You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'The Maslov correction to the wave function is the jump of $-\pi/2$ in the phase when the system passes through a caustic point. This phenomenon is related to the second variation and to the geometry of paths, as conveniently explained in Feynman’s path integral framework. The results can be extended to any system using the semiclassical approximation. The $1$-dimensional harmonic oscillator is used to illustrate the different derivations reviewed here.'
author:
- |
    P. A. Horváthy[^1]\
    Laboratoire de Mathématiques et de Physique Théorique\
    Université de Tours\
    Parc de Grandmont\
    F-37 200 TOURS (France)
title: The Maslov correction in the semiclassical Feynman integral
---

arXiv : `quant-ph/0702236`

Introduction
============

A fascinating fact, first observed at the end of the 19th century [@Gouy], is that, after passing through a focal point, the phase of light jumps by $-\pi/2$. If a light beam is split into two parts one of them passes through a focal point while the other does not, then, when the two partial waves are recombined, a destructive interference is observed.

This curious phenomenon has later been extended to the quantum mechanics of massive particles, where it is referred to as the “Maslov correction” [@Keller; @Maslov; @Arnold]. A simple illustration is provided by the harmonic oscillator [@JMS]. Feynman’s path integral framework [@FeynmanHibbs] is ideally suited to understand how this comes about [@HPA].

The aim of this Review is to derive the Maslov correction and study some if its aspects in the path integral framework.

The Feynman propagator of the oscillator {#Feynmanprop}
========================================

Let us first consider, for simplicity, a problem in $1$ space dimension. Let $x_1$ and $x_2$ two points and let $T>0$ a time interval [^2]. Let us indeed consider the set, denoted by ${\cal P}$ of all curves $\gamma(t)$ between $x_1$ and $x_2$, i. e. such that $\gamma(0)=x_1$ and $\gamma(T)=x_2$. The classical action is a real valued function defined on ${\cal P}$, $$S(\gamma)=\int_0^T\!\! L\big(\gamma(t),\dot{\gamma}(t)\big)dt,
\label{hatas}$$ where $L(x,\dot{x})$ is the Lagrange function of the system. Then the Principle of Least Action (Hamilton’s Principle) tells us that the actual motion which starts in $x_1$ and arrives to $x_{2}$ in $T$ is the $\bar{\gamma}(t)$ that makes $S$ stationary.

Apart of exceptional cases, (see below) the two given points $x_1$ to $x_2$ are joined, in time $T$, by a unique classical motion curve $\bar\gamma(t)$. But in Quantum Mechanics the situation is different. Intuitively, a quantum particle moves not along a unique path, but along [*all paths*]{} which join these points in the given time ! Feynman, in his Thesis, argued indeed that to any path $\gamma$ in $\cP$ is associated a complex number, namely $$\exp\left[\frac{i}{\hbar}S(\gamma)\right],
\label{Feynmanfactor}$$ where $S$ denotes the classical action (\[hatas\]) calculated along $\gamma$. Next, if the “amplitude” (whose square $|\psi|^2$ is the probability) of finding our particle in the instant $t=0$ at the point $x_{1}$ is $\psi(x_{1})$, then the amplitude, $\psi_{T}(x_{2})$, of finding it at $x_{2}$ at time $T$ will be $$U_T\psi(x_2)\equiv\psi_T(x_2)=\int_{-\infty}^{\infty}
K(x_2,T|x_1,0)\psi(x_{1})dx_{1}.
\label{idofej}$$ Here the propagator, $K(x_2,T|x_1,0)$, which describes the transition from $x_{1}$-ből $x_{2}$ in time $T$, is, says Feynman, a “sum” of the contributions (\[Feynmanfactor\]) of all paths, $$K(x_2,T|x_1,0)=\int_\cP\exp\left[\frac{i}{\hbar}S(\gamma)\right]
\cD\gamma.
\label{propag}$$

But all this intuitive, unless we say what “$\cD\gamma$” means here. The definition of the integration measure is indeed the main stumbling block of the whole theory, and, despite many efforts, no fully satisfactory answer is available as yet. The miracle is that the integral can, in some cases, nevertheless evaluated [@FeynmanHibbs]. Below we present one possible method we illustrate on the example of a [*one dimensional harmonic oscillator*]{}.

Let us indeed consider an arbitrary path $\gamma(t)$ in $\cP$ that satisfies the given boundary conditions. Let us assume that $\cP$ contains a unique classical path, $\bar{\gamma}(t)$, and let us decompose $\gamma$ into the sum of $\bar{\gamma}(t)$ and a $$\gamma(t)=\bar{\gamma}(t)+\eta(t),
\qquad\hbox{ahol}\qquad
\eta(0)=\eta(T)=0,
\label{eta}$$ since the end points are kept fixed. The Lagrange function of the oscillator is $
L=\frac{m}{2}\big({\dot{x}}^2-\omega^2x^2\big). 
$ The action along $\gamma(t)$ is therefore $$S(\gamma)=\int_0^T\!{{\scriptstyle{\frac{1}{2}}}}m\big(\dot{\bar{\gamma}}^2
-\omega^2{\bar{\gamma}}^2\big)dt
+
m\int_0^T\!\big(\dot{\bar{\gamma}}\,\dot{\eta}
-\omega^2\bar{\gamma}\,\eta\big) dt
+
{{\scriptstyle{\frac{1}{2}}}}m\int_0^T\!(\dot{\eta}^2
-\omega^2{\eta}^2)dt.
\label{hatkifejtes}$$ For the oscillator, $
\ddot{x}+\omega^2x=0
$, so that $
x(t)=A\sin\omega t +B\cos\omega t,
$ where the constants $A$ and $B$ are determined by the initial conditions, $$B=x_1\qquad\hbox{and}\qquad
A=\frac{x_2-x_1\cos\omega T}{\sin\omega T},$$ provided $\sin\omega T\neq 0$,i.e., if $\omega T\neq N\pi$. Hence [*If $T$ is not integer multiple of the half-period, $$T\neq N\times\frac{\tau}{2},
 \qquad\tau=\frac{2\pi}{\omega},
 \label{nemkauszt}$$ there exists a unique classical path that starts in $x_1$ and arrives, after time $T$, to $x_2$-be*]{}. The action calculated along this path is $$S(\bar{\gamma})=\frac{m\omega}{2\sin\omega T}\left(
(x_1^2+x_2^2)\cos\omega T-2x_1x_2\right).
\label{oszchat}$$

