n particle wave function

[Gerard Westendorp]

Hi,

I'm trying to interpret the time-dependence part of the n-particle
Schroedinger equation, and its wave function solutions.

i d_t PSI(x1,x2,x3...,t)
= -1/2m (d2_x1 + d2_x3 + d2_x3+...+ V) PSI(x1,x2,x3,...,t)

It seems that the more particles you add, the higher the frequency
of the wave function solutions.

w= k1^2 + k2^2 + k3^2 +... -V

I'm having some trouble with this. Is the frequency really higher,
or is this some kind of unobservable thing?


thanks,

Gerard

[Gerard Westendorp]

Arnold Neumaier wrote:
[..]

>> It seems that the more particles you add, the higher the frequency
>> of the wave function solutions.
>>
>> w= k1^2 + k2^2 + k3^2 +... -V
>>
>> I'm having some trouble with this. Is the frequency really higher,
>> or is this some kind of unobservable thing?
>
> The only frequencies that have a physical meaning are the differences
> of eigenvalues of the Hamiltonian, multiplied by Planck's constant.

I thought of something observable: wave-packets.

The group velocity of a wave packet is observable.

v_g = dw/dk = dE/dp

So for the group velocity, it does not matter if the frequency is offset
by a constant, such as mc^2.

(See Wikipedia
http://en.wikipedia.org/wiki/Group_velocity )

This is not really what I was worried about though.
So what about the group velocity of an n-particle wave function?

Fill in a solution in free space:

Psi = exp(ik_1*x_1)*exp(ik_2*x_2)*... exp(iwt)

-> w = (h_bar/2m) * (k1^2+ k2^2+ k3^2 +....)

Aha!

So the frequency is multiplied by N, but

dw/dk_i = h_bar*k_i/m = v_classical

So the wave is traveling at group velocity sqrt(N)*v_classical, but in
N-dimensional space.

I guess the weirdness comes from the N-dimensional (or 3N-dimensional)
space.

Gerard

[Hendrik van Hees]

Arnold Neumaier wrote:

> A 1 MHz radio wave is not an N-particle system but (in the relevant
> approximation) an ensemble of single-particle systems prepared in
> the same state.

A 1 MHz radio wave is not an ensemble of single-particle systems, but
a coherent state of definite frequency. Such a state can have any
average photon number you like (but it's not a state with a definite
photon number).
>
> This explains the lack of oscillations at N*f.

Sure, the state has (approximately) one frequency. After all that's
what a radio station likes to emit ;-).

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/

[Igor Khavkine]

On Jan 15, 3:09 am, Gerard Westendorp <west...@xs4all.nl> wrote:

> But take a 1 MHz radio wave. The 1 MHz can be be observed directly on an
> oscilloscope.
> This radio wave consists of a large number of 1 MHz photons. It seems
> that both the collective of these particles and the individual particles
> have the same frequency, seemingly in contradiction with the idea that N
> particles oscillate at N times the frequency of a single one.

> I find it hard to believe that there is something oscillating at N*f.

Not all frequencies are created equal. Just because "something"
oscillates at frequency N*f does not mean that "everything else" must
oscillate with the same frequency. When you look at a wave form on an
oscilloscope, you are essentially measuring the superposition of the
wave forms of individual photons the many that compose the state of
the electromagnetic field. The frequency that you see on the
oscilloscope is the average frequency, which will be f, no matter what
N is. N is encoded in the amplitude of the field's wave form. To
measure N, you need to calculate the energy contained in the intensity
of the field. Now, energy is not measured in Hz, but if you divide
that energy by hbar, you'll get N*f. Ts expected, since the wave
function frequency N*f measures the total energy of the state
(relative say to the ground state of the field).

Hope this helps.

Igor

[Arnold Neumaier]

Gerard Westendorp schrieb:
> Igor Khavkine wrote:
>
>> A wave function with n sets of independent coordinates represents a
>> collective state of n particles. The frequency of the wave function is
>> then proportional to the total energy of the n-particle system, which
>> is just the sum of the energies of the n particles. The energy of each
>> particle is its kinetic energy (the k^2 term) and its potential energy
>> (that's contained in the V term). So, what you observed is perfectly
>> correct and not in any way unusual.
>
> Thanks.
> This seems indeed to be what the n-particle Schroedinger equation is saying.
>
> But take a 1 MHz radio wave. The 1 MHz can be be observed directly on an
> oscilloscope.
> This radio wave consists of a large number of 1 MHz photons. It seems
> that both the collective of these particles and the individual particles
> have the same frequency, seemingly in contradiction with the idea that N
> particles oscillate at N times the frequency of a single one.

