Hydrogen Molecule Ion

The hydrogen molecule ion consists of an electron orbiting about two protons, and is the simplest imaginable molecule. Let us investigate whether this molecule possesses a bound state: that is, whether it possesses a ground-state whose energy is less than that of a ground-state hydrogen atom plus a free proton. In fact, according to the variational principle (see the previous section), we can prove that the $H_2^{\,+}$ ion possesses a bound state if we can find any trial wavefunction for which the total Hamiltonian of the system has an expectation value that is less than that of a ground-state hydrogen atom plus a free proton.

Suppose that the first and second protons lie at $+{\bf X}/2$ and $-{\bf X}/2$ , respectively. Let ${\bf x}$ be the position vector of the electron. The position vectors of the electron relative to the first and second protons are thus ${\bf x}_1 = {\bf x} - {\bf X}/2$ and ${\bf x}_2={\bf x}+{\bf X}/2$ , respectively. The Hamiltonian of the system is written

$$H = \frac{{\bf p}^{\,2}}{2\,m_e} -\frac{e^{\,2}}{4\pi\,\epsilon_0}\left(\frac{1}{r_1}+ \frac{1}{r_2}\right) + \frac{e^{\,2}}{4\pi\,\epsilon_0\,R},$$ (9.97)

where ${\bf p}$ is the electron momentum, $r_1=\vert{\bf x}_1\vert$ , $r_2=\vert{\bf x}_2\vert$ , and $R = \vert{\bf X}\vert$ . 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 Born-Oppenheimer approximation [13].

The Hamiltonian (9.98) is manifestly invariant under the transformation ${\bf X}\rightarrow -{\bf X}$ , which simply swaps the positions of the two identical protons. This transformation is equivalent to ${\bf x}_1\leftrightarrow {\bf x}_2$ . If $P_{12}$ is the operator that swaps the proton positions then it is clear that $P_{12}^{\,2}=1$ . (See Exercise 1.) Hence, the eigenvalues of $P_{12}$ are $\pm 1$ . Furthermore, the fact that the Hamiltonian is invariant under exchange of proton positions implies that $[P_{12}, H]=0$ . (See Section 9.2.) Thus, the eigenkets of the Hamiltonian are simultaneous eigenkets of $P_{12}$ . Now, it is easily shown that the eigenkets of $P_{12}$ corresponding to the eigenvalues $\pm 1$ are, respectively, even and odd under the transformation ${\bf x}_1\leftrightarrow {\bf x}_2$ . (See Exercise 2.) It follows that the ground-state wavefunction of the $H_2^{\,+}$ ion has the general form

$$\psi_\pm({\bf x}) = A\left[\psi_0({\bf x}_1) \pm \psi_0({\bf x}_2)\right],$$ (9.98)

where $A$ is a complex number.

Let us adopt

$$\psi_0({\bf x}) = \frac{1}{\sqrt{\pi}\,a_0^{\,3/2}}\,{\rm e}^{-r/a_0}$$ (9.99)

as our trial single-proton wavefunction, where $r=\vert{\bf x}\vert$ . Here, $\psi_0({\bf x})$ is a properly normalized hydrogen ground-state wavefunction, and $a_0$ 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

$$\int d^{\,3}{\bf x}\,\vert\psi_\pm\vert^{\,2} = 1.$$ (9.100)

Hence, from Equation (9.99), $A = I^{\,-1/2}$ , where

$$I = \int d^{\,3}{\bf x}\left[\vert\psi_0({\bf x}_1)\vert^{\,2} + ... ...psi_0({\bf x}_2)\vert^{\,2} \pm 2\,\psi_0({\bf x}_1)\,\psi_0({\bf x}_2)\right].$$ (9.101)

