Quantum mechanics
touqra
How can we know its history/past, if QM forbids us to know what is
happening inside the box, unless we take measurements? And by
measurements, we are trying to find out what is happening at the
moment of measurement, not the history of the box.
So, does the history of the box exist prior to measurements?
What if two observers wanted to measure the property of the particle.
Observer A measures first, let say its position. And then, after some
time later, observer B measures its position. Before B made any
measurements, B has to assume that the particle is in a superposition
state. But, A measured the particle already but did not find the
particle in any superposition.
Isn't this a paradox?
Charles Francis
<tou...@yahoo.com> writes
of it (and other qm paradoxes) is to recognise that there is no physical
reality attached to the wave function. It is only a calculational device
used by a particular observer to calculate a probability. There is no
such inconsistency in the calculation of probabilities by different
observers.
Regards
--
Charles Francis
John F
: Consider a particle in a box.
: happening inside the box, unless we take measurements? And by
: measurements, we are trying to find out what is happening at the
: moment of measurement, not the history of the box.
: So, does the history of the box exist prior to measurements?
The history "exists" only so far as it's "encoded" in the current
state. A fossil exists only so far as its impression remains in
a present-day rock. More physically, consider a particle moving
left-to-right in the following potential (I'm not sure the
illustrated well works as claimed, but there's some potential
that'll do it):
o ----->
+--+ +--+
| | | |
---------+ | | +----------
| |
+-------+
Instead of going directly over the well, the particle may temporarily
resonate with it, "bouncing" back and forth a few times before exiting
to the right. So I ask you: after bouncing back and forth for a while,
how does the particle "remember" that it entered from the left and
therefore must (conservation of momentum, assuming no transfer of
momentum to the apparatus creating the well) exit to the right?
: What if two observers wanted to measure the property of the particle.
: Observer A measures first, let say its position. And then, after some
: time later, observer B measures its position. Before B made any
: measurements, B has to assume that the particle is in a superposition
: state. But, A measured the particle already but did not find the
: particle in any superposition. Isn't this a paradox?
Assuming (as I assume you are) that A's measurement leaves the
particle in an eigenstate of the observable measured by B, then
that corresponding eigenvalue will, with probability 1, be the
outcome of B's subsequent measurement, regardless of what B assumes.
That is, in this case B's assumption is simply wrong, or,
more kindly, the initial (after A's measurement and before B's)
"superposition" happens to have weight 1 for the eigenstate produced
by A and weight 0 for all other eigenstates of that observable.
B simply wasn't aware of these initial weights. Moreover, even
after B's single measurement, he remains unaware of the _initial_
weights. (If he made a million measurements and got the same outcome
every time, then he'd probably realize somebody was pulling a fast
one on him.)
--
John Forkosh ( mailto: j...@f.com where j=john and f=forkosh )
John T Lowry
book about the history of a particle modifying its wave function. Anyone
know any more about that?
John.
--
John T Lowry, PhD
Flight Physics
5217 Old Spicewood Springs Rd, #312
Austin, Texas 78731
(512) 231-9391
jlow...@earthlink.net
"Charles Francis" <cha...@clef.demon.co.uk> wrote in message
news:c4sari$c9b$1...@lfa222122.richmond.edu...
Alfred Einstead
> Before B made any
> measurements, B has to assume that the particle is in a superposition
> state.
All states are "superposition states" -- even the ones that "aren't".
Superposition does not mean mixture, which is the notion implicit in
your comment.
Even the "of" in "superposition of" is inappropriate, since this is
not a parts/part of relation. It's symmetric:
a is a superposition involving b
if *and only if*
b is a superposition involving a.
Both relations are characterized by the relation of non-orthogonality
and incompatibility.
Macroscopic observers can't be on incompatible Hilbert space bases
because they are continuously entangled by their mutual interaction
with each other and with everything else. So, the issue does not
arise in the macroscopic world.
In particular, you can't shield an actual cat to even come up with a
Schroedinger cat, to begin with, since (for one thing) its gravity
can't be shielded. The mere existence of any long-range, universal,
unshieldable interaction (which nature, fortunately, provides one
instance of in the actual world), by itself, makes entanglement at
the macroscopic level virtually unavoidable. Among other things,
everything near the Earth is in under continual observation by the
Earth and is permanently and unavoidably entangled with both it
and each other.