> Or better, lets make it a resonant cavity of 1 MHz. Then we might think
> of this system as a quantum oscillator with frequency (f). Every time it
> changes energy, it adsorbs or emits a photon of the same frequency (f).
> According to the Schroedinger equation of the quantum oscillator, the
> frequency of the oscillator is actually N*f.
>
> I find it hard to believe that there is something oscillating at N*f.

A 1 MHz radio wave is not an N-particle system but (in the relevant
approximation) an ensemble of single-particle systems prepared in
the same state.

This explains the lack of oscillations at N*f.


Arnold Neumaier

[Gerard Westendorp]

Hendrik van Hees wrote:
> Arnold Neumaier wrote:
>
>> A 1 MHz radio wave is not an N-particle system but (in the relevant
>> approximation) an ensemble of single-particle systems prepared in
>> the same state.
>
> A 1 MHz radio wave is not an ensemble of single-particle systems, but
> a coherent state of definite frequency. Such a state can have any
> average photon number you like (but it's not a state with a definite
> photon number).

Thanks.
As usual, Wikipedia has a good article on coherent states:
http://en.wikipedia.org/wiki/Coherent_state
So as you say there is not a definite photon number in a coherent state,
but some are more likely than others, and the expectation value is for
photon number is E/(hf).

An important difference between photons and matter Bose particles is
that photons are their own antiparticles, so you can have states that
are superpositions of particles and antiparticles (if you interpret the
complex conjugate of a photon creation operator as the creation of an
antiphoton), such as

(a exp(iwt) + i*a^ exp(-iwt))^n |0>

These states have real-valued and directly observable fields (such as E
and B), but indefinite particle number.

If in the other hand you have matter particles, you would have states like

a^exp(-iwt)^n |0>

These always have complex-valued field strength.
I think this means the field strength is unobservable, but I am not sure
yet.


>> This explains the lack of oscillations at N*f.
>
> Sure, the state has (approximately) one frequency. After all that's
> what a radio station likes to emit ;-).

Even though the electric and magnetic fields oscillate with 1 MHz,
according to the texts I am reading at the moment, the quantum states
have frequencies larger than this. A coherent state would be a
superposition of terms like:

Psi_n = (a^)^n exp(-(inw+1/2)t)

But I can see how the field strength expectation value will end up
oscillating with exp(-iwt),

Field ~ < Psi_n| (a^exp(iwt) + a^exp(-iwt)) | Psi_n>

~exp(-iwt)



Gerard

[Arnold Neumaier]

Hendrik van Hees schrieb:
> Arnold Neumaier wrote:
>
>> A 1 MHz radio wave is not an N-particle system but (in the relevant
>> approximation) an ensemble of single-particle systems prepared in
>> the same state.
>
> A 1 MHz radio wave is not an ensemble of single-particle systems, but
> a coherent state of definite frequency.

Phrasing it this way makes being ''an ensemble of single-particle
systems'' and being ''a coherent state of definite frequency''
look like alternatives excluding each other. But this is not
the case.

A coherent state, interpreted in terms of particles, is the state of
an ensemble of photons arriving at different times. This follows easily
from the fact that the number of photons counted is proportional to
the counting time. If you make your counting time short enough,
you get only single particles.


> Such a state can have any
> average photon number you like (but it's not a state with a definite
> photon number).

It is a stationary state with an average _rate_ of photons, but not
one with an average _number_ of photons. The photons arrive one after
the other. In an N-particle state they would arrive simultaneously.


>> This explains the lack of oscillations at N*f.
>
> Sure, the state has (approximately) one frequency. After all that's
> what a radio station likes to emit ;-).
>
> --
> Hendrik van Hees Institut für Theoretische Physik
> Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
> Fax: +49 641 99-33309 D-35392 Gießen
> http://theory.gsi.de/~vanhees/faq/
>

[Arnold Neumaier]

Hendrik van Hees schrieb:
> Arnold Neumaier wrote:
>
>> A 1 MHz radio wave is not an N-particle system but (in the relevant
>> approximation) an ensemble of single-particle systems prepared in
>> the same state.
>
> A 1 MHz radio wave is not an ensemble of single-particle systems, but
> a coherent state of definite frequency.

Phrasing it this way makes being ''an ensemble of single-particle
systems'' and being ''a coherent state of definite frequency''
look like alternatives excluding each other. But this is not
the case.

A coherent state, interpreted in terms of particles, is the state of
an ensemble of photons arriving at different times. This follows easily
from the fact that the number of photons counted is proportional to
the counting time. If you make your counting time short enough,
you get only single particles.


> Such a state can have any
> average photon number you like (but it's not a state with a definite
> photon number).

It is a stationary state with an average _rate_ of photons, but not
one with an average _number_ of photons. The photons arrive one after
the other. In an N-particle state they would arrive simultaneously.