It follows that

$$I = 2\,(1\pm J),$$ (9.102)

where

$$J = \int d^{\,3}{\bf x}\,\psi_0({\bf x}_1)\,\psi_0({\bf x}_2).$$ (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 ${\bf X}= R\,{\bf e}_z$ . Let ${\bf x}= r\,(\sin\theta\,\cos\varphi,\, \sin\theta\,\sin\varphi,\, \cos\theta)$ be the position vector of the electron. It follows that $r_1=r$ and $r_2=(r^{\,2}+R^{\,2}-2\,r\,R\,\cos\theta)^{1/2}$ . Hence,

$$J = 2\int_0^\infty dx\,x^{\,2}\int_0^\pi d\theta\,\sin\theta\,\exp\left[-x-(x^{\,2}+y^{\,2}-2\,x\,y\,\cos\theta)^{1/2}\right],$$ (9.104)

where $x=r/a_0$ and $y=R/a_0$ . Here, we have already performed the trivial $\varphi$ integral. Let $z(\theta)=(x^{\,2}+y^{\,2}-2\,x\,y\,\cos\theta)^{1/2}$ . It follows that $d(z^{\,2})=2\,z\,dz =2\,x\,y\,\sin\theta\,d\theta$ , giving

$$\int_0^\pi d\theta\,\sin\theta\,{\rm e}^{\,-(x^{\,2}+y^{\,2}-2\,x\,y\,\cos\theta)^{1/2}} = \frac{1}{x\,y}\int_{\vert x-y\vert}^{x+y} dz\,z\,{\rm e}^{\,-z}$$ (9.105)

$$= - \frac{1}{x\,y}\left[{\rm e}^{\,-(x+y)}\,(1+x+y) - {\rm e}^{\,-\vert x-y\vert}\,(1+\vert x-y\vert)\right].$$

Thus,

$$J = - \frac{2}{y}\,{\rm e}^{\,-y}\int_0^y dx\,x\left[{\rm e}^{\,-2\,x}\,(1+y+x)- (1+y-x)\right]$$

$$\phantom{=}-\frac{2}{y}\int_y^\infty dx\,x\,{\rm e}^{\,-2\,x}\left[{\rm e}^{\,-y}\,(1+y+x)- {\rm e}^{\,y}\,(1-y+x)\right],$$ (9.106)

which evaluates to

$$J = {\rm e}^{\,-y}\left(1+y+\frac{y^{\,2}}{3}\right).$$ (9.107)

(See Exercise 20.)

Now, the Hamiltonian of the electron is written

$$H_e = \frac{{\bf p}^{\,2}}{2\,m_e} - \frac{e^{\,2}}{4\pi\,\epsilon_0}\left(\frac{1}{r_1}+\frac{1}{r_2}\right).$$ (9.108)

Note, however, that

$$\left(\frac{{\bf p}^{\,2}}{2\,m_e} - \frac{e^{\,2}}{4\pi\,\epsilon_0\,r_{1,2}}\right)\psi_0({\bf x}_{1,2}) = E_0\,\psi_0({\bf x}_{1,2}),$$ (9.109)

where $E_0$ is the hydrogen ground-state energy, because the $\psi_0({\bf x}_{1,2})$ are hydrogen ground-state wavefunctions. It follows that

$$H_e\,\psi_\pm =A\left[\frac{{\bf p}^{\,2}}{2\,m_e} - \frac{e^{\,2}}{4\pi\,\epsi... ...frac{1}{r_2}\right)\right] \left[\psi_0({\bf x}_1) \pm \psi_0({\bf x}_2)\right]$$

$$= E_0\,\psi_\pm - A\,\left(\frac{e^{\,2}}{4\pi\,\epsilon_0}\right)\left[ \frac{\psi_0({\bf x}_1)}{r_2}\pm \frac{\psi_0({\bf x}_2)}{r_1}\right].$$ (9.110)

Hence,

$$\langle H_e\rangle \equiv \langle \psi_\pm\vert\,H_e\,\vert\psi_\pm\rangle = E_0 + 4\,A^2\,(D\pm E)\,E_0,$$ (9.111)

where

$$D =\left\langle \psi_0({\bf x}_1)\left\vert\frac{a_0}{r_2}\right\vert\psi_0({\bf x}_1)\right\rangle,$$ (9.112)

$$E = \left\langle \psi_0({\bf x}_1)\left\vert\frac{a_0}{r_1}\right\vert\psi_0({\bf x}_2)\right\rangle.$$ (9.113)

Here, use has been made of the fact that $e^{\,2}/4\pi\,\epsilon_0\,a_0 = -2\,E_0$ , as well as the fact that the Hamiltonian is invariant under the transformation ${\bf x}_1\leftrightarrow {\bf x}_2$ .

