Google Groups no longer supports new Usenet posts or subscriptions. Historical content remains viewable.

Dismiss

34 views

Skip to first unread message

Apr 2, 2001, 7:09:15 PM4/2/01

to

When a succession of photons, all of the same energy, E, is fed into a

spectrometer the measured wavelength must surely be equal to the de

Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

spectrometer the measured wavelength must surely be equal to the de

Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

If, using the same laser source, we locate the spectrometer within a

pressure vessel and flood this with a dielectric medium of refractive

index, n, classical wave theory asserts that the measured wavelength is

reduced to b' = b/n. So quantum mechanics seems to be saying that the

momentum of each photon in the medium is increased to p' = h/b' = np.

This counter-intuitive increase agrees with Minkowski"s theoretical

value for the momentum of light. But it conflicts with all other

theoretical values of the momentum of light (Abrahams, Peierls, etc )

that have ever been put forward..

Is there any consensus on the value of p' today? Has the value of b'

ever been checked experimentally?

Phil Gardner <pej...@oznetcom.com.au>

Apr 3, 2001, 6:10:50 PM4/3/01

to

Phil Gardner wrote:

> When a succession of photons, all of the same energy, E, is fed into a

> spectrometer the measured wavelength must surely be equal to the de

> Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

>

> If, using the same laser source, we locate the spectrometer within a

> pressure vessel and flood this with a dielectric medium of refractive

> index, n, classical wave theory asserts that the measured wavelength is

> reduced to b' = b/n. So quantum mechanics seems to be saying that the

> momentum of each photon in the medium is increased to p' = h/b' = np.

> When a succession of photons, all of the same energy, E, is fed into a

> spectrometer the measured wavelength must surely be equal to the de

> Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

>

> If, using the same laser source, we locate the spectrometer within a

> pressure vessel and flood this with a dielectric medium of refractive

> index, n, classical wave theory asserts that the measured wavelength is

> reduced to b' = b/n. So quantum mechanics seems to be saying that the

> momentum of each photon in the medium is increased to p' = h/b' = np.

From the conservation of energy, it seems that the number

of photons passing through the medium is the same as in vacuum,

the energy per photon is the same,

but they are moving slower, and therefor they must be closer

together. The energy density in the medium is greater than in

vacuum. This checks with the fact that the medium has

polarization energy.

The extra energy density would result in a net force, pushing

the medium apart. This checks with the higher momentum flux

inside the medium, which also leads to a net medium-expanding

force.

It seems a bit counterintuitive that slower moving photons

have bigger momenta. But apart from that there seems to be

problem. You could also check the Poynting vector,

P = (c/n)E X B, which should be continuous as you pass from

vacuum into the medium. The momentum density is P/(c/n)= nP/c,

which is also bigger inside the medium than outside. So it

seems that this has nothing to do with quantum mechanics,

it is also there is classical electromagnetism.

Gerard

[Moderator's note: there's nothing wrong with slowly moving

things having more momentum, since the real definition of

momentum for a particle is not p = mv but p = dL/dq'. - jb]

Apr 3, 2001, 1:22:56 PM4/3/01

to

Phil Gardner wrote (in message <3AC958...@oznetcom.com.au>):

>When a succession of photons, all of the same energy, E, is fed into a

>spectrometer the measured wavelength must surely be equal to the de

>Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

>

>If, using the same laser source, we locate the spectrometer within a

>pressure vessel and flood this with a dielectric medium of refractive

>index, n, classical wave theory asserts that the measured wavelength is

>reduced to b' = b/n. So quantum mechanics seems to be saying that the

>momentum of each photon in the medium is increased to p' = h/b' = np.

As far as I understand, it's like this:

The refractive index is a bulk property that only has a meaning when

applied to quantities that are averaged over the microscopic details. A

single photon can thus not be related to the refractive index by p' =

np. Instead, the refractive index is related to the process by which the

photons interact with the electrons in the dielectric medium:

Classically, the light wave incident on the dielectric medium induces a

polarization wave therein. What is propagating at phase velocity c/n

through the medium is not merely a decelerated light wave, but the

excitation of the system of the two coupled oscillators "light" and

"electron polarization". The momentum relation of this wave may thus be

different than that of a pure light wave.

The quantum description is given by quantum field theory of solids: The

approximate Hamiltonian of the coupled system is in terms of the

creation and annihilation operators of light and electron modes with

interaction terms describing absorption of a photon and creation of an

electron mode and vice versa. This Hamiltonian may be diagonalized by a

Bogoliubov transformation replacing photon and electron operators by a

linear combination of them, which may be interpreted as quasi particle

operators of "polaritons", the quanta of the coupled oscillation. The

dispersion relation of the polaritons is a mixture of the dispersion

relations of the light and the polarization wave and is approximately

linear for small wavenumbers / large wavelengths with

\omega \approx ck/n

while the dispersion relation for photons and electrons remains unchanged.

