[SOLVED] The time it takes to emit one photon

Paul Danaher

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>We have ("Mossbauer effect and retarded interactions") a fairly subtle\ndebate going on which an outsider with some mathematics but not nearly\nenough physics feels should be capable of a direct answer. One of the\nquestions posed in this is, how long does it take to emit one photon. I\nwould really welcome a "simple" answer.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>We have ("Mossbauer effect and retarded interactions") a fairly subtle
debate going on which an outsider with some mathematics but not nearly
enough physics feels should be capable of a direct answer. One of the
questions posed in this is, how long does it take to emit one photon. I
would really welcome a "simple" answer.

Eugene Stefanovich

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nPaul Danaher wrote:\n> We have ("Mossbauer effect and retarded interactions") a fairly subtle\n> debate going on which an outsider with some mathematics but not nearly\n> enough physics feels should be capable of a direct answer. One of the\n> questions posed in this is, how long does it take to emit one photon. I\n> would really welcome a "simple" answer.\n\nThat\'s a good question, and I don\'t have a good answer for it.\n\nIn QM we describe the photon emission as a decay process\nA -> B + C\n\nThe system is initially prepared in the (metastable) state A.\nThis is an excited state of the radioactive nucleus in our case.\nAt time t>0 we measure whether the system remained in A or\nchanged to B+C (the ground state of the nucleus plus a photon).\nThen we repeat this preparation-measurement process many times for\neach value of t, and finally get the time-dependent "decay law"\nomega_A(t) which is the probability of finding the system in A at time\nt if it was prepared in A at time t=0.\n\nQuantum mechanics allows us to calculate the function\nomega_A(t). In most cases this is an\nexponential function of t with the decay parameter called\n"the lifetime".\n\nThis parameter has nothing to do with the question you asked, i.e,\n"how long does it take to emit one photon", and the above approaches\n(both experimental and theoretical) do not offer any answer.\n\nTo answer this question experimentally, one needs to perform\nrepetitive measurements on _the same_ metastable system. E.g.,\nprepare system in A at time t=0, measure the state of the system at\nt=1ns, then measure the same system at t=2ns, etc. The result of one\nsuch experimental run may look\nlike this (note that in experiment we can find the system only\nin one of the two states A or B+C, and not somewhere in between):\n\ntime state\n0ns A\n1ns A\n2ns A\n3ns B+C\n4ns B+C\n.....\n\nThis may suggest that the photon emission time is less than 1ns.\nHowever, this "repetitive measurement" experiment is not\ncovered well by the standard QM formalism: you need to worry\nabout how the wave function changes after each measurement (collapse),\nyou meet some weird things like "Zeno effect", etc.\n\nI think that your question boils down to the following:\n"how fast is the collapse of the wave function?" My best guess is that\nit happens instantaneously.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Paul Danaher wrote:
> We have ("Mossbauer effect and retarded interactions") a fairly subtle
> debate going on which an outsider with some mathematics but not nearly
> enough physics feels should be capable of a direct answer. One of the
> questions posed in this is, how long does it take to emit one photon. I
> would really welcome a "simple" answer.

That's a good question, and I don't have a good answer for it.

In QM we describe the photon emission as a decay process
[itex]A -> B + C[/itex]

The system is initially prepared in the (metastable) state A.
This is an excited state of the radioactive nucleus in our case.
At time t>0 we measure whether the system remained in A or
changed to B+C (the ground state of the nucleus plus a photon).
Then we repeat this preparation-measurement process many times for
each value of t, and finally get the time-dependent "decay law"
[itex]\omega_A(t)[/itex] which is the probability of finding the system in A at time
t if it was prepared in A at time t=0.

Quantum mechanics allows us to calculate the function
[itex]\omega_A(t)[/itex]. In most cases this is an
exponential function of t with the decay parameter called
"the lifetime".

This parameter has nothing to do with the question you asked, i.e,
"how long does it take to emit one photon", and the above approaches
(both experimental and theoretical) do not offer any answer.

To answer this question experimentally, one needs to perform
repetitive measurements on [itex]_the same_[/itex] metastable system. E.g.,
prepare system in A at time [itex]t=0,[/itex] measure the state of the system at
[itex]t=1ns,[/itex] then measure the same system [itex]at t=2ns,[/itex] etc. The result of one
such experimental run may look
like this (note that in experiment we can find the system only
in one of the two states A or [itex]B+C,[/itex] and not somewhere in between):

time state
0ns A
1ns A
2ns A
3ns B+C
4ns B+C
[itex].....[/itex]

This may suggest that the photon emission time is less than 1ns.
However, this "repetitive measurement" experiment is not
covered well by the standard QM formalism: you need to worry
about how the wave function changes after each measurement (collapse),
you meet some weird things like "Zeno effect", etc.

