photon's wavefunction

kvblake

What is the coordinate representation of the wavefunction of the
photon in QFT?
According QFT the wavefunction of a single photon is a+/0> where a+ is
the creation operator of the photon and /0> means the vacuum state.
What is then <x/ in QFT in order to make <x/a+/0> which will be the
coordinate representation of the photon's wavefunction?

Arnold Neumaier

kvblake schrieb:
> What is the coordinate representation of the wavefunction of the
> photon in QFT?
> According QFT the wavefunction of a single photon is a+/0> where a+ is
> the creation operator of the photon and /0> means the vacuum state.
> What is then <x/ in QFT in order to make <x/a+/0> which will be the
> coordinate representation of the photon's wavefunction?

A single photon has the same degrees of freedom as a classical
electromagnetic field; its shape is characterized by an arbitrary
4-potential A(x) satisfying (in the Lorentz gauge) the equations
nabla dot nabla A(x) = 0,
nabla dot A(x) = 0,
expressing the zero mass and the transversality of photons.

In momentum space, single photon states have the form
psi = integral d\p^3/p_0 A(\p)|\p>,
where |\p> is a single particle state with definite 3-momentum
\p (fat p), p_0=|\p| is the corresponding photon energy divided by
the speed of light, and A(\p) a polarization 4-vector orthogonal to
the 4-momentum p composed of \p and p_0, obtained by a Fourier
transform of the 4-potential A(x) in the Lorentz gauge. (The wave
equation translates into the condition p_0^2=\p^2, causality requires
p_0>0, and orthogonality p dot A(\p) = 0 expresses the Lorentz gauge
condition. For massless particles, there remains the additional gauge
freedom to shift A(\p) by a multiple of the 4-momentum p, which can
be used to fix A_0=0.)

A(\p) is usually written (in the gauge with time component 0) as a
linear combination of two specific polarization vectors eps^+(p) and
eps^-(p) for circularly polarized light. The term A(\p)|\p>
approximately represents a monochromatic beam with polarization
determined by A(\p).

Thus a general photon is a superposition of monochromatic waves of
arbitrary frequencies and directions, though for experiments one
usually uses nearly monochromatic photons bundled into narrow beams.

A(\p)=<\p|psi> can be regarded as the photon's wave function in
momentum space. Since photons are not localizable. there is no photon's
wave function in coordinate space. One could regard the 4-potential
A(x) as coordinate space wave function, but because of its gauge
dependence, this is not really useful.

Arnold Neumaier

kvblake

Kvblake

Vielen Dank!
Your post is very precious but unfortunately I need some
clarifications about your answer. I haven't seen nothing about my
question in any approved book and I can assure you that I have looked
in a lot. In fact for me such questions are the essence of physics, in
understanding of nature not just calculating possibilities which I
relate to mathemathics.

In fact I have seen a formula for <x/ in a book by the russian
Prohorov (2003) ---- <0/A(x) where one easy calculates for
wavefunction of the photon exp(-ikx)eps(k). I don't know why <x/=<0/
A(x) and I am bothered that the function is a complex function. This
is necessity for de Broglie function but how can one calculate E and M
which have to be real. By the way A is taken to be a hermitian
operator in order A to be real.

==========================
==========================
=
Arnold Neumaier schrieb:

1=2E A single photon has the same degrees of freedom as a classical
electromagnetic field; its shape is characterized by an arbitrary
4-potential A(x) satisfying (in the Lorentz gauge) the equations

nabla dot nabla A(x) = 0,
nabla dot A(x) = 0,

expressing the zero mass and the transversality of photons.
==========================
==========================
=
Kvblake

Do you mean by this that the photon generates(or is equal to,
identical to) A(x) and that two photons may have different A(x)?? Or
if one puts it another way that any E=CC field may in fact can be one
photon? I'm especially impressed by the words 'its shape is
characterized by an arbitrary 4-potential A(x)' which I understand as
above described.
In fact I thought that if you has a box full of EM radiation
(arbitrary A(x)) and if you let the photons to leave one by one at the
end you will stay with a specific A(x) which will be the photons
wavefunction.

==========================
==========================
===
Arnold Neumaier schrieb:
2=2E In momentum space, single photon states have the form
psi = integral d\p^3/p_0 A(\p)|\p>,

where is a single particle state with definite 3-momentum
| is the corresponding photon energy divided by
the speed of light, and a polarization 4-vector orthogonal to
the 4-momentum p composed of p, p0, obtained by a Fourier
transform of the 4-potential A(x) in the Lorentz gauge. (The wave
equation translates into the condition p2=(p0)2 , causality requires
p0>0, and orthogonality p. = 0 expresses the Lorentz gauge
condition. For massless particles, there remains the additional gauge
freedom to shift by a multiple of the 4-momentum p, which can
be used to fix A0=0.)

