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Preprint CAMTP/96-5\\
July 1996\\
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\begin{center}
\large
{\bf WKB to all orders and the accuracy of the semiclassical quantization}\\
\vspace{0.25in}
\normalsize
Marko Robnik$^{(*)}$\footnote{e--mail: robnik@uni-mb.si} and
Luca Salasnich$^{(*)(+)}$\footnote{e--mail: salasnich@math.unipd.it} \\
\vspace{0.2in}
$^{(*)}$ Center for Applied Mathematics and Theoretical Physics,\\
University of Maribor, Krekova 2, SLO--2000 Maribor, Slovenia\\
\vspace{0.2in}
$^{(+)}$ Dipartimento di Matematica Pura ed Applicata \\
Universit\`a di Padova, Via Belzoni 7, I--35131 Padova, Italy \\
and\\
Istituto Nazionale di Fisica Nucleare, Sezione di Padova,\\
Via Marzolo 8, I--35131 Padova, Italy
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\vspace{0.3in}
\normalsize
{\bf Abstract.} We perform a systematic WKB expansion to all orders
for a one--dimensional system with potential $V(x)=U_0/\cos^2{(\alpha x)}$.
We are able to sum the series to the exact energy spectrum.
Then we show that at any finite order the error of the WKB approximation
measured in {\em the natural units of the mean energy level spacing}
does not go to zero when the quantum number goes to infinity.
Therefore we make the general conclusion that the semiclassical
approximations fail to predict the individual energy levels within a
vanishing fraction of the mean energy level spacing.
\vspace{0.6in}
PACS numbers: 03.65.-w, 03.65.Ge, 03.65.Sq \\
Submitted to {\bf Journal of Physics A: Mathematical and General}
\normalsize
\vspace{0.1in}
\newpage
\section{Introduction}
\par
In the last years many studies
have been devoted to the transition from classical mechanics
to quantum mechanics. These studies are motivated by the so--called
quantum chaos (see Ozorio de Almeida 1990,
Gutzwiller 1990, Casati and Chirikov 1995).
An important aspect is the semiclassical quantization formula of the
energy levels for integrable and quasi--integrable systems,
i.e. the torus quantization initiated by Einstein (1917) and
completed by Maslov (1972, 1981).
As is well known, the torus quantization is just the first term
of a certain $\hbar$-expansion, the so--called WKB expansion,
whose higher terms
can be calculated with a recursion formula at least for one degree systems
(Dunham 1932, Bender, Olaussen and Wang 1977, Voros 1983).
\par
Recently it has been observed by Prosen and Robnik (1993) and also
Graffi, Manfredi and Salasnich (1994)
that the leading--order semiclassical approximation fails to predict
the individual energy levels within a vanishing fraction of the mean
energy level spacing. This result has been shown to
be true also for the leading (torus) semiclassical approximation
by Salasnich and Robnik (1996).
\par
In this paper we analyze a simple one--dimensional system
for which we are able to perform a systematic
WKB expansion to all orders resulting in a convergent series whose sum is
identical to the exact spectrum. For this system we show that
any finite order WKB (semiclassical) approximation
fails to predict the individual energy levels within a vanishing
fraction of the mean energy level spacing.
\section{The system and the WKB expansion method}
\par
The Hamiltonian of the system is given by
\beq
H={p^2\over 2m}+ V(x) \; ,
\eeq
where
\beq
V(x)= {U_0\over \cos^2{(\alpha x)} } \; .
\eeq
Of course, the Hamiltonian is a constant
of motion, whose value is equal to the total energy $E$.
To perform the torus quantization it is necessary to introduce
the action variable
\beq
I={1\over 2\pi}\oint p dx = {\sqrt{2m}\over \alpha} (\sqrt{E} - \sqrt{U_0})
\; .
\eeq
The Hamiltonian as a function of the action reads
\beq
H={\alpha^2 \over 2m}I^2 + {2\alpha \sqrt{U_0\over 2m}} I + U_0 \; ,
\eeq
and after the torus quantization
\beq
I = (\nu + {1\over 2})\hbar \; ,
\eeq
where $\nu = 0,1,2,\dots$, the energy spectrum is given by
\beq
E_{\nu}^{tor} = A[(\nu +{1\over 2})+{1\over 2}B]^2 \; ,
\eeq
where $A=\alpha^2 \hbar^2 /(2m)$ and $B=\sqrt{8mU_0}/(\alpha \hbar )$.
