next up previous
Next: Energies in the LCAO Up: lecture_14 Previous: lecture_14

Linear combination of atomic orbitals

Linear combination of atomic orbitals (LCAO) is a simple method of quantum chemistry that yields a qualitative picture of the molecular orbitals (MOs) in a molecule. Let us consider H$_2^+$ again. The approximation embodied in the LCAO approach is based on the notion that when the two protons are very far apart, the electron in its ground state will be a $1s$ orbital of one of the protons. Of course, we do not know which one, so we end up with a Schrödinger cat-like state in which it has some probability to be on one or the other.

As with the HF method, we propose a guess of the true wave function for the electron

\psi_g({\bf r}) = C_A \psi_{1s}^{\rm A}({\bf r}) + C_B\psi_{1s}^{\rm B}({\bf r})

where $\psi_{1s}^{\rm A}({\bf r}) = \psi_{1s}({\bf r}-{\bf R}_A)$ is a $1s$ hydrogen orbital centered on proton A and $\psi_{1s}^{\rm B}({\bf r}) = \psi_{1s}({\bf r}-{\bf R}_B)$ is a $1s$ hydrogen orbital centered on proton B. Recall $\psi_{1s}({\bf r}) = \psi_{100}(r,\phi,\theta)$. The positions ${\bf R}_A$ and ${\bf R}_B$ are given simply by the vectors

{\bf R}_A = (0,0,R/2)\;\;\;\;\;\;\;\;\;\;{\bf R}_B = (0,0,-R/2)

The explicit forms of $\psi_{1s}^{\rm A}({\bf r})$ and $\psi_{1s}^{\rm B}({\bf r})$ are
$\displaystyle \psi_{1s}^{\rm A}({\bf r})$ $\textstyle =$ $\displaystyle {1 \over (\pi a_0^3)^{1/2}}
e^{-\vert{\bf r}-{\bf R}_A\vert/a_0}$  
$\displaystyle \psi_{1s}^{\rm B}({\bf r})$ $\textstyle =$ $\displaystyle {1 \over (\pi a_0^3)^{1/2}}
e^{-\vert{\bf r}-{\bf R}_B\vert/a_0}$  

Now, unlike the HF approach, in which we try to optimize the shape of the orbitals themselves, in the LCAO approach, the shape of the $\psi_{1s}$ orbital is already given. What we try to optimize here are the coefficients $C_A$ and $C_B$ that determine the amplitude for the electron to be found on proton A or proton B.

The guess wave function $\psi_g({\bf r})$ is not normalized as we have written down. Thus, our guess of the ground-state energy is given by

E_g = {\int \psi_g \hat{H}_{\rm elec}\psi_g dV \over \int \psi_g^2 dV}

where $dV$ is the electron's volume element, and $\hat{H}_{\rm elec}$ is the electronic Hamiltonian (minus the nuclear-nuclear repulsion $ke^2/R$:

\hat{H}_{\rm elec} = \hat{K}_e - ke^2\left({1 \over \vert{\b...
...{\bf R}_A\vert}
+ {1 \over \vert{\bf r}-{\bf R}_B\vert}\right)

(we will account for the nuclear-nuclear repulsion later when we consider the energies). Consider the denominator first:

\int \psi_g^2 dV = C_A^2 \int \psi_{1s}^2 ({\bf r}-{\bf R}_A... \psi_{1s}({\bf r}-{\bf R}_A)\psi_{1s}({\bf r}-{\bf R}_B)dV

Now, the $1s$ wave function of hydrogen is normalized so

\int \psi_{1s}^2 ({\bf r}-{\bf R}_A) = \int \psi_{1s}^2({\bf r}-{\bf R}_B) = 1

In the cross term, however, the integral

\int \psi_{1s}({\bf r}-{\bf R}_A)\psi_{1s}({\bf r}-{\bf R}_B)dV

is not 1 because the orbitals are centered on different protons (it is only one if the two protons sit right on top of each other, which is not possible). It is also not 0 unless the protons are very far apart. We can calculate the integral analytically, however, it is not that important to do so since there is no dependence on $C_A$ or $C_B$. Let us just denote this integral as $S$. We know that $0 \leq S \leq 1$ and this is good enough for now. Thus, the denominator is simply