Returning to (\[hatkifejtes\]), the first term is $S(\bar{\gamma})$. Next, integration by parts of the middle term yields $$\int_0^T\!\big(\dot{\bar{\gamma}}\,\dot{\eta}
-\omega^2\bar{\gamma}\,\eta\big) dt=-\int_0^T\!\big(\ddot{\bar{\gamma}}
+\omega^2\bar{\gamma}\big)\,\eta\, dt=0,$$ since $\bar{\gamma}$ satisfies the classical equation of motion $\ddot{x}=-\omega^2x$. This term vanishes therefore, providing us with the propagator $$K(x_2,T|x_1,0)=
\exp\left[\frac{i}{\hbar}S(\bar{\gamma})\right]\times F(T),
\label{oscpropag}$$ where the “reduced propagator”, $F(T)$, is a path integral taken over all variations, $$F(T)=\int \exp\left\{\frac{im}{2\hbar}\int_0^T\!\left[\big(
\dot{\eta}^2-\omega^2\eta^2\right]dt\right\}\cD\eta.
\label{F(T)}$$

But all this is still intuitive, as we still not say what “$\cD\eta$” actually means. To answer this question, Feynman expands the (periodic) variation into a Fourier series, $$\eta(t)=\sum_{k=1}^\infty a_k\sin [k\frac{\pi}{T}t]
\quad\Rightarrow\quad
\int_0^T\!{\eta}^2dt=\frac{T}{2}\sum_ka_k^2,
\qquad 
\int_0^T\!\dot{\eta}^2dt
 =\frac{T}{2}\sum_k\big(\frac{k\pi}{T}\big)^2a_k^2.
 \label{four}$$ Then he argues as follows. Any path is determined by its Fourier coefficients $a_{k}$; instead of “integrating over all paths” let us integrate over all Fourier coefficients, $$\begin{aligned}
F(T)=\lim_{n\to\infty}\,{\cal J}\int_{-\infty}^{\infty}\!
\dots \int_{-\infty}^{\infty} \!
\exp\left\{\sum_{k=1}^ni\,\frac{\lambda_k}{2\hbar}a_k^2
\right\}\times da_1\dots da_n
\label{sokintegral}\end{aligned}$$ where $$\begin{aligned}
\lambda_k=m\left(
\big(\frac{\pi k}{T})^2-\omega^2\right)
\label{eigenvalue}\end{aligned}$$ and where ${\cal J}$ denotes the Jacobian of the (linear) transformation $\cP\to \Big\{\hbox{Fourier coefficients}\Big\}$. Let us observe that ${\cal J}$ is independent of the data ($\omega, m$, etc) of the oscillator and even of $\hbar$.

The classical Fresnel integrals can be evaluated, $$\int_{-\infty}^{\infty}\exp\left[i\frac{\lambda}{2}x^2\right]dx=
\sqrt{\frac{2\pi}{\lambda}}\,e^{i\frac{\pi}{4}}.
\label{efresnel}$$ Then the sum in the exponent in (\[sokintegral\]) becomes a product, $$F(T)=\lim_{n\to\infty} C_{n}
\left(\prod_{k=1}^n\lambda_k\right)^{-1/2},
\label{faktorok}$$ where $C_{n}$ is the product of various (divergent) factors.

The product of eigenvectors can be split into two parts. $$\prod_{k=1}^n\lambda_k=
\prod_{k=1}^nm\frac{k^2\pi^2}{T^2}\times
\prod_{k=1}^n\Big(1-\frac{\omega^2T^2}{k^2\pi^2}\Big).
\label{tenyezok}$$ The Euler formula says now that $$\prod_{k=1}^\infty\left(1-\frac{x^2}{k^2\pi^2}\right)=
\frac{\sin x}{x},
\label{Euler}$$ so that $$F(T)=C\sqrt{\frac{\omega T}{\sin\omega T}},$$ where $C$ denotes the product of all $\omega$-independent factors.

Let us now remember that for $\omega\to 0$ we get a free particle, and therefore $$F^{free}(T)=\sqrt{\frac{m}{2\pi i\hbar T}}.
\label{szabadredprop}$$ This yields $C$, and inserting the action (\[oszchat\]) the propagator becomes, at last, $$K(x_2,T|x_1,0)=\left(\frac{m\omega}{2\pi i\hbar\sin\omega T}\right)^{1/2}\!
\times\exp\left\{\frac{im\omega}{2\hbar\sin\omega T}
\big[(x_1^2+x_2^2)\cos\omega T-2x_1x_2\big]\right\}.
\label{Fprop}$$

The phase correction
====================

Let us observe that if the “$\sin$” in the denominator vanishes, the whole expression becomes meaningless. How should we then continue after such a singular point. The answer comes from the evaluation of the Fresnel integrals : Euler’s formula, (\[efresnel\]), is only valid for $\lambda>0$. But $$\lambda_k>0\qquad \Longleftrightarrow\qquad
0<T<k\frac{\tau}{2},
\qquad$$ where $\tau=2\pi/\omega$ is the period. Before reaching the first half period, $0<T<\tau/2$, all factors under the square root are positive and Feynman’s calculation is correct. After the first half period (but before a full periode), however, i. e. for $\tau/2<T<\tau$, the first factor under the root is negative, while all the other factors remain positive. $$\lambda_1<0,\qquad \lambda_k>0, \quad k\geq2.$$ As a result, the propagator gets multiplied by $$\frac{1}{\sqrt{-1}}=-i=e^{-i\frac{\pi}{2}}.$$ Thus, the [*phase of the propagator (and therefore also of the wave function) jumps by $(-\pi/2)$.*]{}

Similarly, after $N$ but before $(N+1)$, half-periods, i.e. for $$N\times\frac{\tau}{2}<T<(N+1)\times\frac{\tau}{2},
\label{N}$$ the first $N$ factors in (\[faktorok\]) become negative. The phase jumps, therefore, by $N\times(-\pi/2)$. The correct result is hence $$\begin{array}{ll}
K(x_2,T|x_1,0)=
&\left(\displaystyle{\frac{m\omega}{2\pi\hbar|\sin\omega T|}}\right)^{1/2}
\!e^{-\frac{i\pi}{4}} 
\\[16pt]
&\times\exp\left\{\displaystyle{\frac{im\omega}{2\hbar|\sin\omega T|}}
\times[(x_1^2+x_2^2)\cos\omega T-2x_1x_2]
\right\}\times e^{-\frac{i\pi}{2}N}.
\end{array}
\label{Masprop}$$

We must admit that our argument has been somewhat sloopy: each factor $\sqrt{-1}$ could be $i$ or $-i$, and choosing the second one would change every phase jump from $-\pi/2$ to $\pi/2$. Worse: every jump can be chosen independently ! Which square root of $(-1)$ has to be chosen ? During the first half-period, the question is irrelevant, since it is merely a global phase. But repeating it $N$ times it [*is*]{} relevant !