>> This explains the lack of oscillations at N*f.
>
> Sure, the state has (approximately) one frequency. After all that's
> what a radio station likes to emit ;-).
>
> --
> Hendrik van Hees Institut für Theoretische Physik
> Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
> Fax: +49 641 99-33309 D-35392 Gießen
> http://theory.gsi.de/~vanhees/faq/
>

[Hendrik van Hees]

Arnold Neumaier wrote:

> Phrasing it this way makes being ''an ensemble of single-particle
> systems'' and being ''a coherent state of definite frequency''
> look like alternatives excluding each other. But this is not
> the case.

It is the case! An "ensemble of single-particle systems" is, by
definition, a repeated setup of an experiment, where in each experiment
you send precisely one photon into an apparatus to do some measurements
on it, i.e., in each setup you prepare a single-photon state.

A coherent state is not a single-photon state. So you do not have
an "ensemble of single photons" but a superposition of states with an
indefinite photon number.

> It is a stationary state with an average _rate_ of photons, but not
> one with an average _number_ of photons. The photons arrive one after
> the other. In an N-particle state they would arrive simultaneously.

FACK.

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen

[Arnold Neumaier]

Hendrik van Hees wrote (in: n particle wave function)
> Arnold Neumaier wrote:
>
>> Phrasing it this way makes being ''an ensemble of single-particle
>> systems'' and being ''a coherent state of definite frequency''
>> look like alternatives excluding each other. But this is not
>> the case.
>
> It is the case! An "ensemble of single-particle systems" is, by
> definition, a repeated setup of an experiment, where in each experiment
> you send precisely one photon into an apparatus to do some measurements
> on it, i.e., in each setup you prepare a single-photon state.

If you prepare a weak beam of light (or radio waves) in a coherent
state and test the photon contents at the end of the beam by a
photodetector you measure a series of clicks indicating according to
tradition the presence of single photons.

Each click is conventionally regarded as the measurement of a
single photon; hence one measures an ensemble of photons.
Without this interpretation, much of the talk about photons in
quantum optics would not make sense.

> A coherent state is not a single-photon state. So you do not have
> an "ensemble of single photons" but a superposition of states with an
> indefinite photon number.

Technically, and completely precisely, you have an ensemble of photons
in an indefinite photon number state. (Even a superposition of states
describes an ensemble, in the conventional interpretation.)

In a weak coherent state, the multiparticle contents is negligible;
one has essentially a superposition of the vacuum and the single
particle state. Conventially (as for all somewhat rare events),
the vacuum part is ignored - one just restricts attention to the
times where a particle is present. This leaves a single particle state.
Thus, at least for weak coherent states, it is a good approximation
to say that a coherent state of definite frequency is an ensemble
of single-particle systems.

Now I was saying: ''A 1 MHz radio wave is not an N-particle system but
(in the relevant approximation) an ensemble of single-particle systems
prepared in the same state.''

The relevant approximation is the approximation relevant for a receiver
of the radio wave. The receiver integrates the number of impinging
photons within its time interval of response, and cannot distinguish
between two photons arriving simultaneously and two photons arriving
one after the other. Thus, from the point of view of the receiver,
the wave behaves as an ensemble of single photons. Indeed, if you
weaken the signal enough, or shorten the interval enough, the
expected number of photons in the response interval will be much
less than one, and one can measure (in principle) the single
photon events.

>> It is a stationary state with an average _rate_ of photons, but not
>> one with an average _number_ of photons. The photons arrive one after
>> the other. In an N-particle state they would arrive simultaneously.
>
> FACK.

If you accept this, I don't understand why you complain about the
ensemble.

A plane monochromatic radio wave in the form of a coherent state
(three mathematical idealizations involved here) is an endless stream
of infinitely many photons passing with the speed of light through a
particular position on the beam.

How can this _not_ be an ensemble of photons?

But the difficulty of this discussion illustrates the fact that the
particle picture in quantum field theory is of very limited use
and creates more confusion than understanding.

As the Nobel prize winner Willis E. Lamb (the discoverer of the
Lamb shift) maintained, there are no photons. Excitations of a
harmonic oscillator in quantum field theory are conventionally
_called_ particles although the relation to real objects is quite
vague.

For a fuller discussion, see the section
S2f. What is a photon?
in my theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt

Arnold Neumaier

[Gerard Westendorp]

Arnold Neumaier wrote:
[..]

> If you prepare a weak beam of light (or radio waves) in a coherent
> state and test the photon contents at the end of the beam by a
> photodetector you measure a series of clicks indicating according to
> tradition the presence of single photons.
>
> Each click is conventionally regarded as the measurement of a
> single photon; hence one measures an ensemble of photons.
> Without this interpretation, much of the talk about photons in
> quantum optics would not make sense.

What about stimulated emission?
Doe an ensemble ensemble of N photons enhance the rate of photon
emission by atoms by a factor N, just as an N-photon state does?


