OK, I am reviewing my basic quantum mechanics by watching Jim Branson's
QM class on streaming video. There is something he keeps say that
bothers me. He keeps saying that a wave function like exp(ikx) can't be
normalized to "one particle" and so must be a beam of particles?
...Huh?...
I understand what it means to say that exp(ikx) can't be normalized to
one, but that never meant to me anything about the number of particles
to me. It is just an idealized state that is not in technically in the
Hilbert space (we might use "rigged Hilbert spaces" for this kind of
thing). So what is he talking about? What am I missing?

But now this brings up an interesting question. In plain old quantum
mechanics (not QFT), the wave function for a pair of pairticles (moving
in 1D for simplicity) is an L^2 function on R \times R. Thus it seems
that exp(ikx) can't refer to more than one particle in anycase! It is a
generalized eigenfunction of momentum for a single particle (isn't it?).
The wave function describing, say two particles must be a functions of
two variables like say \psi (x_1,x_2).
So wouldn't a beam of many particles have wave functions of many
variable (the number of particles)? And yet we hear that exp(ikx) is a
beam of particles!! Remember I am talking about a QM class here, not a
QFT class (so no Fock space etc).

Thus spake Cyberkatru
OK, I am reviewing my basic quantum mechanics by watching Jim Branson's
QM class on streaming video. There is something he keeps say that
bothers me. He keeps saying that a wave function like exp(ikx) can't be
normalized to "one particle" and so must be a beam of particles?
..Huh?...
I understand what it means to say that exp(ikx) can't be normalized to
one, but that never meant to me anything about the number of particles
to me. It is just an idealized state that is not in technically in the
Hilbert space (we might use "rigged Hilbert spaces" for this kind of
thing). So what is he talking about? What am I missing?

You have it right. A wave function describes a single particle state. If
he is really saying this, ignore him and find a teacher or a book that
understands the subject.

But now this brings up an interesting question. In plain old quantum
mechanics (not QFT), the wave function for a pair of pairticles (moving
in 1D for simplicity) is an L^2 function on R \times R. Thus it seems
that exp(ikx) can't refer to more than one particle in anycase! It is a
generalized eigenfunction of momentum for a single particle (isn't it?).
The wave function describing, say two particles must be a functions of
two variables like say \psi (x_1,x_2).
So wouldn't a beam of many particles have wave functions of many
variable (the number of particles)? And yet we hear that exp(ikx) is a
beam of particles!! Remember I am talking about a QM class here, not a
QFT class (so no Fock space etc).

In a beam such as a laser all particles may be in the same state. Then
you can use one wave function. Otherwise not.


Regards

--
Charles Francis
Please reply by name

"Cyberkatru" a écrit dans le message de news:
%VEtf.59072$4l5.17061@dukeread05...

OK, I am reviewing my basic quantum mechanics by watching Jim Branson's
QM class on streaming video. There is something he keeps say that
bothers me. He keeps saying that a wave function like exp(ikx) can't be
normalized to "one particle" and so must be a beam of particles?
..Huh?...
I understand what it means to say that exp(ikx) can't be normalized to
one, but that never meant to me anything about the number of particles
to me. It is just an idealized state that is not in technically in the
Hilbert space (we might use "rigged Hilbert spaces" for this kind of
thing). So what is he talking about? What am I missing?

A wave function doesn't even need to be normalized. That has no physical
meaning, just some holy water. On the other hand, a wave function always
does represent an infinity of particles, that's why we can derive a
probability from it. The description of one particle is still a mystery,
whatever the mathematical coating. The trick to manage quantum mechanics is
to never confuse the calculus which allows to make predictions, and the
understanding of the underlying physical reality.

In that case, a perfect plane wave obviously can't exist in reality since it
would have an infinite extension, but we can make very precise calculations
using it. That doesn't imply anything in the physical reality.

But now this brings up an interesting question. In plain old quantum
mechanics (not QFT), the wave function for a pair of pairticles (moving
in 1D for simplicity) is an L^2 function on R \times R. Thus it seems
that exp(ikx) can't refer to more than one particle in anycase! It is a
generalized eigenfunction of momentum for a single particle (isn't it?).
The wave function describing, say two particles must be a functions of
two variables like say \psi (x_1,x_2).
So wouldn't a beam of many particles have wave functions of many
variable (the number of particles)? And yet we hear that exp(ikx) is a
beam of particles!! Remember I am talking about a QM class here, not a
QFT class (so no Fock space etc).