\par
The Schr\"odinger equation of the system
\beq
[-{\hbar^2 \over 2 m} {d^2 \over dx^2} + V(x)] \psi (x) = E \psi (x) \; ,
\eeq
can be solved analytically (as shown in Landau and Lifshitz 1973,
Fl\"ugge 1971) and the exact energy spectrum is:
\beq
E_{\nu}^{ex} = A [( \nu + {1\over 2}) +{1\over 2}\sqrt{1+B^2}]^2 \; ,
\eeq
where $\nu = 0,1,2,\dots$.
We see that the torus quantization does not give the correct
energy spectrum, but it is well known that the torus quantization
is just the first term of the WKB expansion.
To calculate all the terms of the WKB expansion we observe that
the wave function can always be written as
\beq
\psi (x) = \exp{ \big( {i\over \hbar} \sigma (x) \big) } \; ,
\eeq
where the phase $\sigma (x)$ is a complex function that satisfies
the differential equation
\beq
\sigma{'}^2(x) + ({\hbar \over i}) \sigma{''}(x) = 2m(E - V(x)) \; .
\eeq
The WKB expansion for the phase is given by
\beq
\sigma (x) = \sum_{k=0}^{\infty} ({\hbar \over i})^k \sigma_k(x) \; .
\eeq
Substituting (11) into (10) and comparing like powers of $\hbar$ gives
the recursion relation ($n>0$)
\beq
\sigma{'}_0^2=2m(E-V(x)), \;\;\;\;
\sum_{k=0}^{n} \sigma{'}_k\sigma{'}_{n-k}
+ \sigma{''}_{n-1}= 0 \; .
\eeq
\par
The quantization condition is obtained by requiring
the single-valuedness of the wave function
\beq
\oint d\sigma =
\sum_{k=0}^{\infty} ({\hbar \over i})^{k} \oint d\sigma_{k}=
2 \pi \hbar \; \nu \;\; ,
\eeq
where $\nu =0,1,2,\dots$ is the quantum number.
\par
The zero order term, which gives the Bohr-Sommerfeld formula,
is given by
\beq
\oint d\sigma_0 = 2 \int dx \sqrt{2m(E - V(x))} = 2\pi \hbar
(\sqrt{E\over A}-{1\over 2}B) \; ,
\eeq
and the first odd term in the series gives the Maslov corrections
(Maslov index is equal to 2)
\beq
({\hbar \over i}) \oint d\sigma_{1} = ({\hbar \over i}) {1\over 4}
\ln{p}|_{contour} = - \pi \hbar \; .
\eeq
The zero and first order terms give the equation (6), which is
the torus quantization formula for the energy levels
(Bohr--Sommerfeld--Maslov).
Here we want to analyze the quantum corrections to this formula.
We observe that all the other odd terms vanish when integrated along the closed
contour because they are exact differentials (Bender, Olaussen and
Wang 1977). So the quantization condition (13) can be written
\beq
\sum_{k=0}^{\infty} ({\hbar \over i})^{2k} \oint d\sigma_{2k} = 2 \pi \hbar
(\nu +{1\over 2}) \; ,
\eeq
thus again a sum over even--numbered terms only.
The next two non--zero terms are (Narimanov 1995, Bender, Olaussen and
Wang 1977, Robnik and Salasnich 1996)
\beq
({\hbar \over i})^{2} \oint d\sigma_2
= - {\hbar^2 \over \sqrt{2m}}{1\over 12} {\partial^2 \over \partial E^2}
\int dx {V{'}^2(x) \over \sqrt{E - V(x)} } \; ,
\eeq
\beq
({\hbar \over i})^{4} \oint d\sigma_4 = {\hbar^4\over (2m)^{3/2}}
[{1\over 120} {\partial^3\over \partial E^3}
\int dx {V{''}^2(x) \over \sqrt{E - V(x)} }
- {1\over 288 } {\partial^4\over \partial E^4}
\int dx {V{'}^2(x) V{''}(x) \over \sqrt{E - V(x)} } ] \; .