Gerard

[Arnold Neumaier]

Gerard Westendorp schrieb:
> Arnold Neumaier wrote:
>
>> If you prepare a weak beam of light (or radio waves) in a coherent
>> state and test the photon contents at the end of the beam by a
>> photodetector you measure a series of clicks indicating according to
>> tradition the presence of single photons.
>>
>> Each click is conventionally regarded as the measurement of a
>> single photon; hence one measures an ensemble of photons.
>> Without this interpretation, much of the talk about photons in
>> quantum optics would not make sense.
>
> What about stimulated emission?
> Does an ensemble ensemble of N photons enhance the rate of photon
> emission by atoms by a factor N, just as an N-photon state does?

The rate of emission is proportional to the intensity
of the incident beam. It is the field of the incident beam that counts;
the talk about photons in the incoming beam is not meaningful and
only blurs the picture; the right language is that of field theory.

Stimulated emission of a single atom creates a wave pulse, not a
coherent state.


Arnold Neumaier

[Hendrik van Hees]

Arnold Neumaier wrote:

> If you prepare a weak beam of light (or radio waves) in a coherent
> state and test the photon contents at the end of the beam by a
> photodetector you measure a series of clicks indicating according to
> tradition the presence of single photons.

I'm a bit surprised that you changed your mind on this issue. Once we
discussed in the German physics newsgroup, and there you convinced me
from the opposite, and as I've learned from a talk by Marlan Scully
at Texas A&M, there is the principle possibility to distinguish a
weak coherent state from a real single-photon state although it's
pretty difficult experimentally:

PRL 96, 010501 (2006)

>
> Each click is conventionally regarded as the measurement of a
> single photon; hence one measures an ensemble of photons.
> Without this interpretation, much of the talk about photons in
> quantum optics would not make sense.

That's clear, and I didn't contradict it. Of course, a coherent state
describes an ensemble of photons, but not an ensemble of single
photons! You can measure the total photon number in each event, and
you'll find fluctuations following a Poisson distribution.
>
>> A coherent state is not a single-photon state. So you do not have
>> an "ensemble of single photons" but a superposition of states with
>> an indefinite photon number.
>
> Technically, and completely precisely, you have an ensemble of
> photons in an indefinite photon number state. (Even a superposition
> of states describes an ensemble, in the conventional
> interpretation.)

Whether a state is a superposition or not depends on the basis and is
physically irrelevant. A coherent state is of course a superposition
of Fock-basis states. Any Hilbert-space vector corresponds to a
(pure) physical state, and a coherent state is such a Hilbert-space
vector.

>
> In a weak coherent state, the multiparticle contents is negligible;
> one has essentially a superposition of the vacuum and the single
> particle state. Conventially (as for all somewhat rare events),
> the vacuum part is ignored - one just restricts attention to the
> times where a particle is present. This leaves a single particle
> state. Thus, at least for weak coherent states, it is a good
> approximation to say that a coherent state of definite frequency is
> an ensemble of single-particle systems.

Well, it's approximately a superposition of the vacuum and a
single-photon state, but not a single-photon state. Both ensembles
are distinguishable altough it's experimentally tough.

> If you accept this, I don't understand why you complain about the
> ensemble.

I don't complain about the ensemble, it is only important to clearly
dinstinguish between coherent states of weak intensity and
single-photon Fock-basis states.
>
> A plane monochromatic radio wave in the form of a coherent state
> (three mathematical idealizations involved here) is an endless
> stream of infinitely many photons passing with the speed of light
> through a particular position on the beam.

The expectation value of the photon number in a coherent state is
finite!
>
> How can this _not_ be an ensemble of photons?

It is an ensemble of photons, but not an ensemble of single photons!
>
> But the difficulty of this discussion illustrates the fact that the
> particle picture in quantum field theory is of very limited use
> and creates more confusion than understanding.

There is no principle problem with the particle picture in QFT. The
notion of a particle is of course different if you look in
relativistic an (approximately) non-relativistic contexts. A massless
particle is always relativistic, and thus one has to adapt the notion
of particle to the ultrarelativistic realm.
>
> As the Nobel prize winner Willis E. Lamb (the discoverer of the
> Lamb shift) maintained, there are no photons. Excitations of a
> harmonic oscillator in quantum field theory are conventionally
> _called_ particles although the relation to real objects is quite
> vague.

I don't believe that physics papers are necessarily good only because
they are written by Nobel Laureats. IMHO this paper is an example for
the opposite and it's disproved experimentally since nowadays the
quantum opticians can prepare single- or multiple photon states with
a definite photon number, e.g., by parametric down conversion.

