\input phyzzx
%\input pstricks

\def\Re{\mathop{\rm Re}\nolimits}
\def\Im{\mathop{\rm Im}\nolimits}
\def\Det{\mathop{\rm Det}\nolimits}
\def\eqdef{\buildrel\rm def\over=}
\def\pathint{\mathop{\intop\mskip-18mu\intop\mskip-18mu\intop}\limits }
\def\modN{{{\rm mod}\,N}}
\def\QED{\hskip 1em $\cal Q.~E.~D.$}

%\postdisplaypenalty=1000
\hsize=6.7in \hoffset=-0.1in
%\vsize=8.5in \voffset=-0in
%\overfullrule=0pt

\centerline{\seventeenrm PATH INTEGRAL for HARMONIC OSCILLATOR}
\smallskip

In class, I have showed how to use path integral formalism to calculate
the partition function of a quantum system.
Formally,
$$
\eqalign{
Z(T)\ &
=\ \Tr\left[ e^{-iT\hat H}\ =\ \hat{\cal U}(T,0) \right]\ =\ \int\
\!dx_0\,U(x_0,T;x_0,0)\cr
&=\int\!\!dx_0 \pathint_{x(0)=x_0}^{x(T)=x_0}\!\!{\cal
D}'[x(t)]\,e^{iS[x(t)]}\cr
&=\pathint^{x(T)=x(0)}\!\!{\cal D}[x(t)]\,e^{iS[x(t)]}\cr
}\eqn\MinkInt
$$
where
$$
S[x(t)]\ = \int_0^T\!\!dt\left(\coeff{M}{2} \dot x^2\,-\,V(x)\right)
\eqn\SMink
$$
is the Lagrangian action functional.
Note the boundary conditions in the last path integral in eq.~\MinkInt:
$x(t)$ is required to be periodic in time, $x(T)=x(0)$ but there are no separate
initial or final conditions.

In class, I evaluated the path integral~\MinkInt\ for the harmonic oscillator,
but I was deliberately ignoring all issues of convergence and hoping that
all the pre-exponential factors would somehow take care of themselves.
In this note I remedy the problem and do the calculation right.

Actually, there are two separate convergence problems.
Formally, the path integral is defined via time discretization according to
$$
\pathint^{x(T)=x(0)}\!\!{\cal D}[x(t)]\,e^{iS[x(t)]}\
=\ \lim_{N\to\infty} \left({MN\over 2\pi iT}\right)^{N/2}
\int\!\!dx_1^{}\,\cdots\int\!\!dx_N^{}\,
\exp\left( iS^{\rm discr}(x_1,x_2,\ldots,,x_N)\right),
\eqn\MultiInt
$$
so there is an obvious convergence problem of the continuum-time limit
of $N\to\infty$.
But even for finite~$N$ there is a separate convergence problem of an
$N$--dimensional integral of a rapidly-oscillating but unimodular
function~$e^{iS}$.
In fact, for $N\ge2$ this integral does not converge, not even conditionally,
so it must be re-defined via analytic continuation.

