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Copyright © 2003 jsd

*   Contents

1  The Quantum / Classical Boundary
2  Bibiliography

1  The Quantum / Classical Boundary

Every day we encounter systems that are completely classical, i.e. systems that behave in accordance with the laws of classical physics. Yet there are other systems that are not classical, and must be described by the laws of quantum mechanics. It is well worth exploring where the boundary lies. In particular, we will investigate systems that are “just barely” quantum-mechanical or “just barely” classical.

To keep things simple, we will concentrate on two-state systems. Most two-state systems are completely classical ... but not all. Examples include:

Ammonia tunnels from one state to the other via the umbrella inversion as described in reference 1. Another example is NHDT, an isotope of ammonia, which is chiral, making the inversion particularly easy to detect. Because the ligands are heavier, the inversion rate is less for NHDT than for ammonia, but it is nonzero.

In the effort to pin down where the quantum / classical boundary lies, it is interesting to consider methylethylamine. It is in some sense homologous to NHDT. Its ligands are larger and heavier, but not wildly so. It behaves classically, unlike ammonia. The rule is that “almost” everything heavier than ammonia is classical. Where is the boundary? Why is there a boundary?

Note: At room temperature, left-handed methylethylamine converts fairly rapidly to the right-handed form and vice versa, but this is due to thermal activation. It is not due to the quantum tunneling we see in ammonia. The inversion ceases for methylethylamine at low temperature, whereas the inversion in ammonia continues just fine down to arbitrarily low temperature.

Let’s be clear: thermal activation is classical. Tunneling is 100% quantum mechanical. The factors that govern the tunneling rate are:

In contrast, thermal activation is only weakly (if at all) sensitive to the width of the barrier, and increasing the dissipation (i.e. the coupling to the heat bath) generally increases the rate of thermally-activated processes.

Let us examine things more closely. We have a two-state system. It has the symmetry group SU(2). Any state of the system can be represented by a point on the SU(2) sphere. The conventional ball-and-stick basis states are states of definite parity. In reference 1 these are called |1⟩ and |2⟩. These are two very particular points on the equator of the SU(2) sphere. Elementary quantum mechanics tells us that points on this sphere near the equator rotate at a rate given by the off-diagonal terms in the Hamiltonian. This rotation carries |1⟩ into |2⟩ and back again. This is exactly what we call tunneling. The stationary states are the poles of the SU(2) sphere. They are denoted |I⟩ and |II⟩. They are the ungerade and gerade combinations of the ball-and-stick basis states, namely

|I⟩ = 
|1⟩ − |2⟩ 
√2
 
|II⟩ = 
|1⟩ + |2⟩) 
√2
             (1)

The rotation (described by the Hamiltonian) leaves the polar points fixed.

In the case of ammonia, elementary QM gives the right answer. The rotation represents the umbrella inversion. If the system-point starts out in state |1⟩ it will rotate to state |2⟩ and back again, visiting in turn all the states on the equator of the SU(2) sphere. If we start out in state |I⟩ or |II⟩ the system will just sit in that state.

In the case of methylethylamine, we would expect the situation to be analogous, except that the rotation rate should be exceedingly slow, because of the greater mass of the ligands.

At first glance you might think that the ultra-slow tunneling rate is the whole story, explaining why methylethylamine is classical. Alas, that is far from sufficient. Elementary theory suggests that any state (i.e. any point on the SU(2) sphere) should be an allowed state of the system, but that’s not what we see. We never see states other than the classical ball-and-stick states. Saying that if we have a classical state it will stay that way is not the whole story, because if we had any other state – a superposition state – it would stay that way also. We need to explain why we never see superposition states.

It is easy to say the rotation-rate is ultra-small. The hard and interesting part is to explain why the rotation stopped exactly at one of the special points representing the classical ball-and-stick states |1⟩ or |2⟩.

The short answer is that it gets "measured to death" by the dissipative coupling to the environment. This is sometimes called the "watched pot effect".

To understand this, consider the following analogy. Imagine a photon that starts out in the X polarization state and then propagates through an optically active medium. After it has traveled a short distance, it will have rotated through a small angle θ. At this point we insert an X-polarizer into the medium. Assuming the photon survives the polarizer, it will emerge in the pure X polarized state. The chance that it won’t survive goes like θ2, so if we keep θ small enough we can neglect this. By putting lots and lots of polarizers into the medium, closely spaced, we can arrest the rotation of the photon. The rotation has been measured to death.

So that’s the real reason why the system-point describing typical chiral molecules doesn’t rotate. The ball-and-stick basis states are the states of definite parity. These are the states that couple to the electromagnetic field (reference 2). These are also the states that couple to the solvent for dissolved molecules. It is the coupling to the dissipative environment that makes the classical states special.

You can actually quantify the contrast between ammonia and methylethylamine in terms of the radiation resistance i.e. the strength of the coupling to the environment. As you increase the coupling, you get a second-order symmetry-breaking phase transition. If the coupling gets too large, the effective mass of the dressed state (the molecule plus all of the electromagnetic field that it needs to drag around with it) becomes infinite, and the rotation-rate of the SU(2) sphere goes to zero. Not just small, but zero. And the rotation doesn’t stop at some arbitrary point; it stops in states that are eigenstates of the measurement operator.

The classical universe emerges from the quantum universe
via a second-order phase transition.
     

You can actually walk back and forth across the quantum / classical boundary in some model systems, such as a two-hole SQUID (superconducting quantum interference device).

See also reference 3 and references therein.

2  Bibiliography

1.
Feynman, Leighton, and Sands The Feynman Lectures on Physics volume III chapter 9 "The Ammonia Maser".

2.
Pfeiffer, unpublished thesis, citation details missing.

3.
W. H. Zurek, "Decoherence and the transition from quantum to classical – REVISITED" Quantum Physics abstract quant-ph/0306072, http://arxiv.org/abs/quant-ph/0306072
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Copyright © 2003 jsd