==========================
==========================
=====================
Kvblake

I haven't also found a monentum representation of the wavefunction of
the photon in any textbook. I would appreciate if you can write where
you found it. I can't understand why it is integrated in dp - psi will
then not depend on p and what is that particle with 3-momentum p.
Actually why not receive the coordinate representation by the reversed
Fourier transform.

Arnold Neumaier schrieb:
==========================
==========================
=======
3=2E is usually written (in the gauge with time component 0) as a
linear combination of two specific polarization vectors eps^+(p) and
eps^-(p) for circularly polarized light. The term
approximately represents a monochromatic beam with polarization
determined by
==========================
==========================
==========
Arnold Neumaier schrieb:

4=2E Thus a general photon is a superposition of monochromatic waves of
arbitrary frequencies and directions, though for experiments one
usually uses nearly monochromatic photons bundled into narrow beams.
==========================
==========================
===========
Kvblake

This here is very very suprising for me. I always tought that a photon
is a particle and wave of 1 direction and 1 frequency. As far as I
know a decay of an atom emits one photon of one frequency. Also if an
atom absorbs EM radiation it happens in 1 frequency. How can we then
regard a photon to be superposition of monochromatic waves of
arbitrary frequencies and directions?

==========================
==========================
==========
Arnold Neumaier schrieb:
5=2E A(x) can be regarded as the photon's wave function in
momentum space. Since photons are not localizable there is no
photon's
wave function in coordinate space. One could regard the 4-potential
A(x) as coordinate space wave function, but because of its
gauge
dependence, this is not really useful.
==========================
==========================
===========
Kvblake

Sorry but I can't also agree with that that photons are not
localizable. For localization of course one can use the photoelectric
effect, Compton effect and generally every interaction with matter. Of
course you will base yourself on the Heisenberg relation but maybe
photons are wavepackets. Also if one calculates the S matrix in
Besetzungszahl representatiion with photons as end particles for a
given process at the end he receives theire angle distribution which
means they are located partially in space nevertheless.
What if one has an excited atom in a box. Soon the atom emits a photon
which is localized in the box (refractive walls b.e.)

Thank you very much for the your consideration.

Kevin Blake

Arnold Neumaier

kvblake schrieb:

> In fact I have seen a formula for <x/ in a book by the russian
> Prohorov (2003) ---- <0/A(x) where one easy calculates for
> wavefunction of the photon exp(-ikx)eps(k).

Yes; this corresponds to my description with the Fourier transform undone.

> I am bothered that the function is a complex function. This
> is necessity for de Broglie function but how can one calculate E and M
> which have to be real.

Read about the complex analytic signal in
L. Mandel and E. Wolf,
Optical Coherence and Quantum Optics,
Cambridge University Press, 1995.
to see, how real E and M are described naturally in terms of complex
quantities. The complex version is very natural for the discussion of
causality issues.


> Arnold Neumaier schrieb:
>> A single photon has the same degrees of freedom as a classical
>> electromagnetic field; its shape is characterized by an arbitrary
>> 4-potential A(x)
>
> Do you mean by this that the photon generates(or is equal to,
> identical to) A(x)

No. Just that there are as many distinct photons as there are distinct
(gauge orbits of) A(x).


> and that two photons may have different A(x)??

Yes.


> Or if one puts it another way that any EM field may in fact can be one
> photon?

No. The electromagnetic field is the expectation of the EM field
operator, whereas the wave functions define the ensemble for which
the expectations are taken.

In an analogy to quantum mechanics, x, A (the photon amplitude),
E(x) (the field operator) correspond to
k, x, and p_k. Thus the coordinate index k is inflated to the spacetime
position x, the argument of the wave function is inflated to a solution
of the free Maxwell equations, the momentum operator is inflated to
a field operator, and the integral over x becomes a functional integral
over photon amplitudes.

As the observed components of the mean momentum, say, are
<p_k> = integral dx psi(x)^* p_k psi(x),
so the observed values of the electromagnetic field are
<E(x)> = integral dA psi(A)^* E(x) psi(A).
Here psi(A) is the most general state vector in Fock space; for a
single photon, psi depends linearly on A,
psi(A) = integral d\p^3/p_0 A(\p)|\p> = |A>.


