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\begin{document}\thispagestyle{empty}

\parbox[b]{4.8in}{
  \footnotesize\url{http://drorbn.net/?title=AKT-14}
  \newline\bf
  \href{\myurl}{Dror Bar-Natan}:
  \href{\myurl/classes/}{Classes}:
  \href{\myurl/classes/\#1314}{2014}:
  \href{http://drorbn.net/?title=AKT-14}{MAT 1350 --- Algebraic Knot Theory}:
}
\hfill\parbox[b]{2in}{
  \null\hfill{\Large\bf Friday Introduction}
}

\begin{multicols}{2}

{\Large\bf What happens to a quantum particle on a pendulum at $T=\frac{\pi}{2}$?}

\vskip 3mm
{\bf Abstract.}
This subject is the best one-hour introduction I know for the mathematical
techniques that appear in quantum mechanics --- in one short lecture we
start with a meaningful question, visit Schr\"odinger's equation, operators
and exponentiation of operators, Fourier analysis, path integrals, the
least action principle, and Gaussian integration, and at the end we land
with a meaningful and interesting answer.

\vskip 3mm
{\footnotesize Based a lecture given by the author in the ``trivial notions'' seminar in Harvard on April 29, 1989. This edition, \today.}

\section{The Question} Let the complex valued function $\psi=\psi(t,x)$ be
a solution of the Schr\"odinger equation
\[ \frac{\partial\psi}{\partial t} =
  -i\left(-\frac12\Delta_x+\frac12x^2\right)\psi
  \qquad \text{with} \qquad
  \psi|_{t=0}=\psi_0.
\]
What is $\psi|_{t=T=\frac{\pi}{2}}$?

In fact, the major part of our discussion will work just as well for the
general Schr\"odinger equation,
\[ \frac{\partial\psi}{\partial t} = -iH\psi, \qquad H=-\frac12\Delta_x+V(x), \]
\[ \psi|_{t=0}=\psi_0, \qquad \text{arbitrary }T, \]
where,
\begin{itemize}
\item $\psi$ is the ``wave function'', with $|\psi(t,x)|^2$
  representing the probability of finding our particle at time $t$ in
  position $x$.
\item $H$ is the ``energy'', or the ``Hamiltonian''.
\item $-\frac12\Delta_x$ is the ``kinetic energy''.
\item $V(x)$ is the ``potential energy at $x$''.
\end{itemize}

\section{The Solution} \label{sec:solution}
The equation $\frac{\partial\psi}{\partial t} = -iH\psi$ with
$\psi|_{t=0}=\psi_0$ formally implies

\[ \psi(T,x)=\left(e^{-iTH}\psi_0\right)(x)
  = \left(e^{i\frac{T}{2}\Delta-iTV}\psi_0\right)(x).
\]
By Lemma~\ref{lem:Trotter} with $n=10^{58}+17$ and setting $x_n=x$ we find that $\psi(T,x)$ is

\[ \left(
    e^{i\frac{T}{2n}\Delta}e^{-i\frac{T}{n}V}
    e^{i\frac{T}{2n}\Delta}e^{-i\frac{T}{n}V}
    \dots
    e^{i\frac{T}{2n}\Delta}e^{-i\frac{T}{n}V}
    \psi_0
  \right)(x_n).
\]
Now using Lemmas~\ref{lem:eV} and~\ref{lem:eDelta} we find that this is: ($c$
denotes the ever-changing universal fixed numerical constant)
\begin{multline*} c\!  \int\!\!
    dx_{n-1}e^{i\frac{(x_n-x_{n-1})^2}{2T/n}}e^{-i\frac{T}{N}V(x_{n-1})}
  \ldots \\
  \int\!\! dx_1e^{i\frac{(x_2-x_1)^2}{2T/n}}e^{-i\frac{T}{N}V(x_1)} \\
  \int\!\! dx_0e^{i\frac{(x_1-x_0)^2}{2T/n}}e^{-i\frac{T}{N}V(x_0)}
  \psi_0(x_0).
\end{multline*}