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen
http://theory.gsi.de/~vanhees/faq/

[Arnold Neumaier]

Hendrik van Hees schrieb:
> Arnold Neumaier wrote:
>
>> If you prepare a weak beam of light (or radio waves) in a coherent
>> state and test the photon contents at the end of the beam by a
>> photodetector you measure a series of clicks indicating according to
>> tradition the presence of single photons.
>
> I'm a bit surprised that you changed your mind on this issue. Once we
> discussed in the German physics newsgroup, and there you convinced me
> from the opposite,

I haven't changed my mind. The real situation is complicated, due to
the very informal usage of the word photon in many contexts; the recent
collection of articles explaining different current views,
The Nature of Light: What Is a Photon?
Optics and Photonics News, October 2003
http://www.osa-opn.org/Content/ViewFile.aspx?Id=3185
give widely diverging definitions of what a photon is....
And a mail should be short.

[[Mod. note -- Sadly, that collection of articles seems to require
a subscription for access, and web searches failed to find an
open-access version. -- jt]]


I was speaking from the main stream view, i.e., ''according to
tradition'', as I explicitly mentioned. readers here did not have the
benefit of our discussion on the German mailing list de.sci.physik;
so I could not assume that, and I didn't what to generate a new long
foundational debate. (What I thought I had convinced you was only that
the clicks don't mean that the arriving photons change into the pure
state determined by the measuring apparatus; but apparently we
misunderstood each other.)

My own views (not mainstream, but consistent with experiment) are:
- that clicks have nothing at all to do with photons, they just are
a stochastic measure of intensity, and arise also if the incindent
field is modelled completely classical;
- that what is typically called a photon is not an arbitrary single
particle state of the electromagnetic field (in particular, never
an approximately plane wave) but a state of the electromagnetic field
that at each time is localized in space, whose energy contents is
that of hbar*omega. Otherwise, the idea of producing photons of
demands makes no sense.
See
http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf
http://www.mat.univie.ac.at/~neum/ms/optslides.pdf


> and as I've learned from a talk by Marlan Scully
> at Texas A&M, there is the principle possibility to distinguish a
> weak coherent state from a real single-photon state although it's
> pretty difficult experimentally:
>
> PRL 96, 010501 (2006)

I don't see where this paper discusses that question, but
yes, it is possible to distinguish the two:

The difference is that a weak coherent state generates an infinitely
long random sequence of Poisson-distributed clicks, while a single
photon (in the above sense of a space-localized field) generates
a single click only (in an ideal detector).



>> Each click is conventionally regarded as the measurement of a
>> single photon; hence one measures an ensemble of photons.
>> Without this interpretation, much of the talk about photons in
>> quantum optics would not make sense.
>
> That's clear, and I didn't contradict it. Of course, a coherent state
> describes an ensemble of photons, but not an ensemble of single
> photons! You can measure the total photon number in each event, and
> you'll find fluctuations following a Poisson distribution.

For a weak coherent state of a stationary monochromatic beam,
psi = (1-eps||0> + eps|1> + O(eps^2), (*)
with eps<<1, and
<n> = <a^*a> = \psi^*(a^*a)psi = eps^2 + O(eps^3)
is not a mean photon number, but a mean rate - the mean intensity.

The events are the clicks, and there is exactly one click per event
in a weak signal (for strong signals, one cannot separate the events).
But the events happen randomly in time, with a rate proportinal to eps.

Thus (*) does not describe _one_single_photon_ but an ensemble of
O(T*eps^2) _single_photons_, where $T$ is the duration of the
experiment.

Note that two photons arriving at different times cannot be considered
as being part of a N-particle state with N>1, since states are
considered at a fixed time! The fact that the weak coherent state has
a negligible contribution of doubly excited states is here completely
irrelevant. Thus one has an ensemble of single photons.




>>> A coherent state is not a single-photon state. So you do not have
>>> an "ensemble of single photons" but a superposition of states with
>>> an indefinite photon number.
>> Technically, and completely precisely, you have an ensemble of
>> photons in an indefinite photon number state. (Even a superposition
>> of states describes an ensemble, in the conventional
>> interpretation.)
>
> Whether a state is a superposition or not depends on the basis and is
> physically irrelevant. A coherent state is of course a superposition
> of Fock-basis states. Any Hilbert-space vector corresponds to a
> (pure) physical state, and a coherent state is such a Hilbert-space
> vector.

Yes, but I don't see how this related to my statement.


>> In a weak coherent state, the multiparticle contents is negligible;
>> one has essentially a superposition of the vacuum and the single
>> particle state. Conventially (as for all somewhat rare events),
>> the vacuum part is ignored - one just restricts attention to the
>> times where a particle is present. This leaves a single particle
>> state. Thus, at least for weak coherent states, it is a good
>> approximation to say that a coherent state of definite frequency is
>> an ensemble of single-particle systems.
>
> Well, it's approximately a superposition of the vacuum and a
> single-photon state, but not a single-photon state. Both ensembles
> are distinguishable altough it's experimentally tough.