The formula that generalizes (\[efresnel\]) can be derived through analytic extension. For any real $\lambda\neq0$, $$\int_{-\infty}^{\infty}\exp\left[i\frac{\lambda}{2}x^2\right]dx
=\left\{
\begin{array}{cc}
\displaystyle(\frac{2\pi}{\lambda})^{1/2}e^{i\frac{\pi}{4}}
&\lambda >0
\\[14pt]
\displaystyle(\frac{2\pi}{-\lambda})^{1/2}e^{-i\frac{\pi}{4}}
&\lambda <0
\end{array}\right..
\label{Fresnel}$$ Euler’s formula, (\[Euler\]), only holds for $x<\pi$, and otherwise it should be replaced by $$\prod_{k=1}^\infty\left|1-\frac{x^2}{k^2\pi^2}\right|=
\frac{|\sin x|}{x}. 
\qquad x>0,
\label{Eulerbis}$$

This confirms the validity of our previous argument : passing through every half-peridod contributes a new negative $\lambda$, and this changes the phase by $\pi/2$.

What happens for $T=N\times\tau/2$ ? The propagator plainly diverges, since $\sin\omega T\to \sin N\pi=0$ in the denominator. To derive the correct result, let us use the (semi-)group property of the time evolution, $$U_{t+t'}=U_t\smallcirc U_{t'}
\qquad\Rightarrow\qquad
U_{N\frac{\tau}{2}}=(U_{\frac{\tau}{4}})^{2N}.$$ According to (\[Fprop\]) $$U_{\frac{\tau}{4}}\psi(x_2)=(\frac{m\omega}{2\pi\hbar})^{1/2}
e^{-i\pi/4}\times
\int e^{-i\frac{m\omega}{\hbar}x_1x_2}\psi(x_1)dx_1,$$ which is, essentially, a Fourier-transformation. But the square of such a Fourier-transformation carries any function into itself up to a change of its argument. Hence $$\psi_{N\frac{\tau}{2}}(x_2)=e^{-i\frac{\pi}{2} N}\psi((-1)^Nx_2),$$ i.e. $$K(x_{2},T|x_{1},0)=\exp\left[-i\frac{\pi}{2}N\right]\times
    \delta\big(x_{1}-(-1)^Nx_{2}\big).
    \label{kausztprop}$$

Going to $D$ dimensions, the phase will jump by $D\times\pi/2$. What happens if the oscillator is not perfectly harmonic ? One has to study higher-order terms [@Schulman].

The result can be extended to several similar situations.

$\bullet$ For a [*forced oscillator*]{} driven by a constant external force [@FeynmanHibbs; @Cheng; @LiangMorandi], the previous calculation can be repeated word-by-word. The Lagrange function is $$L=\frac{m}{2}\big({\dot{x}}^2-\omega^2x^2\big)+fx, 
\label{forcedoszc}$$ where $f=\const$. For $T\neq k\tau/2$ the Hamiltonian action is $$\begin{array}{ll}
S_f(\bar{\gamma})=&\displaystyle\frac{m\omega}{2\sin\omega T}\left(
(x_1^2+x_2^2)\cos\omega T-2x_1x_2\right)
\\[14pt]
&+2f\displaystyle\frac{(1-\cos\omega T)}{m\omega^2}(x_1+x_2)
-f^2\displaystyle\frac{2(1-\cos\omega T)-\omega T\sin\omega T}{m^2\omega^4}
\end{array}
\label{focedoszchat}$$ cf. (\[oszchat\]). The propagator is again (\[Masprop\]) with the only change that the extra terms in (\[focedoszchat\]) should be accounted for. In fact [@LiangMorandi], $$K_f(x_2,T|x_1,0)=\exp\left[i\displaystyle\frac{f^2T}{2m\omega^2\hbar }\right]K_{osc}(x_2-x^*,T|x_1-x^*,0)$$ where $x^*=f/m\omega^2$. This latter formula also holds for $T=N\times\tau/2$.

Derivation from the wave function
=================================

Before further investigating the various aspects and derivations of the Maslov correction from the Feynman integral viewpoint, we would like to show how it can be understood using our knowledge of the solution of the Schrödinger equation. The clue [@LiangMorandi] is to write the propagator as $$K(x_1,x_2;T)=\sum_{n=0}^\infty\exp[-i(n+{{\scriptstyle{\frac{1}{2}}}})\omega T]\psi_n(x_1)\psi_nx_2,
\label{propwf}$$ where the $\psi_n$ s are the normalized harmonic oscillator wavefunctions : $$\psi_n(x)=\left(\frac{1}{2^nn!}\sqrt{\frac{m\omega}{\pi\hbar}}\right)^{1/2}
\,\exp(-{{\scriptstyle{\frac{1}{2}}}}\zeta^2)H_n(\zeta),
\label{owf}$$ where $\zeta=\sqrt{m\omega/\hbar}x$ and the $H_n$ denote the Hermite polynomials. Their property important for us is their behavior under space reflection, $
\psi_n(-x)=(-1)^n\psi_n(x),
$ implied by that of the Hermite polynomials.

(\[propwf\]) can now be evaluated. Let us assume that we are not in caustic, $t\neq N\pi/\omega$, where $N$ is some integer. Then write $ 
\omega t=N\pi+\omega\tau,
$ $0<\tau<\pi/\omega$. $N$ is hence the integer part of $\omega t/\pi$. Now from (\[propwf\]), we infer that $$K=\exp(-{{\scriptstyle{\frac{1}{2}}}}iN\pi)\,\sum_{n=0}^\infty\exp[-inN\pi-(i(n+{{\scriptstyle{\frac{1}{2}}}})\omega\tau]\psi_n(x_1)\psi_n(x_2).$$

$$K(x_1,x_2;T)=\exp(-{{\scriptstyle{\frac{1}{2}}}}i\pi{N})\sum_{n=0}^\infty
\exp[-i(n+{{\scriptstyle{\frac{1}{2}}}})\omega t]\psi_n(x_1)\psi_n((-1)^Nx_2).
\label{propwfallt}$$

