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  How does one quantize the phase-space semiclassically?

+ 7 like - 0 dislike

Often, when people give talks about semiclassical theories they are very shady about how quantization actually works.

Usually they start with talking about a partition of $\hbar$-cells then end up with something like the WKB-wavefunction and shortly thereafter talk about the limit $\hbar\rightarrow 0$.

The quantity that is quantized is usually the Action $\oint p\, \mathrm{d}q$ which is supposed to be a half-integer times $2\pi\hbar$.

What is the curve we integrate over? Is it a trajectory, a periodic orbit or anything like this? And how does this connect to the partition in Planck-cells?

Additionally, what is the significance of the limit $\hbar\rightarrow 0$?

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user user9886
asked Jul 12, 2012 in Theoretical Physics by user9886 (85 points) [ no revision ]
Chapter 2 of these lecture notes could be useful.

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user user10001
Related: physics.stackexchange.com/q/27492/2451

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user Qmechanic
For an expansion in $\hbar$ (more informative than just the classical limit) see ns.math.cas.cz/~englis/38.ps

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user Arnold Neumaier
It's an integer times $2\pi \hbar$, sometimes there is a half-integer offset. The half-integer comes from HO ground state, and has a semiclassical explanation in terms of Maslov indices, but the integer sequence is the main rule.

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user Ron Maimon

3 Answers

+ 6 like - 0 dislike

First, one must appreciate that the phase space is classically parameterized by $x,p$ and coordinates on an ordinary plane commute with each other, $xp=px$. However, in quantum mechanics, this ain't the case. Instead, we have the Heisenberg commutator $$ xp - px = i\hbar. $$ This means that quantum mechanically, the phase space is not an ordinary plane (or higher-dimensional space) but a "noncommutative geometry".

Still, one may use the ordinary plane to parameterize the quantum phase space, too. The operators on the Hilbert space may be canonically associated with ordinary functions $f(x,p)$ of the commuting coordinates via the so-called Wigner transform:


It's important to notice that the nonzero commutator only affects the geometry of the phase space if we study it with a better resolution than $\hbar$. For very large changes of $x$ and $p$, the quantum phase space may be approximated by the classical one.

The $\hbar\to 0$ is the classical limit. In this limit, $xp-px=i\hbar$ is sent to zero, too. So the phase space effectively becomes a classical geometry. If you want to keep $\hbar$ fixed and imagine what the limit means, it means such values of $x,p$ and their changes so that $\Delta x \cdot \Delta p \gg \hbar $. In this limit, the phase space is classical and one may define ordinary contour integrals in it.

In this limit, the number of "cells" confined in such contours is huge, much larger than one, $N\gg 1$. One may calculate the number of cells (i.e. number of basis vectors of an orthonormal basis representing a particle with $x,p$ in the region bordered by the contour) – with precision that only neglects the "somewhat fuzzy boundary" – by taking the area and dividing it by $2\pi \hbar$.

This map may be visualized as the following correspondence between the quantum trace and the classical integral over the phase space: $${\rm Tr} \leftrightarrow \frac{1}{2\pi \hbar}\int dp\,dx $$ This relationship works for any region in the phase space, bounded by any curve. But there are special kinds of curves for which we may say additional things.

We may also give examples explaining why the correspondence above works. Imagine that the space is a circle of radius $R$, i.e. $x$ is identified with $x+2\pi R$, with the circumference. In that case, the momentum $p$ is quantized because the wave function $\exp(ipx/ \hbar)$ must be single-valued on the circle. It follows that $$ p = \frac{N\hbar}{R} $$ The eigenstates of $p$ form a complete basis and we may count how many states are there in a region of the phase space. Imagine the phase space as a thin strip (or, more precisely, a long infinite cylinder: because of the periodic identification of $x$). It is thin in the $x$ direction, the width being $2\pi R$, the circumference of the circle, of course.

