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Matrix Version

We can use matrix notation to re-do the problem above. The time-dependent Schrödinger equation in the original $\left\{ \Phi_1({\bf r}),
\Phi_2({\bf r}) \right\}$ basis becomes
$\displaystyle i \hbar \left[ \begin{array}{c} {\dot c}_1 \\  {\dot c}_2 \end{ar...
...2} \end{array} \right]
\left[ \begin{array}{c} c_1 \\  c_2 \end{array} \right].$     (42)

Now, just as in eq. 3, we can see that if H was a diagonal matrix, then the equations for c1 and c2 would become decoupled. Again, we can make H diagonal by going to the special basis made of the eigenvectors of H. In this new basis, we will denote vector coefficients and matrix elements with tildes as a reminder that the basis set has changed, and we obtain:
 
$\displaystyle i \hbar \left[ \begin{array}{c}
{\dot {\tilde c}}_1 \\  {\dot {\t...
...ght]
\left[ \begin{array}{c}
{\tilde c}_1 \\  {\tilde c}_2 \end{array} \right].$     (43)

Now remember that we've gone to the eigenvector basis (which we'll also denote with tildes to distinguish it from the original basis), so we know that
$\displaystyle {\hat H} {\tilde \Phi}_1({\bf r})$ = $\displaystyle E_1 {\tilde \Phi}_1({\bf r})$ (44)
$\displaystyle {\hat H} {\tilde \Phi}_2({\bf r})$ = $\displaystyle E_2 {\tilde \Phi}_2({\bf r}).$ (45)

Thus we can further simplify the diagonal elements as
$\displaystyle {\tilde H}_{11}$ = $\displaystyle \int {\tilde \Phi}_1^*({\bf r}) {\hat H}
{\tilde \Phi}_1({\bf r}) d{\bf r}$ (46)
  = $\displaystyle E_1 \int {\tilde \Phi}_1^*({\bf r})
{\tilde \Phi}_1({\bf r}) d{\bf r}$  
  = E1  

for normalized basis functions. Likewise of course ${\tilde H}_{22} = E_2$. Hence, we can expand our matrix equation 43 as
$\displaystyle i \hbar {\dot {\tilde c}}_1$ = $\displaystyle {\tilde c}_1 E_1$ (47)
$\displaystyle i \hbar {\dot {\tilde c}}_2$ = $\displaystyle {\tilde c}_2 E_2$ (48)

which is the same thing we got before when we assumed the given functions were orthonormal. The only difference is that here we emphasized the diagonalization of H rather than getting the eigenvectors, but of course it is the same process. These decoupled equations can be solved the same way as before to give
$\displaystyle {\tilde c}_1(t)$ = $\displaystyle {\tilde c}_1(0) e^{- i E_1 t / \hbar}$ (49)
$\displaystyle {\tilde c}_2(t)$ = $\displaystyle {\tilde c}_2(0) e^{- i E_2 t / \hbar},$ (50)

which we could write back in matrix notation (still in the eigenvector or tilde basis) as
$\displaystyle \left[ \begin{array}{c}
{\tilde c}_1(t) \\  {\tilde c}_2(t) \end{array} \right]$ = $\displaystyle \left[ \begin{array}{cc}
e^{- i E_1 t / \hbar} & 0 \\
0 & e^{- i...
...left[ \begin{array}{c} {\tilde c}_1(0) \\  {\tilde c}_2(0) \end{array}\right] .$ (51)

We can identify the matrix as the propagator,
 
$\displaystyle {\mathbf {\tilde G}}(t)$ = $\displaystyle \left[ \begin{array}{cc}
e^{- i E_1 t / \hbar} & 0 \\
0 & e^{- i E_2 t / \hbar} \end{array} \right]$ (52)

in the eigenvector basis.

In the vector/matrix representation, we can go from our original to our tilde coefficients and back as

$\displaystyle \left[ \begin{array}{c} c_1(t) \\  c_2(t) \end{array} \right]$ = $\displaystyle {\bf U} \left[ \begin{array}{c}
{\tilde c}_1(t) \\  {\tilde c}_2(t) \end{array} \right]$ (53)
$\displaystyle \left[ \begin{array}{c}
{\tilde c}_1(t) \\  {\tilde c}_2(t) \end{array} \right]$ = $\displaystyle {\bf U}^{\dagger}
\left[ \begin{array}{c} c_1(t) \\  c_2(t) \end{array} \right],$ (54)

where ${\bf U}$ is the matrix made by making each column an eigenvector of ${\hat H}$ in the original basis. We could transform our propagator ${\mathbf {\tilde G}}(t)$ from the eigenvector basis to the original basis by

\begin{displaymath}{\mathbf G}(t) = {\mathbf U} {\mathbf {\tilde G}}(t) {\mathbf U}^{\dagger}.
\end{displaymath} (55)


next up previous
Next: Dirac Notation Version Up: Decoupling of Equations in Previous: Basis Functions in Coordinate
C. David Sherrill
2000-05-02