\int \psi_g^2 dV = C_A^2 + C_B^2 + 2SC_A C_B

As for the numerator

\int \psi_g\hat{H}_{\rm elec}\psi_g dV = C_A^2 H_{AA} + C_B^2 H_{BB} + 2C_A C_B H_{AB}

where we have defined a bunch of integrals I'm too lazy to do as
$\displaystyle H_{AA}$ $\textstyle =$ $\displaystyle \int \psi_{1s}^{\rm A}({\bf r})\hat{H}_{\rm elec}\psi_{1s}^{\rm A}({\bf r})dV$  
$\displaystyle H_{BB}$ $\textstyle =$ $\displaystyle \int \psi_{1s}^{\rm B}({\bf r})\hat{H}_{\rm elec}\psi_{1s}^{\rm B}({\bf r})dV$  
$\displaystyle H_{AB}$ $\textstyle =$ $\displaystyle \int \psi_{1s}^{\rm A}({\bf r})\hat{H}_{\rm elec}\psi_{1s}^{\rm B}({\bf r})dV$  

Again, these are integrals we can do, but it is not that important, so we will just keep the shorthand notation. Note that since the two nuclei are the same (they are both protons), we expect $H_{AA} = H_{BB}$. Since these are equal, we will just call them both $H_{AA}$. Hence, the guess energy becomes

E_g = {H_{AA}\left(C_A^2 + C_B^2\right) + 2H_{AB}C_A C_B \over C_A^2 + C_B^2 + 2SC_A C_B}

which is just a ratio of two simple polynomials. Since we know that $E_g > E_0$, where $E_0$ is the true ground-state energy, we can minimize $E_g$ with respect to the two coefficients $C_A$ and $C_B$. Thus, we need to take two derivatives and set them both to 0:

{dE_g \over dC_A} = 0\;\;\;\;\;\;\;\;\;\;{dE_g \over dC_B} = 0

Defining the denominator simply as $D$, where $D = C_A^2 + C_B^2 + 2SC_A C_B$, the two derivatives are
$\displaystyle {dE_g \over dC_A}$ $\textstyle =$ $\displaystyle {2C_A H_{AA} + 2C_B H_{AB} \over D}
- {H_{AA}\left(C_A^2 + C_B^2\right) + 2C_A C_B H_{AB} \over D^2}(2C_A + 2SC_B) = 0$  
$\displaystyle {dE_g \over dC_B}$ $\textstyle =$ $\displaystyle {2C_B H_{AA} + 2C_A H_{AB} \over D}
- {H_{AA}\left(C_A^2 + C_B^2\right) + 2C_A C_B H_{AB} \over D^2}(2C_B + 2SC_A) = 0$  

Thus, we have two algebraic equations in two unknowns $C_A$ and $C_B$. In fact, these will not determine $C_A$ and $C_B$ uniquely because they are redundant. However, we also have the normalization of $\psi_g$ as a third condition, so we have enough information to determine the coefficients absolutely.

The equations can be solved as follows: First we write them as

$\displaystyle {2C_A H_{AA} + 2C_B H_{AB} \over D}$ $\textstyle =$ $\displaystyle {H_{AA}\left(C_A^2 + C_B^2\right) + 2C_A C_B H_{AB} \over D^2}(2C_A + 2SC_B)$  
$\displaystyle {2C_B H_{AA} + 2C_A H_{AB} \over D}$ $\textstyle =$ $\displaystyle {H_{AA}\left(C_A^2 + C_B^2\right) + 2C_A C_B H_{AB} \over D^2}(2C_B + 2SC_A)$  

We now divide one equation by the other, which yields

{C_A H_{AA} + C_B H_{AB} \over C_B H_{AA} + C_A H_{AB}}
= {C_A + S C_B \over C_B + S C_A}

Since we cannot solve for $C_A$ and $C_B$ independently, we choose, instead, to solve for the ratio $r = C_A/C_B$. Dividing both sides of the above by equation by $C_B$ yields

{r H_{AA} + H_{AB} \over H_{AA} + r H_{AB}}
= {r+S \over 1+rS}

To solve for $r$, we cross multiply:

\left(r H_{AA} + H_{AB}\right)\left(1+rS\right)
= \left(H_{AA} + r H_{AB}\right)\left(r+S\right)

Multiplying this out, we obtain
$\displaystyle rH_{AA} + r^2SH_{AA} + H_{AB} + rS H_{AB}$ $\textstyle =$ $\displaystyle rH_{AA} + r^2H_{AB} + SH_{AA} + rSH_{AB}$  
$\displaystyle r^2\left(SH_{AA} - H_{AB}\right)$ $\textstyle =$ $\displaystyle SH_{AA} - H_{AB}$  
$\displaystyle r^2$ $\textstyle =$ $\displaystyle 1$  

which implies $r = \pm 1$.

