Simple? fundamental question about EM in cavity
mercury
and L isn't equal to n. lambda/2 will this atom radiate or not?
A) If yes (radiates): There will not be a standing wave in the fiber
as it is contrary stated in any textbook
B) If no: The atom must learn the value of L at every instant of time
immediately!
Another question I don't understand is when there is a standing wave
in a cavity how is the Heisberg like relation to be fulfilled dE.dt>h
If one measures E one must receive an interval (E1,E2) respectively
interval for lambda (lambda1,lambda2). But lambda is only one!!
Tom Roberts
> If an atom emits EM radiation in an optic fiber (cavity) of lenght L
> and L isn't equal to n. lambda/2 will this atom radiate or not?
Sure it will. Ends of the fiber far away can't affect it. And, of
course, one cannot predict in advance whether the emitted light will be
within the acceptance of the fiber, or will leave it through its wall.
Optical fibers are designed so that most of the light
that enters will come out the other end, with minimal
losses due to absorption and scattering. But light
emitted inside the fiber has a large probability of
leaving through the nearest wall.
> A) If yes (radiates): There will not be a standing wave in the fiber
> as it is contrary stated in any textbook
Yes, there won't be a standing wave in the fiber.
> Another question I don't understand is when there is a standing wave
> in a cavity how is the Heisberg like relation to be fulfilled dE.dt>h
> If one measures E one must receive an interval (E1,E2) respectively
> interval for lambda (lambda1,lambda2). But lambda is only one!!
No real cavity is perfect, and has imperfections in its length; any
measurement of its length will have errors as well. So by measuring its
length you cannot determine the wavelength of the light precisely.
Wavelength, of course, is related to the light's momentum, not its
energy, and there can be an additional error involved in converting
between them: uncertainty in the index of refraction of the fiber.
To measure the energy of the light, one could measure its frequency.
Then the Heisenberg relation applies directly: to count cycles with a
higher relative precision you must count more of them, thus requiring a
longer duration of the measurement.
In actual practice, real-world measurement resolutions are usually
considerably larger than Heisenberg's limit. (but not always...)
Tom Roberts
Rich L.
> mercury wrote:
> > If an atom emits EM radiation in an optic fiber (cavity) of lenght L
> > and L isn't equal to �n. lambda/2 � will this atom radiate or not?
I think it might help to consider two representations of waves on a
string. If you have a string of length L with fixed ends, initially
at rest, and then apply a very short impulse to a point on the string
(not at an end), we all know that waves will propagate towards each of
the ends from the point of the impulse.
In a quantum mechanical representation, the modes of the string are
considered independently, and each mode is excited (or not) by the
impulse. The energy in each mode is represented by a quantum (e.g.
photon) and is considered to have an independent existence.
In classical mechanics one is more likely to represent the string as
very small masses separated by springs (using calculus to go to the
limit of uniform mass density and elasticity). In this picture the
effect of the impulse is to displace adjacent mass elements, and thus
the disturbance propagates down the string.
Both these points of view are equivalent. To see that, one must
realize that the pulses that propagate in the time domain (e.i. the
classical mechanics view presented above) is the superposition of many
of the eigen modes of the quantum mechanical model. The phases of the
different modes contribute to them all adding up to discrete
propagating pulses seen classically.
In the question you ask, you are assuming that the atom emits only one
photon at a discrete and precise wavelength. In actual practice there
is a time dependence to the emission that results in a broadening of
the wavelength spectrum. If the fiber is so short, and the emission
by the atom so slow, that there is no overlap of the emission spectrum
by the modes supported by the optical fiber, then the atom will NOT be
able to emit the photon and it will remain in an excited state for a
very long time. In practice this is unlikely due to the generally
fast emission times. Even an emission that is very slow, however,
will have a tail to the wavelength spectrum that will overlap the
fiber modes somewhere, even if with a very low probability, so
eventually the atom will decay and emit something.
I think a very interesting question is one of conservation of energy.
Suppose one excites such an atom with a 2ev photon, then puts it into
a cavity where it can only emit 0.5ev, 1.5ev, 2.5ev, etc.Eventually
one will observe one of these energies being emitted. What happens to
the difference in energy? Where did the extra energy come from, or go
to? This is not a question of the Uncertainty Principle because the
energy difference will persist for a very long time.
Rich L.
mercury
the letter of L.Rich yet and it will take a while to do so. If he is
reading this I can recommend him to keep an eye on the theme
(probably by clicking on the star above). I hope I will be able
to respond in 2-3 days.
