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# John Baez: Ever wonder why they call it ``phase space''?

OK, so let's start with a classical dot racing around in a circle. Projecting onto the x-axis gives simple harmonic motion. We'll use q for the x-coordinate and p for the y-coordinate, following Hamiltonian tradition.
Of course, by ``x-coordinate'' you mean ``position'' and by ``y-coordinate'' you mean ``momentum''. We've got here a point in phase space, oscillating harmonically. And as you note, one key to understanding quantum mechanics is to see this phase space as the complex plane!
Now as complex number, q+ip is that racing dot. Everything flows from q+ip.
Indeed. Ever wonder why they call it ``phase space''? I don't know the history, but here we see a damn good reason: as our point Z = q+ip circles the origin, nothing changes but its phase.

The whole point of coherent states is to see very clearly what happens to this picture when we go into quantum mechanics. Of course, everybody knows that when p and q become operators, we can make Z and its complex conjugate into operators too, which are basically just the creation and annihilation operators:

a = (Z/ sq2)      a* = (Z*/ sq2)               [Reminder: sq2 stands for sqrt(2)]

But the cool part is that there are also states that are quantum analogues of points circling the origin: the coherent states. As you note, these are just the eigenstates of the annihilation operator. But I prefer to visualize them as Gaussian wavefunctions: a kind of blurred-out version of a state in which a particle has a definite position and momentum. If you start with such a state, and evolve it in time, its (expectation value of) position and momentum oscillate just like that of a classical particle, and if I remember correctly, they maintain a basically Gaussian shape, though probably with some funny complex phase factor stuff thrown in...

...nothing races around at all, instead we have a bunch of wavefunctions that just sit there...
Yeah, sure, the eigenstates of the Hamiltonian just sit there, after we ignore the time-dependent phase. But these states are very unlike the classical states we know and love. I like to pose the following puzzle to kids just learning quantum mechanics:

``Take this eraser. [I brandish my eraser threateningly as I stand before the blackboard.] Put it into an eigenstate of the Hamiltonian. Now it's in a stationary state! It doesn't do anything as time passes. It just sits there, except for a time-dependent phase! [I demonstrate an eraser nonchalantly hovering in midair, only its phase wiggling slightly.] So what does this mean, that you can levitate an eraser just by putting it into an eigenstate?''

And of course they eventually get the point: it's not so easy in practice to put anything big into an eigenstate of the Hamiltonian. It's the coherent states that are close to the classical physics we know and love.

Could these utterly different kinematics really come from the same cinematographer?

Sure! Just use coherent states. As D increases continuously, the average energy increases continuously, as does the range of motion of the bump.

Hmm. Guess I should check that last assertion.

That should be true. Remember, the expected value of the Hamiltonian is

<H> = (<p2> + <q2>)/2

As we take our basic Gaussian bump (the ground state) and translate it, <p2> stays the same, since its shape stays the same. <q2>, on the other hand, gets bigger, basically because the average value of q gets bigger (though I'm being a bit rough here). So we can tune <H> to whatever value we want...at least for values bigger than the ``zero-point energy'', which is 1/2.

Next: John Baez: Perhaps the Up: Schmotons Previous: Michael Weiss: Hmm, homework
Michael Weiss

3/10/1998