The Hermite polynomials admit a generating function, namely $${\cal G}(\zeta_1,\zeta_2;z)=
\frac{1}{\sqrt{1-z^2}}\exp\left(\frac{2z\zeta_1\zeta_2-z^2(\zeta_1^2+\zeta_2^2)}{1-z^2}\right).
\label{Hgenfunc}$$ ${\cal G}$ is analytic over the $z$ plane with cuts going from $-\infty$ to $-1$ and from $+1$ to $+\infty$. For $|z|<1$ the generating function admits the expansion $$\begin{aligned}
{\cal G}(\zeta_1,\zeta_2;z)=\sum_{n=0}^\infty\frac{z^n}{n!2^n}\,H_n(\zeta_1)H_n(\zeta_2).
\label{Hgenfuncexp}\end{aligned}$$ The product of the Hermite polynomials is expressed therefore, as $$\begin{aligned}
H_n(\zeta_1)H_N(\zeta_2)=2^n\frac{\ d^n}{dz^n}{\cal G}(\zeta_1,\zeta_2;z)\bigg|_{z=0}.\end{aligned}$$ Then putting $z=\exp(-\alpha i),\, 0<\alpha<\pi$, a lengthy calculation [@LiangMorandi] yields $${\cal G}=\exp[{{\scriptstyle{\frac{1}{2}}}}(i\alpha+\zeta_1^2+(\zeta_2)^2)]\frac{e^{-i\pi/4}}{2\sin\alpha}
\exp\left(\frac{i}{2\sin\alpha}[\zeta_1^2+(\zeta_2)^2\cos\alpha-2\zeta_1\zeta_2]\right).$$ Collecting our formulae, $$K(x_1,x_2,T)=\sqrt{m\omega/\pi\hbar}\exp(-{{\scriptstyle{\frac{1}{2}}}}\pi{N}[\frac{\omega t}{\pi}]
\exp[-{{\scriptstyle{\frac{1}{2}}}}(i\omega\tau+\zeta_1^2+\zeta_2^2)]{\cal G}(\zeta,(-1)^k\zeta_2;e^{-i\omega\tau}).$$ Noting finally that $
\sin(\omega\tau)=|\sin(\omega\tau)|=(-1)^k\sin(\omega t)
$ and $
\cos(\omega\tau)=(-1)^k\cos(\omega t)
$ we get precisely the propagator (\[Fprop\]), valid for times which are not integer multiples of half-periods.

In a caustic i.e. if $t=(\pi/\omega)N$, then, letting $\omega t\to N\pi$ in (\[propwfallt\]) yields [@LiangMorandi], using the completeness of the wave functions, the formula valid in the caustic point, i.e., $$\lim_{\omega t\to N\pi}K(x_1,x_2;t)=
\exp(-{{\scriptstyle{\frac{1}{2}}}}iN\pi)\delta(x_1-(-1)^Nx_2)$$ cf. (\[kausztprop\]).

Let us mention that the Maslov correction can also be derived from a semiclassical analysis of the wave function [@Keller].

How minimal is the “minimal action” ?
=====================================

Let us now return to classical mechanics. The classical hamiltonian action is, as explained in Section \[Feynmanprop\], a real function, $S(\gamma)$, defined on the “infinite dimensional manifold”, $\cP$, of all paths which join $x_{1}$ to $x_{2}$ in times $T$. A variation $\eta$ can be viewed in turn as a “tangent vector” to $\cP$ in $\gamma$ [@HPAU], cf. Fig. 1.

![Those paths which join $x_1$ to $x_2$ in time $T$ form an “infinite dimensional manifold” $\cP$. The the actual motion, $\bar\gamma(t)$, is a critical point of the hamiltonian action, viewed as a real-valued function on $\cP$.](varcurves.pdf "fig:") \[varcurves.pdf\]

$\delta S$, the first variation of the action, is the directional derivative of $S$ in the “point” $\gamma$, $$\delta S_{\gamma}(\eta)=\lim_{s\to0}
    \frac{S(\gamma+s\eta)-S(\gamma)}{s}.
  \label{1vari}$$ $\delta S_{\gamma}$ is, hence a one-form on $\cP$. Just like in finite dimensional calculus, if $\bar{\gamma}$ is an extremal point of $S$, then the directional derivative must vanish in any direction, $$\delta S_{\bar{\gamma}}\equiv\delta S_{\bar{\gamma}}(\eta)=
    \int\left\{\frac{\ d}{dt}
    \Big(\frac{\p L}{\p \dot{x}}\Big)-\frac{\p L}{\p x}\right\}\eta dt=0.
    \label{variegy}$$ This yields the classical (Euler-Lagrange) equations, $
\frac{\ d}{dt}
    \Big(\frac{\p L}{\p \dot{x}}\Big)-\frac{\p L}{\p x}=0,
$ whose solution satisfying the boundary conditions, $\bar{\gamma}$ is, by Hamilton’s Principle, is the actual motion.

But does such a solution always exist, and if it does, is it unique ? The answer is, generically, affirmative. Not always, however. This point can again illustrated by the oscillator. Let us indeed assume that the elapsed time is an integer multiple of the half-period, $$T= N\times\frac{\tau}{2}.
\label{kauszt}$$ Then, independently of the initial velocity, [*all*]{} motions which start in $x_1$ arrive, at time $T$, to $x_2=(-1)^Nx_1$. (cf. Fig. 2.).

![After a half period, all trajectories meat again, independently of the initial velocity, in the same point opposite to the one they started from.](kausztika.pdf)

\[kausztika\]

For this particular time, (i) either i.e. when $x_2\neq(-1)^Nx_1$, there is no classical path at all with the required boundary condition, (ii) or i.e. for $x_2=(-1)^Nx_1$, there are infinitely many of them.

The solutions of the classical equation of motion behave, hence, as light rays that start from one focus of an elliptic mirror: after reflection, they are all collected into the same point, namely the other focus. (By analogy, such point are also called focal points also in variational mechanics).

Apart of this particular situation, there is a unique classical path, $\bar{\gamma}$, in $\cP$. Let us assume that we are in this, generic, situation. Is the action minimal ? Let us emphasized that, just like in finite dimension, (\[variegy\]) is merely a [*necessary*]{} condition for having a minimum. It is not sufficient, though. In other words, $\bar{\gamma}$ is a critical point of $S$, but not necessarily a minimal one.

