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From: brooks@lll-crg.ARPA (Eugene D. Brooks III)
Newsgroups: net.physics
Subject: Quantum Field Theory
Message-ID: <693@lll-crg.ARPA>
Date: Thu, 11-Jul-85 00:39:33 EDT
Article-I.D.: lll-crg.693
Posted: Thu Jul 11 00:39:33 1985
Date-Received: Tue, 16-Jul-85 08:02:12 EDT
References: <368@sri-arpa.ARPA>
Organization: Lawrence Livermore Labs, CRG group
Lines: 161

> Not true. Let us not confuse between the classical and quantum electromagnetic
> fields. The quantum EM fields are as much unobservable as the wave function.

I think you need to seriously study quantum field theory before you make this
statement.  The field operator in a quantized electromagnetic field is an
observable.  When you measure it you get a value of the "classical" field.
The measurement begets eigenvalues phi(x) of the field operator PHI(x) just
as the position operator X of the harmonic oscillator can be measured to
obtain the eigenvalue x.  What you have to be careful not to confuse is
the operator X of the harmonic oscillator with the scalar index x of the quantum
field.

To understand this clearly consider the harmonic oscillator. 

You can measure the operator X (the position operator) and get a value x
(one of the eigenvalues of the operator X) as a result.  The state becomes
an eigenstate of the operator X, with eigenvalue x, as a (I think, desireable)
side effect of the measurement.  This is natural, what else should happen to
a state when you measure it?  Notice that I avoid the use of the term "wave
function collapse" here as it creates a misunderstanding of what is going on
and makes you want to worry about relativity and all that.  It does not apply
for harmonic oscillator anyway.  The classical physics underlying the harmonic
oscillator is Newtonian.

The wavefunction is not something that is distributed in space.  If you use
the momentum (P) or number (N) representation to represent the state, the notion
of a wavefunction that is spread in space does not even arise.  The wavefunction
is now spread in momentum space or in a discrete but infinitely long vector
respectively.  You can have it anyway you want it.  (Signals going out to the
wavefunction that is distributed in space to tell it to collapse???  Gag me
with a spoon!  Try transforming to the P or N representation and working there
for a while to get rid of the tendancy for such thinking.  Where are your
signals propagating now?)

Note that the operator X does not commute with H and as a result if you measure
X at a later time you will get a new and potentially different eigenvalue
"x prime", you can compute the probability of various values of x prime using
QM.  The particle has not moved in the classical sense that you could have
followed it along a path.  First is was here and then it was there.  Thats
all there is to it.  Using the Newtonian equations of motion which the the
harmonic oscillator is based on the mass can be found rather far away rather
quickly.  There just isn't much chance of it.

Now suppose that you instead measure N, the number operator, and get the
eigenvalue n.  This is a positive integer.  N commutes with the Hamiltonian
H and if you measure N at a later time you will still get n.  Voila!, we are
measuring without disturbing a darn thing.  Stick that under your quantum
mechanical hat for a while.  It will lead you down the road to the QND work
at Caltech.

Now what does all this have to do with Quantum Field theory?  In Quantum
Field theory the classical field (the value of the field at a given point
in space) is the thing that corresponds to the the position of the mass in
a harmonic oscillator.  There is an operator PHI that you can measure at each
point in space x, which is no longer an operator as was the position X of the
mass in the harmonic oscillator.  x is simply the index that determines which
field operator PHI(x) you are going to measure.  When you measure PHI(x),
and you might indeed only measure the field PHI at only one point in space
x say 0 for instance, you toss the state into an eigenstate of the operator
PHI(0) with eigen value phi(0).  The "wave function" is the probability
amplitude associated with each possible eigenvalue of the field operator PHI(0).

You see, the wave function is not spread out in space at all!  Its spread out
over the possible eigenvalues of the field operator PHI(0).  Now of course there
is an operator associated with each value of the index x which turns out to be
continous in this case.  Thats a whole lot of operators!  The complete
wavefunction for the whole field everwhere (in x) is the outer product of the
wavefunctions associated with the field at each value of the index x.  The
wavefunctions are functions of phi(x) and not of x.