> I'm especially impressed by the words 'its shape is
> characterized by an arbitrary 4-potential A(x)' which I understand as
> above described.

If you have a coherent state whose single photon part has A(x) as its
wave function then it generates a classical coherent electromagnetic
field with 4-potential A(x). If you have superpositons or mixtures of
states whose whose single photon part have different A(x), you get
instead some kind of average field.


> In fact I thought that if you has a box full of EM radiation
> (arbitrary A(x)) and if you let the photons to leave one by one at the
> end you will stay with a specific A(x) which will be the photons
> wavefunction.

Each photon is the whole ray - a mode of the electromagnetic field, and
not an object that leaves the source one by one...

Yes, this is somewhat counterintuitive, but that's what it is.
Not everything in quantum mechanics has an easy intuition.


>> Arnold Neumaier schrieb:
>> In momentum space, single photon states have the form
>> psi = integral d\p^3/p_0 A(\p)|\p>,
>
> I haven't also found a monentum representation of the wavefunction of
> the photon in any textbook. I would appreciate if you can write where
> you found it. I can't understand why it is integrated in dp - psi will
> then not depend on p

This is how things always look in second quantization. Even a harmonic
oscillator. The wave function psi(x) or psi(p) in standard (first
quantized) quantum mechanics becomes the state vector
psi = integral dx psi(x) |x> or integral dp psi(p) |p>
in Fock space; the wave function at x or p turns into the coefficient
of |x> or |p>.


> Arnold Neumaier schrieb:
>> 4=2E Thus a general photon is a superposition of monochromatic waves of
>> arbitrary frequencies and directions, though for experiments one
>> usually uses nearly monochromatic photons bundled into narrow beams.
>
> This here is very very suprising for me. I always thought that a photon
> is a particle and wave of 1 direction and 1 frequency.

This may be the basis of your difficulties with the concept of a photon.
Only a monochromatic beam has a fixed frequency and direction.


> As far as I
> know a decay of an atom emits one photon of one frequency.

But already arbitrary direction - since its shape is a radial wave and
not a plane wave. Written in terms of plane waves, it is a
superposition.


> How can we then
> regard a photon to be superposition of monochromatic waves of
> arbitrary frequencies and directions?

It just means that some photons are harder to produce than others.
Atomic decays only produce special kinds of photons.


> Arnold Neumaier schrieb:
> >A(x) can be regarded as the photon's wave function in
>> momentum space. Since photons are not localizable there is no photon's
>> wave function in coordinate space. One could regard the 4-potential
>> A(x) as coordinate space wave function, but because of its gauge
>> dependence, this is not really useful.
>
> Sorry but I can't also agree with that that photons are not
> localizable. For localization of course one can use the photoelectric
> effect,

Of course they are approximately localizable, otherwise we couldn't do
experiments with them. But unlike electrons, they cannot be localized
arbitrarily well. This is because they lack the longitudinal part of
a general wave function.

You can read about all this in the standard reference for quantum optics,
L. Mandel and E. Wolf,
Optical Coherence and Quantum Optics,
Cambridge University Press, 1995.


Arnold Neumaier

kvblake

Arnold Neumaier schrieb:
A single photon has the same degrees of freedom as a classical
electromagnetic field; its shape is characterized by an arbitrary 4-
potential A(x)
kvblake schrieb: Do you mean by this that the photon generates (or is
equal to, identical to) A(x)
Arnold Neumaier schrieb:
No. Just that there are as many distinct photons as there are distinct
(gauge orbits of) A(x).

kvblake
How can one understand that two A(x) represent different photons? --if
they are not connected by a gauge transformations A+df (where d is
partial derivative and f an arbitrary function?But if so EM field is
just arbitrary - what about its intensity which should be proportial
to the number of photons?
==========================================
Arnold Neumaier schrieb:

Each photon is the whole ray - a mode of the electromagnetic field,
and not an object that leaves the source one by one...

kvblake
I feel some lack of competence here.
Can you put it as simple as possible.
Let's have a box and concentrate on one mode only - a standing cosine
wave of EM of the lowest basic frequency f (it has of course a
specific vector potential A when one fixes the gauge) or two running
in opposite directions waves. I think there is a quantization here
because one has thrown away many EM configurations.
Now the cosine wave has an energy proportional to E.E. One then simply
divide by hf and find the number of photons in that mode.
Contrariwise can one now find E like proportional to square root of
1.hf and calculate A(x) - classical for 1 photon.
Now if one uses A /0> one would receive A(x)/0>
But then A(x) should be a very real function living inside the box.
One will has for the EM tensor the 4-rot of A(x).
If this is right how is this formula
<E(x)> = integral dA psi(A)^* E(x) psi(A).
Going to reproduce this result.
I would like to stress once more that I not very sure in that and
that I would appreciate your critique on the subject. Please keep in
mind that I have no problems with math but with the physical problems
for II quantization.
===============================================
kvblake schrieb As far as I know a decay of an atom emits one photon
of one frequency.
Arnold Neumaier schrieb:
But already arbitrary direction - since its shape is a radial wave and
not a plane wave. Written in terms of plane waves, it is a
superposition.