Therefore I'd say one never measures a photon with \omega = ck/n for n

\neq 1 but one could measure polaritons with this relation.

Apr 9, 2001, 10:12:57 PM4/9/01

to

Gerard Westendorp wrote:

>

> Phil Gardner wrote:

>

> > When a succession of photons, all of the same energy, E, is fed into a

> > spectrometer the measured wavelength must surely be equal to the de

> > Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

> >

> > If, using the same laser source, we locate the spectrometer within a

> > pressure vessel and flood this with a dielectric medium of refractive

> > index, n, classical wave theory asserts that the measured wavelength is

> > reduced to b' = b/n. So quantum mechanics seems to be saying that the

> > momentum of each photon in the medium is increased to p' = h/b' = np.

>

> From the conservation of energy, it seems that the number

> of photons passing through the medium is the same as in vacuum,

> the energy per photon is the same,

> but they are moving slower, and therefor they must be closer

> together. The energy density in the medium is greater than in

> vacuum. This checks with the fact that the medium has

> polarization energy.

[...]>

> Phil Gardner wrote:

>

> > When a succession of photons, all of the same energy, E, is fed into a

> > spectrometer the measured wavelength must surely be equal to the de

> > Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

> >

> > If, using the same laser source, we locate the spectrometer within a

> > pressure vessel and flood this with a dielectric medium of refractive

> > index, n, classical wave theory asserts that the measured wavelength is

> > reduced to b' = b/n. So quantum mechanics seems to be saying that the

> > momentum of each photon in the medium is increased to p' = h/b' = np.

>

> From the conservation of energy, it seems that the number

> of photons passing through the medium is the same as in vacuum,

> the energy per photon is the same,

> but they are moving slower, and therefor they must be closer

> together. The energy density in the medium is greater than in

> vacuum. This checks with the fact that the medium has

> polarization energy.

If, as a photon goes from vacuum into a medium of refractive index n,

its energy, E, stays constant and its momentum, p, increases to nE/c,

what do we do about the energy equation, E^2 = m^2 c^4 + c^2 p^2 ?

Does the photon acquire an imaginary mass, or should the equation be

changed to read E^2 = m^2 c^4 + v^2 p^2, where v = c/n, the mean

velocity of the photon?

Phil Gardner <pej...@oznetcom.com.au>

[Moderator's note: Quoted text trimmed. -MM]

Apr 10, 2001, 8:38:47 PM4/10/01

to

Phil Gardner wrote (in message <3ACFD8...@oznetcom.com.au>):

>If, as a photon goes from vacuum into a medium of refractive index n,

>its energy, E, stays constant and its momentum, p, increases to nE/c,

>what do we do about the energy equation, E^2 = m^2 c^4 + c^2 p^2 ?

>Does the photon acquire an imaginary mass, or should the equation be

>changed to read E^2 = m^2 c^4 + v^2 p^2, where v = c/n, the mean

>velocity of the photon?

QFT seems to teach that it is a misconception to consider a

photon with a momentum different from E/c. Look at the Hamiltonian

of the free photon field, it reads

E = \sum_k \hbar \omega_k N_k

= \sum_k \hbar c |k| N_k

and the total momentum observable reads

P = \sum_k \hbar |k| N_k

(with N_k being the usual operator measuring the number of photons

in mode k).

Thus if photons are what is counted by N_k, then their energy and

momentum will always satisfy p_k = E_k/c and p = E/c.

If now these photons are incident on some medium with refractive index

n, we classically observe the wave number of the radiation to increase

by a factor n. On the QFT level the free Hamiltonian of the photons

does not change, but it is accompanied by an interaction term that

describes annihilation of photons and excitiation of electrons (of the

medium) and vice versa. It turns out that one can, by a Bogoliubov like

transformation, introduce new operators N' such that the total

Hamiltonian, consisting of free photons, free electrons and the interaction,

may be concisely written in the fashion of a single free field

E_total = \sum_k n c |k| N'_k

(in the low energy limitt).

The quanta counted by N' are called "polaritons". These quasi particles

represent the quantized oscillations of the two coupled oscillators

"EM field" and "electron field". They do have momentum nck. This is

the QFT model of the classical polarization of the medium.