DickT
> of it (and other qm paradoxes) is to recognise that there is no physical
> reality attached to the wave function. It is only a calculational device
> used by a particular observer to calculate a probability. There is no
> such inconsistency in the calculation of probabilities by different
> observers.
>
>
> Regards
So on this view, not having an observer doesn't keep physics from
happening, it just keeps us from knowing (by calculation) what the
details are?
Arkadiusz Jadczyk
wrote:
When A measured the position, the wave function has jumped to a
different state (localized around the place where the detector was).
Then B measures position. The wave function jumps again.
What exists is a sequence of recordings and, perhaps, a sequence
of wave functions. While we can relatively easily know the history of
recordings, we can only guess the history of wave functions.
But I see no paradox here.
ark
--
Arkadiusz Jadczyk
http://www.cassiopaea.org/quantum_future/homepage.htm
--
John Baez
John F <jo...@SeeSigForAddress.com> wrote:
>[...] consider a particle moving
>left-to-right in the following potential (I'm not sure the
>illustrated well works as claimed, but there's some potential
>that'll do it):
>
> o ----->
>
> +--+ +--+
> | | | |
>---------+ | | +----------
> | |
> +-------+
>
>Instead of going directly over the well, the particle may temporarily
>resonate with it, "bouncing" back and forth a few times before exiting
>to the right. So I ask you: after bouncing back and forth for a while,
>how does the particle "remember" that it entered from the left and
>therefore must (conservation of momentum, assuming no transfer of
>momentum to the apparatus creating the well) exit to the right?
I'm not sure what the point of this example is. If you have
a particle moving in a fixed potential well, its momentum will
typically *not* be conserved. Physically, this is because it
typically *will* transfer momentum to the apparatus creating
potential well. So, in the above situation, the quantum particle
can exit either to the left or to the right. In other words,
there's a chance that it'll bounce back, and a chance that it'll
tunnel all the way through.
Maybe I'm misunderstanding your point.
chronon
tou...@yahoo.com (touqra) wrote in message news:<4a5d59d9.04033...@posting.google.com>...
> What if two observers wanted to measure the property of the particle.
> Observer A measures first, let say its position. And then, after some
> time later, observer B measures its position. Before B made any
> measurements, B has to assume that the particle is in a superposition
> state. But, A measured the particle already but did not find the
> particle in any superposition.
>
> Isn't this a paradox?
B's lack of knowledge of the system means that he ascribes to it a
*mixed* state, which is not the same as a superposition. Making this
change means that there is then nothing specifically quantum about the
question - the same would apply in classical mechanics, for which
mixed states are also defined.
There is some question about what mixed states actually *mean* in
quantum mechanics, since they can't be interpreted as simply not
knowing what the *real* state is, but I'm not sure what the resolution
of that should be.
Stephen Lee
www.chronon.org
John F
John Baez <ba...@galaxy.ucr.edu> wrote:
Yeah, right :) -- I'm pretty sure you understand it, but maybe
I didn't make its intent clear (or maybe it's just wrong).
Touqra (the original poster) asked about the present
"existence" of the "history/past", and I chose to interpret the
past as existing only so far as it's encoded in the present state
(which definition you and/or Touqra may or may not agree with).
The above example was intended to illustrate this.
If you'll imagine the idealized case where no momentum is
transferred between particle and apparatus, then the
particle must "remember" its history/momentum while it's
resonating above the well, so that it knows to exit to
the right. How does it do this? That *past* information must remain
encoded in the particle's *present* (quasi-)stationary resonant state.
And it would be to that extent that the particle's past "exists".
Even more generally, I suppose information representing
the value of any conserved quantity is never lost from the
time-dependent wavefunction. (And I'd guess this is either wrong or
else there are many references for it.) Conversely, you might argue
that information which isn't lost always represents some "conserved"
physical quantity/observable. In that case, getting back to Touqra's
original question, it's only such "conserved quantities" from the past
that continue to exist in the present.
To ramble further, then the present state would consist of
these conserved quantities plus some "inexplicable" initial conditions.
