\question
{Shotaro Maruyama}       
{2/3/12}          
{shotaro.maruyama@cmpt.phys.tohoku.ac.jp}  
{2-7}   
    {Solve the Schr\"{o}dinger equation for the H molecule for the case that $s=\left\langle \Psi_1|\Psi_2 \right\rangle$ is not zero. Obtain both eigenvalues and wave functions.}
{\\
We consider the Hamiltonian $H=H_{0}+V$ where $H_{0}$ and $V$ mean, respectively,  unperturbed Hamiltonian of isolated H atoms and interactions between one hydrogen atom and another hydrogen atom. $H_0$ satisfy the following equation.
\begin{eqnarray}
H_0|\Psi_i\rangle=E_s|\Psi_i\rangle\,(i=1,2),
\end{eqnarray}
where $E_s$ and $|\Psi_i \rangle$ are eigenvalue and eigen vector of each H atom, respectively. We express the eigenstate of the molecule by taking linear combination of atomic orbitals as follows:
\begin{eqnarray}
|\Psi \rangle=C_1|\Psi_1 \rangle+C_2|\Psi_2 \rangle,
\end{eqnarray}
where $C_1$ and $C_2$ are coefficients to be solved. The Schr\"{o}dinger equation\, $H|\Psi\rangle=E|\Psi \rangle$ gives
\begin{eqnarray}
C_1H|\Psi_1\rangle+C_2H|\Psi_2\rangle=E(C_1|\Psi_1 \rangle+C_2|\Psi_2 \rangle).
\end{eqnarray}
Operating $\langle \Psi_1|$ from left side in Eq.\,(2-7-3), we get 
\begin{eqnarray}
C_1E_s^{'}+C_2E_sS+C_2V_0&=&C_1E+C_2ES\nonumber\\
(E_s^{'}-E)C_1+(E_sS+V_0-ES)C_2&=&0,
\end{eqnarray}
where $E_s^{'}=\left\langle\Psi_i|H|\Psi_i\right\rangle$, 
$V_0= \left\langle\Psi_1|V|\Psi_2\right\rangle= \left\langle\Psi_2|V|\Psi_1\right\rangle$, and $S=\left\langle\Psi_1|\Psi_2\right\rangle = \left\langle\Psi_2|\Psi_1\right\rangle$. Similarly, we get the following equation by operating $\langle\Psi_2|$ from left side in Eq.\,(2-7-3):
\begin{eqnarray}
(E_sS+V_0-ES)C_1+(E_s^{'}-E)C_2=0.
\end{eqnarray}
In matrix form, Eqs.\,(2-7-5) and (2-7-6) can be written as:
\begin{eqnarray}
\left(\begin{array}{ccc}
E_s-E&\epsilon -ES \\ \epsilon -ES&E_s^{'}-E 
 \end{array}\right)
\left(\begin{array}{ccc}
C_1 \\ C_2
\end{array}\right)
=\left(\begin{array}{ccc}0\\0 \end{array}\right),
\end{eqnarray}
where $\epsilon =E_sS+V_0$. In order to get not trivial solution for $\left(\begin{array}{ccc}
C_1 \\ C_2
\end{array}\right)$, we get the following secular equation.
\begin{eqnarray}
\det
\left(\begin{array}{ccc}
E_s^{'}-E&\epsilon-ES \\ \epsilon-ES&E_s^{'}-E
 \end{array}\right)=0.
\end{eqnarray}

We obtain $E$ by using Eq.\,(2-7-7) as follows:
\begin{eqnarray}
(E_s^{'}-E)^2-(\epsilon-ES)^2&=&0\nonumber\\
(1-S^2)E^2+2(\epsilon-E_s^{'})E+E_s^{'2}-\epsilon^{2}&=&0\nonumber\\
(1-S^2)\left\{E-\frac{E_{s}^{'}+\epsilon}{1+S}\right\}\left\{E-\frac{E_{s}^{'}-\epsilon}{1-S}\right\}&=&0\,,\nonumber
\end{eqnarray}
which gives
\begin{eqnarray}
E_{\pm} &=& \frac{E_s^{'}\pm \epsilon}{1\pm S}.
\end{eqnarray}
In the case of $E=E_{+}$, Eq.\,(2-7-6) is written as:
\begin{eqnarray}
\left(
    \begin{array}{ccc}
E_s^{'}-E_{+}&\epsilon-E_{+}S \\ \epsilon-E_{+}S&E_s^{'}-E_{+}
 \end{array}
  \right)
\left(
    \begin{array}{c}
C_1 \\ C_2
 \end{array}
  \right)
=\left(\begin{array}{ccc}
0 \\ 0
\end{array}\right).
\end{eqnarray}
Eq.\,(2-7-9) gives 
\begin{eqnarray}
C_1:C_2=\epsilon-E_{+}S:E_{+}-E_s^{'}=1:1.
\end{eqnarray}

Normalizing $|\Psi \rangle$, we determine the coefficients of Eq.\,(2-7-2) as follows:
\begin{eqnarray}
1=\left\langle \Psi|\Psi \right\rangle\nonumber&=&|C_1|^2(\left\langle \Psi_1|\Psi_1 \right\rangle+\left\langle \Psi_1|\Psi_2 \right\rangle+\left\langle \Psi_2|\Psi_1 \right\rangle+\left\langle \Psi_2|\Psi_2 \right\rangle)\nonumber \\
&=&
2|C_1|^2(1+S).
\end{eqnarray}
Using Eq.\,(2-7-11), $C_1$ and $C_2$ are written as:
\begin{eqnarray}
C_1=C_2=\frac{1}{\sqrt{2(1+S)}}.
\end{eqnarray}

Thus we get the eigenstate of the molecule:
\begin{eqnarray}
|\Psi_{+} \rangle=\frac{1}{\sqrt{2(1+S)}}(|\Psi_1 \rangle+|\Psi_2 \rangle).
\end{eqnarray}
Similarly, we get the eigenstate of the molecule for the case of $E=E_{-}$ as follows:
\begin{eqnarray}
|\Psi_{-} \rangle= \frac{1}{\sqrt{2(1-S)}} (|\Psi_1 \rangle-|\Psi_2 \rangle).
\end{eqnarray}
$|\Psi_{+} \rangle and |\Psi_{-} \rangle$ are, respectively, bonding and anti-bonding states.
}