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*From*: "Fayngold, Moses" <fayngold@ADM.NJIT.EDU>*Date*: Sat, 17 Dec 2005 12:16:40 -0500

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=

=20

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=20

unpolarized light, and, accordingly, non-zero entropy. But there is=

=20

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=20

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=20

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=

=20

obtained the zero entropy using the density matrix, which only=20

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=

=20

system of more than 1 particles. No surprise we always get the zero=

=20

entropy when formally applying corresponding definition to a single

particle.=20

That a single particle may be in (in fact, can always be=20

represented as) a superposition of different states, does not change=

=20

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=20

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.=

=20

On the other hand, this state (Psi(x) =3D const exp{ikx}) is a=20

superposition of the infinite number of eigenstates of the position

operator: xFi(x',x)=3D x'Fi(x',x), with Fi(x',x) =3D delta(x-x'). It=

=20

tells us that there is equal chance to find the photon in any of=20

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.=20

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=20

non-scalar quantities depend, say, on the state of motion of the=20

observer, which is an objective physical distinction between two

observers, while John's entropy depends on the state of our=20

knowledge, which may be different for two observers in the same Lab.

But in the above two examples the situation is even worse still,=

=20

since the entropy of a single photon is either zero or not, may even=

=20

be zero and infinite for THE SAME observer depending on what=20

representation he chooses to describe a quantum state!

John then quotes my statement that the entropy is the objective=20

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=20

energy of a system to take on different values for the same=20

observer.

Moses Fayngold,

NJIT

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