Most photon states produced in the laboratory are superposition with
the vacuum, and still people speak of photons.
Entanglemaent studies are typically made with squeezed states,
which differ from coherent states only in thast they have instead
of (*) a representation
psi = (1-eps||0> + eps|2> + O(eps^2), (**)
and everyone refers to (**) as an ensemble of 2-photon states!


>> If you accept this, I don't understand why you complain about the
>> ensemble.
>
> I don't complain about the ensemble, it is only important to clearly
> dinstinguish between coherent states of weak intensity and
> single-photon Fock-basis states.

I wasn't claiming that a cohernent state is a Fock state!!!


>> A plane monochromatic radio wave in the form of a coherent state
>> (three mathematical idealizations involved here) is an endless
>> stream of infinitely many photons passing with the speed of light
>> through a particular position on the beam.
>
> The expectation value of the photon number in a coherent state is
> finite!

But as discussed above, it does not give an absolute photon number but
an intensity, which translates into a rate of photons!


>> How can this _not_ be an ensemble of photons?
>
> It is an ensemble of photons, but not an ensemble of single photons!

I think you mean: It is an ensemble of photons, but not an ensemble
of Fock states.

Indeed, it is an ensemble of single photons but not an ensemble of
Fock states.


>> But the difficulty of this discussion illustrates the fact that the
>> particle picture in quantum field theory is of very limited use
>> and creates more confusion than understanding.
>
> nowadays the
> quantum opticians can prepare single- or multiple photon states with
> a definite photon number, e.g., by parametric down conversion.

But their understanding of these states is something different from
yours.

Yes, parametric down conversion is well-known to produce an
ensemble of 2-photon states, but if one looks closer at the models
one finds that they actually produce states of the form (**)
that produce endless streams of photon pairs.

[Gerard Westendorp]

There is a nice demo of a coherent state:

http://web.ift.uib.no/AMOS/MOV/HO/


Note that the higher components of the Energy eigenstates *have* to
oscillate at higher frequencies, to reproduce this nice dispersion-free
hump.

Gerard

[Hendrik van Hees]

Arnold Neumaier wrote:

> My own views (not mainstream, but consistent with experiment) are:
> - that clicks have nothing at all to do with photons, they just are
> a stochastic measure of intensity, and arise also if the incindent
> field is modelled completely classical;

I don't know what you mean by "photon". However within QED it is very
easy to define what one means by "one photon". It's defined as a
one-particle Hilbert-space vector. Of course plane waves never are a
state, but a generalized state. So a single-photon state, written in
terms of generalized single-photon momentum-helicity eigenstates (you
can take any generalized single-photon basis for this) is given by

|Psi_{one photon}>=\sum_{lambda=\pm 1} \int d^3 p A(p,lambda)
|\hat{a}^{dagger}(p,lambda),

where A(p,lambda) should be square integrable functions.

> - that what is typically called a photon is not an arbitrary single
> particle state of the electromagnetic field (in particular, never
> an approximately plane wave) but a state of the electromagnetic
> field that at each time is localized in space, whose energy
> contents is that of hbar*omega. Otherwise, the idea of producing
> photons of demands makes no sense.

I don't know, what you call a photon, but the above given state is what
I call a single-photon state.
> See
> http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf
> http://www.mat.univie.ac.at/~neum/ms/optslides.pdf
>
>
>> and as I've learned from a talk by Marlan Scully
>> at Texas A&M, there is the principle possibility to distinguish a
>> weak coherent state from a real single-photon state although it's
>> pretty difficult experimentally:
>>
>> PRL 96, 010501 (2006)
>
> I don't see where this paper discusses that question, but
> yes, it is possible to distinguish the two:
>
> The difference is that a weak coherent state generates an infinitely
> long random sequence of Poisson-distributed clicks, while a single
> photon (in the above sense of a space-localized field) generates
> a single click only (in an ideal detector).

That's not true. A simple example is a laser which you switch on at a
certain time and then switch off after a finite amount of time. It
emits a coherent state of finite extension in both, space and time.
There is no such thing as a plane-wave state in nature, as is well
reflected by quantum theoretical formalism since plane-wave states
belong not to Hilbert space but to the dual of a smaller dense subspace
(see, e.g., Ballentine, Quantum Mechanics for this formulation of
quantum theory in terms of "rigged Hilbert spaces").

> Most photon states produced in the laboratory are superposition with
> the vacuum, and still people speak of photons.
> Entanglemaent studies are typically made with squeezed states,
> which differ from coherent states only in thast they have instead
> of (*) a representation
> psi = (1-eps||0> + eps|2> + O(eps^2), (**)
> and everyone refers to (**) as an ensemble of 2-photon states!