Now,

$$D = 2\int_0^\infty dx\,x^{\,2}\int_0^\pi d\theta\,\sin\theta\, \frac{{\rm e}^{\,-2\,x}}{(x^{\,2}+y^{\,2}-2\,x\,y\,\cos\theta)^{1/2}},$$ (9.114)

where $x=r/a_0$ and $y=R/a_0$ , which reduces to

$$D =\frac{4}{y}\,\int_0^y dx\,x^{\,2}\,{\rm e}^{\,-2\,x} + 4\int_y^\infty dx\,x\,{\rm e}^{\,-2\,x},$$ (9.115)

giving

$$D = \frac{1}{y} \left[ 1-(1+y)\,{\rm e}^{\,-2\,y}\right].$$ (9.116)

(See Exercise 21.) Furthermore,

$$E = 2\int_0^\infty dx\,x\int_0^\pi d\theta\,\sin\theta\,\exp\left[-x-(x^{\,2}+y^{\,2}-2\,x\,y\,\cos\theta)^{1/2}\right],$$ (9.117)

which reduces to

$$E = - \frac{2}{y}\,{\rm e}^{\,-y}\int_0^y dx \left[{\rm e}^{\,-2\,x}\,(1+y+x)- (1+y-x)\right]$$

$$\phantom{=}-\frac{2}{y}\int_y^\infty dx\,{\rm e}^{-2\,x}\left[{\rm e}^{\,-y}\,(1+y+x)- {\rm e}^{\,y}\,(1-y+x)\right],$$ (9.118)

yielding

$$E = (1+y)\,{\rm e}^{\,-y}.$$ (9.119)

(See Exercise 22.)

Figure: The functions $F_+(y)$ (solid curve) and $F_-(y)$ (short-dashed curve), where $y=R/a_0$ . The long-dash-dotted curve shows the improved $F_+(y)$ curve obtained from Exercise 23. The dots indicate the minimum values of $F_+(y)$ .

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

$$\langle H_e\rangle = \left[1+ 2\,\frac{(D\pm E)}{(1\pm J)}\right] E_0,$$ (9.120)

where $J$ , $D$ , and $E$ are specified as functions of $y=R/a_0$ 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,

$$E_{\rm total} = \langle H_e\rangle + \frac{e^{\,2}}{4\pi\,\epsilon_0\,R} = \langle H_e\rangle - \frac{2}{y}\,E_0,$$ (9.121)

because $E_0=-e^{\,2}/(8\pi\,\epsilon_0\,a_0)$ . Hence, we can write

$$E_{\rm total} = - F_\pm(R/a_0)\,E_0,$$ (9.122)

where

$$F_\pm(y) = -1 + \frac{2}{y}\left[\frac{(1+y)\,{\rm e}^{\,-2\,y}\pm(1-2\,y^{\,2}/3) \,{\rm e}^{\,-y}}{1\pm (1+y+y^{\,2}/3)\,{\rm e}^{\,-y}}\right].$$ (9.123)