Kvblake

But the atom becames a certain impulse (which can be measured) and
hence the photon has also the opposite directed impulse. Then this
radial wave immidiately collapses. I don't think it is necessary to
measure at all.
The same happens if one register the photon. I don't believe one ever
had measured a radial wave as a whole object.

Thanks in advance for the considerations.

Kevin Blake

Arnold Neumaier

kvblake schrieb:
> Arnold Neumaier schrieb:
> A single photon has the same degrees of freedom as a classical
> electromagnetic field; its shape is characterized by an arbitrary 4-
> potential A(x)
> kvblake schrieb: Do you mean by this that the photon generates (or is
> equal to, identical to) A(x)
> Arnold Neumaier schrieb:
> No. Just that there are as many distinct photons as there are distinct
> (gauge orbits of) A(x).
>
> kvblake
> How can one understand that two A(x) represent different photons? --if
> they are not connected by a gauge transformations A+df (where d is
> partial derivative and f an arbitrary function?

Because the response of a measurement device to it will be different.


> But if so EM field is
> just arbitrary - what about its intensity which should be proportial
> to the number of photons?

For a monochromatic beam of light, it is proportional to the expectation
of the photon number operator. In general, the measured intensity also
depends on the position of the detector (which selects a particular
direction from the EM field - only beams in this direction will
contribute) and to the spectral sensitivity of the detector (which
typically responds in different degrees to different frequencies).
This shows that direction and frequency (which, together are in 1-1
correspondence with the momentum vector \p) are part of the beam's
(and hence the photon's) identity.


> ==========================================
> Arnold Neumaier schrieb:
>
> Each photon is the whole ray - a mode of the electromagnetic field,
> and not an object that leaves the source one by one...
>
> kvblake
> I feel some lack of competence here.
> Can you put it as simple as possible.
> Let's have a box and concentrate on one mode only - a standing cosine
> wave of EM of the lowest basic frequency f (it has of course a
> specific vector potential A when one fixes the gauge) or two running
> in opposite directions waves.

> I think there is a quantization here
> because one has thrown away many EM configurations.

No; this has nothing to do with quantization.

> Now the cosine wave has an energy proportional to E.E. One then simply
> divide by hf and find the number of photons in that mode.

No. This gives an intensity. The number of photons is also proportional
to the time of measurement since, when taking expectations, you need to
integrate over some time interval.


> Contrariwise can one now find E like proportional to square root of
> 1.hf and calculate A(x) - classical for 1 photon.
> Now if one uses A /0> one would receive A(x)/0>
> But then A(x) should be a very real function living inside the box.

For a pure 1-photon state, the derivatives of A(x) _are_ very real
objects inside the box, since they agree with the expectations of the
corresponding field operators. For multiphoton states or mixed states,
these expectations are still real, but the individual contributions of
the photons are not.


> ===============================================
> kvblake schrieb As far as I know a decay of an atom emits one photon
> of one frequency.
> Arnold Neumaier schrieb:
> But already arbitrary direction - since its shape is a radial wave and
> not a plane wave. Written in terms of plane waves, it is a
> superposition.
>
> Kvblake
>
> But the atom becames a certain impulse (which can be measured) and
> hence the photon has also the opposite directed impulse. Then this
> radial wave immidiately collapses.

Look at the derivations, and you'll find out that radial waves are
produced in the rest frame of the particle (typically assumed to be
inside some crystal. The decay of an atom has no preferred direction.
Even if it was excited from a preferred direction.


> The same happens if one register the photon. I don't believe one ever
> had measured a radial wave as a whole object.

Of course not, because all measurements take place at a particular
position.


Arnold Neumaier

kvblake

In an article "photon's wavefunction"
in sci.physics.research Mr.Neumaier wrote:

For a pure 1-photon state, the derivatives of A(x) _are_ very real
objects inside the box, since they agree with the expectations of the
corresponding field operators. For multiphoton states or mixed
states,
these expectations are still real, but the individual contributions
of
the photons are not.