On 14 Окт, 04:30, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
> mercury wrote:
> > If an atom emits EM radiation in an optic fiber (cavity) of lenght L
> > and L isn't equal to n. lambda/2 will this atom radiate or not?
>
> Sure it will. Ends of the fiber far away can't affect it. And, of
> course, one cannot predict in advance whether the emitted light will be
> within the acceptance of the fiber, or will leave it through its wall.
>
> Optical fibers are designed so that most of the light
> that enters will come out the other end, with minimal
> losses due to absorption and scattering. But light
> emitted inside the fiber has a large probability of
> leaving through the nearest wall.
>
> > A) If yes (radiates): There will not be a standing wave in the fiber
> > as it is contrary stated in any textbook
>
> Yes, there won't be a standing wave in the fiber.
>
Thank you very much for your reply.
I believe you misunderstood my post but it’s my fault because
I thought it is self understood what I have in mind.
I never had in mind an open fiber. I was thinking about a closed
fiber
with a 100 % reflectivity at the end. And a perfect wall with no
roughnesses. I think a crystal with its crystal grid can serve
practically
well but if no then with certain aid from nanotechnology such a wall
can be achieved – I think at least in the future.
This condition is always used in order to quantize the EM field in any
QED textbook. I only consider an optic fiber because I want to make
the
cavity extremely long. But I also think it is very possible to
realize such
an arrangement in a real experiment.
I am almost(100%) sure there won’t be radiation in a cavity with L
not
approximately (in the limits of Heisenberg relation) equal to n.lambda/
2.
This is exactly the Planck black body radiation condition, where only
those
oscillators in the walls can radiate for which this equation holds
true.
Suppose the atom in the wall radiates at 600 nm but the length of the
cavity
(closed fiber) is (300. 10^30 + 100) nm. Of course there must be a
huge gap
between 600 nm and the two surrounding frequencies v=c/lambda of the
atom.
Nevertheless the atom has been put in an excited state in won’t
radiate
(of course the imperfection of the reflective wall should be smaller
then 200 nm).
So now if one changes the position of the reflective end (b.e. using a
piezocrystal)
so that it is in the reach of the Heisenberg principle for 600 nm
(that is L=300.10^30) the atom would have the possibility to radiate
IMMEDIATELY. Of course it is not sure it will radiate immediately but
surely this makes it possible for interchanging signals over
indefinite
distances with a speed greater than admitted by SR.
Am curious about how is this ‘gedanken experiment’ to be disproved?
But as I think the model of the photon accepted in QED implies
exactly this. May be the model is not right in admitting the creation
of a standing wave mode in a cavity (the photon itself) as a whole
thing
at once – e.g. immediately in every point within a cavity.
> > Another question I don't understand is when there is a standing wave
> > in a cavity how is the Heisberg like relation to be fulfilled dE.dt>h
> > If one measures E one must receive an interval (E1,E2) respectively
> > interval for lambda (lambda1,lambda2). But lambda is only one!!
>
> No real cavity is perfect, and has imperfections in its length; any
> measurement of its length will have errors as well. So by measuring its
> length you cannot determine the wavelength of the light precisely.
OK. If I admit this then it follows that HUP is not an intrinsic
property
of the quantum system (as I think) but follows from coupling to the
environment. In this sense if there is not an atom anywhere in a
straight
line from the source, the source would not been able to radiate.
Here also applies again the argument I described above – the wall can
be made not to be rough – suppose we use just one atom as a reflector
in a one dimensional world or maybe 5 atoms which move in synchrony
to stay on a line perpendicular to the radiation.
> Wavelength, of course, is related to the light's momentum, not its
> energy, and there can be an additional error involved in converting
> between them: uncertainty in the index of refraction of the fiber.
>
I don’t agree. E=hv ; lambda. v = c so lambda = ch/E e.g an
absolute
constant multiplied be 1/E and so the uncertainty in lambda and the
energy are the same (this is true for v and momentum as well).
> To measure the energy of the light, one could measure its frequency.
> Then the Heisenberg relation applies directly: to count cycles with a
> higher relative precision you must count more of them, thus requiring a
> longer duration of the measurement.
If dE is 0 the time interval doesnt matter dE.dt = 0
Is there is just one lambda and one v dE should be 0.
>
> In actual practice, real-world measurement resolutions are usually
> considerably larger than Heisenberg's limit. (but not always...)
>
> Tom Roberts
I think that HUP follows as an intrinsic property of the atom and is
not
due to our experimental abilities. I think the atom always emits a
superposition of waves with different wavelengths which let as
register
a continuum of energies even if we can determine E with arbitrary
accuracy.