Let us mention, at this point, that the variational calculus is in fact local, i.e., it is a differential calculus in the neighborhood of a path $\bar\gamma$. When calculating the first variation, we compare in fact the value of $S$ calculated along paths which can be smoothly deformed to $\bar\gamma$. All such paths belong to a single path component of $\cP$. If the latter has more then one path components, as it happens in the, one has to perform the variational calculus in each path-connected sector separately.

In technical terms, a “point” (i.e. a path in a manifold $M$) $\gamma$ belongs to the path-component of $\bar\gamma$ if $\gamma$ can joined to the “point” $\bar\gamma$ in $\cP$ if there exists a continuous “path” of “points” $\gamma_s$ such that $ 
\gamma_1=\gamma,
\,
\gamma_0=\bar\gamma.
$ But this means precisely that $\bar\gamma$ can be deformed into $\bar\gamma$ by a homotopy. The path components of $\cP$ are therefore labeled with the homotopy classes of the underlying space, $
\pi_0(\cP)=\pi_1(M).
$

An illustration is provided by the Aharonov-Bohm experiment [@AB], where $M$ is the punctured plane $\IR^2\setminus\{0\}$, whose $\pi_1$ is $\IZ$. The (free) action has, hence, a minimum in two homotopy classes, namely in those which pass near the solenoid (assumed infinitely thin) on either side. In all other path components, i. e. for paths which turn around the origin, the action has no critical point.

Returning to the critical points of the action, they behave just like those of a function defined over a finite ($D$) dimensional space : $\vx_0$ is a critical point of a function $V(\vx)$ if $$\delta V(\vx_0)=0
\qquad\hbox{i.e.}\qquad
\p_iV(\vx_0)=0\qquad\forall \ i.$$ Whether a given critical point $\vx_0$ is a minimum, a maximum or a saddle point, can be determined by looking at the matrix formed from second-order partial derivatives, $$\delta^2 V\equiv \delta^2 V(\vx_0)=
\frac{\p^2V}{\p x^i\p x^j}(\vx_0).$$ Is the matrix of $\delta^2 V$ definite ? Being symmetric, $\delta^2 V$ will have $D$ (real) eigenvalues, $$\delta^2 V\,e_a=\lambda_a\,e_a,
\qquad
a=1,\dots, D.$$ Then $\vx_0$ is $$\left\{\begin{array}{lll}
\hbox{a minimum if}\;&\lambda_a>0\,&\forall a
\\[8pt]
\hbox{a maximum if}\;&\lambda_a<0\,&\forall a
\\[8pt]
\hbox{saddle point if}\qquad &\lambda_a>0\quad\lambda_b<0\;&\hbox{for some}\; a, b.
\end{array}\right.$$

In an analogous way, to see if a given classical path $\bar\gamma$ makes the action minimal one or not, we have to calculate the second variation, $\delta^2S_{\bar{\gamma}}$. This is a quadratic form on the “tangent space” of the “variations”. To be a minimum, the second variation has to be positive definite, $$\delta^2S_{\bar{\gamma}}(\eta,\eta)>0
    \qquad\forall \eta.
    \label{minfelt}$$ If, however, $$\delta^2S_{\bar{\gamma}}(\eta,\eta)<0
    \qquad\hbox{but}\qquad
    \delta^2S_{\bar{\gamma}}(\eta',\eta')>0
    \label{nyeregfelt}$$ for suitable variations $\eta$ and $\eta'$, then $\bar{\gamma}$ is a saddle point.

A variation $\eta$ such that $$\delta^2S_{\gamma}(\eta,\eta)<0$$ is a negative mode, and if $${{\scriptstyle{\frac{1}{2}}}}\delta^2S(\eta,\eta')=0
\qquad\forall\eta',
\label{nullter}$$ it is a zero-mode. In the direction of a zero mode the function is, in the first approximation, invariant. In this case, the nature of the critical point depends on the higher variations.

Let us examine the second variation in some detail. $${{\scriptstyle{\frac{1}{2}}}}\delta^2S\equiv{{\scriptstyle{\frac{1}{2}}}}\delta^2S_{\bar{\gamma}}(\eta,\eta)
={{\scriptstyle{\frac{1}{2}}}}\int_{0}^T\!\left\{
\frac{\p^2L}{\p x^2}\eta^2+
2\frac{\p^2L}{\p x\p \dot{x}}\eta\dot{\eta}+
\frac{\p^2L}{\p x^2}\eta^2+
\frac{\p^2L}{\p \dot{x}^2}{\dot{\eta}^2}
\right\}dt,$$ where the integration is along the (assumed unique) classical path $\bar\gamma$. After partial integration, $$\begin{aligned}
&{{\scriptstyle{\frac{1}{2}}}}\delta^2S=\int_{0}^T\!\big(\eta,\Lambda\eta\big)dt,\nonumber
\\[10pt]
&\Lambda=
    -\displaystyle\frac{\ d}{dt}\left(
    \displaystyle\frac{\p^2L}{\p \dot{x}^2}\displaystyle\frac{\ d}{dt}+
    \displaystyle\frac{\p^2L}{\p x\p \dot{x}}\right)
    +
    \left(\displaystyle\frac{\p^2L}{\p x\p \dot{x}}
    \displaystyle\frac{\ d}{dt}+\displaystyle\frac{\p^2L}{\p x^2}
    \right).
    \label{Lambda}\end{aligned}$$ where $\Lambda$ is the operator of the second variation. ${{\scriptstyle{\frac{1}{2}}}}\delta^2S$ is hence positive definite if all eigenvalues of the quadratic form $\Lambda$ are positive, $\lambda>0$, for all $$\Lambda\,\eta=\lambda\eta,
    \qquad
    \eta(0)=\eta(T)=0.
    \label{sajert}$$

$\bullet$ In the simplest possible case of a one-dimensional massive particle in a potential, $$L=\frac{m{\dot{x}}^2}{2}-V(x)
    \qquad\Rightarrow\qquad
    \Lambda=-m\left(\frac{\ d^2}{dt^2}+\frac{d^2V}{dx^2}\right).
    \label{Lambdabis}$$ For the $1D$ oscillator, e.g., $$\Lambda_{osc}=-m\left(\frac{\ d^2}{dt^2}+\omega^2\right).
    \label{oscLambda}$$ Hence, taking into account the boundary conditions, $$\begin{aligned}
m(\ddot{\eta}+\omega^2\eta\big)=-\lambda\eta
\quad\Rightarrow\quad
\eta(t)=\sin[\big(\omega^2+\frac{\lambda}{m}\big)^{1/2}\,t].\end{aligned}$$ But $T\sqrt{\omega^2+\frac{\lambda}{m}}=k\pi,\; k=0,\pm1,\dots$ due to periodicity, so that $$\lambda_{k}=m\left(\big(\frac{k\pi}{T}\big)^2-\omega^2\right)
    \qquad\hbox{and}\qquad
    \eta_{k}(t)=\sin(\frac{k\pi}{T}t),
    \label{oscsajert}$$ cf. (\[four\]) and (\[eigenvalue\]). The integer $N>0$ in (\[N\]) $
N\tau/2<T<(N+1)\tau/2,
$ counts hence the negative eigenvalues. Thus, the oscillator trajectory is a minimum of the action during the first half period. For time beyond $N\tau/2$, it becomes a saddle point with $N$ negative modes !