> ===============================================
> kvblake schrieb As far as I know a decay of an atom emits one photon
> of one frequency.
> Arnold Neumaier schrieb:
> But already arbitrary direction - since its shape is a radial wave and
> not a plane wave. Written in terms of plane waves, it is a
> superposition.

> Kvblake


> But the atom becames a certain impulse (which can be measured) and
> hence the photon has also the opposite directed impulse. Then this
> radial wave immidiately collapses.



Look at the derivations, and you'll find out that radial waves are
produced in the rest frame of the particle (typically assumed to be
inside some crystal. The decay of an atom has no preferred direction.
Even if it was excited from a preferred direction.


> The same happens if one register the photon. I don't believe one ever
> had measured a radial wave as a whole object.


Of course not, because all measurements take place at a particular
position.

Arnold Neumaier

======================================================================

Sorry for the great delay for my answer.

I've read your post a month or so before but I couldn't say anything
though I didn't agree (or understand) much of it. I just waited for
Mandel and Wolf.

But at last I've got Mandel and Wolf you have recommend to me. I've
found there
the q-representation (Schroedinger wave functon) of the photon(s). I
was
glad to see that it is (1) not complex and (2) looks very much like
the
wave function of the quantum oscillator which the field is proven to
be
equivalent to and (3) is not arbitrary (gauge fixed).

But meanwhile I found in Beresteckii and Lifshitz book on QFT that
the
Schoedinger wave function of the photon is the plane wave taking part
in
the expansion of the operator of the vector-potential and standing in
front of the anihilation operator.
Again now the vector-potential is complex A=sqrt(4PI/
2w)eps(alfa)exp(ikr)
where eps(alfa) is the polarization. There is absolutely no speech
about
the complex signals.
I read in Mandel, Wolf a bit about the complex signal and as much as
I
understood its just a representation (in the means of some kind of
shorthand
writting) - the vector potential is nevertheless real. One should
take
A and A* in order to recover the original quantity.

So I am now in a big amazement what is than true? - Mandel's
Hermitian
polynomials (H0=q for one photon) or the complex plane wave exp(ikq).
I dont see any connection between these two.
And your notion about arbitrarity of the shape of the wave function of
the
photons is completely obscure to me. Maybe you mean it depends on the
boundery
conditions which can be arbitrary. Even so I cant agree - as a photon
one understands
always one and only one frequency. An arbitrary function may be
represented
only by many frequencies and hence many photons.

Arnold Neumaier

kvblake schrieb:
> In an article "photon's wavefunction"
> in sci.physics.research Mr.Neumaier wrote:
>
> For a pure 1-photon state, the derivatives of A(x) _are_ very real
> objects inside the box, since they agree with the expectations of the
> corresponding field operators.

Let me be precise and quote from the theoretical physics FAQ at
http://www.mat.univie.ac.at/~neum/physics-faq.txt
where you can find all the details (S4k. What is a photon?):

In momentum space (frequently but not always the appropriate choice),
single photon states have the form
|A> = integral d\p^3/p_0 A(\p)|\p>,
where |\p> is a single particle state with definite 3-momentum
\p (fat p), p_0=|\p| is the corresponding photon energy divided by c,
and the photon amplitide A(\p) is a polarization 4-vector.
Thus a general photon is a superposition of monochromatic waves with
arbitrary polarizations, frequencies and directions.
(The Fourier transform of A(\p) is the so-called analytic signal
A^(+)(x), and by adding its complex conjugate one gets the real
4-potential A(x) in the Lorentz gauge.)
[*** I just added this in view of your remarks below]


> For multiphoton states or mixed states, these expectations are still real,
> but the individual contributions of the photons are not.


> ======================================================================
>
> Sorry for the great delay for my answer.
>
> I've read your post a month or so before but I couldn't say anything
> though I didn't agree (or understand) much of it. I just waited for
> Mandel and Wolf.
>
> But at last I've got Mandel and Wolf you have recommend to me. I've
> found there
> the q-representation (Schroedinger wave functon) of the photon(s). I
> was
> glad to see that it is (1) not complex and (2) looks very much like
> the
> wave function of the quantum oscillator which the field is proven to
> be
> equivalent to and (3) is not arbitrary (gauge fixed).
>
> But meanwhile I found in Beresteckii and Lifshitz book on QFT that
> the
> Schoedinger wave function of the photon is the plane wave taking part
> in
> the expansion of the operator of the vector-potential and standing in
> front of the anihilation operator.
> Again now the vector-potential is complex A=sqrt(4PI/
> 2w)eps(alfa)exp(ikr)
> where eps(alfa) is the polarization. There is absolutely no speech
> about
> the complex signals.