Thus, we have two solutions: $C_A = C_B$ or $C_A = -C_B$. So, to simplify the notation, Let $C_A = C$ (drop the A subscript). We, therefore, have two possible guess wave functions:

$\displaystyle \psi_+({\bf r})$ $\textstyle =$ $\displaystyle C\left[\psi_{1s}^{\rm A}({\bf r}) + \psi_{1s}^{\rm B}({\bf r})\right]$  
$\displaystyle \psi_-({\bf r})$ $\textstyle =$ $\displaystyle C\left[\psi_{1s}^{\rm A}({\bf r}) - \psi_{1s}^{\rm B}({\bf r})\right]$  

The constant $C$ can be determined for each guess wave function by imposing the normalization condition. For $\psi_+$, we have
$\displaystyle \int \psi_+^2({\bf r})dV$ $\textstyle =$ $\displaystyle C^2 \left[\int \psi_{1s}^2({\bf r}-{\bf R}_A)dV + \int \psi_{1s}^...
...}_B)dV +
2\psi_{1s}({\bf r}-{\bf R}_A)\psi_{1s}({\bf r}-{\bf R}_B)dV\right] = 1$  
    $\displaystyle C^2 \left[1+1+2S\right] = 1$  
    $\displaystyle C = {1 \over \sqrt{2(1+S)}}$  

Similarly, one can show that for $\psi_-$, $C = 1\sqrt{2(1-S)}$.

Let us now see how the two approximate solutions lead to bonding and antibonding orbitals. Consider first $\psi_+({\bf r})$. We will add these two wave functions pictorally, keeping in mind that $\psi_{1s}$ is a spherically symmetric orbital with an exponential decay. Thus,

$\displaystyle \psi_{1s}^{\rm A}({\bf r})$ $\textstyle \propto$ $\displaystyle e^{-\vert{\bf r}-{\bf R}_A\vert/a_0}$  
$\displaystyle \psi_{1s}^{\rm B}({\bf r})$ $\textstyle \propto$ $\displaystyle e^{-\vert{\bf r}-{\bf R}_B\vert/a_0}$  

The figure below shows the orbitals that result from the addition or subtraction for $\psi_+$ or $\psi_-$.
Figure: LCAO approximation to the ground ($1\sigma _g$) and first excited ($1\sigma _u^*$) orbitals in H$_2^+$.
In the bottom left panel, we show the two $1s$ orbitals centered on protons A and B in $\psi_+$. In the center on the left, the result of adding them together is shown. It should be clear from the figure that $\psi_+$ is a bonding orbital that is an approximation to the true $1\sigma _g$ orbital that is the true ground state. This approximation is often denoted $\sigma_{g1s}$. On the other hand, the orbital $\psi_-$ is the difference between $\psi_{1s}({\bf r}-{\bf r}_A)$ and $\psi_{1s}({\bf r}-{\bf r}_B)$, which is depicted in the top left panel. This orbital has a node between the two nuclei and the amplitude between the two nuclei is generally low. This is clearly an anti-bonding orbital corresponding to $1\sigma _u^*$. This approximate orbital is often denoted $\sigma_{u1s}^*$. Since this orbital has a node, we expect that its energy will be higher than the approximate ground-state energy $\sigma_{g1s}$.

Concerning the energies, the figure below shows the two lowest energies curves as functions of $R$, comparing the results from the exact solution for H$_2^+$ and the LCAO approximation. Note, in particular, that, in accordance with the variational principle, the exact energy curve for the ground state is always lower than the LCAO approximation.


next up previous
Next: Energies in the LCAO Up: lecture_14 Previous: lecture_14
Mark E. Tuckerman 2011-11-21