What happens in the focal points ? For $T=N\times \tau/2$, as seen above, either we have no classical path at all or we have infinitely many between our fixed endpoints. Let $x_2=(-1)^Nx_1$ and let us consider some $\bar{\gamma}$. Then all the other classical paths can be viewed as a variation of $\bar{\gamma}$, labeled by a parameter $s$, $\gamma=\gamma_s$. The action calculated for any such classical path $\gamma=\gamma_s$ is the same. Varying $s$ we get, hence, a “curve” in $\cP$ inscribed onto the “level surface” $S=\const$. The derivative w.r.t. $s$ of $S(\gamma_s)$ is therefore zero, so that $$\eta(t)=\frac{\ d}{ds}\gamma_s(t)\bigg|_{s=0}$$ is a null mode at $\bar\gamma$.

Minimum or not : what are the physical consequences ? At the purely classical level, nothing at all : (\[variegy\]) yields the correct equations of motion in all cases. In Quantum Mechanics, however, the consequence is precisely the Maslov phase jump, as we explain it below.

Let us first consider another example.

$\bullet$ A phase jump similar to the one found for the oscillator is observed for a charged particle moving perpendicularly to the induction lines of a constant magnetic field [@FeynmanHibbs]. Classically, the particle rotates uniformly with Larmor frequency ${eB}/{m}$ i.e. with period $\tau=2\pi m/eB$. The Lagrange function is $$L_B=
\frac{m}{2}\left(\dot{x}^2+\dot{y}^2+
2\omega\big(x\dot{y}-y\dot{x}\big)\right),
\qquad
\omega=\frac{eB}{2m}.$$

If $T$ is not an integer multiple of a (full) period, $
T\neq N\times\tau=\frac{2N\pi m}{eB},
$ then there is a unique classical trajectory that links any two points $(x_1,y_1)$ and $(x_2,y_2)$ in time $T$. The action calculated for it is $$S_B=\frac{m\omega}{2}
    \Big[\big[(x_{2}-x_{1})^2+(y_{2}-y_{1})^2\big]
    \cot(\omega T)
    +2(x_{1}y_{2}-x_{2}y_{1})\Big].
\label{Bhatas}$$ The matrix of the second variation is now $$\Lambda_{B}=m
    \left(\begin{array}{cc}
    \displaystyle\frac{\ \ d^2}{dt^2}&-2\omega\displaystyle\frac{\ d}{dt}
    \\[8pt]
    2\omega\displaystyle\frac{\ d}{dt}
    &\displaystyle\frac{\ \ d^2}{dt^2}
    \end{array}
    \right).
    \label{BLambda}$$ Our task is to solve the eigenvalue problem $$\left\{\begin{array}{ll}
m\ddot{\eta}_x-2m\omega\dot{\eta}_y=-\lambda\eta_x
\\[8pt]
m\ddot{\eta}_y+2m\omega\dot{\eta}_x=-\lambda\eta_y
\end{array}\right.,
\qquad
\eta_x(0)=\eta_y(0)=0=\eta_x(T)=\eta_y(T),
\label{Beigen}$$

The solutions are readily derived by separating the equations (\[Beigen\]) by applying a time-dependent rotation, $$\begin{aligned}
\left\{\begin{array}{ll}
\eta_x=\ \ \cos\omega t\,\xi+\sin\omega t\,\zeta
\\[8pt]
\eta_y=-\sin\omega t\,\xi+\cos\omega t\,\zeta
\end{array}\right.
\quad\Rightarrow\quad
\left\{\begin{array}{ll}
m\ddot{\xi}+(m\omega^2+\lambda)\xi=0
\\[8pt]
m\ddot{\zeta}+(m\omega^2+\lambda)\zeta=0
\end{array}\right..\end{aligned}$$ By periodicity, $
\xi,\zeta\propto\sin\big(\sqrt{\omega^2+\lambda}\,t\big)
=\sin\frac{k\pi}{T}t,
$ where $k$ is some integer. The eigenvalues are, therefore, doubly degenerate, and are identical to those in the oscillator problem : $$\lambda_{k}=m\left(\big(\frac{k\pi}{T}\big)^2-\omega^2\right),
    \qquad
    k=0,1,\dots.
    \label{Bsajert}$$ The [*reduced propagator is hence identical to that of a planar oscillator whose frequency is half of the Larmor value,* ]{} $\omega=eB/2m$.

If $N\tau<T<(N+1)\tau$, then $$\begin{array}{ll}
    K(x_2,y_2,T|x_1,y_1 0)=\displaystyle
    \frac{m\omega}{2\pi i\hbar|\sin\omega T|}\times
    \\[18pt]
    \exp
    \left\{\displaystyle\frac{im\omega}{2\hbar}
    \Big[\big[(x_{2}-x_{1})^2+(y_{2}-y_{1})^2\big]\cot(\omega T)
    +2(x_{1}y_{2}-x_{2}y_{1})\Big]\right\}(-1)^N,
    \quad\omega=\displaystyle\frac{eB}{2m}.
    \end{array}
    \label{Bprop}$$ cf. [@Cheng2; @LiangMorandi]. After $N$ full periods i.e. at $
T=\frac{2N\pi m}{eB}
$ \[which corresponds to $N$ half-oscillator-periods\], all classical motions meet again in the point they started from. According to our previous results, taking into account the dimension of the problem, the propagator is again a Dirac-delta with a sign change : $$K(x_2,y_2,\tau|x_1,y_1, 0)=(-1)^N\delta\big(x_1-x_2,y_1-y_2\big).$$

It is worth mentioning that this calculation should turn out to be useful to explain the Sagnac effect [@Sagnac]. This experiment, originally proposed and performed with light and later been repeated with massive particles, amounts to perform a two-slit type interference experiment, when the whole apparatus is fixed on a turntable. The clue is that the inertial force due to rotation behaves exactly as a fictious magnetic field, with twice the mass, $2m$, replacing the electric charge.