There are different ways to derive certain things.
The complex analytic signal is just very convenient, since it
covers the part that contains the annihilation operators.
But one does not _have_ to use it.


> I read in Mandel, Wolf a bit about the complex signal and as much as I
> understood its just a representation (in the means of some kind of shorthand
> writting) - the vector potential is nevertheless real. One should take
> A and A* in order to recover the original quantity.

Yes.

He writes A for the real potential and A^(+) for the analytic signal.
It just contains the positive frequency part, and A is obtained by
adding its adjoint, which he denotes as A^(-).


> So I am now in a big amazement what is than true? - Mandel's
> Hermitian
> polynomials (H0=q for one photon) or the complex plane wave exp(ikq).
> I dont see any connection between these two.

The q-representation is an equivalent way of writing quantum optics.
It is useful for calculations and drawing pictures but is not
fundamental. The fundamental representation is that in terms of the
vector potential in momentum space.


> And your notion about arbitrarity of the shape of the wave function of
> the
> photons is completely obscure to me. Maybe you mean it depends on the
> boundery
> conditions which can be arbitrary. Even so I cant agree - as a photon
> one understands
> always one and only one frequency.

No. A photon can be in an arbitrary mode, which is a superposition of
modes of fixed frequency. Any linear combination of 1-photon states is
again a 1-photon state.


> An arbitrary function may be represented
> only by many frequencies and hence many photons.

No. Multiple photons are obtained by taking tensor products, not by
superposition.


Arnold Neumaier

kvblake

>>Kvblake schrieb:
>>But at last I've got Mandel and Wolf you have recommend to me. I've
>>found there the q-representation (Schroedinger wave functon) of the
>>photon(s). I was glad to see that it is (1) not complex and (2) looks
>>very much like the wave function of the quantum oscillator which the
>>field is proven to be equivalent to and (3) is not arbitrary (gauge
>>fixed).

>>But meanwhile I found in Beresteckii and Lifshitz's book on QFT that
>>the Schoedinger wave function of the photon is the plane wave taking
>>part in the expansion of the operator of the vector-potential and
>>standing in front of the anihilation operator. Again now the vector-
>>potential is complex A=sqrt(4PI/ 2w)eps(alfa)exp(ikr) where eps(alfa)
>>is the polarization. There is absolutely no speech about the complex
>>signals.

>Neumaier schrieb:

>There are different ways to derive certain things.

As I see it these are not different ways but different results.

>The q-representation is an equivalent way of writing quantum optics.
>It is useful for calculations and drawing pictures but is not
>fundamental. The fundamental representation is that in terms of the
>vector potential in momentum space.

OK. But what is its connection (of q-representation) to
A=sqrt(4PI/ 2w)eps(alfa)exp(ikr) . That's the question?

>Neumaier schrieb:

>A photon can be in an arbitrary mode, which is a superposition of
>modes of fixed frequency. Any linear combination of 1-photon states is
>again a 1-photon state.

I think your definition of the photon is a third alternative to the
above two.
As it concerns polarization states I am familier with this but not
with frequencies and hence energies. What experiments proves this?

What is the meaning of the coefficients ? - like in polarization?


>>Kvblake schrieb:

>>An arbitrary function may be represented only by many frequencies and hence many photons.

>No. Multiple photons are obtained by taking tensor products, not by
superposition.

What I am talking about is the q-representation which I understand as
the result from acting by the fileld operator on the state vector -
the eigenvalue of the field operator A. This has to be the vector
potential from which one gets the EM and calulates the Lorentz force.
I think what you mean is the state vector (in Fock space) itself which
is tensor product of all photon's state vectors.
So what I have in mind is the Fourier expansion of this 4-vector
A(mu).


O.K. Let I assume the photon can be in an arbitrary mode, which is a
superposition of modes of fixed frequency. But what are than its
energy and impulse - every mode will have its fixed frequency. So
which one will be taken in the Planck formulae.
>From the wave equation for A it follows strictly Omega = modul (k)

On the other hand I read somewhere that a wavepacket dissipate with
time in vacuum (I think it
was tried by de Broigle or during his time).


Regards: Kevin Blake

P.S
I'm also concerned about an incosistancy in Mandel. In formulae
10.4.12. they insist that the state vector
is a product of the modes, and than costruct the state of the
pseudolocalized photon using a sum of
modes k (the formulae just before 10.4.18.)

