Variational methods in quantum mechanics

Based on a lecture given by me to the PH3520 class on March 25, 2011. (Updated November 4, 2015 for another lecture!)


The variational method is a commonly used to set bounds on the ground state (and possibly first excited state) in several problems in quantum mechanics. We shall illustrate the general idea using an example that also shows the limitations of this approach. The system of interest is one involving two electrons (possibly) bound to a nucleus of charge $Z$. Thus, $Z=1$ corresponds to $H^-$ (hydrogen ion), $Z=2$ corresponds to $He$ (Helium) and $Z=3$ corresponds to $Li^+$ (Lithium ion). The Hamiltonian for the system without the centre of mass degrees of freedom is one with two degrees of freedom which we take to be the coordinates ($\mathbf{r}_1$ and $\mathbf{r}_2$) of the two electrons.

(1)
\begin{align} H = \Big[\frac{\mathbf{p}_1^2}{2m} -\frac{Ze^2}{r_1} \Big]+ \Big[\frac{\mathbf{p}_2^2}{2m} -\frac{Ze^2}{r_2} \Big]+ \frac{e^2}{r_{12}}\ , \end{align}

in notation where the subscripts $1,\ 2$ are for the two electrons and $r_{12}=|\mathbf{r}_1-\mathbf{r}_2|$ and the mass $m$ is the reduced mass of the electrons. The last term in the above Hamiltonian is the inter-electronic repulsion. If this term were not present, the Hamiltonian would be equivalent to decoupled (i.e., commuting) Hydrogen-like hamiltonians: $h_i= \frac{\mathbf{p}_i^2}{2m} -\frac{Ze^2}{r_i}$ for $i=1,2.$ The ground state for such an Hamiltonian is given by the $1s$ wavefunction

(2)
\begin{align} \phi_{1s}^{Z}(\mathbf{r}) = \left(\frac{Z^3}{\pi a_0^3}\right)^{1/2} \ e^{-\frac{Zr}{a_0}}\ , \end{align}

where $a_0=\hbar^2/me^2$ is the Bohr radius. The energy eigenvalue of this wavefunction is $-Z^2 R$ where $R =me^4/2\hbar^2$ is the Rydberg constant. It is useful to work in units where length is measured in terms of the Bohr radius and energy is measured in Rydbergs and $\hbar=1.$ In such coordinates, the Hamiltonian simplifies as follows1:

(3)
\begin{align} H = \Big[\mathbf{p}_1^2 -\frac{2Z}{r_1} \Big]+ \Big[\mathbf{p}_2^2 -\frac{2Z}{r_2} \Big]+ \frac{2}{r_{12}}\ , \end{align}

and the $1s$ wavefunction becomes

(4)
\begin{align} \phi_{1s}^{Z}(\mathbf{r}) = \left(\frac{Z^3}{\pi }\right)^{1/2} \ e^{-Zr}\ . \end{align}

The ansatz for the variational wavefunction that we choose is the following2

(5)
\begin{align} \boxed{ \psi^{\widetilde{Z}}(\mathbf{r}_1,\mathbf{r}_2) = \phi_{1s}^{\widetilde{Z}}(\mathbf{r}_1)\times \phi_{1s}^{\widetilde{Z}}(\mathbf{r}_2)\ , } \end{align}

where we treat the parameter $\widetilde{Z}$ as a variational parameter. Note that without the inter-atomic repulsion, the above would be an eigenfunction with $\widetilde{Z}=Z.$ We will compute the expectation value of $H$ in for this wavefunction as a function of $\widetilde{Z}$ and choose its value such that the expectation values takes its smallest value.

A non-trivial computation (details given below) leads to the following result

(6)
\begin{align} \boxed{\langle \psi^{\widetilde{Z}}\left|\ H\ \right| \psi^{\widetilde{Z}} \rangle = 2 \widetilde{Z}^2 - 4 Z \widetilde{Z} +\frac54 \widetilde{Z}\ , } \end{align}

whose minimum occurs at $\widetilde{Z}= (Z -\tfrac5{16}).$ Thus the lower bound on the ground state energy $E_0$ obtained for the ansatz is

(7)
\begin{align} E_0 \leq -2R \Big(Z -\tfrac5{16}\Big)^2 \end{align}

We need to compare the above number with the energy where one electron is bound to the nucleus and the second electron is unbound and at infinity. The energy in this case is given by $-Z^2R$. Thus, we can use the above bound to see whether it predicts the existence of a two-electron bound state. This would follow if the bound implies energy less than $-Z^2R$. In the table below, we carry out the comparison for $Z=1,2,3$ by studying the ratio $2\left(Z-\tfrac5{16}\right)^2\big /Z^2$:

Z $\mathbf{ -2\left(Z-\tfrac5{16}\right)^2\big /Z^2}$ Bound state? Ratio in [1]
1 $-\tfrac{121}{128}\sim -0.95$ Can't decide $-1.05$
2 $-\tfrac{729}{512}\sim -1.42$ Yes $-1.45$
3 $-\tfrac{1849}{1152}\sim -1.61$ Yes $-1.62$

So our variational ansatz shows that both $He$ and $Li^+$ can occur as bound states. It turns out that $H^-$ also occurs and an improved variational ansatz[1] gives a ratio of $-1.05$. The original result is due to Hans Bethe[2]. Further, it has been shown (by Hill[3]) that $H^-$ has no more bound states! See the paper by Fontenelle et. al. linked below for further details.