The Semiclassical approximation {#szemi}
===============================

Returning to the general case, let us assume that $\cP$ has a unique classical path, $\bar{\gamma}$, and let us develop the action to second order [@LeSmi] : $$S(\gamma)=S(\bar{\gamma})+\delta_{\bar{\gamma}}(\eta)+
{{\scriptstyle{\frac{1}{2}}}}\delta^2S_{\bar{\gamma}}(\eta,\eta)+\dots
\label{hatkif}$$ where the “dots, $\dots$”, denote all higher-order variations.

Taking the semiclassical approximation amounts of dropping all these terms. According to Hamilton’s Principle $\delta S_{\bar{\gamma}}=0$. The semiclassical propagator is, hence, $$K(\vec{x}_2,T|\vec{x}_1,0)=
\exp\left[\frac{i}{\hbar}S(\bar{\gamma})\right]\times
F(T),
\qquad 
F(T)=\int\!\exp\left\{\frac{i}{2\hbar}
\delta^2_{\bar{\gamma}}S(\eta,\eta)\right\}\cD\eta.
\label{SCpropag}$$ cf. (\[oscpropag\])-(\[F(T)\]).

The reduced propagator can be determined diagonalizing the second variation. Let us assume, for simplicity, that the system is $1$-dimensional. The quadratic form $\Lambda$ in the eigenvalue equation (\[sajert\]) is a self-adjoint Sturm-Liouville operator on the space of all “tangent vectors” $\eta$. The eigenvalues, $\lambda_n$, are therefore all real and form a complete orthonormal system w.r.t. the usual scalar product, $
(\eta_{n},\eta_{m})=
T^{-1}\displaystyle\int\eta_{n}\eta_{m}dt=\delta_{nm}.
$ Expanding the variation $\eta$ as $
\eta=\sum_{n}a_{n}\eta_{n}$ yields $
\Big(\eta,\Lambda\eta\Big)=\sum_{k} a_{k}^2\lambda_{k}.
$ The reduced propagator is, therefore, once again (\[sokintegral\]), with the $\lambda_{k}$ denoting the eigenvectors of $\Lambda$ in (\[sajert\]). Hence $$F(T)=C\sqrt{\prod_k\frac{2i\pi\hbar}{\lambda_k}}\ .
\label{Jredprop}$$ As explained before, the Jacobian is independent of the dynamics, so that (\[Jredprop\]) holds also for the free factor, $$F(T)^{free}=C\sqrt{\prod_k\frac{2i\pi\hbar}{\lambda_k^{free}}}\, ,
\label{szabJredprop}$$ where $F(T)^{free}$ and $\lambda_k^{free}$ are the free propagator and the eigenvalues, respectively. Dividing (\[Jredprop\]) by (\[szabJredprop\]) $C$ drops out, $$F(T)=F(T)^{free}\times\sqrt{
{\displaystyle\prod_k\lambda_k^{free}}/{\displaystyle\prod_k\lambda_k\ \ \ \  }}\ \ .
$$ The semiclassical propagator is, therefore, $$K=e^{iS(\bar{\gamma})}\left(\frac{m}{2\pi i\hbar T}\right)^{1/2}\times\sqrt{
\frac{\prod_k\lambda_k^{free}}{\,\prod_k\lambda_k\ \ \ \  }}\ ,
\label{szemiklprop}$$ since, in $D$ dimensions, $
F(T)^{free}=\left(\frac{m}{2\pi i\hbar T}\right)^{D/2},
$ cf. (\[szabadredprop\]).

$\bullet$ For a $1D$ oscillator, we recover the previous result : by (\[tenyezok\]), the quotient of the products of the eignevalues under the root is exactly the infinite product we determined using in the Euler formula, since $\lambda_k^{free}=m\pi^2 k^2/T^2$.

. Our result can also be presented using the [*Van Vleck matrix*]{} [@LeSmi]. Let us indeed remember that, still assuming the uniqueness of the classical path, $\bar{\gamma}$, between the to given points $\vec{x}_{1}$-et $\vec{x}_{2}$ in time $T$, the action can be viewed as function of the end points. $$S(\vec{x}_{1},\vec{x}_{2})=S(\bar{\gamma})$$ is in fact Hamilton’s Principal Function. The determinant of the ${D}\!\times\!{D}$ matrix $$\left[\frac{\p^2 S}{\p \vec{x}_1\p \vec{x}_2}\right]=
    \left[\frac{\p^2 S}{\p x_{1}^i\p x_{2}^j}\right]$$ is called the [*Van Vleck determinant*]{}. Then

([@LeSmi]) : In $D$ [*dimensions, the semiclassical propagator is $$K(\vec{x}_{2},T|\vec{x}_{1},0)
    =
    \left(\frac{1}{2\pi i\hbar}\right)^{1/2}
    \left|{\rm det}
    \frac{\p^2 S}{\p \vec{x}_1\p \vec{x}_2}\right|^{D/2}\!
    \exp\left[-iDN\frac{\pi}{2}\right]\times
    \exp\left[\frac{i}{\hbar}S(\bar{\gamma})\right].
\label{semiclassprop}$$ where $\vec{x}_1=(x_1^i)$ and $\vec{x}_2=(x_2^j)$ are the initial and final point, respectively.*]{}

$\bullet$ For the $1D$ oscillator $$\frac{\p^2 S}{\p x_{1}\p x_{2}}=-\frac{m\omega}{\sin\omega t},
\label{oscivV}$$ so that (\[semiclassprop\]) yields, once again, the previous oscillator propagator formula.

$\bullet$ Let us consider a charged particle in a [*constant magnetic field*]{}, and let us assume that $T\neq\tau$. Hamilton’s Principal function is now (\[Bhatas\]), and the absolute value of the Van Vleck determinant reads $$\big|{\rm det}\,\frac{\p^2 S}{\p \vec{x}_{1}\p \vec{x}_{2}}\big|=
    m^2\omega^2\times{\rm det}\,\left(
    \begin{array}{cc}
    -\cot\omega T&1
    \\[6pt]
    -1&-\cot\omega T
    \end{array}\right)=
    \frac{m^2\omega^2}{\sin^2\omega T}.
    \label{BvVdet}$$ From (\[semiclassprop\]) we get (\[Bprop\]) once again.

