The trial wavefunction: a linear combination of atomic orbitals

Physically, when the two protons are far apart, and the electron is close to one or the other proton, the ground state wavefunction of the system should resemble that of a $1s$ orbital of hydrogen centered on one of the protons. When the electron is proton p$_1$, its wavefunction should approximately be

$$\psi_0({\bf r}) = \left({1 \over \pi a_0^3}\right)^{1/2}e^{-... ..._0} \equiv \psi_1({\bf r}) = \langle {\bf r}\vert\psi_1\rangle$$

Similarly, when it is on proton p$_2$, it should be

$$\psi_0({\bf r}) = \left({1 \over \pi a_0^3}\right)^{1/2}e^{-... ..._0} \equiv \psi_2({\bf r}) = \langle {\bf r}\vert\psi_2\rangle$$

This suggests that we might construct a trial wavefunction from an arbitrary linear combination of these two atomic orbitals according to

$$\vert\psi\rangle = C_1\vert\psi_1\rangle + C_2 \vert\psi_2\rangle$$

$$\psi({\bf r};C_1,C_2) = C_1\psi_1({\bf r}) + C_2\psi_2({\bf r})$$

and treat the coefficients $C_1$ and $C_2$ as variational parameters. Such a trial wave function is called a linear combination of atomic orbitals or LCAO wave function. Notice that $\vert\psi_1\rangle$ and $\vert\psi_2\rangle$ are not orthogonal.

In accordance with the variational procecure, we construct the function

$$E(C_1,C_2) = {\langle \psi\vert H\vert\psi\rangle \over \langle \psi\vert\psi\rangle }$$

Then,

$$\langle \psi\vert\psi\rangle = \left[C_1^*\langle \psi_1\vert + C_2^*\langle \psi_2\vert\right] \left[C_1\vert\psi_1\rangle + C_2\vert\psi_2\rangle \right]$$

$$= \vert C_1\vert^2 + \vert C_2\vert^2 + C_1C_2^*\langle \psi_2\vert\psi_1\rangle + C_1^*C_2\langle \psi_1\vert\psi_2\rangle$$

Define the overlap between $\vert\psi_1\rangle$ and $\vert\psi_2\rangle$ by

$$S_{12} = \langle \psi_1\vert\psi_2\rangle$$

$$S_{21} = S_{12}^*$$

Then,

$$\langle \psi\vert\psi\rangle = \vert C_1\vert^2 + \vert C_2\vert^2 + C_1C_2^*S_{21} C_1^*C_2S_{12}$$

Similarly,

$$\langle \psi\vert H\vert\psi\rangle = \vert C_1\vert^2\langl... ...si_1\rangle +C_1^*C_2\langle \psi_1\vert H\vert\psi_2\rangle$$

Define the matrix elements of $H$ by

$$H_{ij} = \langle \psi_i\vert H\vert\psi_j\rangle$$

we find

$$\langle \psi\vert H\vert\psi\rangle = \vert C_1\vert^2\langl... ...t C_2\vert^2\langle H_{22} + C_1C_2^* H_{21} + C_1^*C_2 H_{12}$$

so that the energy becomes

$$E(C_1,C_2) = { \vert C_1\vert^2\langle H_{11} + \vert C_2\... ..._1\vert^2 + \vert C_2\vert^2 + C_1C_2^*S_{21} C_1^*C_2S_{12} }$$

Now we perform the variation $\partial E/\partial \langle \psi\vert$, which is equivalent to the conditions:

$${\partial E \over \partial C_1^*}=0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; {\partial E \over \partial C_2^*}=0$$