Ilian
Rich L.
> This a reply to the post of Tom Roberts only. I haven’t read
> the letter of L.Rich yet and it will take a while to do so. If he is
> reading this I can recommend him to keep an eye on the theme
> (probably by clicking on the star above). I hope I will be able
> to respond in 2-3 days.
>
> On 14 Окт, 04:30, Tom Roberts <tjroberts...@sbcglobal.net> wrote:
>
> Ilian
I think there are two misunderstandings here. First, for a very long
fiber (long cavity) the modes will be closely spaced and many modes
will overlap the emission band of the atom. To do what you are trying
to do you will want a fairly short cavity so the modes are fairly
widely spaced. Of course the meaning of "widely spaced" depends on
the emission rate (decay time constant) of the atom. An atom that has
a fast decay will have a wider emission bandwidth than an atom with a
slower decay.
The other problem is the relation between time domain and frequency
domain representation of the problem. If only one mode is being
excited, then by definition there can be no time dependence! This may
seem counter intuitive, but consider the more familiar case of an
impulse exciting a wave on a string. In the time domain we will see
the impulse at a position x=a produce two pulses propagating left and
right from x=a. In the frequency domain we will see several of the
allowed modes excited with various phases. In the frequency domain
representation it appears that the entire mode excitation happens at
once, and in the mathematical sense it does. But due to the
superposition of all those modes with the appropriate phases, they all
add up to the two pulses propagating away from the x=a point as seen
in the time domain representation. In particular, far from where
these pulses appear, the modes will add up distructively resulting in
no disturbance to the string. If you do the quantum mechanics
carefully you will see exactly the same thing. If only one mode is
being excited, you will find that the decay time is very, very long
because of the narrow overlap of the mode spectrum and the emission
spectrum. If the atom can excite many modes at once, then you will
see something a little closer to the pulse propagating left and right
from the atom (of course it will be more like a damped sinusoid than
an impulse, but a similar idea.).
In either case, if you do the math carefully you will not violate SR.
Rich L.
mercury
>
> - Показване на цитирания текст -
Thank you for that consideration. I've just got the case a little bit
before reading your post. Look at the end of the post for comment.
NEVERTHELESS there still stands the case of a long wavelength WL in
the radiofrequencies say 100 m. If we put a filter in the cavity which
would pass 90-110 m WL s. Then the cavity would be transparent for
90-110 m and by moving the reflecting end of the fiber maybe it would
be possible to violate SR?
I would highly appreciate your opinion about it?
What am worried about is that way of QED in creating the field in a
cavity at once in the whole volume of the cavity? If so there must be
a way to brake SR.
=========================================================
Comment for the post of Rich
I've just got the case. I think you mean the same.
If we have a long fiber (cavity) and lambda L1 doesnt fit well in
L=N.L1/2+k where k<L1/2 then the WL which would fit would be
L2/2= L/N=L1/2+k/N. So L2=L1+2k/N
If N is great L2 will be well within the reach of Heisenberg and the
atom radiates on L2 (very close to L1).
I even developed an every day example to better grasp it.
If there are 2 swimmers and the one has to beat the other by a pool
length: he has to be very quick if they must swim just one poollength.
But if they have to swimm kilometers (e.g.100 pools L) he has to be
just a little better. He have to outswim the second not by 50 m pro
poollength but just by 1m.
So he has to be just a little better.
Ilian
Rich L.
...
> NEVERTHELESS there still stands the case of a long wavelength WL in
> the radiofrequencies say 100 m. If we put a filter in the cavity which
> would pass 90-110 m WL s. Then the cavity would be transparent for
> 90-110 m and by moving the reflecting end of the fiber maybe it would
> be possible to violate SR?
> I would highly appreciate your opinion about it?
> What am worried about is that way of QED in creating the field in a
> cavity at once in the whole volume of the cavity? If so there must be
> a way to brake SR.
> Ilian
If the cavity fundamental wavelength is 100m (i.e. a 50m long cavity),
the frequency is 3MHz, and that is also the spacing between cavity
modes. In terms of energy this is 1.2E-8 electron volts, a very very
small energy! If this is a good "Q" cavity (i.e. it has low losses;
the modes do not decay rapidly on their own) then the natural decay
time of this mode will be on the order of 33 microseconds. This means
there is an uncertainty in the exact frequency of this mode on the
order of 30KHz, which allows this mode to excite/decay in times on the
order of 33 microseconds. This time is much longer than the cavity
period (0.33 microseconds), so you would be hard pressed to observe an
excitation at one location and a mode response at another and be sure
that the causal effect is exceeding the speed of light.