> Yes, parametric down conversion is well-known to produce an
> ensemble of 2-photon states, but if one looks closer at the models
> one finds that they actually produce states of the form (**)
> that produce endless streams of photon pairs.

But it's very easy to produce single-photon states from this. If your
entangled photon pair is well separated in space, you can trigger on
one of the photons. Then you are sure that there is this photon and due
to the entanglement that there is also its partner at the other end.
Then you consider only events where you've seen the trigger photon and
do experiments with the other one. That's then an ensemble of single
photons and not a coherent state or a superposition of the vacuum and a
single photon.

--
Hendrik van Hees Institut für Theoretische Physik
Phone: +49 641 99-33342 Justus-Liebig-Universität Gießen
Fax: +49 641 99-33309 D-35392 Gießen

[Gerard Westendorp]

Arnold Neumaier wrote:
[..]

>> What about stimulated emission?
>> Does an ensemble ensemble of N photons enhance the rate of photon
>> emission by atoms by a factor N, just as an N-photon state does?
>
> The rate of emission is proportional to the intensity
> of the incident beam. It is the field of the incident beam that counts;
> the talk about photons in the incoming beam is not meaningful and
> only blurs the picture; the right language is that of field theory.

It is interesting to note, as you say, that the rate of emission is
proportional to the classical field strength.

However, the qualifications "right language" versus "not meaningful" are
going a bit far. Quantum mechanics should give correct answers, even if
a problem is more conveniently put in classical terms.

Comparing an "ensemble of N photons" to an "ensemble of N
*distinguishable* particles", it would seem that the ensemble of photons
would need to be symmetric with respect to exchange of photons.
Otherwise, the difference between identical and non-identical particles
could not be expressed in this ensemble picture. But a "symmetrised
ensemble" is suspiciously like an "N particle wave function". What is
the difference?

Gerard

[p.kinsler@ic.ac.uk]

Hendrik van Hees <Hendrik.vanHees@theo.physik.uni-giessen.de> wrote:
> I don't know what you mean by "photon". However within QED it is very
> easy to define what one means by "one photon". It's defined as a
> one-particle Hilbert-space vector.

Hmm. Within QED I might start with a mode containing photons (perhaps
a number state, perhaps coherent, perhaps something else), and then
ADD ONE PHOTON. IF the initial state was an N-photon number state,
the final state is a N+1 photon number state -- where is the
one-particle Hilbert-space vector? If it was an initial coherent state,
I get a coherent superposition between that and a single photon
number state.

Of course it's easy to argue that once the extra photon is in the mode
I cant tell one from the other, so shouldn't talk about single photons
there any more.

But if I rephrase and say that a photon is a single excitation, I can
talk clearly about the "extra" single photon for whatever explanatory
purpose, and most will understand what I mean (Even if strictly I'm
not allowed to distingish photons in the same mode).



--
---------------------------------+---------------------------------
Dr. Paul Kinsler
Blackett Laboratory (PHOT) (ph) +44-20-759-47734 (fax) 47714
Imperial College London, Dr.Paul.Kinsler@physics.org

[Arnold Neumaier]

Hendrik van Hees schrieb:
> Arnold Neumaier wrote:
>
>> My own views (not mainstream, but consistent with experiment) are:
>> - that clicks have nothing at all to do with photons, they just are
>> a stochastic measure of intensity, and arise also if the incindent
>> field is modelled completely classical;
>
> I don't know what you mean by "photon".

Different people mean different and often quite vague things by this
notion, if they bother to spell out the meaning in some detail
(which is usually not done). This can be seen from the diverging
explanations given in the special issue already cited.

But a theoretical model of a photo-detector excited by an external
classical monochromatic e/m field responds by clicking randomly
according to a Poisson statistics, although the model contains
no photons. Thus the precise meaning of ''photon'' is not
needed to defend my statement.

> However within QED it is very
> easy to define what one means by "one photon". It's defined as a
> one-particle Hilbert-space vector.

This only defines a ''one-photon state'' on paper, but not ''one
photon'' in an experiment. The relation between the two is quite
indirect, and there is no agreement in the literature on the precise
relation.

> Of course plane waves never are a
> state, but a generalized state. So a single-photon state, written in
> terms of generalized single-photon momentum-helicity eigenstates (you
> can take any generalized single-photon basis for this) is given by
>
> |Psi_{one photon}>=\sum_{lambda=\pm 1} \int d^3 p A(p,lambda)
> |\hat{a}^{dagger}(p,lambda),
>
> where A(p,lambda) should be square integrable functions.

Yes, this is a general 1-photon state. The question is how you relate
this state to what you actually prepare in a beam of light.

What does it mean in experimental terms to have prepared _one_ photon
in this state?

