You might also want to see David Mermin's Commentary column in
the July 2012 issue of Physics Today, at
http://www.physicstoday.org/resource/1/phtoad/v65/i7/p8_s1?bypassSSO=1
In fact, I posted a similarly-minded question to spr back then...
To quote the part [of Mermin's commentary] I'm asking a question about,
QBism immediately disposes of the paradox of "Wigner's friend."
The friend makes a measurement in a closed laboratory, notes
the outcome, and assigns a state corresponding to that outcome.
Wigner, outside the door, doesn't know the outcome and assigns
the friend, the apparatus, and the system an entangled state
that superposes all possible outcomes. Who is right?
For the QBist, both are right: The friend assigns a state
incorporating her experience; Wigner assigns a state incorporating
his. Quantum state assignments, like probability assignments,
are relative to the agent who makes them.
I'm sure Mermin's right and I'm wrong, but the above seems wrong
to me: the friend's objectively right and Wigner's wrong, as follows.
Suppose the friend makes a p(rojection)vm that throws the system
into a nice eigenstate of that observable. Then we know a second
measurement will result in the same outcome with probablility 1.
And the friend's state assignment correctly predicts that, whereas
Wigner's doesn't. So isn't the friend right and Wigner wrong?
Apparently not, since I doubt Mermin could've missed that.
So I must be missing something: how is it that "both are right"?
Nobody [else] answered that, but Stephen Parrott subsequently replied...
I was surprised that no one answered your question, so I will try.
At first I intended to send this just to you, but then I decided that
its main point, though elementary, is rather fundamental and
might possibly have wider interest.
Neither Wigner nor his friend are "right" about what is the (quantum)
state. They have different states corresponding to their different
information about the system.
This is essentially a classical effect which has nothing to
do with quantum theory. It has to hold in any sensible probabilistic
theory in which not all observers have complete information.
To give an explicit example, suppose that both Wigner and his
friend know that the state of the system in the laboratory is initially
the *pure* state ( | e > + | f > ) / \sqrt{2} where | e >, | f > is
an orthonormal basis of quantum states.
Suppose the friend measures in this basis obtaining state | e >
with probability 1/2 and | f > with probability 1/2.
Suppose the friend tells Wigner that he has made that measurement,
but he does not reveal the result. Then the friend's state
is either | e > or | f >, but Wigner's state is the *mixed* state which
is | e > with probability 1/2 and | f > with probability 1/2.
The same would hold for *any* probabilistic theory
(not just quantum theory) in which the laboratory measurement had
to result in just one of two possibilities..
I read Mermin's comment (and basically agree) as pointing out that
Bayesian probabilistic ideas have to be a fundamental part of quantum
theory (and indeed of any nontrivial probabilistic theory, though he
does not point this out).
Hope that helps, Stephen Parrott
--
John Forkosh ( mailto:
j...@f.com where j=john and f=forkosh )