so this isn’t too bad.

1.4. The variational method, take 2: an improved trial wavefunction

We content ourselves with the simplest possible application of the variational method where we take the variational wavefunction to have the same form as (13) but we replace the nuclear charge {Z} with {Z_\text{eff} = Z-\sigma}. The physical motivation here is that each electron is partially screened from the nucleus by the presence of the other. Thus our trial wavefunction is

\displaystyle \Psi(\mathbf{r}_1, \mathbf{r}_2; \sigma) = \frac{(Z-\sigma)^{3}}{\pi a^{3}}e^{-\frac{(Z-\sigma)(r_1+r_2)}{a}}. \ \ \ \ \ (16)

If {\sigma} is left arbitrary then the energy expectation value is

\displaystyle E(\sigma) = \langle H \rangle = -\frac{e^2}{a}\left(Z^2 - \frac{5}{8}Z + \frac58 \sigma - \sigma^2\right). \ \ \ \ \ (17)

Exercise: work out this result. The minimum of {E(\sigma)} is obtained for {\sigma = 5/16}, which corresponds to a screening effect where one electron reduces the nuclear charge by approximately one-third of an electron charge. The minimum value is given by

\displaystyle E(5/16) = -\left(Z - \frac6{16}\right)\frac{e^2}{a}. \ \ \ \ \ (18)

Exercise: compare this with the observed value. How much of an improvement over the crude estimate is there?

1.5. Conclusions

The agreement between theory and experiment can be improved by using a better trial function than the simple products we employed here. The design of such trial functions has occupied researchers since the early days of quantum mechanics. The helium atom is a testbed problem and offers a basic challenge because the two-electron system is mathematically tractable, although solutions in closed form cannot be found. The test of an improved trial state is measured by how closely the variational minimum lies above the precisely measured energy.


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