Repackaging, we get
\begin{multline*} c\int dx_0\ldots dx_{n-1} \\ \exp
  \left(
    i\frac{T}{2n}\sum_{k=1}^n\left(\frac{x_k-x_{k-1}}{T/n}\right)^2
    - i\frac{T}{n}\sum_{k=0}^{n-1}V(x_k)
  \right)\\
  \psi_0(x_0).
\end{multline*}

Now comes the novelty. keeping in mind the picture

\resizebox{\columnwidth}{!}{\input{PolyPath.pdf_t}}

and replacing Riemann sums by integrals, we can write
\begin{multline*} \psi(T,x) =
  c\int dx_0 \int_{W_{x_0x_n}}{\mathcal D}x \\
  \exp\left(
    i\int_0^Tdt\left(\frac12\dot{x}^2(t)-V(x(t))\right)
  \right) \psi_0(x_0),
\end{multline*}
where $W_{x_0x_n}$ denotes the space of paths that begin at $x_0$ and end
at $x_n$,
\[ W_{x_0x_n} =
  \left\{x:[0,T]\to{\mathbb R}:\ x(0)=x_0,\,x(T)=x_n\right\},
\]
and ${\mathcal D}x$ is the formal ``path integral measure''.

This is a good time to introduce the ``action'' ${\mathcal L}$:
\[ {\mathcal L}(x):=\int_0^Tdt\left(\frac12\dot{x}^2(t)-V(x(t))\right). \]
With this notation,
\[ \psi(T,x) =
  c\int dx_0 \psi_0(x_0)\int_{W_{x_0x_n}}{\mathcal D}x
  e^{i{\mathcal L}(x)}.
\]

Let $x_c$ denote the path on which ${\mathcal L}(x)$ attains its minimum
value, write $x=x_c+x_q$ with $x_q\in W_{00}$, and get
\[ \psi(T,x) =
  c\int dx_0 \psi_0(x_0)\int_{W_{00}}{\mathcal D}x_q
  e^{i{\mathcal L}(x_c+x_q)}.
\]

In our particular case ${\mathcal L}$ is quadratic in $x$, and therefore
${\mathcal L}(x_c+x_q)={\mathcal L}(x_c)+{\mathcal L}(x_q)$ (this uses the
fact that $x_c$ is an extremal of ${\mathcal L}$, of course). Plugging this
into what we already have, we get
\begin{multline*} \psi(T,x) =
  c\int dx_0 \psi_0(x_0)\int_{W_{00}}{\mathcal D}x_q
  e^{i{\mathcal L}(x_c)+i{\mathcal L}(x_q)} \\
  = c\int dx_0 \psi_0(x_0)e^{i{\mathcal L}(x_c)}
    \int_{W_{00}}{\mathcal D}x_qe^{i{\mathcal L}(x_q)}.
\end{multline*}

Now this is excellent news, because the remaining path integral over
$W_{00}$ does not depend on $x_0$ or $x_n$, and hence it is a constant!
Allowing $c$ to change its value from line to line, we get
\[ \psi(T,x) = c\int dx_0 \psi_0(x_0)e^{i{\mathcal L}(x_c)}. \]

Lemma~\ref{lem:EL} now shows us that $x_c(t)=x_0\cos t+x_n\sin t$. An
easy explicit computation gives ${\mathcal L}(x_c)=-x_0x_n$, and we arrive
at our final result,
\[ \psi(\frac{\pi}{2},x) = c\int dx_0 \psi_0(x_0)e^{-ix_0x_n}. \]

Notice that this is precisely the formula for the Fourier transform of
$\psi_0$!  That is, the answer to the question in the title of this
document is ``the particle gets Fourier transformed'', whatever that may
mean.