How many states are there? If the length of the strip/cylinder in the $p$ direction is $\Delta p$, the number of states in it is $\Delta p\cdot R/ \hbar$ because the spacing is $\hbar/R$. And because the width of the strip is $2\pi R$, the number of states per unit area of the phase space is $$ \frac{\Delta p\cdot R}{\hbar} \cdot \frac{1}{\Delta p} \cdot \frac{1}{2\pi R} = \frac{1}{2\pi\hbar} $$ which exactly agrees with the claim that $2\pi \hbar$ is the area of the single cell. This result would hold true not only for the compactification on the circle but for the infinite $x,p$ as well.

The WKB approximation uses these insights to derive many other things. In particular, the most interesting contours for $\int p\cdot dx$ in the WKB approximation are contours connecting points with $H(x,p)=H_0$, i.e. contours of a fixed energy. These are places of the phase space where an energy eigenstate may be imagined to be localized. If you draw many contours of this form for various values of $H_0$ and you make sure that the area in between them is $2\pi\hbar$ for each adjacent pair, you may claim that the inter-contour "annulus" represents the region of the phase space associated with a particular eigenstate.

For example, you may pick the harmonic oscillator. In the right units chosen for the graph, $H(x,p)=H_0$ are circles for each value of $H_0$ and the inter-contour regions are genuine annuli. Their width will go like $1/\sqrt{n}$ and their radius will go like $\sqrt{n}$ where $n$ is the integer labeling the energy level. You should also choose the internal contour, surrounding the origin, to encircle the appropriate area of the phase space as well.

The classical limit is only accurate enough if the spacing of the contours is dense enough. However, even though it is dense enough, one may quantify what the density actually is – by using the methods of the limit only. To calculate such things, we're really not using just the "classical limit"; we're using some information about the "first quantum correction" to the classical physics, too.

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user Luboš Motl
answered Jul 12, 2012 by Luboš Motl (10,278 points) [ no revision ]
+ 6 like - 0 dislike

Don't be intimidated, semiclassical quantization is very simple, and it can be straightforwardly understood from a few examples which lead to the general case.

Consider a particle in a box. The classical motions are reflections off the wall. These make a box in phase space, as the particle goes left, hits the wall, goes right, and hits the other wall. If the particle has momentum p and the length of the box is L, the area enclosed by this motion in phase space is

$$ p L $$

and the condition is that this is an integer multiple of $h=2\pi\hbar$. This gives the momentum quantization condition from quantum mechanics.

For a 1 dimesional system, the rule is that

$$ \int p dx = n h$$

With a possible offset, so that the right-hand side might be $(n+1/2)h$, or $(n+3/4)h$, as appropriate, but the spacing between levels is given by this rule to leading order in h. This rule can be understood from deBroglie's relation--- the momentum at any x is the wavenumber, or the rate of change of the phase of the wavefunction. The condition (in natural units where $h=2\pi$ ) is saying that the phase change as you follow a classical orbit should be an integer multiple of $2\pi$, i.e. that the wave should form a standing wave.

This formula is not exact, because the quantum wave doesn't follow the classical trajectory, but the WKB approximation just takes this as a starting point, and makes a wave whose phase is given by the value of this integral, and whose amplitude is the reciprocal of the square root of the classical velocity.

The reason this works was known already before quantum theory was fully formulated. But to understand it requires familiarity with action-angle variables

Action-angle variables

Consider an orbit of a particle in one-dimension, with position x and momentum p. You call the area in phase space enclosed by the orbit J, and this is the action. J is only a function of H and it is constant in time (by definition).

The conjugate variable to J is a variable which distinguishes the points of the orbit, and this is called $\theta$. Now you notice that the area in phase space is invariant under canonical transformations (for infinitesimal canonical transformations this is Liouville's theorem), so that the area between the orbits at J and J+dJ is the same as the area in x-p coordinates between J and J+dJ, which is just dJ because that's the definition of J. But this area in J,$\theta$ coordinates is dJ times the period of $\theta$, so $\theta$ has the same period for all J, which I will take to be $2\pi$.

