The de Broglie wavelength of a photon

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Phil Gardner

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

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>

Gerard Westendorp

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

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]

Urs Schreiber

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


--
Urs.Sc...@uni-essen.de

Phil Gardner

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Apr 9, 2001, 10:12:57 PM4/9/01
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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.
[...]

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]

Urs Schreiber

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

--
Urs.Sc...@uni-essen.de

Neil

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Apr 17, 2001, 4:07:44 PM4/17/01
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"Urs Schreiber" <Urs.Sc...@uni-essen.de> wrote in message
news:d6AA6.1740$FY5.1...@www.newsranger.com...
<snip>

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

Jonathan Scott

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

Phil Gardner

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

Squark

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Apr 25, 2001, 4:22:04 PM4/25/01
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On Wed, 25 Apr 2001 04:19:29 GMT, Phil Gardner wrote (in
<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

Gerard Westendorp

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

Urs Schreiber

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

--
Urs.Sc...@uni-essen.de

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