I think that your question boils down to the following:
"how fast is the collapse of the wave function?" My best guess is that
it happens instantaneously.

Eugene.

Marcel LeBel

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Paul Danaher wrote:\n> We have ("Mossbauer effect and retarded interactions") a fairly subtle\n> debate going on which an outsider with some mathematics but not nearly\n> enough physics feels should be capable of a direct answer. One of the\n> questions posed in this is, how long does it take to emit one photon. I\n> would really welcome a "simple" answer.\n>\n\nHow long does it take for a photon to be absorbed ? This amount of time\nfor a light photon is very short, and consequently, usually rounded up\nto "instantaneous". But if you try to apply this idea to a long wave\nradio photon of 100km wavelength .. it does not strike me as too\nconvincing. The radio photon is absorbed by a radio antenna by inducing\nelectron oscillation in the antenna....not by collapsing on it\ninstantaneously. The photon is a quantum of action h packaged in a fixed\namount of time, the period. The photon is something that can happen, but\nhappen only in the time of the package. The photon is a fixed specific\npower. Once it is absorbed, we factor/integrate the total work done and\ncall it Energy. And this is the problem. We loose sight of what we\nstudy. What is real?\n\n*-The calculation, or the event? The event is real and our calculations\nare not. Everything we put on paper is one order of time\nover-integration. We integrate and measure energy, but the dimension of\nthe real event happening is "power". We integrate and measure time, but\nthe dimension of the real event happening is "the passage of time". We\nintegrate and measure the fall distance of an object, but the real event\nis an object falling.. or its existence vs time. Take a light photon\nabsorbed by chlorophyll in plants. It eventually degrade down to\ninfra-red heat photons. The quantum of action h is still the same. The\nonly difference is that the time package is now bigger. The power of\nthe heat photon is lower than the power of a light photon.-*\n\nThe photon is a quantum of action h packaged in a specific amount of\ntime. It will be absorbed by oscillating structures tuned to receiving\nthe quantum just at the same rate. The same happen for emission. How\nlong does it take for a 100km long radio photon to be emitted? Exactly\none period T, the event time. Same for light photons and all other\nphotons of the EM spectrum.\n\nlebel@muontailpig.com remove particle\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Paul Danaher wrote:
> We have ("Mossbauer effect and retarded interactions") a fairly subtle
> debate going on which an outsider with some mathematics but not nearly
> enough physics feels should be capable of a direct answer. One of the
> questions posed in this is, how long does it take to emit one photon. I
> would really welcome a "simple" answer.
>

How long does it take for a photon to be absorbed ? This amount of time
for a light photon is very short, and consequently, usually rounded up
to "instantaneous". But if you try to apply this idea to a long wave
radio photon of 100km wavelength .. it does not strike me as too
convincing. The radio photon is absorbed by a radio antenna by inducing
electron oscillation in the antenna....not by collapsing on it
instantaneously. The photon is a quantum of action h packaged in a fixed
amount of time, the period. The photon is something that can happen, but
happen only in the time of the package. The photon is a fixed specific
power. Once it is absorbed, we factor/integrate the total work done and
call it Energy. And this is the problem. We loose sight of what we
study. What is real?

[itex]*-The[/itex] calculation, or the event? The event is real and our calculations
are not. Everything we put on paper is one order of time
over-integration. We integrate and measure energy, but the dimension of
the real event happening is "power". We integrate and measure time, but
the dimension of the real event happening is "the passage of time". We
integrate and measure the fall distance of an object, but the real event
is an object falling.. or its existence vs time. Take a light photon
absorbed by chlorophyll in plants. It eventually degrade down to
infra-red heat photons. The quantum of action h is still the same. The
only difference is that the time package is now bigger. The power of
the heat photon is lower than the power of a light photon.-*

The photon is a quantum of action h packaged in a specific amount of
time. It will be absorbed by oscillating structures tuned to receiving
the quantum just at the same rate. The same happen for emission. How
long does it take for a 100km long radio photon to be emitted? Exactly
one period T, the event time. Same for light photons and all other
photons of the EM spectrum.

