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Re: quantized measurements of photons -- or not!



I think this thread originates in a fundamental misconception.

It is not true that the spin (or helicity) of the electromagnetic
field is always quantized as +1 or -1.

For that matter, it is not true that the electromagnetic field
is always quantized in photons.

Here's the real story:

If you measure the EM field with a photon counter, you will see
photons. The measuring apparatus forces the field into a state
of good photon number and tells you the number. It will be an
integer.

In contrast, if you measure the EM field with a voltmeter, you
will see a voltage. The measuring apparatus forces the field
into a state of near-definite voltage and tells you the number.
The voltage will !!not!! be quantized as an integer.

There are some things in quantum mechanics that are discrete,
and some that are not. Position is not quantized in general.
Momentum is not quantized in general. Voltage is not quantized
in general.

Even when you have something that is naturally discrete, e.g.
the angular momentum around the X-axis, you can have superposition
states that do not have a definite value of that variable. For
example if you start with X-angular-momentum of +3 and rotate
the system slightly around the Y-axis, you no longer have a state
of definite X-angular-momentum. It won't be a stationary state,
but who cares? The universe is full of non-stationary states.

To repeat: Much of the quantization in quantum mechanics comes
from the choice of measuring apparatus, not from the equations
of motion of the object being measured. If you choose a different
measuring apparatus, you will see different things.

================

Tangential remark:

Consider the inversion of the ammonia molecule as discussed in
Feynman volume III chapter 9. The conventional basis states |1>
and |2> are states of definite parity. In contrast, the stationary
states (i.e. eigenstates of the Hamiltonian) are called |I> and |II>
and they are the ungerade and gerade combinations of |1> and |2>.

The states of definite parity are not states of definite energy.
And vice versa.

You would expect at sufficiently low temperature to find most
ammonia molecules in the lowest-energy state.

Now.... If we pursue this a little farther, we come to some very deep
issues that were not understood until rather recently. What Feynman
says about NH3 is true for NH3, ND3, NHDT, and a few other things
like that ... but it is not true for almost any larger molecules!
NHDT is chiral, but it will instantly racemize by means of the
umbrella inversion. Larger chiral molecules will not racemize in
this way.

When chemists make ball-and-stick models of molecules, the models
obviously do not undergo umbrella inversion. These models are
incorrect for ammonia, but are OK for larger molecules.

You should expect at any reasonable temperature to find a small
molecule such as alanine in a state of definite parity, |1> or |2>.
You will never see it in, say, the gerade combination of |1> and
|2>, even though that is a stationary state of the isolated
molecule, and you might think it has lower energy.

The trick here is that the molecule is never really isolated.
If it is in a solvent, it interacts strongly with the solvent.
Even if there is no solvent, the molecule interacts so strongly
with the electromagnetic field that it is misleading to consider
the isolated ("undressed") molecule; we must consider the "dressed"
states which include the molecule and its interactions with the
field. These interactions (which tend to force the molecule
into the nearest state of definite parity) totally swamp the
bare-molecule's Hamiltonian (which otherwise would cause transitions
from |1> to |2>, changing the parity).

I mention this because essentially the molecule is being measured
to death. A watched pot never boils.

You need to know what is the coupling to the measurement
apparatus before you know what variable will take on definite
"classical" values. If the apparatus couples to the energy, you
will get states of definite energy (example: ammonia; see Feynman
figure III-9-3). If the apparatus couples to states of definite
parity, you will get states of definite parity (example: alanine).