The functions $F_+(y)$ and $F_-(y)$ are plotted in Fig. 9.1. Recall that, in order for the $H_2^{\,+}$ ion possess a bound state, it must have a lower energy than a hydrogen atom and a free proton. In other words, $E_{\rm total}< E_0$ . It follows, from Equation (9.123), that a bound state corresponds to $F_\pm < -1$ . It is clear, from the figure, that the even trial wavefunction, $\psi_+$ , possesses a bound state, whereas the odd trial wavefunction, $\psi_-$ , 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 $H_2^{\,+}$ ion is defined as minus the difference between its energy and that of a hydrogen atom plus a free proton: that is,

$$E_{\rm bind} = E_0- E_{\rm total}= - (F_+ +1)\,\vert E_0\vert.$$ (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 $F_+$ .) This maximum occurs when $y= y_0=2.493$ and $F_+=F_+(y_0)= -1.130$ . Thus, our estimates for the equilibrium separation between the two protons, and the binding energy of the molecule, are $R_0 = 2.493\,a_0 = 1.32\times 10^{-10}\,{\rm m}$ and $E_{\rm bind} = 0.130 \,\vert E_0\vert = 1.77$ eV, respectively. The experimentally determined values are $R_0=1.06\times 10^{-10}$ m, and $E_{\rm bind}=2.8$ 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 $H_2^{\,+}$ 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 $R$ , as moving in an electric potential $V(R)=F_+(R/a_0)\,\vert E_0\vert$ due to the combination of their electrostatic repulsion and the binding action of the consistent electron. Moreover, close to the equilibrium separation, $R_0$ , we have $V(R)\simeq V(R_0) + \left[(1/2)\,F_0''(y_0)/a_0^{\,2}\right]x^{\,2}+ {\cal O}(x^{\,3})$ , where $x=R-R_0$ . Thus, in the center of mass frame [50], the protons' Hamiltonian takes the form

$$H_p\simeq \frac{{\bf p}_x^{\,2}}{2\,\mu} + \frac{1}{2}\,\mu\,\omega^{\,2}\,x^{\,2},$$ (9.125)

where ${\bf p}_x$ is the momentum conjugate to $x$ , $\mu=2\,m_p$ the reduced mass [50], $m_p$ the proton mass, $\omega^{\,2} = F_0''(y_0)/(a_0^{\,2}\,\mu)$ , and we have neglected an unimportant constant term. Here, $'$ denotes $d/dy$ . 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

$$E_n = \left(n+\frac{1}{2}\right)\hbar\,\omega,$$ (9.126)

where $n$ is a non-negative integer. In particular, there is a non-zero lowest oscillation energy--the so-called zero-point energy [40],

$$E_{\rm zero} = \frac{1}{2}\,\hbar\,\omega= \sqrt{F_+''(y_0)}\left(\frac{m_e}{m_p}\right)^{1/2}\vert E_0\vert=0.11\,{\rm eV},$$ (9.127)

that must be subtracted from the previously determined electric binding energy of $1.77\,{\rm eV}$ to give the true binding energy. In calculating, $E_{\rm zero}$ we have made use of the numerically determined value $F_0''(y_0)= 0.1257$ [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

$$H_{\rm rot} = \frac{L^{\,2}}{2\,I_\perp},$$ (9.128)

where ${\bf L}$ is the angular momentum of rotation, and $I_\perp=m_p\,R_0^{\,2}/3$ 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

$$E_l = l\,(l+1)\,\frac{\hbar^{\,2}}{2\,I_\perp},$$ (9.129)

where the eigenvalues of $L^{\,2}$ are $l\,(l+1)\,\hbar^{\,2}$ , and $l$ is a non-negative integer. It follows that

$$E_l = l\,(l+1)\,\frac{3}{y_0^{\,2}}\,\frac{m_e}{m_p}\,\vert E_0\vert=l\,(l+1)\,E_r,$$ (9.130)

where

$$E_r= \frac{3}{y_0^{\,2}}\,\frac{m_e}{m_p}\,\vert E_0\vert=3.58\times 10^{-3}\,{\rm eV}.$$ (9.131)

Note that our estimate for the electric binding energy of the hydrogen molecule ion, $1.77\,{\rm eV}$ , is significantly greater than our estimate for the zero-point oscillation energy, $0.11\,{\rm eV}$ , which, in turn, is very much greater than a typical rotational energy, $\sim 7\times 10^{-3}\,{\rm eV}$ . 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.