# 13. Adiabatic Invariants and Action-Angle Variables

*Michael Fowler*

## Adiabatic Invariants

Imagine a particle in one dimension oscillating back and
forth in some potential. The potential doesn't have to be harmonic, but it must
be such as to trap the particle, which is executing periodic motion with period. Now suppose we *gradually* change
the potential, but keeping the particle trapped. That is, the potential
depends on some parameter, which we change
gradually, meaning over a time much greater than the time of oscillation:

A crude demonstration is a simple pendulum with a string of variable length, for example (see figure) one hanging from a fixed support, but the string passing through a small loop that can be moved vertically to change the effective length.

If were fixed, the system would have constant energy and period As is gradually changed from outside, there will be energy exchange in general, we'll write the Hamiltonian the energy of the system will be (Of course,also depends on the initial energy before the variation began.) Remember now that from Hamilton's equations so during the variation

It's clear from the diagram that the energy fed into the system as the ring moves slowly down varies throughout the cycle -- for example, when the pendulum is close to vertically down, its energy will be almost unaffected by moving the ring.

Moving slowly down meansvaries very little in one cycle of the system, we can average over a cycle:

where

Now Hamilton's equation means that we can replace with , so the time for going round one complete cycle is

(This won't integrate to zero, because on the return leg both and will be negative.)

Therefore, replacing in as well,

Now, we assumeare varying slowly enough that they are close to constant over one cycle, meaning that at a given point on the circuit, the momentum can be written, regarding as constant and independent parameters. (We can always adjust at fixedby giving the pendulum a little push!)

If we now partially differentiate with respect to keeping constant (appropriate infinitesimal pushes required!), we get, at pointon the circuit,

which is the integrand in the numerator of our expression for , so

In the denominator, we've replaced by

Rearranging,

This can be written

is an *adiabatic invariant*:
That means it stays constant when the parameters of the system change
gradually, even though the system's energy changes.

*Important*! The partial derivative with respect to
energy determines the period of the motion:

(*Note*: here is another connection with quantum
mechanics. If the system is connected to the outside world, for example if the
orbiting particle is charged, as it usually is, and can therefore emit
radiation, since in quantum mechanics successive action numbers differ by integers, and the quantum of action
is , the energy radiated per quantum drop in
action is . This is of course in the classical
limit of high quantum numbers.)

Notice that *is the area of
phase space enclosed by the integral,*

For the SHO, it's easy to check from the area of the ellipse that :

Take

The phase space elliptical orbit has semi-axes with lengths , so the area enclosed is

The bottom line is that as we *gradually* change the
spring strength (or, for that matter, the mass) of an oscillator (not
necessarily harmonic), the *energy changes proportionally with the frequency***.**

## Adiabatic Invariance and Quantum Mechanics

This finding, the invariance of for slow variation of the potential strength in a simple harmonic oscillator, connects directly with quantum mechanics, as was first pointed out be Einstein in 1911. Suppose the (quantum mechanical) oscillator is in the energy eigenstate with Then the spatial wave function has zeros. If the potential is changed slowly enough (meaning little change over one cycle of oscillation) the oscillator will not jump to another eigenstate (or, more precisely, the probability will go to zero with the speed of change). The wave function will gradually stretch (or compress) but the number of zeroes will not change. Therefore the energy will stay at and track with Of course, the classical system is a little different: the quantum system is "locked in" to a particular state if the perturbation has vanishingly small frequency components corresponding to the energy differences to available states. The classical system, on the other hand, can move to states arbitrarily close in energy. Landau gives a nontrivial analysis of the classical system, concluding that the change in the adiabatic "invariant" is of order for an external change acting over a time

## Action-Angle Variables

For a closed one-dimensional system undergoing finite motion (essentially a bound state), the equations of motion can be reformulated using the action variable in place of the energy is a function of energy alone in a closed one-dimensional system, and vice versa.

We're
visualizing here a particle moving back and forth in a one-dimensional well
with potential zero at the origin, and the potential never decreasing on going
out from the origin to infinity. Obviously, if a potential has *two *low
points, local bound states can arise in different places, and the relationship is complicated, with
different branches, possibly coming together at high energies.

*Important!* Notice the integral sign in the
expression for the action variable issignifying an integral around a *closed
path*, a circuit. *Don't* confuse this integral with the abbreviated
action integral, which has the same integrand, but is an integral along a contour from a fixed starting
point, say the origin, to the endpoint , not
going around a closed path. (Apologies for using the same letter for the
differential and the endpoint, just following Landau.)

In the spirit of the discussion of constants of motion above, we make a canonical transformation toas the new "momentum", using as generating function the abbreviated action

The original momentum

The new "coordinate" conjugate to the momentumwill be

This is called an *angle variable*,is the *action variable*, they are
canonical.

To find Hamilton's equations in the transformed variables, since there is no time-dependence in the transformation, and the system is closed, the energy remains constant. Also, the energy is a function of (meaning not of)

Hence

so the angle is a linear function of time:

One further point about the action variable and the action: since we define the action as

it follows that if we track the change in this integral as time goes on and the system moves round and round the circuit in phase space, an additional term will be added to the action for each time round, so the action is multi-valued.

## *Kepler Orbit Action-Angle Variables

We have not yet covered Kepler orbits, so skip this section for now: it's here to refer back to later. It's from Landau, p 167.

For motion confined to a plane, we can take the central potential analysis with and , the angular momentum, so the Hamiltonian is

The Hamilton-Jacobi equation is therefore

So, following the previous analysis of separation of variables for motion in a central potential, here

The action variable for the angular motion is just the angular momentum itself,

And the radial action variable, with potential , is

(Details on doing the integral are given in the Appendix, *Mathematica*
can do it too.)

So the energy is

The motion is *degenerate*: the two
fundamental frequencies coincide, This has major
consequences in quantum mechanics: the actions are all quantized in units of
Planck's constant, for the hydrogen atom, from the formula above, the energy
depends only on the *sum* of the quantum numbers: above the ground state,
energy levels are degenerate, which is why the energy spectrum has the
deceptively simple form so successfully explained by the Bohr model.

The orbital parameters, semi-latus rectum and eccentricity, fromand , are

Recall the semi-major axis is given by and from the above expression

in the hydrogen atom quantum number notation.

### Appendix: Doing the Integral for The Radial Action *I*_{r}

The integral can be put in the form

which can be integrated by taking a contour encircling the cut from to The integral will have a contribution from the pole at the origin equal to , and another from the circle at infinity, which is

.

Equating coefficients (multiplying the term inside the square root by )

So the contribution from the origin gives the the circle at infinity .

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