The usual analytic continuation keeps the $x_n=x(t_n)$ real but makes the
time itself imaginary, $t=-it_E$ which runs from $0$ to $T=-i\beta$.
In field theory, $t_E$ is called the Euclidean time because the 4D spacetime
spanned by $(x_1,x_2,x_3,x_4=t_E)$ is Euclidean rather tan Minkowski.
Going from the real Minkowski time $t$ to the real Euclidean time $t_E=it$
turns the oscillating phase function
$$
e^{iS^{\rm discr}}\ =\ \exp\left[{iT\over N}\sum_{n=1}^N\left(
{M\over2}\,\left({x_n-x_{n-1}\over (T/N)}\right)^2\ -\ V(x_n)\right)\right]
\eqno\eq
$$
of the (discretized) path integral into a real narrowly peaked function
$\exp(-S_E^{\rm discr})$ where the discretized Euclidean action
$$
S_E^{\rm discr}=\ {\beta\over N}\sum_{n=1}^N\left(
{M\over2}\,\left({x_n-x_{n-1}\over (\beta/N)}\right)^2\ +\ V(x_n)\right)
\eqno\eq
$$
receives positive contribution from both the Kinetic and the potential energy
terms.
In the continuous Euclidean time limit
$$
S_E[x(t_E)]\ =\int_0^\beta\!dt_E\,\left[
 {M\over2}\,\left({dx\over dt_E}\right)^2\
 +\ V(x)\right]
\eqn\EuclAction
$$
and the Euclidean path integral becomes
$$
Z_E(\beta)\ =\ \Tr\left[ e^{-\beta \hat H}\right]\
=\pathint^{x(\beta)=x(0)}\!{\cal D}[x(t_E)]\,e^{-S_E[x(t_E)]}.
\eqn\EuclZ
$$
Note that this Euclidean partition function is precisely the partition function
of Statistical Mechanics, so it is well worth calculating in its own right.

Unlike the Minkowski-time path Integral, the Euclidean path integral is well
defined because for each finite $N$ we have an absolutely convergent
integral
$$
\left({MN\over 2\pi\beta}\right)^{N/2}
\int\!\!dx_1^{}\,\cdots\int\!\!dx_N^{}\,
\exp\left( -S_E^{\rm discr}(x_1,x_2,\ldots,,x_N)\right),
\eqn\MultiIntE
$$
and the continuous-Euclidean-time limit $N\to\infty$ usually behaves well.
Consequently, the technical definition of the Minkowski-time path integral
is nothing but the analytic continuation of the Euclidean-time PI back to
Minkowski time $t=it_E$.

\bigskip
\centerline{\fourteenpoint $\star\quad\star\quad\star$}
\medskip

So, after all these preliminaries, let is calculate the Euclidean path
integral for the harmonic oscillator.
The Euclidean action of the oscillator
$$
S_E\ =\int_0^\beta\!\!dt_E\,{M\over2}\left[
 \left({dx\over dt_E}\right)^2\ +\ \omega^2x^2\right]
\eqno\eq
$$
discretizes to
$$
S_E^{\rm discr}(x_1,\ldots,x_N)\
=\ {NM\over2\beta}\sum_{n=1}^N\left[ (x_n-x_{n-1})^2\
+\ {\omega^2\beta^2\over N^2} x_n^2\right] .
\eqn\DiscrAct
$$
which is a quadratic function of the integration variables $x_1,\ldots,x_N$.
Consequently, the discretized path integral
$$
Z(\beta,N)\ =\ \left({MN\over 2\pi\beta}\right)^{N/2}
\int\!d^Nx\,\exp\bigl( -S_E^{\rm discr}(x_1,\ldots,x_N)\bigr)
\eqn\DiscrInt
$$
is Gaussian and may be evaluated exactly.
Unfortunately, the determinant of the quadratic form~\DiscrAct\ is rather
formidable, so the best way to evaluate the integral~\DiscrInt\ is to
diagonalize the action as a quadratic form.