Arnold Neumaier

kvblake schrieb:
>>> Kvblake schrieb:
>>> But at last I've got Mandel and Wolf you have recommend to me. I've
>>> found there the q-representation (Schroedinger wave functon) of the
>>> photon(s). I was glad to see that it is (1) not complex and (2) looks
>>> very much like the wave function of the quantum oscillator which the
>>> field is proven to be equivalent to and (3) is not arbitrary (gauge
>>> fixed).
>
>>> But meanwhile I found in Beresteckii and Lifshitz's book on QFT that
>>> the Schoedinger wave function of the photon is the plane wave taking
>>> part in the expansion of the operator of the vector-potential and
>>> standing in front of the anihilation operator. Again now the vector-
>>> potential is complex A=sqrt(4PI/ 2w)eps(alfa)exp(ikr) where eps(alfa)
>>> is the polarization. There is absolutely no speech about the complex
>>> signals.
>
>> Neumaier schrieb:
>
>> There are different ways to derive certain things.
>
> As I see it these are not different ways but different results.
>
>> The q-representation is an equivalent way of writing quantum optics.
>> It is useful for calculations and drawing pictures but is not
>> fundamental. The fundamental representation is that in terms of the
>> vector potential in momentum space.
>
> OK. But what is its connection (of q-representation) to
> A=sqrt(4PI/ 2w)eps(alfa)exp(ikr) . That's the question?

The two are related by a Wigner transform. I don't have the book ready
to check the details, but the q-representation is derived in Mandel and
Wolf from the canonical field quantization, so you should be able to
figure it out for yourself. It is also derived in Gardiner's book on
quantum noise (''Phase space methods'')


>> Neumaier schrieb:
>
>> A photon can be in an arbitrary mode, which is a superposition of
>> modes of fixed frequency. Any linear combination of 1-photon states is
>> again a 1-photon state.
>
> I think your definition of the photon is a third alternative to the
> above two.

It is the definition of quantum electrodynamics. Quantum optics is a
subfield, obtained by making various useful assumptions and
approximations, in particular the assumption of quasi-monochromaticity
and the paraxial approximation


> As it concerns polarization states I am familier with this but not
> with frequencies and hence energies. What experiments proves this?

QED is proved in thousands of everyday applications...

Have you ever seen light of different colors? They are made up of
different frequencies, and by E = h*nu = hbar * omega related to
different energies.


> What is the meaning of the coefficients ? - like in polarization?

They tell something about the frequencies of the electromagnetic waves
and the directions of the wavefronts. You need to look at a classic
optics book (or corresponding sections of a treatise on theoretical
physics) to get the details; I don't want to do this by email...


>>> Kvblake schrieb:
>
>>> An arbitrary function may be represented only by many frequencies and hence many photons.
>
>> No. Multiple photons are obtained by taking tensor products, not by
> superposition.
>
> What I am talking about

Single and multiple photons _are_ states, not operators, so what you are
talking about does not make sense in terms of operators.


> is the q-representation which I understand as
> the result from acting by the field operator on the state vector -
> the eigenvalue of the field operator A. This has to be the vector
> potential from which one gets the EM and calulates the Lorentz force.
> I think what you mean is the state vector (in Fock space) itself which
> is tensor product of all photon's state vectors.
> So what I have in mind is the Fourier expansion of this 4-vector
> A(mu).

The Fourier expansion of A has nothing to do with photon numbers but
simply expands a general single photon state into a superposition of
plane wave single photon states.


> O.K. Let I assume the photon can be in an arbitrary mode, which is a
> superposition of modes of fixed frequency. But what are than its
> energy and impulse - every mode will have its fixed frequency.

Yes, and the superposition has not a well-defined energy but an
energy density in space-time.


> I'm also concerned about an incosistancy in Mandel. In formulae
> 10.4.12. they insist that the state vector
> is a product of the modes,

I don't have the book here, but if they do it, it is not the general case...

> and than costruct the state of the
> pseudolocalized photon using a sum of
> modes k (the formulae just before 10.4.18.)

Arnold Neumaier

kvblake

kvblake schrieb:

>> OK. But what is its connection (of q-representation) to
>> A=sqrt(4PI/ 2w)eps(alfa)exp(ikr) . That's the question?

>The two are related by a Wigner transform.

