" 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=
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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:
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.