Reading the details of preparation schemes for photons on demands,
as discussed in the two pdf files quoted, one finds that no clear
answer can be given to this question. The closest explanation is
what I have said.

>> - that what is typically called a photon is not an arbitrary single
>> particle state of the electromagnetic field (in particular, never
>> an approximately plane wave) but a state of the electromagnetic
>> field that at each time is localized in space, whose energy
>> contents is that of hbar*omega. Otherwise, the idea of producing
>> photons of demands makes no sense.
>
> I don't know, what you call a photon, but the above given state is what
> I call a single-photon state.

You define a theoretical object, but fail to say how it relates to
reality. Thus your explanation is incomplete.

>> See
>> http://www.mat.univie.ac.at/~neum/ms/lightslides.pdf
>> http://www.mat.univie.ac.at/~neum/ms/optslides.pdf
>>
>>
>>> and as I've learned from a talk by Marlan Scully
>>> at Texas A&M, there is the principle possibility to distinguish a
>>> weak coherent state from a real single-photon state although it's
>>> pretty difficult experimentally:
>>>
>>> PRL 96, 010501 (2006)
>> I don't see where this paper discusses that question, but
>> yes, it is possible to distinguish the two:
>>
>> The difference is that a weak coherent state generates an infinitely
>> long random sequence of Poisson-distributed clicks, while a single
>> photon (in the above sense of a space-localized field) generates
>> a single click only (in an ideal detector).
>
> That's not true. A simple example is a laser which you switch on at a
> certain time and then switch off after a finite amount of time. It
> emits a coherent state of finite extension in both, space and time.

I don't see how your statement could imply your wrong claim that
my statement is not true.

Let us look at this setting in more detail. The time that the laser
is switched on is a variable T that we can choose at will.

Each coherent state has a mode A=A(p); the modes are in
1-1 correspondence with creation operators a^*(A). They create,
in field theory language, one photon in this mode. So far, these
photons are only constructs on paper, used to be able to write down
multiparticle states, and have not yet an obervable meaning.
An N-particle state of mode A is defined recursively from the vacuum
state by
|1,A> := a^*(A)|vac>, |N,A>: = a^*(A)|N-1,,A> for N>0,
and coherent states with mode A have the form
|z,A>> := const* sum_N z^N/sqrt{N!} |N,A>
with a complex amplitude z. and satisfies
a(A)|z,A>> = z|z,A>>.
The mean photon number associated with the coherent state is
Nbar := <N> = <a^*(A)a(A)> = <<z,A|a^*(A)a(A)|z,A>>
= <<z,A|z^*z|z,A>> = z^*z <<z,A|z,A>> = z^*z,
hence
Nbar = |z|^2,
independent of the time T.

On the other hand, the number of observable photons (in the sense
of detector clicks) is proportional to T. This shows that the
formal photon number operator in Fock space has nothing to do
with the photon number as defined by the number of clicks.

The fact that the model is an approximation only and that observation
is bounded in space and time does not change the results of
this analysis.

>> Most photon states produced in the laboratory are superposition with
>> the vacuum, and still people speak of photons.
>> Entanglemaent studies are typically made with squeezed states,
>> which differ from coherent states only in thast they have instead
>> of (*) a representation
>> psi = (1-eps||0> + eps|2> + O(eps^2), (**)
>> and everyone refers to (**) as an ensemble of 2-photon states!
>
>> Yes, parametric down conversion is well-known to produce an
>> ensemble of 2-photon states, but if one looks closer at the models
>> one finds that they actually produce states of the form (**)
>> that produce endless streams of photon pairs.
>
> But it's very easy to produce single-photon states from this. If your
> entangled photon pair is well separated in space, you can trigger on
> one of the photons. Then you are sure that there is this photon and due
> to the entanglement that there is also its partner at the other end.

This is precisely the mechanism that I was referring to, which
eliminates the vacuum part. You restrict your attention to the
2-photon sector of (**) by ignoring the times where nothing but the
vacuum part is observed, and focussing on the times when something
- and then by the form of (**) the 2-photon part - is observed.
In this sense you can interpret (**) as an ensemble of 2-photon states,
of which you observe the part in one beam, to know when a photon
is present in the other beam.

If you do exactly the same with an ordinary coherent state (**),
you get the interpretation of a weak coherent state as an ensemble
of 1-photon states.

If one does not accept this, one has no reason left to interpret
the case of (**) in the above way.

> Then you consider only events where you've seen the trigger photon and
> do experiments with the other one. That's then an ensemble of single
> photons and not a coherent state or a superposition of the vacuum and a
> single photon.

One still has the superposition with the vacuum, except that you choose
to ignore the times where nothing happens to get rid of the vacuum.
One is allowed to do this, of course; but if one accepts this liberty,
one has the same liberty for a simple coherent state....

Arnold Neumaier