Hydrogen Molecule Ion

Suppose that the first and second protons lie at and , respectively. Let be the position vector of the electron. The position vectors of the electron relative to the first and second protons are thus and , respectively. The Hamiltonian of the system is written

where is the electron momentum, , , and . Here, we are treating the protons as essentially stationary, which is a reasonable approximation because the electron's motion is much more rapid than that of the protons. Of course, this is the case because the electron mass is very much less than the proton mass. Incidentally, the neglect of nuclear motion when calculating the electronic structure of a molecule is known as the

The Hamiltonian (9.98) is manifestly invariant under the transformation , which simply swaps the positions of the two identical protons. This transformation is equivalent to . If is the operator that swaps the proton positions then it is clear that . (See Exercise 1.) Hence, the eigenvalues of are . Furthermore, the fact that the Hamiltonian is invariant under exchange of proton positions implies that . (See Section 9.2.) Thus, the eigenkets of the Hamiltonian are simultaneous eigenkets of . Now, it is easily shown that the eigenkets of corresponding to the eigenvalues are, respectively, even and odd under the transformation . (See Exercise 2.) It follows that the ground-state wavefunction of the ion has the general form

where is a complex number.

Let us adopt

as our trial single-proton wavefunction, where . Here, is a properly normalized hydrogen ground-state wavefunction, and is the Bohr radius. Thus, our trial molecular wavefunction, which is specified in Equations (9.99) and (9.100), is simply a linear combination of hydrogen ground-state wavefunctions centered on each proton [18].

Our first task is to normalize the trial wavefunction. We require that

(9.100) |

Hence, from Equation (9.99), , where

(9.101) |

It follows that

(9.102) |

where

(9.103) |

Without loss of generality, we can perform the previous integral using a modified coordinate system in which the first proton lies at the origin, and the second at . Let be the position vector of the electron. It follows that and . Hence,

where and . Here, we have already performed the trivial integral. Let . It follows that , giving

(9.105) | ||

Thus,

which evaluates to

(See Exercise 20.)

Now, the Hamiltonian of the electron is written

(9.108) |

Note, however, that

(9.109) |

where is the hydrogen ground-state energy, because the are hydrogen ground-state wavefunctions. It follows that

(9.110) |

Hence,

(9.111) |

where

(9.112) | ||

(9.113) |

Here, use has been made of the fact that , as well as the fact that the Hamiltonian is invariant under the transformation .

Now,

where and , which reduces to

giving

(See Exercise 21.) Furthermore,

which reduces to

yielding

(See Exercise 22.)

Our final expression for the expectation value of the electron Hamiltonian is

(9.120) |

where , , and are specified as functions of in Equations (9.108), (9.117), and (9.120), respectively. In order to obtain the total energy of the molecule, we must add the potential energy of the two protons to this expectation value. Thus,

(9.121) |

because . Hence, we can write

where

The functions and are plotted in Fig. 9.1. Recall that, in order for the ion possess a bound state, it must have a lower energy than a hydrogen atom and a free proton. In other words, . It follows, from Equation (9.123), that a bound state corresponds to . It is clear, from the figure, that the even trial wavefunction, , possesses a bound state, whereas the odd trial wavefunction, , does not. [See Equation (9.99).] This is hardly surprising, because the even wavefunction maximizes the electron probability density between the two protons, thereby reducing their mutual electrostatic repulsion. On the other hand, the odd wavefunction does exactly the opposite. The binding energy of the ion is defined as minus the difference between its energy and that of a hydrogen atom plus a free proton: that is,

(9.124) |

According to the variational principle, the binding energy is greater than, or equal to, the maximum binding energy that can be inferred from Figure 9.1. (A maximum in the binding energy corresponds to a minimum of .) This maximum occurs when and . Thus, our estimates for the equilibrium separation between the two protons, and the binding energy of the molecule, are and eV, respectively. The experimentally determined values are m, and eV, respectively [53]. Clearly, the previous estimates are not particularly accurate. However, our calculation does establish, beyond any doubt, the existence of a bound state of the ion, which is all that we set out to achieve. (See Exercise 23 for a somewhat improved calculation.)

We can think of the two constituent protons of the hydrogen molecule ion, whose separation is , as moving in an electric potential due to the combination of their electrostatic repulsion and the binding action of the consistent electron. Moreover, close to the equilibrium separation, , we have , where . Thus, in the center of mass frame [50], the protons' Hamiltonian takes the form

(9.125) |

where is the momentum conjugate to , the reduced mass [50], the proton mass, , and we have neglected an unimportant constant term. Here, denotes . It can be seen, by comparison with Exercise 3, that the previous Hamiltonian is identical to that of a harmonic oscillator. Thus, the allowed radial oscillation energies of the molecule are

(9.126) |

where is a non-negative integer. In particular, there is a non-zero lowest oscillation energy--the so-called

(9.127) |

that must be subtracted from the previously determined electric binding energy of to give the true binding energy. In calculating, we have made use of the numerically determined value [derived from Equation (9.124.)]

Of course, the hydrogen molecule ion can rotate, as well as vibrate. According to Exercise 7, the rotational component of the molecular Hamiltonian can be written

(9.128) |

where is the angular momentum of rotation, and is the molecular moment of inertia [about a perpendicular (to the line joining the two protons) axis passing through the center of mass]. In calculating the moment of inertia, we have neglected the electron mass, and have treated the protons as point particles. The analysis of Exercise 7 reveals that the rotational energy levels are

(9.129) |

where the eigenvalues of are , and is a non-negative integer. It follows that

(9.130) |

where

(9.131) |

Note that our estimate for the electric binding energy of the hydrogen molecule ion, , is significantly greater than our estimate for the zero-point oscillation energy, , which, in turn, is very much greater than a typical rotational energy, . This separation in energy scales lies at the heart of the previously mentioned Born-Oppenheimer approximation [13], according to which the electric, vibrational, and rotational energy levels of molecules can all be calculated independently of one another.