Defining the denominator as $D$

$$D = \vert C_1\vert^2 + \vert C_2\vert^2 + C_1C_2^*S_{21} C_1^*C_2S_{12}$$

these conditions yield two equations for $C_1$ and $C_2$:

$${\partial E \over \partial C_1^*} = {C_1 H_{11} + C_2 H_{12} \over D} - {E(C_1,C_2) \over D} \left(C_1 + C_2S_{12}\right) =0$$

$${\partial E \over \partial C_2^*} = {C_2 H_{22} + C_2 H_{21} \over D} - {E(C_1,C_2) \over D} \left(C_2 + C_1S_{21}\right) =0$$

or, since $D\neq 0$,

$$\left(H_{11}-E(C_1,C_2)\right)C_1 + \left(H_{12}-E(C_1,C_2)S_{12}\right)C_2 = 0$$

$$\left(H_{22}-E(C_1,C_2)\right)C_2 + \left(H_{21}-E(C_1,C_2)S_{21}\right)C_1 = 0$$

These may be written as a matrix equation:

$$\left(\matrix{H_{11} & H_{12} \cr H_{21} & H_{22}}\right) \l... ...{12} \cr S_{21} & 1}\right) \left(\matrix{C_1 \cr C_2}\right)$$

which is called a generalized eigenvalue equation. In matrix notation is becomes

$${\rm H}{\bf C} = E{\rm S}{\bf C}$$

where ${\rm H}$ and ${\rm S}$ are the Hamiltonian and overlap matrices, respectively.

We may, therefore, regard $E$ as an eigenvalue and solve the above eigenvalue equation. This requires that we solve the determinant:

$$\left\vert\matrix{H_{11}-E & H_{12}-ES_{12} \cr H_{21}-ES_{21} & H_{22}-E}\right\vert=0$$

Recognizing that $\langle {\bf r}\vert\psi_1\rangle$ and $\langle {\bf r}\vert\psi_2\rangle$ are both real, it is clear that $S_{12}=S_{21}\equiv S$. Similarly, by symmetry $H_{11}=H_{22}\equiv H_{11}$ and $H_{12}=H_{21}\equiv H_{12}$. Thus, the determinant simplifies to

$$\left\vert\matrix{H_{11}-E & H_{12}-ES \cr H_{12}-ES & H_{11}-E}\right\vert=0$$

which yields the condition

$$(H_{11}-E)^2 = (H_{12}-ES)^2$$

or

$$(H_{11}-E) = \pm (H_{12}-ES)$$

which yields two solutions. For the case of $+$, we have

$$E_+ = {H_{11}-H_{12} \over 1-S}$$

and for $-$,

$$E_- = {H_{11}+H_{12} \over 1+S}$$

The overlap, $S$, and Hamiltonian matrix elements will now be computed explicitly. For the overlap, $S$, the integral that needs to be performed is

$$S = \langle \psi_1\vert\psi_2\rangle = \int\;d{\bf r}\; \psi_1({\bf r})\psi_2({\bf r}) = {1 \over \pi a_0^3}\int\;d{\bf r}\; e^{-\left\vert{\bf r} + {R \o... ...right\vert/a_0}e^{-\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert/a_0}$$

$$= {1 \over \pi a_0^3}\int\;d{\bf r}\; e^{(\left\vert{\bf r} + {R \o... ...bf z}}\right\vert+\left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert)/a_0}$$

This is integral is most easily performed in the confocal elliptic coordinate system using

$$\mu = {\left\vert{\bf r} + {R \over 2}\hat{{\bf z}}\right\vert+ \left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert\over R}$$

$$\nu = {\left\vert{\bf r} + {R \over 2}\hat{{\bf z}}\right\vert- \left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert\over R}$$

$$J = {\partial (x,y,z) \over \partial (\mu,\nu,\phi)} = {R^3 \over 8}(\mu^2-\nu^2)$$