The rate at which $\theta$ increases with time is given by Hamilton's equations

$$ \dot{\theta} = {\partial H\over \partial J} = H'(J) $$

And this is constant over the entire orbit, because H is constant, and so is J. So you learn that $\theta$ increase monotonically at a constant rate at each J, and the time period of $\theta$ is:

$$ T = {2\pi\over H'(J)} $$

Semiclassical quantization

Suppose you weakly couple this one-dimensional system to electromagnetism. The classical orbital frequency is going to be the frequency of the emitted photons (and double this frequency, and three times this frequency), so that if you want to have discrete photon-emission transitions, you must ensure that emitting a photon of frequency $f={1\over T}$, and taking away energy $hf$ leaves you with a quantum state to fall to. So if there is a quantum state corresponding to a classical motion with one value of J, at energy H(J), there must be another quantum state with energy

$$ H(J) - {2\pi h\over T} = H(J) - H'(J)h \approx H(J-h) $$

in other words, the quantum states must be spaced evenly in J. To this order, this means that there are states at J-h,J-2h,J-3h and so on, and transitions to these states have to reproduce the classical radiation harmonics produced when you weakly couple the thing to electromagnetism.

So the quantization rule is $J=nh$, up to a possible offset. The derivation makes it clear that it is only true to leading order in h. This was Bohr's correspondence argument for the quantization condition.

When you have more than one degree of freedom, and the system is integrable, you have action variables $J_1,J_2...J_n$ and conjugate angle variables periodic with period $2\pi$ each. You can couple any of the degrees of freedom to electromagnetism weakly, and each classical period of the $\theta$ variable in time is

$$T_k = {\partial H \over \partial J_k}$$

so the statement is that for each orbit, each J variable is quantized according to the Bohr rule.

$$ J_k = nh $$

The $J_k$ variable is the area enclosed in the one dimensional projection of the motion in those coordinates where the motion separates into multiperiodic motion (this is the torus of Bar Moshe's answer). This is Sommerfeld's extension of Bohr quantization.

So the integral $\int p dq$ is taken with p and q any conjugate variables which make a period motion. In 1d, there is nothing to do, in multiple dimensions, you just choose variables which separately execute a 1d motion, and in general, you have to find J variables. This procedure doesn't work for classically chaotic systems.

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user Ron Maimon
answered Jul 13, 2012 by Ron Maimon (7,730 points) [ no revision ]
+ 4 like - 0 dislike

Many symplectic manifolds (phase spaces of mechanical systems) admit a coordinate system where the symplectic two form can be written locally as:

$\omega = \sum_i dp_i \wedge dq_i + \sum_j dI_j \wedge d\theta_j$

Where $ p_i, q_i$ are linear coordinates $ I_j$ are radial coordinates and $\theta_j$ are angular coordinates.

The submanifold parameterized by $\theta_j$ is a torus and a result due to Snyatycki states that the system can be quantized by means of a Hilbert space of wave functions (distributions) supported only on points whose coordinates $ I_j$ satisfy:

$ 2 I_j = m \hbar$

One of the simplest examples admitting such a quantization is the two sphere whose symplectic form is its area form

$A = r \mathrm{sin} \theta d\theta \wedge d\phi = dz \wedge d\phi $

where $\theta, \phi$ are the spherical coordinates and $z$ is the coordinate along the axis.

In this case the Bohr-Sommerfeld condition is given by:

$ z =\frac{ m}{2} \hbar$,

which is the spin projection quantization condition.

In a more sophisticated mathematical language one says that an open dense subset of he symplectic manifold is foliated by lagrangian tori and the integration is over the generating cycles of the tori.

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user David Bar Moshe
answered Jul 12, 2012 by David Bar Moshe (4,355 points) [ no revision ]
Thank You for the answer. I will unfortunately have to chew on it for quite a while in order to understand it, as I have no idea what the symplectic two-form is and what the hat-product does. Additionally this is the first time I hear about Snyatycki states.

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user user9886
@user9886: the symplectic structure of the cotangent-bundle (ie momentum-space) is responsible for the $-$ in Hamilton's equations (and the Poisson brackets as well); however, Hamilton's equations only take their simple form if you use canonical coordinates - if you allow arbitrary coordinate transformations, you'll get more complex coefficients - in particular, the components of an anti-symmetric rank 2 tensor, the symplectic form; in fact, a symplectic form on an arbitrary manifold is enough to do Hamiltonian mechanics (eg you can forget about the vector-bundle structure)

This post imported from StackExchange Physics at 2017-03-13 12:20 (UTC), posted by SE-user Christoph

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