lebel@muontailpig.com remove particle

Oz

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich <eugenev@synopsys.com> writes\n>\n>\n>In QM we describe the photon emission as a decay process\n>A -> B + C\n>\n>Quantum mechanics allows us to calculate the function\n>omega_A(t). In most cases this is an\n>exponential function of t with the decay parameter called\n>"the lifetime".\n>\n>This parameter has nothing to do with the question you asked, i.e,\n>"how long does it take to emit one photon", and the above approaches\n>(both experimental and theoretical) do not offer any answer.\n\nNB I am an ignorant amateur.\n\nI am not happy with the assumption that there are two states, existed\nand emitted and nothing in between.\n\nWe can have a superposition of \'emitted\' and \'excited\' which seems to me\nto be as good a description of \'partly emitted\' as one could wish to\nhope for.\n\n>To answer this question experimentally, one needs to perform\n>repetitive measurements on _the same_ metastable system. E.g.,\n>prepare system in A at time t=0, measure the state of the system at\n>t=1ns, then measure the same system at t=2ns, etc. The result of one\n>such experimental run may look\n>like this\n\n\n>(note that in experiment we can find the system only\n>in one of the two states A or B+C, and not somewhere in between):\n\nNote carefully the above sentence. In essence one has FORCED the system\ninto state A or (B+C). Not unsurprisingly one sees ONLY A or (B+C). This\nsays nothing about the state of the system before you enforced a\nmeasurement. It says a LOT about the measuring equipment\'s interaction\nwith the state.\n\n>time state\n>0ns A\n>1ns A\n>2ns A\n>3ns B+C\n>4ns B+C\n>....\n>\n>This may suggest that the photon emission time is less than 1ns.\n>However, this "repetitive measurement" experiment is not\n>covered well by the standard QM formalism: you need to worry\n>about how the wave function changes after each measurement (collapse),\n>you meet some weird things like "Zeno effect", etc.\n\nThe above says more than that.\nIt says that the measuring apparatus can force the system into one state\nor the other in less than 1ns. It says absolutely nothing about how fast\nA -> (A+B) happens in the absence of the measuring apparatus.\n\nNow how about the following scenario.\nHere we measure 100,000 changes of A -> (B+C) ensuring A is always in\nthe same state before we start. This apparatus can force the system to\nbe in either A or (B+C) in 1/10ns. We get:\n\n%A %(B+C)\n2.0ns 100 0\n2.1 99 1\n2.2 97 3\n2.3 85 15\n2.4 60 40\n2.5 50 50 (OK its convenient...)\n2.6 40 60\n2.7 15 85\n2.8 3 97\n2.9 1 99\n3.0 0 100\n\nNow you can decide what you like, but I would say that this system takes\nabout 0.6 ns to emit a photon to first order. Mind you IMHO it would be\na slightly odd system that behaved like the above.\n\n>I think that your question boils down to the following:\n>"how fast is the collapse of the wave function?" My best guess is that\n>it happens instantaneously.\n\nMy guess is that it happens typically rather slowly and isn\'t a collapse\nat all but a gradual change via complex superposition of states.\n\nOf course if you measure it very fast, using an apparatus that enforces\nvery fast transitions, then the superposition includes that of the\nwavefunction of the very fast detector. Not unsurprisingly the\ntransition is as fast as the measuring equipment will deliver.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich <eugenev@synopsys.com> writes
>
>
>In QM we describe the photon emission as a decay process
[itex]>A -> B + C[/itex]
>
>Quantum mechanics allows us to calculate the function
[itex]>\omega_A(t)[/itex]. In most cases this is an
>exponential function of t with the decay parameter called
>"the lifetime".
>
>This parameter has nothing to do with the question you asked, i.e,
>"how long does it take to emit one photon", and the above approaches
>(both experimental and theoretical) do not offer any answer.

NB I am an ignorant amateur.

I am not happy with the assumption that there are two states, existed
and emitted and nothing in between.

We can have a superposition of 'emitted' and 'excited' which seems to me
to be as good a description of 'partly emitted' as one could wish to
hope for.

>To answer this question experimentally, one needs to perform
>repetitive measurements on [itex]_the same_[/itex] metastable system. E.g.,
>prepare system in A at time [itex]t=0,[/itex] measure the state of the system at
[itex]>t=1ns,[/itex] then measure the same system [itex]at t=2ns,[/itex] etc. The result of one
>such experimental run may look
>like this

>(note that in experiment we can find the system only
>in one of the two states A or [itex]B+C,[/itex] and not somewhere in between):

Note carefully the above sentence. In essence one has FORCED the system
into state A or [itex](B+C)[/itex]. Not unsurprisingly one sees ONLY A or [itex](B+C).[/itex] This
says nothing about the state of the system before you enforced a
measurement[itex]. It[/itex] says a LOT about the measuring equipment's interaction
with the state.