\section{The Lemmas}

\begin{lemma} \label{lem:Trotter} For any two matrices $A$ and $B$,
\[ e^{A+B} = \lim_{n\to\infty}\left(e^{A/n}e^{B/n}\right)^n. \]
\end{lemma}

\begin{proof} (sketch)
Using Taylor expansions, we see that $e^{\frac{A+B}{n}}$ and
$e^{A/n}e^{B/n}$ differ by terms at most proportional to $c/n^2$. Raising
to the $n$th power, the two sides differ by at most $O(1/n)$, and thus
\[ e^{A+B} = \lim_{n\to\infty} \left(e^{\frac{A+B}{n}}\right)^n =
  \lim_{n\to\infty}\left(e^{A/n}e^{B/n}\right)^n,
\]
as required.
\end{proof}

\begin{lemma} \label{lem:eV}
\[ \left(e^{itV}\psi_0\right)(x) = e^{itV(x)}\psi_0(x). \]
\qed
\end{lemma}

\begin{lemma} \label{lem:eDelta}
\[
  \left(e^{i\frac{t}{2}\Delta}\psi_0\right)(x)
  = c\int dx' e^{i\frac{(x-x')^2}{2t}}\psi_0(x').
\]
\end{lemma}

\begin{proof} In fact, the left hand side of this equality is just a
solution $\psi(t,x)$ of Schr\"odinger's equation with $V=0$:
\[ \frac{\partial\psi}{\partial t}=\frac{i}{2}\Delta_x\psi,
   \qquad \psi|_{t=0}=\psi_0.
\]
Taking the Fourier transform $\tilde\psi(t,p)=\frac{1}{\sqrt{2\pi}}\int
e^{-ipx}\psi(t,x)dx$, we get the equation
\[ \frac{\partial\tilde\psi}{\partial t} = -i\frac{p^2}{2}\tilde\psi,
  \qquad \tilde\psi|_{t=0}=\tilde\psi_0.
\]
For a fixed $p$, this is a simple first order linear differential equation
with respect to $t$, and thus,
\[ \tilde\psi(t,p) = e^{-i\frac{tp^2}{2}}\tilde\psi_0(p). \]

Taking the inverse Fourier transform, which takes products to convolutions
and Gaussians to other Gaussians, we get what we wanted to prove.
\end{proof}

\begin{lemma} \label{lem:EL} With the notation of Section~\ref{sec:solution}
and at the specific case of $V(x)=\frac12x^2$ and $T=\frac{\pi}{2}$, we
have
\[ x_c(t) = x_0\cos t + x_n\sin t. \]
\end{lemma}

\begin{proof}
If $x_c$ is a critical point of ${\mathcal L}$ on $W_{x_0x_n}$, then for
any $x_q\in W_{00}$ there should be no term in ${\mathcal L}(x_c+\epsilon
x_q)$ which is linear in $\epsilon$. Now recall that
\[ {\mathcal L}(x) = \int_0^T dt\left(\frac12\dot{x}^2(t) - V(x(t))\right), \]
so using $V(x_c+\epsilon x_q)\sim V(x_c)+\epsilon x_qV'(x_c)$ we find that
the linear term in $\epsilon$ in ${\mathcal L}(x_c+\epsilon x_q)$ is
\[ \int_0^Tdt\left(\dot{x}_c\dot{x}_q-V'(x_c)x_q\right). \]
Integrating by parts and using $x_q(0)=x_q(T)=0$, this becomes
\[ \int_0^Tdt\left(-\ddot{x}_c-V'(x_c)\right)x_q. \]

For this integral to vanish independently of $x_q$, we must have
$-\ddot{x}_c-V'(x_c)\equiv 0$, or
\[ \ddot{x}_c = -V'(x_c). \qquad\left(\parbox{2in}{
  This is the famous $F=ma$ of Newton's, and we have just rediscovered
  the principle of least action!
} \right) \]
In our particular case this boils down to the equation
\[ \ddot{x}_c = -x_c,\qquad x_c(0)=x_0,\qquad x_c(\pi/2)=x_n, \]
whose unique solution is displayed in the statement of this lemma.
\end{proof}

\end{multicols}

\end{document}
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