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[Phys-L] Re: entropy and electricity



Brian Whatcott wrote:

" Hmmmm.... Moses talks of a single photon, and John responds with
properties of an ensemble of photons expressed as an average value
per photon. Is that it? "

Exactly! I think, John has confused the issue by substituting the
word "photon" in my example with the word "light" in his. A single
photon is always in some definite polarization state, be it linear,
circular, or elliptical polarization. Light (an ensemble of photons)
may (if it is a pure ensemble) or may not (if it is a mixture) be in
a definite polarization state. In the latter case we have the
unpolarized light, and, accordingly, non-zero entropy. But there is
no such thing as an unpolarized photon.

I have to reinstate here some points in our discusison.

John wrote:

"If you know you have a particular state with 64 photons, the
entropy is zero"

Correct. This would be the case of a pure ensemble mentioned above.

"If you have a single photon that is equally likely to be in any
of 64 states, the entropy is 6 bits."

Wrong. Apparently, John confuses here the state itself with its
possible different representations. The same state can be
represented as a single eigenstate of a certain operator or as a
superposition of distinct eigenstates of some other operator.
Using a possible case from John's own example, if the photon is
known to be in one of 64 states, this one state itself can always
be represented as a superposition of eigenstates of another operator.
It follows then, according to John, that the entropy of this photon
is zero and non-zero at the same time.
In fact, there are no "two extreme cases" in my example with a
single photon. There is only one case, for which John himself has
obtained the zero entropy using the density matrix, which only
confirms my original statement, namely, that a single particle has
the zero entropy. Strictly speaking, the property called entropy, as
well as temperature, being statistical by nature, only emerges for a
system of more than 1 particles. No surprise we always get the zero
entropy when formally applying corresponding definition to a single
particle.
That a single particle may be in (in fact, can always be
represented as) a superposition of different states, does not change
this result, since a superposition of states is again a definite
state that can be considered as an eigenstate of some operator.
Of course, this is if you apply the conventional definition of
entropy, - not John's definition.
Here is another example. Consider a photon with definite momentum.
Its entropy, according to one part of John's definition, is zero.
On the other hand, this state (Psi(x) = const exp{ikx}) is a
superposition of the infinite number of eigenstates of the position
operator: xFi(x',x)= x'Fi(x',x), with Fi(x',x) = delta(x-x'). It
tells us that there is equal chance to find the photon in any of
these eigenstates upon the appropriate measurement. The entropy
of the very same photon must then be infinite, according to another
part of John's definition.
So who is ambiguous?
According to John's definition, the entropy of the same object may
be different for different observers (which is wrong if only the
entropy, just as temperature, is a scalar.) Worse, the known
non-scalar quantities depend, say, on the state of motion of the
observer, which is an objective physical distinction between two
observers, while John's entropy depends on the state of our
knowledge, which may be different for two observers in the same Lab.
But in the above two examples the situation is even worse still,
since the entropy of a single photon is either zero or not, may even
be zero and infinite for THE SAME observer depending on what
representation he chooses to describe a quantum state!

John then quotes my statement that the entropy is the objective
state of a system, and concludes:

"False, for reasons explained in detail at
http://www.av8n.com/physics/thermo-laws.htm#sec-s-context.";

Reference to a sourse is not the proof, even less if the sourse
itself falls short of giving a rigorous proof.

"So why bring it up again?"

It was not I who has brought up again the arbitrary definition
of entropy. In fact, I have shown in the previous discussion
that this definition would inevitably cause also the internal
energy of a system to take on different values for the same
observer.

Moses Fayngold,
NJIT
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