It all depends on what you want to do. If you measure only one-particle
observables, a one-particle wave function is the way to go since it
describes an infinity of particles in the same state. You need
multi-particle wave function only if you measure correlations between
particle. If the particles are charged, they interact with one another, and
a multi-particle wave function may be needed. However, there are
mathematical methods enabling to introduce a mean field and to neglect the
correlations (Hartree-Fock for instance). In the non-relativistic case,
that leads to a non-linear Schrödinger equation.

--
~~~~ clmasse on free F-country
Liberty, Equality, Profitability.

Charles Francis wrote:
Thus spake Cyberkatru
OK, I am reviewing my basic quantum mechanics by watching Jim Branson's
QM class on streaming video. There is something he keeps say that
bothers me. He keeps saying that a wave function like exp(ikx) can't be
normalized to "one particle" and so must be a beam of particles?
..Huh?...
I understand what it means to say that exp(ikx) can't be normalized to
one, but that never meant to me anything about the number of particles
to me. It is just an idealized state that is not in technically in the
Hilbert space (we might use "rigged Hilbert spaces" for this kind of
thing). So what is he talking about? What am I missing?

You have it right. A wave function describes a single particle state. If
he is really saying this, ignore him and find a teacher or a book that
understands the subject.

A wave function describes a single-particle state in the case of
fermions.

In the case of photons, the state is a product of the mode function
and the wave function, and the wave function depends on the number
of photons. It has to, because the mode function knows nothing about
photon number, and the wave function is the only place such information
can be stored.

The photon case is a generalised version (in a sense) of the quantised
simple harmonic oscillator, and even a brief read over that subject
in a sutable text clearly shows that the allowed wave functions are the
Hermite polynomials H_n, and that H_n is the wavefunction for an n-excitation
state of the QSHO.

The reason "a wave function like exp(ikx) can't be normalized to
one particle" is that the integral of its norm over all space is
not finite. Thus either we (a) box-normalise, and say there is one
particle per box (of sides L), and hence have some sort of beam;
or we (b) restrict the domain (e.g. to between -L/2 to +L/2) so the
the integral is finite, and hence normalizeable (although then
exp(ikx) is unlikely to be a good wavefn).

--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714
Imperial College London,
SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/

Thus either we (a) box-normalise, and say there is one
particle per box (of sides L), and hence have some sort of beam;
or we (b) restrict the domain (e.g. to between -L/2 to +L/2) so the
the integral is finite, and hence normalizeable (although then
exp(ikx) is unlikely to be a good wavefn).

--

I am sorry to say that this just doesn't make sense to me. I sounds like
hand waving. One particle per box? Again, if there a several fermions then
there sould be many variables in the wave function- an xyz triple for every
particle. That what the "official formalism" states.
I still don't get it.
Now as for your "infinity of particles" since we are talking probability I
think (see Ballentines discussion here) the "infinite" isn't the number of
particles in the experiment but rather an "ensemble of potential instances
of the experiement"-- it is a sort of imaginary infinity. Like me rolling a
die one time, implicitly refers to me being able to repeat the experiment
many times (potentially infinitely many times). But I roll once! Likewise, a
wave function \psi (x,y,z) refers to one real partical and an idea
potentiality of many repetitions of the same experiemnt (always with one
particle).
I still think I am right about this but I am still listening.

Cl.Massé writes

A wave function doesn't even need to be normalized. That has no physical
meaning, just some holy water. On the other hand, a wave function always
does represent an infinity of particles, that's why we can derive a
probability from it. The description of one particle is still a mystery,
whatever the mathematical coating. The trick to manage quantum mechanics is
to never confuse the calculus which allows to make predictions, and the
understanding of the underlying physical reality.

In that case, a perfect plane wave obviously can't exist in reality since it
would have an infinite extension, but we can make very precise calculations
using it. That doesn't imply anything in the physical reality.

That should be clearly said in all elementary (and probably less
elementary) QM lectures.

Unfortunately there is a corollary that is somewhat uncomfortable, which
is that energy-momentum is defined using these non-existent plane waves.
Its implication is, in effect, that energy-momentum is never quite
perfectly conserved.

Personally I don't have a problem with this, I see QM existing in what
is in effect a very noisy environment with h measuring the level of
noise in some sense.

--
Oz
This post is worth absolutely nothing and is probably fallacious.

Use functions].
