\subhead   My notes   \endsubhead

$$i\hbar{d\Psi(\r,t)\over d t}=H \Psi(\r,t)$$

Assuming $\Psi(\r,t)=\psi(\r)\chi(t)$ won't help ($\Psi$ is
non-separable).  However, the above can be understood as an equation
in the Hilbert space of functions from coordinate space (${\Bbb R}^2$)
to $\Bbb C$; so taking $\Psi(\r,t)\equiv\left[\fPsi(t)\right](\r)$ we
get

$$ i\hbar{d\fPsi(t)\over d t} = H \fPsi(t) $$
$$ \fPsi(t) = e^{-iHt/\hbar}\; \fPsi(0) $$

(Simple insertion shows that this is the result we are looking for.)

Taking $\epsilon=t/N$, we get

$$\eqalign{
\fPsi(t) &= \left(e^{-iH\epsilon/\hbar}\right)^N \fPsi(0) \cr
 &\approx \left(e^{-iV\epsilon/2\hbar}\;e^{-iK\epsilon/\hbar}\;
          e^{-iV\epsilon/2\hbar}\right)^N \fPsi(0)\cr
 &= \left(e^{-iV\epsilon\over 2\hbar}\;e^{-iK\epsilon\over \hbar}\;
          e^{-iV\epsilon\over 2\hbar}\right)
    \left(e^{-iV\epsilon\over 2\hbar}\;e^{-iK\epsilon\over \hbar}\;
          e^{-iV\epsilon\over 2\hbar}\right) \ldots
    \left(e^{-iV\epsilon\over 2\hbar}\;e^{-iK\epsilon\over \hbar}\;
          e^{-iV\epsilon\over 2\hbar}\right) \fPsi(0)\cr
 &= e^{-iV\epsilon\over 2\hbar}\;e^{-iK\epsilon\over \hbar}\;
    e^{-iV\epsilon\over \hbar}\; e^{-iK\epsilon\over \hbar}\ldots\;
    e^{-iK\epsilon\over \hbar}\; e^{-iV\epsilon\over 2\hbar} \fPsi(0)\cr
}$$

where $e^{-iVt/2\hbar}$, $e^{-iKt/\hbar}$ and $e^{-iVt/\hbar}$ can be
calculated by

$$e^{-iAt/\hbar} = \sum_{n=0}^\infty {\left(-iAt/\hbar\right)^n \over n!}$$

All the operators are functions from $N$-space to $N$-space, and could
in a general discrete form be described by $2N$-dimensional
``matrices''.  However, in coordinate form $\hat V$ is just a
multiplication of each element by $V(\r)$ corresponding to its
position, so $e^{-i\hat Vt/2\hbar}$ and $e^{-i\hat Vt/\hbar}$ have
that form as well.  Analogously, $\hat K$ is in its {\it momentum}
form is a multiplication of each element by its $K(\p)$, and so is
$e^{-i\hat Kt/\hbar}$.

\centerline{\vbox{\hrule width 2in}}

Taylor sez:

$$\eqalign{
\Psi(\r,t) &= (2\pi)^{-{3\over2}} \int_{V} \Phi(\k,t)
		\exp i \,\k\!\cdot\!\r\, \d\k_x \d\k_y \d\k_z \cr
\Phi(\k,t) &= (2\pi)^{-{3\over2}} \int_{V} \Psi(\r,t)
		\exp -i \,\k\!\cdot\!\r\, \d x \d y \d z \cr
}$$

Numerical Recipes are, of course, geared towards evil electrical
engineers, so they use non-angular frequencies, which corresopond to
spatial frequenties (and not wave vector).  We have $\k\equiv2\pi\f$,
so

$$\eqalign{
\Psi(\r,t) &= (2\pi)^{+{3\over2}} \int_{V} \Phi(\f,t)
		\,e^{2\pi i\,\f\cdot\r} \d\f_x \d\f_y \d\f_z \cr
\Phi(\f,t) &= (2\pi)^{-{3\over2}} \int_{V} \Psi(\r,t)
		\,e^{-2\pi i\,\f\cdot\r} \d x \d y \d z \cr
}$$

