Natural Science Forum / Physics / General Physics / July 2008

On 25 Jul, 11:59, frank_k_shel...@yahoo.co.uk wrote:

Yes and no. I would like to give you not an alternative theory but
another way of looking at it.

Let us start with a single particle in a box. It has a nice set of
discrete quantum states. We add another particle, we get some more
discrete states. Because of the interference of one particle with
another we gat a complex structure. As we add more and more particles
we get an exponentially increasing number of states.

Because of Perturbation Theory therefore Quantum Theory cannot refer
to a single particle but to the ensemble. An ensemble that has more
and more states the more particles there are. A Hamiltonian tells us
that we have mixed states if you refer to a particle, but that is not
really the right way to look at it.

If classical thermodynamics the temperature reflects increase in
Entropy with energy. Entropy is simply number of states. A classical
system will have long since passed the point where individual states
can be identified.

 - Ian Parker

An ensemble is an abstract collection of possible states of a system; it
does not mean that there are actually several copies of the system in real
life.

And looking at the Wikipedia article you may have read:

http://en.wikipedia.org/wiki/Mixed_state

"The need for a statistical description via density matrices arises when
one considers either an ensemble of systems, or one system when its
preparation history is uncertain and one does not know with 100% certainty
which pure quantum state the system is in."

"If the given system is closed, then one can think of a mixed state as
representing a single system with an uncertain preparation history, as
explicitly detailed above; or we can regard the mixed state as representing
an ensemble of systems, i.e. large number of copies of the system in
question, where pj is the proportion of the ensemble being in the state

On Jul 25, 4:59 am, frank_k_shel...@yahoo.co.uk wrote:

The mathematical construction used for mixed states is
called a "density matrix".

Look at the Wikipedia entry for "Density matrix". It points out
(correctly) that one of the cases in which one wants to use a
density matrix is when describing the state of a subsystem
of a system. Even if the whole system is in a pure state, the
state of a subsystem may be mixed. The subsystem can be
a single particle. So, in fact, mixed states can be used to
describe either ensembles or single particles.

Here's one way to think about it. A quantum theory provides you
a recipe for computing probabilities of observations on a system
given knowledge about the state of a system. Actually, a
non-quantum theory does that too. If I perform some experiment
on my system, and there is a measurement made, and I give
some criterion for counting it as a "success", there is some
probability p that it will be a success, which depends upon the
kind of observation O I am making and my knowledge K about
the initial state of the system. So we can write it as p(K,O).

The notion of mixing relevant here can be defined in such terms.
If I have three possible states of knowledge K, K1 and K2 and
for every possible observation O we have

  p(K, O) = q * p(K1, O) + (1-q) * p(K2, O)

then we can say that K is a "mixture" of K1 and K2. It's a
kind of average of them. The most obvious way to have a
mixture is as has already been described in various ways,
simple garden-variety uncertainty about what is happening.
If K1 and K2 correspond to two experimental set-ups, and
somebody flips a coin where you can't see it to decide
whether to set the system up as in K1 or as in K2, then your
knowledge of the system can be represented as just a
equal mixture of K1 and K2. Alternatively, you can have an
ensemble of systems, half of which are as in K1 and half
of which are as in K2; what you know about a randomly
selected example from the ensemble can be represented
as a mixture of K1 and K2 as well.

A "pure" state by contrast is one that isn't a mixture of two
distinct other ones.

Using a term like "mixed state" might then seem a bit
misleading. When someone says "state", I think many of us
tend to expect it to mean what is actually going on in the
system, a veritable description of how the system is. But
a "mixed state" is not necessarily that. It can be just a
description of some uncertain knowledge we have about it.

The distinction between describing the state of a system as
it actually is and describing merely what we know about it
is a philosophically thorny one. Hidden variable theories
amount to ways of taking what was in one theory a "pure"
state and representing it as a mixture of states that have
no direct counterpart in the earlier theory, because one
proceeds as if one knew additional information about the
system that wasn't previously treated as well-defined (such
as the actual particle positions in Bohm mechanics).

The case of entanglement is important because it's the case
where one treats a subsystem of a system using a mixed
state, even though one doesn't necessarily consider it a
matter of your not having complete knowledge of the state
of the system. The temptation to eliminate all mixtures
leads to problems that cause most people who've tried it
to go back to theories in which subsystems can be in
mixed states for "objective" reasons, not just as a way of
describing our lack of knowledge about the system.

It's possible to prepare a pair of photons (particles of light)
in such a way that if you pass one of them through a polarizing
filter, it will pass with probability 1/2 (assuming ideal conditions),
but the second one will pass or not pass through a filter with the
same orientation according as the first one did or didn't. In such
a state, the state of one individual photon is represented as a
mixed state. It can be represented as a mixture of a state in which
the photon is vertically polarized and one in which the photon is
horizontally polarized.

According to some interpretations of quantum physics, once a
(discrete) observation has been made, the system reduces to
a state in which the value of the observable is as observed. In
that case, if we pass one photon through a vertically polarizing
filter, then the other one becomes either vertically or horizontally
polarized, with equal probability. These interpretations would say
that the second photon is to be described as being in a mixed
state merely due to your not knowing yet the results of the
observation made on the first photon.

The probabilities of observations made on the second photon are
not, however, affected by merely whether the first photon is
observed first or second. So if we are people who either don't know
or just don't care about the first photon, for our purposes the state
of the second photon is a mixture. Whether it's a mixture because
it has an actual well-defined polarization that we happen not to
know, or because the polarization is entangled with the state of
something else is irrelevant to people working only with the second
photon. Perhaps at some point during our experiment the people
working with the first photon do something that, according to these
interpretations of quantum physics, switch the "mixed" state of our
photon from being mixed due to objective conditions to being mixed
due to our mere ignorance. The only way we can tell is by going out
and checking with the outside world, to see whether knowledge of
the polarization is already available to someone else.

It's important to distinguish between mixing and superposition.
Quantum interference effects are explained by superposition;
mixing doesn't entail interference effects. That's the essential
reason why we don't just explain all quantum uncertainties as
being merely the result of our lack of knowledge of the state
of the system.

Keith Ramsay

On Jul 25, 2:59 am, frank_k_shel...@yahoo.co.uk wrote:

Bohr said stateless state.

Einstein knew that was bunk.

Mitch Raemsch