"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
equally likely to be found in one of two distinct states,
namely, either along the x- or along the y-direction,
after we will have passed it through, say, a polarizing
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
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
is in the eigenstate, the same logics tells us that its
entropy is zero. Since the entropy cannot depend on
representation of a state, the contradiciton shows that the
above definition of the entropy cannot be correct.
Second, let us forget about the (x',y') basis, and just
measure the polarization in the (x,y) basis. We will find
the photon either x-polarized or y-polarized, which would
mean that the measurement has reduced the photon's entropy
from 1 bit to zero. Once the photon has not been absorbed,
the entropy of the measuring apparatus did not change, so
the net entropy of the isolated system (photon + apparatus)
has spontaneously decreased. We cannot say that this is
just a fluctuation, since the fluctuations are random, while
the described outcome under given conditions is one-way
predictable. Such situation would violate the second law.
Third, whose "not knowing" the entropy is? Knowing or not
knowing is the state of the observer, while the entropy is
the objective state of a system. This has been already
discussed pretty extensively on this Forum.