Of course, we'd like *additional* past information to predict these
"initial conditions" as much as possible. But since, by my defintion,
no additional past information would be available (we've already
described it all), there's an inevitable unpredictablilty of the
present state due to information loss that arises from nonconserved
quantities.
And even more speculatively, maybe this inevitable unpredictability
might correspond to quantum measurement uncertainty. That is,
nonconserved quantities means measurement uncertainties. It doesn't
seem too daunting to formalize at least some of this, to avoid seeming
too crackpot, but I haven't tried. Already been done? Obviously wrong
and not worth doing?
Frank Hellmann
How do we know the history of any physical system? We always have to
apply time evolution, i.e. theory to infer it.
Take the time evolution operator, let's call it U(t) giving you the
evolution of your system by the amount of time t. In classical
mechanics it's easy enough, you know where your particle is after
meassurement apply U(-t) to infer where it was. Now in QM things are
different, a meassurement in general means that you get
macroscopically distinct states. We have a system with two states: |a>
and |b>, and a pointer which has three states |I> for initial, |A> for
measuring position |a> and |B> for |b> now our time evolution looks
like this during the meassurement process:
U(T)|a>|I> = |a>|A>
U(T)|b>|I> = |b>|B>
Now use linearity:
U(T)(c|a> + d|b>)|I> =
U(T)(c|a>|I> + d|b>|I>) =
cU(T)|a>|I> + dU(T)|b>|I> =
c|a>|A> + d|b>|B>
so the system actually is in a superposition of states here.
This looks to be rather bad because of course that's not what we see,
we don't see an apparatus in a superposition. However, as |A> and |B>
are very different microscopically because they are macroscopically
distinct we can see that both of these terms evolve individually if we
look at the density matrix of things:
<A|B> = 0 because they are so different, they have a different support
in Hilbert space.
rho = (c|a>|A> + d|b>|B>)(c<a|<A| + d<b|<B|)
to find the evolution of the subsystem trace out the enviroment:
tr_meas=<A|rho|A> + <B|rho|B> = c^2 |a><a| + d^2 |b><b| this reduced
density matrix governs how these subsystems evolve and you see that as
there are no diagonal terms this evolution does not correspond to a
superposition in Hilbert space. The sub system has decohered.
It obeys a classical statistic not a quantum statistic.
Now this is still not classical as we don't see classical
superposition of particles either.
However there is another thing to remember, we are part of the system.
We are part of |I> before the measurement, and evolve into |A> and |B>
those are different, different support in Hilbert space, no
interference the local subsystems that interact with them (|a> and
|b>) are no longer in superposition, so the version of us in |A> has
meassured |a> the version of us in |B> has meassured |b>.
This is the origin of many world theories. And pertaining to your
question, the information of the history is now split between the two
worlds. Each of the physicists can deduce the time evolution U(-t) for
their part of the system but because their systems don't interact they
can't know about the whole history aas they can't know about the whole
system.
This however does not resolve the issue either because if you look at
the formulas the probability of finding a particle is encoded in the
intensity of the system. However we always get two systems, if both
exist and it's just by chance that we are in this one then we would
expect a probability of 0.5 for finding both instead of c^2 and d^2
respectively.
The very linearity of the evolution that produces the split makes it
unaware of the intensities.
So this does not resolve the problem of meassurement, but it does tell
us why we have no measurable violations of the superpositiopn
principle. This principle hides itself.
I sketched over many details I did not remember properly, some of the
stuff here is a matter of opinion as well. Read up on decoherence why
quantum mechanics can be at the same time practically enormously
succesfull and conceptually so unsatisfying.
This are BTW features retained in QFT as well. It's simply that U is
much more complicated. MUCH more complicated.
Have fun,
Frank Hellmann
Alfred Einstead
> The mere existence of any long-range, universal,
> unshieldable interaction (which nature, fortunately, provides one
> instance of in the actual world), by itself, makes entanglement at
> the macroscopic level virtually unavoidable. Among other things,
> everything near the Earth is in under continual observation by the
> Earth and is permanently and unavoidably entangled with both it
> and each other.
Nobody asked the obvious question: what about in deep space?