Note that the range of $\mu$ is $[1,\infty]$, while that of $\nu$ is $[-1,1]$. In addition, there is the integral over $\phi$ from $[0,2\pi]$. Thus, transforming the integral, one obtains

$$S = {1 \over \pi a_0^3} \int_1^{\infty}\;d\mu\;\int_{-1}^1\;... ... \int_0^{2\pi}\;d\phi\;{R^3 \over 8}(\mu^2-\nu^2)e^{-\mu \rho}$$

where $\rho=R/a_0$. The integral can be performed straightforwardly yielding

$$S = e^{-\rho}\left[1+\rho + {1 \over 3}\rho^2\right]$$

In order to evaluate the Hamiltonian matrix elements, let us look at the structure of $H_{11}$:

$$H = {P^2 \over 2m} - {e^2 \over r_1} - {e^2 \over r_2} + {e^2 \over R}$$

$$H_{11} = \langle \psi_1\vert H\vert\psi_1\rangle = \left<\psi_1\left\vert{... ...er 2m}-{e^2 \over r_1} - {e^2 \over r_2} +{e^2 \over R}\right\vert\psi_2\right>$$

Noting that $\vert\psi_1\rangle$ is an eigenvector of the operator $P^2/2m - e^2/r_1$ with eigenvalue $-e^2/2a_0$:

$$\left({P^2 \over 2m}-{e^2 \over r_1}\right)\vert\psi_1\rangle = -{e^2 \over 2a_0} \vert\psi_1\rangle$$

it can be seen that $H_{11}$ becomes

$$H_{11} = -{e^2 \over 2a_0} + {e^2 \over R} - \left<\psi_1\left\vert {e^2 \over r_2}\right\vert\psi_2\right>$$

$$H_{11} = -{e^2 \over 2a_0} + {e^2 \over R} - C$$

where the last term, $C$, is called the Coulomb integral and is given by

$$C = \left<\psi_1\left\vert {e^2 \over r_2}\right\vert\psi_2\right> = \int\;d{\bf r}\;\psi_1^2({\bf r}){e^2 \over r_2}$$

Explicitly, the Coulomb integral is

$$C = {e^2 \over \pi a_0^3}\int\;d{\bf r}\; e^{-2\left\vert{\b... ...\over \left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert}$$

which, again, is most easily evaluated using the confocal elliptic coordinates. Transforming into this coordinate system gives

$$C = {e^2 \over \pi a_0^3}{R^3 \over 8} \int_1^{\infty}\;d\mu\;\int_{-... ...nu\;\int_0^{2\pi}\;d\phi\; e^{-(\mu+\nu)\rho}{2 \over R(\mu-\nu)}(\mu^2 -\nu^2)$$

$$= {2e^2 \over a_0^3}{R^3 \over 8}{2 \over R} \int_1^{\infty}\;d\mu\;\int_{-1}^1\;d\nu\; e^{-(\mu+\nu)\rho}(\mu+\nu)$$

where the fact that $(\mu^2-\nu^2)=(\mu+\nu)(\mu-\nu)$ has been used. Thus, the integral can be performed straightforwardly, to yield:

$$H_{11} = -{e^2 \over 2a_0} + {e^2 \over R} - {e^2 \over 2a_0}{2 \over \rho}\left[1-e^{-2\rho}(1+\rho)\right]$$

Note that

$$\lim_{R\rightarrow\infty} H_{11}=-{e^2 \over 2a_0}$$

as expected.

The off-diagonal matrix element $H_{12}$ can be evaluated in a similar manner.

$$H_{12} = \left<\psi_1\left\vert{P^2 \over 2m}-{e^2 \over r_1} -{e^2 \over r_2}+{e^2 \over R}\right\vert\psi_2\right>$$

$$= \left<\psi_1\left\vert{P^2 \over 2m}-{e^2 \over r_1}\right\vert\p... ...t\psi_2\rangle - \left<\psi_1\left\vert{e^2 \over r_1} \right\vert\psi_2\right>$$

$$= -{e^2 \over 2a_0}\langle \psi_1\vert\psi_2\rangle + {e^2 \over R}... ...rt\psi_2\rangle - \left<\psi_1\left\vert{e^2 \over r_1}\right\vert\psi_2\right>$$

$$= \left(-{e^2 \over 2a_0} + {e^2 \over R}\right)S - \left<\psi_1\left\vert{e^2 \over r_1}\right\vert\psi_2\right>$$