Morse theory [@Morse]
=====================

The methods of variational calculus allow us to further deepen our understanding. Let us chose a classical motion $\bar{\gamma}$ \[in $D$-dimensions\], and let us focus our attention to the second variation. As seen in (\[Lambda\]) $${{\scriptstyle{\frac{1}{2}}}}\delta^2S(\eta,\eta)=\int_{0}^T\!\big(\eta,\Lambda\eta\big)dt,$$ where $\Lambda$ is the matrix of the second variation. A [*Jacobi field*]{} along $\bar{\gamma}$ is a $D$-dimensional vector $\vec{\xi}(t)=(\xi_i(t))$ such that $$\Lambda\,\vec{\xi}=0\qquad\hbox{i.e.}\qquad
\Lambda_{ij}\,\xi_j=0
\quad\forall\, i.
\label{Jacobifield}$$

This second-order differential equation in $2D$ dimensions has $2D$ independent solutions. Generally, none of them belongs to the tangent space of $\cP$ at $\bar{\gamma}$, since $\vec{\xi}(0)$ and/or $\vec{\xi}(T)$ may not vanish. Two points $p=\bar{\gamma}(t_1)$ and $q=\bar{\gamma}(t_1)$ of a curve $\bar{\gamma}$, $0\leq t_1<t_2\leq T$ are called conjugate point, if there exists a Jacobi field such that $$\vec{\xi}(t_1)=\vec{\xi}(t_2)=0,$$ i.e., if it belongs to the tangent space. The number of such independent fields is called the multiplicity of the conjugate points $p$ and $q$.

The null space of the second variation is the set of all tangent vectors $\eta$ for which $
{{\scriptstyle{\frac{1}{2}}}}\delta^2S(\eta,\eta')=0
$ for all $\eta'$. The dimension, $\nu$, of the null space is thus the number of independent Jacobi fields that also belong to the tangent space. The null-space is, hence, non-trivial, if the starting and the ending points of $\bar{\gamma}$, $p=x_1=\bar{\gamma}(0)$ and $q=x_2=\bar{\gamma}(T)$, are conjugate. $\nu$ is finite and it is readily seen that $\nu<D$ [@Morse].

It is easy to see that $2D$ independent Jacobi fields can be constructed using classical paths. Let us indeed consider a one-paramater family of classical paths \[i.e. solutions of the Euler-Lagrange equations\] $\beta(u),\ 0\leq u\leq 1$, such that $
\beta(0)=\bar{\gamma}.
$ These paths are not required to have the same end points as $\bar{\gamma}$. Then it is easy to see that $$\vec{\xi}=\frac{\p\beta}{\p u}\bigg|_{u=0}$$ is a Jacobi field along $\bar{\gamma}$. The general solution of the Euler-Lagrange equations depends on $2D$ parameters. Deriving w.r.t. these parameters yields $2D$ independent Jacobi fields.

Let us now define the [*Morse index*]{} of $\bar{\gamma}$ as the maximal dimension, $\mu$, of those subspaces of the tangent space upon which the restriction of the second variation is negative definite. $$\delta^2S_{\gamma}(\eta,\eta)<0.$$ Then we have :

(Morse) [@Morse] : [ *Morse index is the number of those points of $\bar{\gamma}$, each counted with its multiplicity, which are conjugate to $x_1=\bar{\gamma}(0)$.*]{}

The essence of the proof is to show that the number of negative eigenvalues is precisely the Morse index $\mu=\mu(T)$. This is proved by showing that the Morse index, $\mu(t)$, calculated on the segment from $t=0$ to $t$ of the curve is a monotonic function of $t$ which, for sufficiently small $t$, is continuous from the left, $
\mu(t-\epsilon)=\mu(t)
$ for $\epsilon>0$ sufficiently small. From the right, instead, $$\mu(t+\epsilon)=\mu(t)+\nu,\label{mujump}$$ where $\nu$ is, as before, the dimension of the null space of the second variation. The $\mu(t)$ as a function of time is hence, at first, constant, and then, upon crossing a conjugate point, it jumps by $\nu$.

The Morse index is finite, so that our curve contains a finite number of points conjugate to the starting point.

Let us again illustrate the general theory on example.

$\bullet$ For the $1D$ oscillator and for $\Lambda$ in (\[oscLambda\]), the solutions of the Jacobi equation (\[Jacobifield\]) are $
a\sin\omega t+b\cos\omega t,
$ which vanish at $t=0$ and at $t=T$ if $b=0$ and $T\omega=N\pi$. A point $q$ of the classical trajectory is conjugate to $x_1=\bar{\gamma}(0)$ if $$q=q_k=\bar{\gamma}(k\frac{\tau}{2}),
\qquad
k=1,\dots$$ Consistently with our general theorem, the null space of $\delta^2S$ can be also obtained by considering the classical motion which starts from $x_1$ with velocity $v$, $$\frac{v}{\omega}\sin\omega t+ x_1\cos\omega t\quad\Rightarrow\quad
\xi(t)=\frac{\p x(t)}{\p v}=\frac{1}{\omega}\sin\omega t.$$ For $t=N\tau/2$, we get manifestly a null-mode of the second variation.

For a $D$-dimensional oscillator, each conjugate point contributes $D$ null-modes. When $T$ crosses $2N/\tau$ the dimension of the null space, i.e., the Morse index jumps by $D$.

Let us mention that, on compact manifolds, the Morse index is related to the topology of the infinite dimensional manifold of paths [@Morse].

Conclusion
==========

Is the phase observable ? The naive answer to this question is negative: when the wave function is multiplied by any phase factor, the absolute value of the amplitude, and hence its square, the probability, are unchanged. This is only true for one wave function, though. If the system is decomposed into parts and the partial waves are multiplied by different phase factors, recombination will yield an interference.

Similar phase jumps are found in molecular [@molek], nuclear [@nukl], and heavy ion [@heavyion] collisions.

After the original ideas reviewed here have been put forward, several developments took place. See, e.g. [@Tsutsui; @noncons]. Related questions are also discussed in some textbooks [@books].

It is worth calling the attention to that all path integrals studied in this paper can be calculated [@Inomata] by transforming the system into the free form using a “non-relativistic conformal transformation” [@Niederer; @DHconf]. For the harmonic oscillator, for example, every half-period can be conformally related to a full free motion. Then the Maslov correction can be recovered by fitting together the wave function at half-periods [@DHconf]. This latter paper also provides a complete catalog of all possibilities.

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[^2]: It is tacitly assumed that the system is conservative. Were this not the case, one should consider two instants $t_1$ and $t_2>t_1$ instead of $T=t_2-t_1$.