You can measure all of the field operators PHI(x) as the same time and get a
whole set of eigenvalues phi(x).  Note that you are obtaining a value for
the classical field everywhere in this case.  Its perfectly measureable.
It turns out that the operators PHI(x) do not commute with the Hamiltonian H
and if you later measure all of the operators PHI(x) again you will get
different values phi(x) prime just as for the harmonic oscillator.

Now then, where are those darned photons in all of this.  Suppose you transform
your description of the classical field phi(x) to momentum space p.  This is a
transformation on the classical description BEFORE you quantize.  Instead of
talking about phi(x) you talk about phi~(p) which are the expansion coefficients
of the classical function phi(x) into a set of continous functions corresponding
to differnt wavelengths p, sines and cosines of (x) if you will, to over
simplify it a bit.  Now you quantize this description of the field.  You can
think of the operator PHI~(p) and its associated momentum, in the quantum
mechanical sense, operator XHI~(p) which corresponds to the operator P for
the harmonic oscillator.  You can also transform in the quantum mechanical
domain to the number representation N(p).

When you measure N(p) you get an eigenvalue n(p) which corresponds to the
number of photons of momentum p.  In the case of a free field, this operator
commutes with the Hamiltonian H and if you measure it again you get the
same eigenvalue n(p).  Again you are measuring without disturbing a darn
thing, except perhaps during the first measurement.  n(p) is them photons
and they correspond exactly to the number of quanta n in the simple old harmonic
oscillator.  Its just that there is a whole lot of oscillators each denoted by
the value of the index p.

You can also talk of the number (photon) representation in x space instead of
p space.  This set of operators N(x) does not commute with the hamiltonian and
if you measure the number of photons at a given position x and get the
eigenvalue n(x) and then try measuring later on you will NOT get the same
eigen value n(x).  This is in exact analogy with the harmonic oscillator
measurements of the operator X.  You are measuring an operator which does not
commute with Hamiltonian, and therefore if you remeasure later you won't get
the same results.

Suppose you measure N(x) everywhere and get 1 at x=0 and 0 everywhere else.
It turns out that the sum of the operator N(x) over x does commute with the
Hamiltonian H and if you measure N(x) everywhere at a later time you will get
1 for some x prime, perhaps not equal to 0 and 0 everywhere else.  The total
number of photons will not change.  One might be tempted to say that the photon
moved from x to x prime and you might even hazard a discussion on what path
was taken.  This is not what happed, nothing "moved" in the sense that you think
of a ball that has been thrown by a baseball pitcher.  The photon simply fired
a phototube at position x at time t and then fired another phototube at position
x prime at time t prime.  If you didn't have your phototubes on inbetween you
don't have any information about where the photon was inbetween t and t prime
and there is no sense in even talking about it.

Of course there are the domain limits that would be set by relativity, and
enforce by the correct relativistic equations of motion for the FIELD.  If
the phototubes had been on during the interval t to t prime the "path" would
have not gone out of this domain.  This would be a uniquely different experiment
which "kicks" the hell out of the photon before it gets to x prime and the
probability of getting there as a result will be quite different.  Put a
phototube at each slit of an interference experiment and see what happens
to the interference pattern.   I assume that you have a mythical phototube that
fires without absorbing the energy of the photon.

All of the quantum mechanical equations of motion, wavefunctions and stuff are
"in your head" just as the newtonian equations of motion are "in your head".
The only reality is the firing of the phototubes.  The quantum mechanical wave
function is a computational entity that allows you to make some probabilistic
predictions about when and where.  I hope that I have case some more light on
this point above but if you still don't understand then I suggest a decade
(or more in most cases) of serious study.










I think that the above discussion should appear in a book sometime somewhere
but who really knows when the phototube will fire!   After all, GOD really
does play dice with universe.








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