Re: confused w/ decoherence
- From: Stephen Parrott <sp@xxxxxxxxx>
- Date: Sat, 31 Mar 2007 06:57:14 +0000 (UTC)
[mahdiarnt@xxxxxxxxx writes:]
1) So you say that in the "frictionless case," (4) is correct for a
pure state. Then it means that the results of successive experiments
are independent of the previous ones (or am I wrong?) Then there is no
wave function reduction at this level at all, whatever interpretation
we choose. This is weird (and perhaps measurable), though I don't know
whether those traditional experiments have really tested this.
Recall that equation (4) stated that if a system is in state S,
before a binary measurement (possible results 0 corresponding to projector P
and 1 corresponding to (1-P) ), then according to the rules of elementary
quantum mechanics, after the measurement it is in state
(4) PSP + (1-P)S(1-P) .
Here S is a (possibly pure) mixed state (represented by a positive Hermitian
operator of trace 1). This is correct whether or not S is pure,
within the context of your original question
(involving how a measurement can change a pure state to a mixed state).
The expression (4) is usually a mixed state.
(It is a pure state if and only if S is a pure state
represented by a projector which commutes with P.)
But you are reading more into my answer than was intended.
The state after the measurement depends on whether or not we look
at the result of the measurement. I didn't mention this because I assumed
you already understood the formalism, given that you proposed (4) as
the post-measurement state.
Let's look more closely at what a measurement entails.
To have a concrete situation in mind, suppose we are measuring the
spin of a particle in a particular direction (the z-axis say). This is a
binary measurement with possible outcomes, spin "up" or "down".
The Hilbert space of pure states is a two-dimensional complex Hilbert space
of vectors [a, b], which I write as row vectors because I can't make
column vectors look right in ASCII. The vector [1,0] represents a state
in which the particle is sure to be measured with spin "up", and [0,1]
represents a state in which spin "down" is sure to be measured.
Consider two physical scenarios:
Scenario I: We measure the spin when the (pure) state is represented by
a given vector [a,b] with |a|^2 + |b|^2 = 1,
and record the result ("up" or "down").
The rules of elementary quantum mechanics state that
the probability that we get spin "up" is |a|^2,
and the probability of "down" is |b|^2.
To avoid degenerate cases, suppose here and below that
neither a nor b is zero.
The rules of elementary quantum mechanics also state that
if the result is "up", then the post-measurement state is [1,0],
and if the result is "down" then it is [0,1].
Note that the post-measurement state has to be either [1,0]
or [0,1], no matter what the original state was,
*assuming that we know the result of the measurement*.
This is "reduction of the wave function".
This is not the same as the prediction of (4),
so your puzzlement is understandable in the context
of unfamiliarity with the rules of elementary quantum mechanics.
Scenario II: We measure the spin, but don't look at the result.
In case this seems artificial, imagine that you measure the
spin in your laboratory, and you tell me that you have done this,
but you don't tell me the result. I know that
your measurement has changed the original "wave function" [a,b],
but only you know *how* it has changed.
All I know is that there is probability |a|^2 that you got "up"
and probability |b|^2 that you got "down". So from my point of view,
the system is now in the mixed state in which it is
[1,0] with probability |a|^2 and [0,1] with probability |b|^2.
This is the mixed state (4) with
1 0
P = (that's supposed to represent a 2x2 matrix),
0 0
and the pure state [a,b] represented by the density matrix
|a|^2 a(b*)
S := , where * denotes complex conjugate.
(a*)b |b|^2
S is the projector on [a,b], and
|a|^2 0
PSP + (1-P)S(1-P) =
0 |b|^2
This is a non-pure state unless a=0 or b=0
because it is a convex linear combination of different states (exercise).
If you don't want to do the exercise,
then a quick and dirty way to see that it is a non-pure state
is to note that it is an invertible matrix, which can't represent a
pure state because its range is not one-dimensional.
In summary, from my point of view, the final state is a mixed state.
From your point of view, it is a pure state, either [1,0]
or [0,1], depending on whether you got "up" or "down".
I won't attempt to answer your other questions because they are not
clear to me.
[End of answer]
________________________________________________________________________
Remarks:
(i) I've never seen an elementary quantum mechanics textbook
which explains this well, but there must be some.
I am a mathematician who learned quantum mechanics decades ago
before finite-dimensional quantum mechanics (e.g., quantum
computation) became important. The above was hardly mentioned in
the usual graduate texts of that day (e.g., Schiff).
They concentrated on analytic aspects of infinite-dimensional
quantum mechanics such as solving the Schroedinger equation.
I'm not familiar with current texts.
(I learned these things from reading papers
and books like Nielsen/Chuang.)
Can some physicist reader recommend a good elementary
to intermediate text for mahdiarnt@xxxxxxxxx which carefully
explains the rules of elementary quantum mechanics,
mixed states, and the mathematics of measurement?
I wouldn't recommend Nielsen/Chuang for him or her
because it only briefly reviews the above
under the assumption that the reader has learned it
in the past. I fear that it would only frustrate
a beginner.
[I suspect that some of the negative reviews
of this book on Amazon.com stem from
frustration of undergraduates at failure to
understand a book that is
optimistically advertised as
"accessible to anyone with
a good undergraduate background in math,
computer science, or physical sciences".]
(ii) While I'm writing, I'd like to correct a couple of minor
inaccuracies in my previous post. I wrote:
"Also, the full (mixed) state space of the
larger system is the tensor product of the individual (mixed)
state spaces."
That's not quite right, though it doesn't make any difference
for the post's analysis.
The larger system has a Hilbert space which is a
tensor product of the Hilbert spaces or two smaller systems.
A mixed state of the larger system is a positive Hermitian
operator of trace 1 on the larger Hilbert space. This is
(isomorphic to) a subset of the tensor product of
the spaces of all operators on the two smaller systems,
but is not itself a tensor product
(because it is not a vector space, e.g.,
you can't multiply a positive operator by a negative scalar
and get another positive operator).
(iii) The URL for my website is www.math.umb.edu/~sp;
the URL given in the post omitted the "math".
.
- References:
- confused w/ decoherence
- From: mahdiarnt
- Re: confused w/ decoherence
- From: Stephen Parrott
- Re: confused w/ decoherence
- From: mahdiarnt
- confused w/ decoherence
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