"Single quantum isn't the issue. Single state is the issue.
If you know you have a particular state with 64 photons,
the entropy is zero.
If you have a single photon that is equally likely to be in
any of 64 states, the entropy is 6 bits.
Entropy is not knowing."
Now, suppose we have a single photon whose polarization is=20
equally likely to be found in one of two distinct states,=20
namely, either along the x- or along the y-direction,
after we will have passed it through, say, a polarizing=20
beam-splitter.
Does this mean that before that the photon's entropy is
1 bit?
I see three objections to this.
First, the initial condition means simply that in another
basis (x',y') that is rotated through 45 degress with=20
respect to (x,y), the photon is, say, in the eigenstate x'.
This basis is as good as (x,y), but since in it the photon=20
is in the eigenstate, the same logics tells us that its=20
entropy is zero. Since the entropy cannot depend on=20
representation of a state, the contradiciton shows that the=20
above definition of the entropy cannot be correct. =20
Second, let us forget about the (x',y') basis, and just=20
measure the polarization in the (x,y) basis. We will find=20
the photon either x-polarized or y-polarized, which would=20
mean that the measurement has reduced the photon's entropy=20
=66rom 1 bit to zero. Once the photon has not been absorbed,
the entropy of the measuring apparatus did not change, so=20
the net entropy of the isolated system (photon + apparatus)
has spontaneously decreased. We cannot say that this is=20
just a fluctuation, since the fluctuations are random, while
the described outcome under given conditions is one-way=20
predictable. Such situation would violate the second law.
Third, whose "not knowing" the entropy is? Knowing or not=20
knowing is the state of the observer, while the entropy is
the objective state of a system. This has been already=20
discussed pretty extensively on this Forum.