My reference for this is

H. Haken: "Quantenfeldtheorie des Festkoerpers", Teubner, (1973)

Urs Schreiber

Apr 17, 2001, 4:07:44 PM4/17/01

to

"Urs Schreiber" <Urs.Sc...@uni-essen.de> wrote in message

news:d6AA6.1740$FY5.1...@www.newsranger.com...<snip>

news:d6AA6.1740$FY5.1...@www.newsranger.com...

Interesting discussion here, but let's not forget the following

important consequence of wavelength, given momentum: the effect on

fixing the position of a particle that the wave is interacting

with. One way to illustrate the uncertainty principle is the

"Heisenberg microscope": as our ability to fix position with

shorter wavelengths increases, the disturbance to the particle by

recoil momentum increases, as per the usual delta x delta p > h

(in this case.)

The problem I have with the photon momentum staying E/c is this:

the photon energy stays the same in a refractive medium, as it

must for proper absorption and energy effects. If the wavelength

decreases in the medium, but the momentum remains E/c rather than

En/c, where n is refractive index, then we could cheat the

Heisenberg microscope by putting non-bound particles (which could

be bits of material, not fundamental particles) in a liquid or

gaseous medium with a high refractive index. We could measure the

particle's position better with the shorter wavelength (as is

actually done with oil immersion microscopes) but the particle

would be subject to the same recoil momentum.

Then again, if everything involves coupling with the medium, all

these quantities are hard to define anyway. What is the story on

effective recoil momentum in a medium, and its implications for

the uncertainty principle? How does this play out in the case of

bizarre condensed states with such high effective refractive index

that light is moving at the speed of a car? Who if anyone asked

this question, and answered it, before, in the literature,

discussions, etc?

Neil Bates

Apr 18, 2001, 3:28:36 PM4/18/01

to

[Note to moderator: I originally tried to post an earlier version of

this on 4th April via my news server but as it hasn't appeared after

two weeks, I'm trying the e-mail route instead]

Phil Gardner wrote:

> When a succession of photons, all of the same energy, E, is fed into a

> spectrometer the measured wavelength must surely be equal to the de

> Broglie wavelength, b = h/p = hc/E (in vacuum), of these photons.

>

> If, using the same laser source, we locate the spectrometer within a

> pressure vessel and flood this with a dielectric medium of refractive

> index, n, classical wave theory asserts that the measured wavelength is

> reduced to b' = b/n. So quantum mechanics seems to be saying that the

> momentum of each photon in the medium is increased to p' = h/b' = np.

According to Special Relativity and classical wave theory, this isn't

correct because this simplified formula mixes up two different uses

of c which are only the same for a vacuum.

In any medium, the relationship between frequency and wavelength is

given by frequency = v / lambda where v is the wave speed in that

medium. Frequency is of course unaffected by entering a different

medium (provided it is effectively at rest).

In special relativity, the momentum of something with energy E and

speed v is given by E v / c^2.

If we substitute h times frequency for E from the wavelength formula,

we get that the momentum is h/lambda v^2/c^2. This is the same as

h/lambda provided that the speed of travel v in both parts of the

equation is c. However, if the speed of travel is less than c,

the factor of v^2/c^2 is required.

This makes sense when a photon enters a medium, in that it

temporarily decreases in momentum, imparting a little momentum to

the medium, but if it comes out the other side it increases in

momentum, giving a kick back on the medium again. (Imagine the

photons to be bouncing around elastically in the medium, so the

overall speed and average momentum is decreased, but if they come

out the other side the last bounce re-imparts the original

momentum and speed. I know that this isn't realistic, but I find

mental models like these help get signs right etc.).

I'm puzzled that you assert that "quantum mechanics" leaves out

this additional factor. I'd guess that the formula you are

using does not apply to this more general case. If you are sure

that the simplified formula is supposed to be valid in this case,

please can you provide some specific reference to where you got

this information?

Jonathan Scott

Apr 25, 2001, 12:19:29 AM4/25/01

to

All the posts to this thread, at least to date, seem to agree that when

a photon goes from vacuum into a medium of refractive index, n:

a photon goes from vacuum into a medium of refractive index, n:

Its energy, E, is unchanged. The measured wavelength (given enough

photons, all of energy, E) is decreased in the ratio 1/n. This should

be equal to the de Broglie wavelength, b = h/p.

They disagree on what the photon momentum, p, changes to from its vacuum

value of E/c. We have:

(1) p = nE/c Gerard Westendorp <wes...@xs4all.nl>

(2) p = E/c Urs Schreiber <Urs.Sc...@uni-essen.de>

(3) p = E/nc Jonathan Scott

Of the three, (3) makes the best sense for anyone who, like me, believes

that the speed of the photon is reduced to c/n. This of course is

strongly denied by proponents of QED's messy and complicated model of

photon motion in a medium (see thread "The speed of a photon" in this

newsgroup ).