O.K. I’ll check for this trasform. But I can’t see how at all a plane
wave exp(-ikq)
Can be equivalent to q-representation -- q.exp(-q.q). The first is
complex and not localized
Whereas the second is not and is almost localized. I suspect that q-
rep and A are different
things (see at the end of the post).

kvblake schrieb:

>>As it concerns polarization states I am familier with this but not
>>with frequencies and hence energies. What experiments proves this?

>QED is proved in thousands of everyday applications...

I suppose that if a photon is in an arbitrary mode, which is
superposition of modes
of fixed frequencies, that it should be entangled via energy
conservation with an emitting
electron (atom) – similarly to the entanglement between photons (EPR)
via angular momentum
conservation. May be than it will be possible to realize some
experiments similar to the
teleportation of polarization (or spin).

>Have you ever seen light of different colors? They are made up of
>different frequencies, and by E = h*nu = hbar * omega related to
>different energies.

How can I be sure that these are not many photons of different
energies? I think they are. Do you think white light is made up of
white light photons – superposition
of different energies?


>The Fourier expansion of A has nothing to do with photon numbers but
>simply expands a general single photon state into a superposition of
>plane wave single photon states.

I think the field is which is expanded. At least it says so in
textbooks (may be they
Are old?)

>Yes, and the superposition has not a well-defined energy but an
>energy density in space-time.

Isn’t it a spectral density or is not spectral density involved? But
if the photon is closed in a box
of total reflecting walls and thus becomes isolated – then the energy
must become well defined
and has just one value (not spectrum of energies for one photon).

Is there any notebook where it stands in straight speech that one
photon has many frequencies
and energies and impulses?
I look now in Beresteckii, Lifshitz. There it stands in chapter 1
“�Photons��:
“These formulae [E=sum(N_ka.w) and P=sum(N_ka.k) let us introduce the
basic for all
QED notion of light quanta or photons. Namely one can consider the
free EM field as an aggregate
of particles , EACH one having energy (hw) and impulse (k=hw/c).

I think in Mandel they use wave packets but manage somehow to keep one
energy (though many
k spread around a central impulse).
In fact I have seen in Itzykson, Zuber QFT they use states in Fock
space a+_k |0> to form packets
a+_f |0> = intergal dk (f_k a+_k |0>) where f_k are subsidiary
functions |f_k|^2<infinity,
They are strongly localized around k.
Then N a+_f |0>=1. a+_f |0> and H a+_f |0> = E. a+_f |
0>
These states are called photons.


I think polarization states are quite different from a wave packet of
modes with exp(ikq). If one lets a X polarized light pass through 45
polarizer then he can see Y polarized photon later.
This could not be made for energy (frequencies) packets.


======================================================================= =======
ON 2 FEBRUARY 2007
kvblake schrieb:

What is the coordinate representation of the wavefunction of the
photon in QFT?
According QFT the wavefunction of a single photon is a+/0> where a+
is
the creation operator of the photon and /0> means the vacuum state.
What is then <x/ in QFT in order to make <x/a+/0> which will be the
coordinate representation of the photon's wavefunction?
======================================================================= ========
With this I wanted to understand

what happens to the classical vector – potential A (and EM) in QED if
one has one photon.

which (A) I regarded as the analog to Schroedinger psi for the
electron.
It’s been made (A) an operator and then simply disappears from
considerations in the textbooks. In QM if a quantity is made an
operator it then has a spectrum of the classical variable: by
example p = -id/dx ïƒ* then for a state |psi> one has many
measurable values of p – p_i after expanding |psi>= sum alfa_i .
p_i .
When one makes A(x,t) an operator then one must have similar spectrum
A_i(x,t) in each point. So I tried to undderstand what is A(x,t) for 1
photon (or 1 photon state).
(I read that A(x,t) can be regarded as WF of the photon like psi for
the electron.)

I think q-rep in Mandel is not this! I think q-rep is the possibility
to register 1 photon in point q. May be this is connected to how
A(x,t) looks like in QFT for 1 photon. (But what be the physical
meaning q-rep of the vacuum |0>?? ) which they (Mandel) also derive –
registration of the vacuum in q , t ?)

Recently I realized that there can not be one value for A(x,t) but in
fact there must be spectrum A_i(x,t) with probabilities alfa_i when
the field is excited in one photon state |1> or intergal dk (f_k a+_k |
0> . That’s because A doesn’t commute with N.

This is very strange from the point of view that the field is the
primary object and the photons are excitations. (I had the notion of
something near to phoNons – which is used in Zuber, Itzykson QFT)

Nevertheless maybe there must be a mean (expectation ?) value for a
set of one photon states which can be thought of as WF in QFT of
photons.

Regards: Kevin