BTOPENWORLD address has ceased. DEMON address has ceased.

Cyberkatru wrote:
Thus either we (a) box-normalise, and say there is one

particle per box (of sides L), and hence have some sort of beam;
or we (b) restrict the domain (e.g. to between -L/2 to +L/2) so the
the integral is finite, and hence normalizeable (although then
exp(ikx) is unlikely to be a good wavefn).

--


I am sorry to say that this just doesn't make sense to me. I sounds like
hand waving. One particle per box? Again, if there a several fermions then
there sould be many variables in the wave function- an xyz triple for every
particle. That what the "official formalism" states.

It sounds like hand-waving but has a completely rigorous formal basis.
The claim was about Bosons, not fermions.

The physical state described by a typical laser beam is a state with
an indeterminate number of photons, since it is not an eigenstate of
the number operator. (This essentially means that a certain number
of photons cannot be meaningfully asserted.)
Thus the traditional N-particle picture does not apply.

Instead one has to work in a suitable Fock space. There, each mode
function (solution of the free Maxwell equation) A(x) gives rise to
a coherent state ||A in the Maxwell Fock space, an associated
annihilation operator a(A) = integral A(x) a(x) dx, and the
corresponding creation operator a^*(A) = a(A)^*.

These produce a single-mode Fock subspace consisting of all |A,psi,
where psi is the unnormalized wave function of a harmonic oscillator;
|psi|^2 is the intensity of the beam.
The coherent state itself corresponds to the normalized vacuum state
of the harmonic oscillator, ||A = |A,vac.

The Maxwell Fock space is the closure of the space spanned by all
the ||A,psi together. [It is not a direct product, though.]
This space is the pure electromagnetic field sector of QED,
describing a physical vacuum, i.e., a region of the universe
where matter is absent.

In optics experiments, laser beams are often idealized by ignoring their
extension perpendicular to the transmission direction. Then each beam
can be described by some ||A,psi. A coherent pair of laser
beams obtained by splitting is described by a superposition
||A_1,psi_1 + ||A_2,psi_2 of the two beams.

Beams of thermal light (such as that from the sun) and pairs of
beams created by independent sources, cannot be described by wave
functions alone, but need a density formulation. A single light beam
is then described (in the same idealization) by a mode A and a density
matrix rho in a single-mode Fock space, while k light beams are
described by k modes A and a density matrix rho in a k-mode Fock space.

In many treatments, the modes are left implicit, so that one works
only in the k-mode Fock space. This simplifies the presentation, but
hides the connection to the more fundamental QED picture.
For a thorough study of the latter, see the bible on quantum optics,
L. Mandel and E. Wolf,
Optical Coherence and Quantum Optics,
Cambridge University Press, 1995.


Arnold Neumaier

Although you got already some answers to your question, I'll give
another one, because I think the matter is much simpler than considered
so far in this thread.

Cyberkatru wrote:

OK, I am reviewing my basic quantum mechanics by watching Jim
Branson's QM class on streaming video. There is something he keeps say
that bothers me. He keeps saying that a wave function like exp(ikx)
can't be normalized to "one particle" and so must be a beam of
particles? ..Huh?...

I'd say, that's a healthy reaction ;-). A wave function

psi(x)=x|psi

describes a single particle and not many particles. A many-particle
state is described by superpositions of product states, i.e., in
position representation these are always wave functions with N position
vectors.

I understand what it means to say that exp(ikx) can't be normalized to
one, but that never meant to me anything about the number of particles
to me. It is just an idealized state that is not in technically in the
Hilbert space (we might use "rigged Hilbert spaces" for this kind of
thing). So what is he talking about? What am I missing?

But now this brings up an interesting question. In plain old quantum
mechanics (not QFT), the wave function for a pair of pairticles
(moving in 1D for simplicity) is an L^2 function on R \times R. Thus
it seems that exp(ikx) can't refer to more than one particle in
anycase! It is a generalized eigenfunction of momentum for a single
particle (isn't it?). The wave function describing, say two particles
must be a functions of two variables like say \psi (x_1,x_2).
So wouldn't a beam of many particles have wave functions of many
variable (the number of particles)? And yet we hear that exp(ikx) is a
beam of particles!! Remember I am talking about a QM class here, not a
QFT class (so no Fock space etc).

You are right. I'd sharpen the argument: a plane wave is not the
representation of a quantum state at all, because it is not in the
Hilbert space.

Within the formulation in the rigged Hilbert space formulation it is a
generalized function, i.e., a member of the dual space of the "nuclear
space". More physically it is a generalized "eigen state" of the
momentum operator, which reads in position representation

\hat{p}=1/i \nabla (\hbar=1 from now on)

\hat{p} is defined on a dense subspace in Hilbert space (or the