Both (2) and (3) require us to accept the fact that the de Broglie

equation, b = h/p, needs changes before it can be applied to photons

moving in a dielectric medium.

Phil Gardner <pej...@oznetcom.com.au>

Apr 25, 2001, 4:22:04 PM4/25/01

to

On Wed, 25 Apr 2001 04:19:29 GMT, Phil Gardner wrote (in

<3AE32B...@oznetcom.com.au>):

<3AE32B...@oznetcom.com.au>):

>All the posts to this thread, at least to date, seem to agree that when

>a photon goes from vacuum into a medium of refractive index, n:

>

>Its energy, E, is unchanged. The measured wavelength (given enough

>photons, all of energy, E) is decreased in the ratio 1/n.

You are forgetting the photon vs. quasi-photon issue. It seems to me the

photon retains its usual wavelength and momentum*, while the quasi-photon

(the quasiparticle born out of the photon-atoms system) has different

wavelength, but violates E = pc (because it is non-relativistic).

*It might be not so because some of the momentum may pass to the matter upon

entry. In any case, the de Broglie relation is preserved.

Best regards,

squark.

--------------------------------------------------------------------------------

Write to me at:

[Note: the fourth letter of the English alphabet is used in the later

exclusively as anti-spam]

dSdqudarkd_...@excite.com

May 2, 2001, 10:19:41 PM5/2/01

to

I think Einstein once said something like:

"Nowadays everyone thinks they know what a photon is,

but they are mistaken."

I think this applies here in a sense. It is hard to get

a mental picture of a photon that is consistent with what

we know.

On the one hand there is a picture in which photons are

bullets that shoot through the medium. In this picture

it seems to make sense that the photons continue to move

at the speed of light when they are not interacting with

atoms. But the bullets are weird, they can be in many

places at the same time, and can interfere with themselves.

Another picture is of a wave that engulfs thousands of

molecules in each wavelength. The electric fields of the

photon polarize the atoms, increasing the net momentum

density of the wave. The wave is also weird, because its

energy comes in quanta.

However, I think the wave picture is a pretty good

approximation here, presuming we are talking for example

about visible light in glass. In this case it makes

sense to talk about a photon as a classical em. wave,

whose energy density, momentum density and momentum flux

can be determined by the laws of classical electromagnetism.

Gerard

May 7, 2001, 8:36:29 PM5/7/01

to

Neil Bates wrote:

>The problem I have with the photon momentum staying E/c is this:

>the photon energy stays the same in a refractive medium, as it

>must for proper absorption and energy effects. If the wavelength

>decreases in the medium, but the momentum remains E/c rather than

>En/c, where n is refractive index, then we could cheat the

>Heisenberg microscope by putting non-bound particles (which could

A photon is a quantum of excitation of a pure EM-wave. The

wave that passes through a medium when light passes through is

not a pure EM-wave, but a coupled oscillation of the EM-field

and the solid (electrons). This coupled oscillation also has

quantized excitation, called polaritons. They are

superpositions of photons and electron excitations and *do*

have the classically expected dispersion relation p = nE/c. So

the quanta of the light wave of wavenumber nk inside the solid

are not photons, they are polaritons. This simple fact may be

hidden by the classical use of D and H instead of E and B. It

should be noted that the classical quantity D does also

incorporate electric field *and* polarization. A D-wave is not

an E-wave.

Gerard Westendorp wrote:

>

> I think Einstein once said something like:

>

> "Nowadays everyone thinks they know what a photon is,

> but they are mistaken."

Photons are the quanta of excitation of the EM-field. We

probably do not really know what *that* means, but it suffices

to distiguish photons from quanta of other fields. So we at

least know what photons are *not*. E.g. if a quantum has a

dispersion relation different from p=E/c it cannot be a

photon. I believe this is rather a question of technical terms

than of philosophy.

>The argument about the field Hamiltonian I don't follow. The

>Hamiltonian should be different from the vacuum Hamiltonian.

It is, but by an additive interaction term. The point is that

this does not change the energy coefficient of the

photon-number operator.

I brought up the QFT Hamiltonian to illustrate the technical

meaning of the technical term "photon". It appears that the

discussion suffers from different usages of the term "photon"

being used.

Urs Schreiber

0 new messages

Search

Clear search

Close search

Google apps

Main menu