\eeq
A straightforward calculation of these terms gives (see the Appendix)
\beq
({\hbar \over i})^{2} \oint d\sigma_2 = - {2\pi \hbar \over 4 B} \; ,
\eeq
and
\beq
({\hbar \over i})^{4} \oint d\sigma_4 = {2\pi \hbar \over 16 B^3} \; .
\eeq
Up to the fourth order in $\hbar \sim B^{-1}$ the quantization condition reads
\beq
E_{\nu}^{(4)} = A[(\nu
+{1\over 2})+{1\over 2}B + {1\over 4 B} - {1\over 16 B^3}]^2
\; .
\eeq
The first two terms on the right side give the torus quantization formula,
and the other two terms are quantum corrections.
Higher--order quantum corrections quickly increase in complexity but in this
specific case they can be calculated.
We first verify by induction, following Bender, Olaussen and Wang (1977),
that the solution to (12) has the general form
\beq
\sigma_n^{'}(x) = (\sigma_0^{'})^{1-3n}P_n(\cos{(\alpha x)})
\sin^{f(n)}{(\alpha x)} \; ,
\eeq
where $f(n)=0$ for $n$ even and $f(n)=1$ for $n$ odd, and $P_n$ is
a polynomial given by
\beq
P_n(\cos{(\alpha x)}) = \sum_{l=0}^{g(n)} C_{n,l}\cos^{2l-3n}{(\alpha x)}
\; ,
\eeq
with $g(n)=(3n-2)/2$ for $n$ even and $g(n)=(3n-3)/2$ for $n$ odd. \\
The integrals in (16) are performed by substituting $z=\tan{(\alpha x)}$.
In this way the $2k$-term reduces to
\beq
({\hbar \over i})^{2k} \oint d\sigma_{2k} =
({\hbar \over i})^{2k} {(2m)^{1/2-3k}\over \alpha}
\sum_{l=0}^{3k-1} C_{2k,l} \oint dz { (1+z^2)^{3k-l-1}\over
(E-U_0-U_0z^2)^{3k -1/2} } \; .
\eeq
We observe that
$$
\oint dz { (1+z^2)^{3k-l-1}\over (E-U_0-U_0z^2)^{3k -1/2} } =
$$
\beq
=(-1)^{3k-1} {\Gamma ({1\over 2})\over \Gamma (3k -{1\over 2})}
{\partial^{3k -1}\over \partial E^{3k-1} }
\oint dz { (1+z^2)^{3k-l-1}\over (E-U_0-U_0z^2)^{1/2} } \; ,
\eeq
so the only non--zero term is for $l=0$
$$
{\partial^{3k -1}\over \partial E^{3k-1} }
\oint dz { (1+z^2)^{3k-1}\over (E-U_0-U_0z^2)^{1/2} } =
{2^{6k-1}\over U_0^{1/2}} {\Gamma (3k -1/2)^2\over \Gamma (6k -1)}
{\partial^{3k -1}\over \partial E^{3k-1} } \beta^{3k-1} \; =
$$
\beq
= {2^{6k-1}\over U_0^{1/2}}{\Gamma (3k -1/2)^2\over \Gamma (6k -1)}
\Gamma (3k ) {1\over U_0^{3k-1/2}} 2\pi \; ,
\eeq
where $\beta = (E-U_0)/U_0$. At this stage we obtain
\beq
({\hbar \over i})^{2k} \oint d\sigma_{2k} = (-1)^{5k-1}
\hbar^{2k} {(2m)^{1/2-3k}\over \alpha} C_{2k,0} {1\over U_0^{3k-1/2}} 2\pi
\; .
\eeq
\par
Now we need to find the coefficient $C_{2k,0}$ explicitly.
By inserting (22) with (23) in the recursion relation (12) we obtain
\beq
\sum_{k=0}^n C_{k,0}C_{n-k,0} -(2mU_0\alpha ) C_{n-1,0} =
\sum_{k=1}^{n-1} C_{k,0}C_{n-k,0} + 2C_{n,0} -(2mU_0\alpha ) C_{n-1,0} = 0
\; ,
\eeq
from which we have
\beq
C_{k,0}={1\over 2}\big[ (2m\alpha U_0)C_{k-1,0}-\sum_{j=1}^{k-1}
C_{j,0} C_{k-j,0} \big] , \;\;\; C_{0,0}=1 \; .