The continuum-time Euclidean action is diagonalized via Fourier transform
$$
\eqalign{
x(t_E)\ &
=\sum_{k=-\infty}^{+\infty}\,\beta^{-1/2}\,e^{-2\pi ikt_e/\beta} y_k\,,\cr
S_E[x]\ &
=\ {M\over2}\sum_k\left( \omega^2\,+\,{(2\pi k)^2\over\beta^2} \right)\,|y_k|^2
\colon\cr
}\eqn\SEFT
$$
note that the frequencies here are discrete because the Euclidean time is
periodic;
also, $y_k^*=y_{-k}^{}$.
For the discretized action~\DiscrAct\ however, we need the discrete Fourier
transform
$$
x_n\ =\ {1\over\sqrt{N}}\sum_{k=1}^N e^{-2\pi i kn/N} y_k
\eqn\DFT
$$
where the discrete frequencies $k$ are defined modulo $N$, \ie\ $y_0\equiv y_N$,
$y_{-k}\equiv y_{N-k}$, \etc, \etc; again, the frequency modes $y_k$
are complex, but the complete set of $y_1,\ldots y_N$ is self-conjugate as
$y_k^*=y_{-k}^{}$.
The key formula of the discrete Fourier transform is
$$
\sum_n e^{-2\pi i (k-\ell)n/N}\ =\ N\delta^\modN(k-\ell).
\eqn\DFTkey
$$
Consequently,
$$
\sum_n^\modN\,x_n^2\ =\sum_n^\modN\,x_n^* x_n^{}\ =\sum_k^\modN y_k^* y_k^{}
\eqno\eq
$$
and likewise
$$
\sum_n^\modN (x_n-x_{n-1})^2\
=\sum_k^\modN \left|1-e^{2\pi ik/N}\right|^2\,y^*_ky^{}_k
\eqno\eq
$$
where the latter follows from
$$
x_n-x_{n-1}\
=\ \sum_k^\modN N^{-1/2} e^{-2\pi i kn/N}\left(1-e^{2\pi ik/N}\right) y_k\,.
$$

Thus
$$
S_E^{\rm discr}[y_k]\
=\ {MN\over2\beta}\sum_k^\modN\left( 4\sin^2{\pi k\over N}\,
        +\,{\omega^2\beta^2\over N^2}\right)\,|y_k|^2,
\eqn\SEDFT
$$
and therefore
$$
\eqalign{
Z(beta,\omega,N)\ &
=\ \left({MN\over 2\pi \beta}\right)^{N/2}\times J(N)\times
 \int\!\!d^Ny\,e^{-S_E^{\rm discr}(y)}\cr
&=\ J(N)\times\prod_k^\modN \left(
        4\sin^2{\pi k\over N}\,+\,{\omega^2\beta^2\over N^2}
        \right)^{-1/2} \cr
}\eqn\Zdiscr
$$
where $J(N)$ is the Jacobian of the discrete Fourier
transform~\DFT.
%
To evaluate this Jacobian, we perform the Fourier transform twice:
$$
y_k\ =\sum_m^\modN N^{-1/2} e^{-2\pi imk/N} z_m,\qquad
x_n\ =\sum_k^\modN N^{-1/2} e^{-2\pi ikn/N} y_k\ =\ (-1)^n z_n\,,
\eqno\eq
$$
which immediately tells us that
$$
\left[\det\partder{x_n}{y_k}\right]^2\ =\ \det\partder{x_n}{z_m}\ =\ \pm1
$$
and hence $J=|\det(\partial x_n/\partial y_k)|=1$
%Naively, this implies unit Jacobian $J=1$
%but there is a subtlety associated with
%the complex conjugacy $y_k^*=y_{N-k}^{}$ while $x_n^*=x_n^{}$.
%Consequently, there is an additional factor of $\half$ in the Jacobian
%(\ie, $d^Nx=\half d^Ny$ for any $N$)
and therefore
$$
Z(\beta,\omega,N)\
=\ \prod_k^\modN\left(4\sin^2{\pi k\over N}\,+\,{\omega^2\beta^2\over
N^2}\right)^{-1/2} .
\eqn\ZZdisk
$$
At this point, let me use without proof a somewhat obscure mathematical formula
$$
\prod_{k=1}^{N-1}\left(2\sin{\pi k\over N}\right)\ =\ N,
\eqn\MATH
$$
which allows me to re-write the discretized partition function as
$$
\eqalign{
Z(\beta,\omega,N)\ &
=\ {N\over\omega\beta}\times \prod_{k=1}^{N-1}
    \left(4\sin^2{\pi k\over N}\,+\,{\omega^2\beta^2\over N^2}\right)^{-1/2}\cr
&=\ {1\over\omega\beta}\times \prod_{k=1}^{N-1}
    \left(1\,+\,{\omega^2\beta^2\over 4N^2\sin^2{\pi k\over N}}\right)^{-1/2} 
\cr
}\eqn\Zfinite
$$

