Helium Atom

Consider the helium atom, which is a good example of a two-electron system. The Hamiltonian is written

$$H = \frac{{\bf p}_1^{\,2}}{2\,m_e} + \frac{{\bf p}_2^{\,2}}{2\,m_... ...{Z\,e^{\,2}}{4\pi\,\epsilon_0\,r_2} + \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{12}},$$ (9.51)

where $Z\,e$ is the nuclear charge, $Z=2$ , $r_1=\vert{\bf x}_1\vert$ , $r_2=\vert{\bf x}_2\vert$ , and $r_{12} = \vert{\bf x}_1-{\bf x}_2\vert$ . Here, ${\bf p}_1$ and ${\bf x}_1$ are the momentum and position of the first electron, et cetera. Suppose that the final term on the right-hand side of the previous expression were absent. In this case, in which the two electrons do not interact with one another, the overall spatial wavefunction can be formed from products of hydrogen atom wavefunctions calculated with $Z=2$ , instead of $Z=1$ . Each of these wavefunctions is characterized by the usual triplet of quantum numbers, $n$ , $l$ , and $m$ . (See Chapter 4.) Now, the total spin of the system is a constant of the motion (because ${\bf S}$ obviously commutes with the Hamiltonian), so the overall spin state is either the singlet or the triplet state. (See the previous section.) The corresponding spatial wavefunction is symmetric in the former case, and antisymmetric in the latter. Suppose that one electron has the quantum numbers $n$ , $l$ , $m$ , whereas the other has the quantum numbers $n'$ ,$l'$ , $m'$ . The corresponding spatial wavefunction is

$$\phi({\bf x}_1,{\bf x}_2)=\frac{1}{\sqrt{2}}\left[\psi_{nlm}({\bf... ..._{n'l'm'}({\bf x}_2)\pm \psi_{nlm}({\bf x}_2)\,\psi_{n'l'm'}({\bf x}_1)\right],$$ (9.52)

where the plus and minus signs correspond to the singlet and triplet spin states, respectively. Here, $\psi_{nlm}({\bf x})$ is a standard hydrogen atom wavefunction (calculated with $Z=2$ ). For the special case in which the two sets of spatial quantum numbers, $n$ , $l$ , $m$ and $n'$ , $l'$ , $m'$ , are the same, the triplet spin state does not exist (because the associated spatial wavefunction is null). Hence, only singlet spin state is allowed, and the spatial wavefunction reduces to

$$\phi({\bf x}_1,{\bf x}_2)=\psi_{nlm}({\bf x}_1)\,\psi_{nlm}({\bf x}_2).$$ (9.53)

In particular, the ground state ($n=n'=1$ , $l=l'=0$ , $m=m'=0$ ) can only exist as a singlet spin state (i.e., a state of overall spin 0), and has the spatial wavefunction

$$\phi({\bf x}_1,{\bf x}_2) =\psi_{100}({\bf x}_1)\,\psi_{100}({\bf... ...2)= \frac{Z^{\,3}}{\pi\,a_0^{\,3}}\,\exp\left[\frac{-Z\,(r_1+r_2)}{a_0}\right],$$ (9.54)

where $a_0$ is the Bohr radius. This follows because [see Equation (4.95) and Exercise 8]

$$\psi_{100}({\bf x})= \frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_0}\right)^{3/2}\exp\left(\frac{-Z\,r}{a_0}\right).$$ (9.55)

The energy of this state is

$$E = 2\,Z^{\,2}\,E_0=8\,E_0 = -108.8\,{\rm eV},$$ (9.56)

because $Z=2$ . (See Exercise 8.) Here, $E_0=-13.6\,{\rm eV}$ is the ground-state energy of a hydrogen atom. In the previous expression, the factor of $2$ (before the factor $Z^{\,2}$ ) is present because there are two electrons in a helium atom.

The previous estimate for the ground-state energy of a helium atom completely ignores the final term on the right-hand side of Equation (9.52), which describes the mutual electrostatic repulsion of the two electrons. We can obtain a better estimate for the ground-state energy by treating Equation (9.55) as the unperturbed wavefunction, and $e^{\,2}/(4\pi\,\epsilon_0\,r_{12})$ as a perturbation. According to standard first-order perturbation theory (see Chapter 7), the correction to the ground-state energy is

$${\mit\Delta} E = \left\langle \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{12}}\right\rangle =2\left\langle \frac{a_0}{r_{12}}\right\rangle \vert E_0\vert,$$ (9.57)

because $E_0=-e^{\,2}/(8\pi\,\epsilon_0\,a_0)$ . Here, the expectation value is calculated using the wavefunction (9.55). The resulting expression for ${\mit\Delta}E$ can be written

$$\frac{{\mit\Delta E}}{\vert E_0\vert} = \frac{2\,Z^{\,6}}{\pi^{\,... ...}{a_0^{\,3}}\,\exp\left[-\frac{2\,Z\,(r_1+r_2)}{a_0}\right] \frac{a_0}{r_{12}}.$$ (9.58)

Now,

$$\frac{1}{r_{12}} = \frac{1}{(r_1^{\,2}+r_2^{\,2}-2\,r_1\,r_2\,\co... ...mma)^{1/2} }= \sum_{l=0,\infty} \frac{r_<^{\,l}}{r_>^{\,l+1}}\,P_l(\cos\gamma),$$ (9.59)

where $r_>$ ($r_<$ ) is the larger (smaller) of $r_1$ and $r_2$ , $\gamma$ the angle subtended between ${\bf x}_1$ and ${\bf x}_2$ , and $P_l(x)$ a Legendre polynomial [50]. Moreover, the so-called addition theorem for spherical harmonics states that [67]

$$P_l(\cos\gamma) = \frac{4\pi}{2\,l+1}\sum_{m=-l,+l} Y_{l\,m}^{\,\ast}(\theta_1,\varphi_1)\,Y_{l\,m}(\theta_2,\varphi_2).$$ (9.60)

Here, ${\bf x}_1 = r_1\,(\sin\theta_1\,\cos\varphi_1,\,\sin\theta_1\,\sin\varphi_1,\,\cos\theta_1)$ , et cetera. However [see Equations (4.94) and (4.95)],

$$\oint d{\mit\Omega} \,Y_{l\,m}(\theta,\phi) = \sqrt{4\pi}\,\delta_{l\,0}\,\delta_{m\,0},$$ (9.61)

so, combining the previous four equations, we obtain

$$\frac{{\mit\Delta E}}{\vert E_0\vert} = 32\,Z\int_0^\infty d x_1\,x_1\left(\int_0^{x_1} d x_2\,x_2^{\,2... ...2\,(x_1+x_2)} +\int_{x_1}^\infty dx_2\,x_1\,x_2\,{\rm e}^{-2\,(x_1+x_2)}\right)$$

$$= \frac{5\,Z}{4}.$$ (9.62)

(See Exercise 10.) Here, $x_1=Z\,r_1/a_0$ and $x_2=Z\,r_2/a_0$ . Thus, our improved estimate for the ground-state energy of the helium atom is [113]

$$E = 2\,Z^{\,2}\,E_0 + {\mit\Delta} E = \left(2\,Z^{\,2}-\frac{5\,Z}{4}\right)E_0 = \left(8 - \frac{5}{2}\right)E_0 = \frac{11}{2}\,E_0$$

$$= -74.8\,{\rm eV},$$ (9.63)

because $Z=2$ and $E_0=-13.6\,{\rm eV}$ . This is much closer to the experimental value of $-78.98\,{\rm eV}$ [75] than our previous estimate, $-108.8\,{\rm eV}$ . [See Equation (9.57).]