I think some of the confusion is the result of the idealized
calculations we often do in physics. When talking about cavities we
often consider ideal cavities with no spontaneous decay modes and
infinitesimal line widths. In reality there is always a finite line
width and an associated time constant. The MATH may imply that the
mode responds everywhere instantaneously, but when the real system
with many frequencies allowed (either many modes or the linewidth of a
single mode) then more gradual changes are calculated, and they
propagate through space without violating SR.
On a somewhat related issue, I am skeptical of the concept of
"instantaneous collapse of the wavefunction" as a real, physical,
thing that is part of the Copenhagen interpretation of QM. I believe
this is the result of misinterpreting the mathematics of an idealized,
unrealistic system. I think when we properly understand QM and SR
(and GR) that the weird ideas like this and the multiverse concept
will be seen to be either nonsense or at most unnecessary.
Rich
mercury
I don’t see any principle objections to the issue I raised in my post
of 19 Oct.
If the energy is low for detection one may use multiphoton excitation
(I am not
a specialist in laser technology but if not possible now then maybe in
the future).
I can’t see why is it impossible to use a bandpass filter of 10 kHz
around 30 MHz – is it principally forbidden by QM or SR??).
At last there is no need to move the reflective end of the 50 m fiber.
There should be two reflective surfaces at the appropriate positions
but the first may be driven transparent electronically (on the
principle of liquid crystals).
I see that this is hard to be realized but I am interested if such a
scheme is possible or principally inadmissible by the laws of QM or
SR.
> I think some of the confusion is the result of the idealized
> calculations we often do in physics. When talking about cavities we
> often consider ideal cavities with no spontaneous decay modes and
> infinitesimal line widths. In reality there is always a finite line
> width and an associated time constant. The MATH may imply that the
> mode responds everywhere instantaneously, but when the real system
> with many frequencies allowed (either many modes or the linewidth of a
> single mode) then more gradual changes are calculated, and they
> propagate through space without violating SR.
As I see your notion of the photon is an EM soliton. This doesn’t
correlate with the point like nature of the photon. Also a wave packet
of EM waves can’t stay compact for long. Also when one photon is
absorbed (e.g. one hv) where do the other frequencies around this v
go?
The wave packet of different frequencies is as I think in fact a
quantum superposition /not a classical one/ . In the sense of the
interpretation of Everett they exist in different universes. I am not
a fan of this interpretation but cite it to be clearer that all
frequencies(modes) exist somehow potentially. Moreover each mode is
also a QM superposition of the oscillators in the different points of
the cavity as far as I understand.
>
> On a somewhat related issue, I am skeptical of the concept of
> "instantaneous collapse of the wavefunction" as a real, physical,
> thing that is part of the Copenhagen interpretation of QM.
In fact I thought that "instantaneous collapse of the wavefunction" as
a real, physical thing has been proven by the teleportation of Bennett
et al. But yesterday I stuck on an article of him himself where he
gives another statistical? interpretation. (in arxiv)
I believed that Copenhagen interpretation of QM is that WF is not
real. (the Born statistical meaning of knowledge).
> I believe this is the result of misinterpreting the mathematics of an idealized,
> unrealistic system. I think when we properly understand QM and SR
> (and GR) that the weird ideas like this and the multiverse concept
> will be seen to be either nonsense or at most unnecessary.
>
> Rich
Best regards: Ilian
mercury
Sorry the article is not of Bennett but of
B. C. Sanctuary
On the existence of Einstein-Podolsky-Rosen Channels
Ilian
Rich L.
> On 22 Окт, 17:45, "Rich L." <ralivings...@sbcglobal.net> wrote:
> > On Oct 21, 9:48 am, mercury <il...@abv.bg> wrote:
...
> around 30 MHz – is it principally forbidden by QM or SR??).
There is no problem with such a narrow bandpass filter in either QM or
SR. A narrow bandpass DOES, however, imply a slow response time, on
the order of the inverse of the bandpass, so a 10KHz bandpass filter
will not respond to an impulse in much less than 100 microseconds,
which is on the order of the time it would take the light to traverse
the cavity.
...
> As I see your notion of the photon is an EM soliton. This doesn’t
> correlate with the point like nature of the photon. Also a wave packet
> of EM waves can’t stay compact for long. Also when one photon is
> absorbed (e.g. one hv) where do the other frequencies around this v
> go?