\eeq
>From this equation one shows $C_{1,0}=m \alpha U_0$. Further,
it easy to show that all higher odd coefficients vanish,
$C_{2k+3,0}=0$ for $k=0,1,2,\dots$.
The solution of this equation for the remaining nonzero
even coefficients is given by
\beq
C_{2k,0}= (-1)^k (2mU_0\alpha )^{2k} 2^{-2k} {{1\over 2}\choose k} \; ,
\eeq
which can be verified by direct substitution in equation (29)
resulting in an identity for half integer binomial coefficients.
Then the integral (27) can be written
\beq
({\hbar \over i})^{2k} \oint d\sigma_{2k} = (-1) \hbar^{2k} 2\pi
\alpha^{2k-1} (2m)^{1/2-k} 2^{-2k} {{1\over 2}\choose k}U_0^{k-1/2}
= - {1\over 2}{{1\over 2}\choose k}{2\pi \hbar \over B^{2k-1}} \; .
\eeq
In conclusion, the WKB quantization to all orders (16) is
\beq
E_{\nu}^{(\infty )} =
A\big[ (\nu +{1\over 2})+ {1\over 2} \sum_{k=0}^{\infty} {{1\over 2}\choose k}
{1\over B^{2k -1}} \big]^2 \; .
\eeq
Because $\sum_{k=0}^{\infty} {{1\over 2}\choose k}
B^{1-2k} = \sqrt{1+B^2}$ we have
$E^{ex}_{\nu}=E^{(\infty )}_{\nu}$, i.e. the WKB series converges
to the exact result (8).
\par
Now we can calculate the error in units of the mean level spacing
$\Delta E_{\nu}=E^{ex}_{\nu +1}-E^{ex}_{\nu}$
between the exact level $E^{ex}_{\nu}$
and its WKB approximation $E^{(N)}_{\nu}$ to $N$th order:
\beq
{E^{ex}_{\nu}-E^{(N)}_{\nu} \over \Delta E_{\nu}}=
{1\over 2} \sum_{k=N+1}^{\infty}
{{1\over 2}\choose k} {1\over B^{2k-1}}
\; , \;\;\;\;\;\; \hbox{for} \;\; \nu \to \infty \; .
\eeq
The limit clearly shows
that even for arbitrarily small but finite $\hbar$ ($1<**From which we obtain
\beq
{\partial^3\over \partial E^3}
\int dx {V{''}^2(x) \over \sqrt{E - V(x)} } =
{135 \pi \alpha^3 \sqrt{U_0}\over 2 U_0^2} \; .
\eeq
For the last integral we have
\beqa
\int dx {V{'}^2(x) V{''}(x) \over \sqrt{E - V(x)} } & = &
{8 \alpha^3 U_0^2\over \sqrt{U_0}}
\int_{-\sqrt{\beta}}^{\sqrt{\beta}} dz
{z^2(3z^2+1)(z^2+1)^2\over \sqrt{\beta - z^2}} =
\nonumber
\\
& = & {8 \alpha U_0^2\over \sqrt{U_0}} (105 \beta^4 + 280 \beta^3 +
240 \beta^2 + 60 \beta ){\pi\over 128} \; ,
\eeqa
from which we obtain
\beq
{\partial^4\over \partial E^4}
\int dx {V{'}^2(x) V{''}(x) \over \sqrt{E - V(x)} }
={315 \pi \alpha^3 \sqrt{U_0}\over 2 U_0^2} \; .
\eeq
In conclusion we have
\beqa
({\hbar \over i})^4 \oint d\sigma_4 & = & {\hbar^4 \over (2m)^{3/2}}
{\alpha^3 \sqrt{U_0}\over U_0^2} \big[
{1\over 120} {135 \over 2} - {1\over 288} {315 \over 2} \big]=
\nonumber
\\
& = & {\hbar^4 \alpha^3 \pi
\sqrt{U_0}\over 64 (2m)^{3/2}U_0^2} = {2\pi \hbar \over 16 B^3} \; .
\eeqa
\newpage
\section*{References}
\parindent=0. pt
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\\\\
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\\\\
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\\\\
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\\\\
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\\\\
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\end{document}
**