References

  1. M.T. Fontenelle, M.R. Gallas and J.A.C. Gallas, The wavefunction for the ground state of the hydrogen ion, Journal of Physics B: Atomic and Molecular Physics, 19, L639(1986).
  2. H. Bethe, Berechnung der Elektronenaffinitaet des Wasserstoffs (Calculating the electron affinity of Hydrogen) Z. Phys., 57, 815 (1929).
  3. R. N. Hill, Proof that the H- Ion Has Only One Bound State, Phys. Rev. Lett. 38, 643–646 (1977).
  4. Rau A. R. P. The Negative Ion of Hydrogen, J. Astrophys. Astr. 17, 113 (1996). pdf file provided as the link appears broken.

Details of the computation


The devil is in the details.


First rewrite the Hamiltonian as $H=H_0+H_1+H_2$ where

(8)
\begin{align} H_0&=\Big[\mathbf{p}_1^2 -\frac{2\widetilde{Z}}{r_1} \Big]+ \Big[\mathbf{p}_2^2 -\frac{2\widetilde{Z}}{r_2} \Big] \\ H_1 &= \Big[ \frac{2(\widetilde{Z}-Z)}{r_1} + \frac{2(\widetilde{Z}-Z)}{r_2} \Big]\\ H_2&= \frac{2}{r_{12}}\ , \end{align}

This separation makes it easy to compute the expectation value of $H_0$ as the variational wave function is the ground state of $H_0$ with energy $-2 \widetilde{Z}^2$ — the two arising from the fact that there are two electrons. The virial theorem can be used to compute the answer for $H_1$. One obtains

(9)
\begin{align} \langle \psi^{\widetilde{Z}}\left|\ H_1\ \right| \psi^{\widetilde{Z}} \rangle = 4(\widetilde{Z}-Z) \left\langle \frac1r \right\rangle_{n=0}= 4(\widetilde{Z}-Z)\widetilde{Z} \ , \end{align}

where $\langle \frac1r \rangle_{n=0}=\widetilde{Z}$ is the expectation value in the ground state of a Hydrogen-like atom with atomic number $\widetilde{Z}$ in units where length is measured in units of the Bohr radius. So all one needs to do is to compute

(10)
\begin{align} I(\widetilde{Z}):=\langle \psi^{\widetilde{Z}}\left|\ H_2\ \right| \psi^{\widetilde{Z}} \rangle=\frac54 \widetilde{Z}\ . \end{align}

The required integral is

(11)
\begin{align} I(\widetilde{Z})&=\left(\frac{\widetilde{Z}^3}{\pi}\right)^2\int dr_1r_1^2d\Omega_1 dr_2r_2^2d\Omega_2\ e^{-2\widetilde{Z}(r_1+r_2)}\ \frac1{r_{12}} \\ &=\widetilde{Z}\ I(1)\ , \end{align}

where in the second line we extract the dependence of the integral on $\widetilde{Z}$ by scaling out the dependence on $\widetilde{Z}$. In order to compute $I(1)$, we begin with the following expression for $1/r_{12}$ which can obtained, for instance, from the generating function for the Legendre polynomials.

(12)
\begin{align} \frac{1}{r_{12}} = \sum_{\ell=0}^\infty \frac{(r_<)^\ell}{(r_>)^{\ell+1}} \ P_\ell(\cos\theta)\ , \end{align}

where $r_>=\text{max}(r_1,r_2)$, $r_<=\text{min}(r_1,r_2)$ and $\theta$ is the angle between the two vectors $\mathbf{r}_1$ and $\mathbf{r}_2$.

(13)
\begin{align} I(1)=\frac2{\pi^2}\int dr_1r_1^2d\Omega_1 dr_2r_2^2d\Omega_2\ e^{-2(r_1+r_2)}\ \sum_{\ell=0}^\infty \frac{(r_<)^\ell}{(r_>)^{\ell+1}} \ P_\ell(\cos\theta) \end{align}

With no loss of generality, assume that $r_1>r_2$ — the other situation is handled by using the $\mathbf{r}_1\leftrightarrow \mathbf{r}_2$ symmetry and putting an overall factor of 2. Further, assume that $\mathbf{r}_1$ is along the positive $z$-axis. The angular integrals can be simplified as follows as the integrand only depends on the angle between the two vectors $\mathbf{r}_1$ and $\mathbf{r}_2$ which is chosen to equal to $\theta_2$.

(14)
\begin{align} \int d\Omega_1 \int d\Omega_2 P_\ell(\cos\theta_2) = 4\pi \times 2\pi \int_0^\pi d\theta_2\sin \theta_2 P_\ell(\cos\theta_2) =(4\pi)^2\ \delta_{\ell,0}\ . \end{align}

We thus obtain after carrying out the summation over $\ell$

(15)
\begin{align} I(1)= 2 \times \frac2{\pi^2}\ \int_0^\infty dr_1^2 r_1^2 \int_0^{r_1}dr_2 r_2^2 \ e^{-2(r_1+r_2)}\ \frac{ (4\pi)^2}{r_1} = \frac54\ . \end{align}

This completes the computation.

Remark: After looking through the Bethe computation for the first time, I notice that he does the computation in a different way. He also works out an improved ansatz which shows the existence of the bound state for the hydrogen ion. The article is in German but the equations make complete sense even if you don't know German.

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