corresponding nuclear space of the G'elfand triple of the rigged
Hilbert space formulation) and has a continuous spectrum.

Physically all boils down to the fact that each state (a true member of
H) can be written in the spectral representation:

|psi=\int dp |pp|psi

For the position representation this reads

psi(x)=x|psi=\int dp x|pp|psi
=\int dp exp(i p x)/(2 pi)^3 p|psi
=\int dp exp(i p x)/(2 pi)^3 psi(p)

First, of course the integral is only defined on the dense subset, where
\hat{p} is defined, but the transformation from momentum representation
to position representation (which is a Fourier transformation, as shown
above) can be unitarily extended to all Hilbert-space vectors. That was
roughly the mathematics of the situation (presented in a very
inrigorous physical manner of course ;-)).

Now to the physics: According to standard quantum theory a single
particle is described by a true Hilbert space vector (or more precisely
by a ray in Hilbert space), and it only describes statistical
properties of the particle: If a particle is prepared in such a way
that its state is described by a normalized Hilbert-space vector |psi,
the probability to find it in another state, represented by a
normalized Hilbert-space vector |phi, is given by

W_{psi}(phi)=|phi|psi|^2

phi| can also be the member of the dual space of the nuclear space,
i.e., it can be also a "generalized eigenstate" in the above mentioned
sense. For instance it could be a generalized eigenstate of the
position operator. Then the probability *distribution* to find the
particle at position x is given by

W_{psi}(x)=|x|psi|^2

Since now QT describes only probabilities, i.e., makes only statistical
predictions about the outcome of measurements, the most simple (and for
all practical purposes sufficient!) interpretation is the "minimal
statistical interpretation", which is due to Ballentine, who wrote a
very nice textbook on quantum mechanics:

L. Ballentine, Quantum Mechanics

According to this interpretation quantum mechanics describes ensembles
of independent identically prepared systems, in our case an ensemble of
single particles prepared such that this ensemble is described by the
state |psi or (equivalently) the wave function psi(x).

In this sense Branson may not be completely wrong (I have not seen the
videos, so I cannot really judge his QM class): The single-particle
state indeed describes "a beam" of particles, i.e., a source of single
particles which delives an ensemble prepared in the state |psi.
Nevertheless this holds true for any quantum state, not only for
"generalized eigenstates" of selfadjoint operators.


--
Hendrik van Hees Texas A&M University
Phone: +1 979/845-1411 Cyclotron Institute, MS-3366
Fax: +1 979/845-1899 College Station, TX 77843-3366
http://theory.gsi.de/~vanhees/

Oz wrote:
Cl.Mass? writes, in part
In that case, a perfect plane wave obviously can't exist in reality ...

Unfortunately there is a corollary that is somewhat uncomfortable, which
is that energy-momentum is defined using these non-existent plane waves.
Its implication is, in effect, that energy-momentum is never quite
perfectly conserved.

It does not matter that we might never manage to generate a perfect
plane wave. But we can still use them (where appropriate) as a basis
to describe physical situations, and get good answers.

Personally I don't have a problem with this, I see QM existing in what
is in effect a very noisy environment with h measuring the level of
noise in some sense.

This is sort of true, as long as you make a number of assumptions
which are not infrequently valid.

--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (QOLS) (ph) +44-20-759-47520 (fax) 47714
Imperial College London,
SW7 2BW, United Kingdom. http://www.qols.ph.ic.ac.uk/~kinsle/

Hendrik van Hees wrote:

Although you got already some answers to your question, I'll give
another one, because I think the matter is much simpler than considered
so far in this thread.

Cyberkatru wrote:

OK, I am reviewing my basic quantum mechanics by watching Jim
Branson's QM class on streaming video. There is something he keeps say
that bothers me. He keeps saying that a wave function like exp(ikx)
can't be normalized to "one particle" and so must be a beam of
particles? ..Huh?...


I'd say, that's a healthy reaction ;-). A wave function

psi(x)=x|psi

describes a single particle and not many particles. A many-particle
state is described by superpositions of product states, i.e., in
position representation these are always wave functions with N position
vectors.

But an even healthier reaction is to try to understand why a competent
physicists can make such a statement with a good conscience.

Physicists are not bound by a single formalism; the usual communication
language is quite flexible and allows many formalizations with different
degrees of rigor. And in essentially the whole theory of the laser, and
in many quantum optics experiments, whole beams are indeed modelled by
single particles, although it is clear that a many-particle picture
should be used for a ''fully correct'' detailed description.

Thus what is correct (i.e., acceptable) in a coarse model becomes
incorrect on a more detailed level. This does not invalidate the
description on the course level, however: Otherwise we'd have to give
up all classical physics...


Arnold Neumaier