Where the fact that $\vert\psi_2\rangle$ is an eigenvector of the operator $P^2/2m - e^2/r_2$ with eigenvalue $-e^2/2a_0$ has been used. The last term

$$\left<\psi_1\left\vert{e^2 \over r_1}\right\vert\psi_2\right... ...=\int\;d{\bf r}\;\psi_1({\bf r}){e^2 \over r_1}\psi_2({\bf r})$$

is called the exchange or resonance integral. Substituting in the atomic wave functions, the integral becomes

$$A = \left({e^2 \over \pi a_0^3}\right) \int\;d{\bf r}\;{1 \o... ... \left\vert{\bf r} - {R \over 2}\hat{{\bf z}}\right\vert)/a_0}$$

Transforming to confocal elliptic coordinates, the integral becomes

$$A = \left({e^2 \over \pi a_0^3}\right){R^3 \over 8} \int_1^{\infty}\;... ...1}^1\;d\nu\;\int_0^{2\pi}\;d\phi\; (\mu^2-\nu^2){2e^{-\mu\rho} \over (\mu+\nu)}$$

$$= 2\pi\left({e^2 \over \pi a_0^3}\right){R^3 \over 8}{2 \over R} \int_1^{\infty}\;d\mu\;\int_{-1}^1\;d\nu\; e^{-\mu\rho}(\mu-\nu)$$

which can be integrated straightforwardly. Thus, the off-diagonal matrix element becomes

$$H_{12} = \left(-{e^2 \over 2a_0} + {e^2 \over R}\right)S - {e^2 \over 2a_0}2e^{-\rho}(1+\rho)$$

Using these expressions, the energies

$$E_+ = {H_{11}-H_{12} \over 1-S} \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; E_- = {H_{11}+H_{12} \over 1+S}$$

can be determined. It is useful, however, to define $\Delta E_{\pm}$ relative to the energy at $R=\infty$, namely $-e^2/2a_0$. This will be given by

$$\Delta E_{\pm} = E_{\pm}+{e^2 \over 2a_0} = -{e^2 \over 2a_... ...er 1 \mp e^{-\rho}\left(1+\rho+{\rho^2 \over 3}\right)}\right]$$

A plot of $\Delta E_{\pm}(\rho)$ is sketched below.

図 2:

It can be seen that $\Delta E_+(\rho)>\Delta E_-(\rho)$ for all $\rho$. Also, $\Delta E_-(\rho)$ has a minimum at a particular value of $\rho$, which corresponds to the equilibrium bond length while $\Delta E_+(\rho)$ exhibits no such minimum. The location of the minimum and depth of the well will be the prediction of the equilibrium bond length and binding energy within this approximation, respectively. Before reporting these values, however, a few points are worth noting.

First, the presence of the minimum in $\Delta E_-$ is due primarily to the contribution of $H_{12}$, the off-diagonal matrix element of $H$. The reason for this is evident. In order for a chemical bond to form, there needs to be a significant overlap between the two atomic orbitals, which can only happen if the distance $R$ between the two protons is not too large.

Second, the fact that $\Delta E_+$ has no minimum means that in the state corresponding to this energy, the molecule will be likely to dissociate and not form a stable H$_2^+$ molecule.

The values obtained at this level of approximation from the location of the minimum of $\Delta E_-$ and the depth of the minimum are:

$${\rm bond\ length}\; = \;2.50 a_0$$

$${\rm binding energy}\; = \;1.76\;{\rm eV}$$

The exact values are $2.0 a_0$ and 2.76 eV, respectively. Therefore, the result is only qualitatively correct. We will explore the problem of how to improve the current approximation by adding more variational parameters to the trial wave function, however, let us first see what other physical insights can be gained from the simple LCAO picture.