> The wave packet of different frequencies is as I think in fact a
> quantum superposition /not a classical one/ . In the sense of the
> interpretation of Everett they exist in different universes. I am not
> a fan of this interpretation but cite it to be clearer that all
> frequencies(modes) exist somehow potentially. Moreover each mode is
> also a QM superposition of the oscillators in the different points of
> the cavity as far as I understand.
It is incorrect to consider the EM waves as a "soliton solution" in
this case especially as the medium is assumed linear. EM fields in QM
are represented by the wave function, a probability wave. It is
improper to consider these to be "real" in the sense that the photon
is spread over space like the EM fields. Instead the amplitudes
represent the probability of a photon being detected in that
location. It actually has the same effect, you only detect photons
where the EM wave amplitudes are significant, therefore you won't
detect a photon where SR would be violated, because the wave
amplitudes do not violate SR either.
Rich L.
mercury
> On Oct 26, 5:55 am, mercury <il...@abv.bg> wrote:
>
> > On 22 Окт, 17:45, "Rich L." <ralivings...@sbcglobal.net> wrote:
> > > On Oct 21, 9:48 am, mercury <il...@abv.bg> wrote:
> ...
> > I can’t see why is it impossible to use a bandpass filter of 10 kHz
> > around 30 MHz – is it principally forbidden by QM or SR??).
>
> There is no problem with such a narrow bandpass filter in either QM or
> SR. A narrow bandpass DOES, however, imply a slow response time, on
> the order of the inverse of the bandpass, so a 10KHz bandpass filter
> will not respond to an impulse in much less than 100 microseconds,
> which is on the order of the time it would take the light to traverse
> the cavity.
>
> ...
Do you mean that the photons of say 3,05 MHz would pass through that
filter which is intended to stop them because of the cavity length?
Does this not imply an instantaneous dependence of the filter’s
properties from the position of the reflective surface at the cavity’s
end?
>
> > As I see your notion of the photon is an EM soliton. This doesn’t
> > correlate with the point like nature of the photon. Also a wave packet
> > of EM waves can’t stay compact for long. Also when one photon is
> > absorbed (e.g. one hv) where do the other frequencies around this v
> > go?
> > The wave packet of different frequencies is as I think in fact a
> > quantum superposition /not a classical one/ . In the sense of the
> > interpretation of Everett they exist in different universes. I am not
> > a fan of this interpretation but cite it to be clearer that all
> > frequencies(modes) exist somehow potentially. Moreover each mode is
> > also a QM superposition of the oscillators in the different points of
> > the cavity as far as I understand.
>
> It is incorrect to consider the EM waves as a "soliton solution" in
> this case especially as the medium is assumed linear. EM fields in QM
> are represented by the wave function, a probability wave. It is
> improper to consider these to be "real" in the sense that the photon
> is spread over space like the EM fields. Instead the amplitudes
> represent the probability of a photon being detected in that
> location. It actually has the same effect, you only detect photons
> where the EM wave amplitudes are significant, therefore you won't
> detect a photon where SR would be violated, because the wave
> amplitudes do not violate SR either.
>
> Rich L.
I think the point of view of the EM waves as probability waves
coincides with the linear superposition of QM ‘potentially existing’
oscillators of the EM tensor in each point of space in the cavity.
I think we agree on that point.
I can accept also that the wall can’t be perfect and there are in
fact a packet of waves of near frequencies corresponding to the
different lengths of the 'rough' reflective surface.
But then as I think you are implying (in order not to violate SR) the
probability amplitudes wouldn’t be zero in a small area (nevertheless
the Fourier decomposition implies non-zero for the separate
frequencies) and this area will travel trough the cavity at a speed of
light so not violating SR.
But then ONE SHOULD OBSERVE A CLASSICAL PICTURE when measuring the EM
field in the cavity immerging from this traveling probability – equal
in the whole cavity(expect at the walls where it should be higher).
But you know that in a cavity there are observed exactly standing
waves with zero EM field on the walls – the quantum picture.
This doesn’t agree with the wavepacket not braking SR – do you think?
I think the point of view of QED is that the photon is delocalized in
the whole cavity and is created in the whole cavity at once as an
intact object of some probabilistic nature (osccupying
probabiistically the whole cavity) and can be found (e.g. put to
existence) at the nodes most probably. This is how QED incorporates
waves and paricles into one object. So when measuring in a cavity one
will observe a standing EM wave - which he does.
Best regards: Ilian