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Dirac Notation Version

Let us do the same problem yet again in Dirac or Bracket notation. For this version, let's go ahead and assume that we expand our state function $\vert \Psi \rangle$ directly in terms of the Hamiltonian eigenvectors

\vert \Psi(t) \rangle = c_1(t) \vert \Phi_1 \rangle + c_2(t) \vert \Phi_2 \rangle.
\end{displaymath} (56)

When substituted into the time-dependent Schrödinger equation, this gives
$\displaystyle i \hbar
\left( {\dot c}_1(t) \vert \Phi_1 \rangle + {\dot c}_2(t) \vert \Phi_2 \rangle \right)$ = $\displaystyle {\hat H} \left( c_1(t) \vert \Phi_1 \rangle + c_2(t) \vert \Phi_2 \rangle \right)$ (57)
  = $\displaystyle E_1 c_1(t) \vert \Phi_1 \rangle + E_2 c_2(t) \vert \Phi_2 \rangle.$  

Now multiply on the left by $\langle \Phi_1 \vert$ and $\langle \Phi_2 \vert$, respectively, and use $\langle \Phi_i \vert \Phi_j \rangle = \delta_{ij}$ to obtain
$\displaystyle i \hbar {\dot c}_1(t)$ = E1 c1(t) (58)
$\displaystyle i \hbar {\dot c}_2(t)$ = E2 c2(t), (59)

just as before, with solutions once again
c1(t) = $\displaystyle c_1(0) e^{- i E_1 t / \hbar}$ (60)
c2(t) = $\displaystyle c_2(0) e^{- i E_2 t / \hbar}.$ (61)

But what if we are given $\vert \Psi(0) \rangle$ in a form that looks different from that of eq. 56? Since the eigenvector basis must be complete (although it will usually have more than two basis vectors, as in this example!), we can always rewrite our state vector in this form, and the coefficients can always be computed as

\begin{displaymath}c_i(0) = \langle \Phi_i \vert \Psi(0) \rangle.
\end{displaymath} (62)

Note that in this subsection we aren't assuming anything about whether we are working in coordinate (${\bf r}$) space or momentum (${\bf p}$) space or some other space. However, if we were working in coordinate space, we could insert the resolution of the identity

\begin{displaymath}{\hat I} = \int d{\bf r} \vert \mathbf r \rangle \langle \mathbf r \vert
\end{displaymath} (63)

to obtain
ci(0) = $\displaystyle \int \langle \Phi_i \vert {\mathbf r} \rangle
\langle \mathbf r \vert \Psi(0) \rangle d{\bf r}$  
  = $\displaystyle \int \Phi_i^*({\bf r}) \Psi({\bf r}, 0) d{\bf r},$ (64)

completely consistent with everything above. The propagator may be written as

\begin{displaymath}{\hat G}(t) = \vert \Phi_1 \rangle \langle \Phi_1 \vert e^{-i...
...vert \Phi_2 \rangle \langle \Phi_2 \vert e^{-i E_2 t / \hbar}
\end{displaymath} (65)

again with $\vert \Phi_1 \rangle$ and $\vert \Phi_2 \rangle$ here representing eigenfunctions of ${\hat H}$ with eigenvalues E1 and E2, respectively. Note the similarity between this propagator and that from the classical example in eq. 27. The only real difference is that there, we chose to work with cosines as a phase factor, and here we are using the more general exponential. Evidently the frequency here is represented by $E_i / \hbar$. You can verify that in the eigenvector basis, this operator becomes ${\tilde
{\mathbf G}}(t)$ from eq. 52 in the previous section. Hopefully you can also see that in the eigenvector basis this definition is equivalent to the more general form 31.

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Next: About this document ... Up: Decoupling of Equations in Previous: Matrix Version
C. David Sherrill