To evaluate the large $N$ limit of this partition function
 (physically, the continuous time limit),
we approximate
$4N^2\sin^2(\pi k/N)\approx (2\pi k)^2$ for $k\ll N$,
and likewise $4N^2\sin^2(\pi k/N)\approx (2\pi(N-k))^2$ for $(N-k)\ll N$,
while for the remaining modes $\sin^2(\pi k/N)=O(1)$ and hence
$$
1\ +\ { \omega^2\beta^2\over 4N^2\sin^2{\pi k\over N} }\ \approx 1.
$$
Consequently,
$$
\eqalign{
Z(\beta,\omega,N)\ &
\becomes{N\gg1}\ {1\over\omega\beta}\times \prod_{1\le k\ll N}
        \left(1\,+\,{\omega^2\beta^2\over (2\pi k)^2}\right)^{-1/2}
    \times \prod_{1\le (N-k)\ll N}
        \left(1\,+\,{\omega^2\beta^2\over (2\pi(N-k))^2}\right)^{-1/2}\cr
&\becomes{N\to\infty}\ {1\over\omega\beta}\times
    \prod_{k=1}^{\infty}\left(
 1\,+\,{\omega^2\beta^2\over (2\pi k)^2}
 \right)^{-1} .\cr
}\eqn\Partition
$$

It remains to evaluate the infinite product in the last formula.
Consider $Z(\omega\beta)$ as an analytic function of a complex argument.
Whenever any factor of on the right hand side has a zero in the complex
$(\omega\beta)$
plane, $Z(\omega\beta)$ has a zero and ditto for the poles.
Also, the product converges, so these are the only poles and zeroes of the
$Z(\omega\beta)$
The individual factors at hand are $1/(\omega\beta)$ and
$$
{1\over 1\,+\,{\omega^2\beta^2\over (2\pi k)^2}}\
=\ {(2\pi k)^2\over(\omega\beta+2\pi k i)\times (\omega\beta-2\pi k i)}
$$
for $k=1,2,3,\ldots$.
Thus, the  $Z(\omega\beta)$ function has no zeroes and it has poles at
$\omega\beta=2\pi k i$ for all integers $k$ (positive, negative and zero).
In other words, it has the same poles and zeroes as the $1/\sinh(\omega\beta/2)$
function and indeed, there is a well known formula
$$
\sinh(z)\ =\ z\prod_{k=1}^{+\infty}\left(1+{z^2\over(\pi k)^2}\right).
$$

Thus, at the end of the long path-integral calculation, we arrive at a rather
simple formula
$$
Z_E(\beta)\ =\ {1\over2\sinh(\omega\beta/2)}
\eqn\ZE
$$
in Euclidean time, and by analytic continuation to Minkowski time
$$
Z_M(T)\ =\ {1\over 2i\sin(\omega T/2)}\,.
\eqn\ZM
$$
Expanding the latter partition function into a sum of $e^{-iET}$ phases, we
have
$$
{1\over 2i\sin(\omega T/2)}\ =\ {e^{-i\omega T/2}\over 1-e^{-i\omega T}}\
=\,\sum_{n=0}^{\infty} 1\times\exp\bigl(-iT\times(n+\half)\omega\bigr),
\eqn\\Spectrum
$$
which immediately tells us that the harmonic oscillator has non-degenerate
energy spectrum with eigenvalues $E_n=(n+\half)\omega$.
Of course, we new {\it that} long before this calculation, but it confirms that
(properly applied) path-integral formalism does yield the correct spectrum. 


\bye