To simplify life a bit, we can change to $\Phi'\equiv(2\pi)^{3\over2}\Phi$:

$$\eqalign{
\Psi(\r,t) &= \int_{V} \Phi'(\f,t)
		\,e^{2\pi i\,\f\cdot\r} \d\f_x \d\f_y \d\f_z \cr
\Phi'(\f,t) &= \int_{V} \Psi(\r,t)
		\,e^{-2\pi i\,\f\cdot\r} \d x \d y \d z \cr
}$$

For discrete $r$ and $\f$, this becomes

$$\eqalign{
\Psi(\r,t) &= \sum_{k_n=0}^{N_n-1}\cdots\sum_{k_1=0}^{N_1-1} \Phi'(\f,t)
		\,e^{2\pi i\,\f\cdot\r} \d\f_x \d\f_y \d\f_z \cr
\Phi'(\f,t) &= {1\over N^n}\sum_{k_n=0}^{N_n-1}\cdots\sum_{k_1=0}^{N_1-1}
		\Psi(\r,t) \,e^{-2\pi i\,\f\cdot\r} \d x \d y \d z \cr
}$$

\centerline{\vbox{\hrule width 2in}}

Let us assume we're sampling the positions at intervals $\Delta$, i.e.

$$ \r_{mnp} = m\Delta\vi+n\Delta\vj+p\Delta\vk
     \hskip1cm m,n,p=\ldots,-1,0,1,\ldots $$

Then the spatial frequencies at points of Fourier transform will be

$$ \f_{mnp} = {m\over N\Delta}\vi+{n\over N\Delta}\vj+{p\over N\Delta}\vk
     \hskip1cm m,n,p=\ldots,-1,0,1,\ldots $$

$$ \k_{mnp} = 2\pi\f_{mnp} $$
$$ \left(\p_{mnp} = \hbar\,\k_{mnp}\right) $$


\centerline{\vbox{\hrule width 2in}}

$$\eqalign{
K(\p) &= {|\p|^2\over2m} \cr
      &= {\p_x^2+\p_y^2+\p_z^2 \over 2m} \cr
      &= {(2\pi\hbar)^2\over2m}{m^2+n^2+p^2\over\Delta^2 N^2} \cr
e^{-iKt/\hbar} &= \exp\left( -it{\hbar\over\Delta^2}{2\pi^2\over m N^2}
	(m^2+n^2+p^2) \right)\cr
}$$

\centerline{\vbox{\hrule width 2in}}

For a harmonic oscillator potential:

$$\eqalign{
V(\r) &= {1\over2} k|\r|^2 \cr
      &= {k\over2}(\r_x^2+\r_y^2+\r_z^2) \cr
      &= {k\over2}\Delta^2(m^2+n^2+p^2) \cr
e^{-iVt/\hbar} &=
	\exp\left( -it{k\over2}{\Delta^2\over\hbar}(m^2+n^2+p^2) \right)\cr
}$$

\centerline{\vbox{\hrule width 2in}}

The maximum values of $Vt/\hbar$ and $Kt/\hbar$ for a $d$-dimensional
matrix, $N\times\ldots\times N$, are

$$t\Delta^2{k\over2\hbar} {d N^2 \over 4}
\,\,\,\hbox{and}\,\,\,
{t\over\Delta^2}{2\hbar\pi^2\over m N^2} {d N^2\over 4}$$

For a ``good'' calculation, the exponentials should stay close to
linear, thus the exponents should be $< \mu \ll 1$.  Still working
with the harmonic oscillator potential, we get

$$ t^2{k\pi^2 \over m} \left(d N \over 4\right)^2 < \mu^2 $$
$$ t < {4 \mu \over d N \pi} \sqrt{m \over k}
     = {2 \mu \over d N \pi^2} T_c $$

where $T_c = 2\pi\sqrt{m/k}$ is the classical period of a harmonic
oscillator.

\centerline{\vbox{\hrule width 2in}}

% $$\eqalign{
% m\ddot x + k x &= 0 \cr
% -m\omega^2 + k &= 0 \cr
% \omega &= \sqrt{k \over m} \cr
% T &= {2\pi \over \omega} = 2\pi \sqrt{m \over k} \cr
% }$$