>time state
>0ns A
>1ns A
>2ns A
>3ns B+C
>4ns B+C
>....
>
>This may suggest that the photon emission time is less than 1ns.
>However, this "repetitive measurement" experiment is not
>covered well by the standard QM formalism: you need to worry
>about how the wave function changes after each measurement (collapse),
>you meet some weird things like "Zeno effect", etc.

The above says more than that.
It says that the measuring apparatus can force the system into one state
or the other in less than 1ns. It says absolutely nothing about how fast
[itex]A -> (A+B)[/itex] happens in the absence of the measuring apparatus.

Now how about the following scenario.
Here we measure 100,000 changes of [itex]A -> (B+C)[/itex] ensuring A is always in
the same state before we start. This apparatus can force the system to
be in either A or [itex](B+C)[/itex] in [itex]1/10ns[/itex]. We get:

[itex]%A %(B+C)[/itex]
2.0ns 100
2.1 99 1
2.2 97 3
2.3 85 15
2.4 60 40
2.5 50 50 (OK its convenient...)
2.6 40 60
2.7 15 85
2.8 3 97
2.9 1 99
3. 100

Now you can decide what you like, but I would say that this system takes
about .6 ns to emit a photon to first order. Mind you IMHO it would be
a slightly odd system that behaved like the above.

>I think that your question boils down to the following:
>"how fast is the collapse of the wave function?" My best guess is that
>it happens instantaneously.

My guess is that it happens typically rather slowly and isn't a collapse
at all but a gradual change via complex superposition of states.

Of course if you measure it very fast, using an apparatus that enforces
very fast transitions, then the superposition includes that of the
wavefunction of the very fast detector. Not unsurprisingly the
transition is as fast as the measuring equipment will deliver.

--
Oz
This post is worth absolutely nothing and is probably fallacious.

Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.

DM Roberts

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Since we cannot measure the state partway through the process of\nemitting a photon (i.e. some half-extruded photon still attached to the\nnucleus - I speak as a fool), there is really not much meaning to the\nquestion "how long does it take?"\n\nOne factor which may influence the answer, however, is the possibility\nof virtual particles (read on, I\'m not proposing anything wierd) -\nunmeasurable, but "existing" for a time less than the upper bound set\nby Heisenberg uncertainty (the more energy, incl mass, the particle\nhas, the less time it could "exist").\n\nWe might have to consider QFT effects, although if we are dealing with\na nucleus the corrections due to higher order diagrams will be small\n(in the sense that we can design lasers properly using QM only).\n\nAlso I agree with Eugene about the measuring of the same state\nrepeatedly to see at which time it collapses - How precisely does one\nmeasure the energy of the nucleus (or whatever) with out disturbing it\nfrom the metastable state?\n\nMore rigorously, the state consisting of the metastable nucleus evolves\ninto a mixture of states, one of which is the undecayed nucleus, the\nother is the decayed nucleus + a photon. Measurement projects the mixed\nstate vector onto one or the other. If we _never_ measure the system,\nwe can never actually say that the decay has taken place! Note that the\nevolution of the state into the mixture starts immediately, so the\ndecay could "take place" immediately. I say "take place", as all we can\nreally tell is that the decay took place sometime before the\nmeasurement, unless we look for extra data like how far the photon has\ntravelled.\n\n[ Mod. note: I believe the poster meant "superposition" rather than\n"mixture". "Mixture" has a different meaning in quantum statistical\nmechanics. -ik ]\n\nJust a question - what\'s the Zeno effect? I\'ve moved off into pure\nmaths and so have lost that little reality-connecting thread I once\nhad.\n\nDavid\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Since we cannot measure the state partway through the process of
emitting a photon (i.e. some half-extruded photon still attached to the
nucleus - I speak as a fool), there is really not much meaning to the
question "how long does it take?"

One factor which may influence the answer, however, is the possibility
of virtual particles (read on, I'm not proposing anything wierd) -
unmeasurable, but "existing" for a time less than the upper bound set
by Heisenberg uncertainty (the more energy, incl mass, the particle
has, the less time it could "exist").

We might have to consider QFT effects, although if we are dealing with
a nucleus the corrections due to higher order diagrams will be small
(in the sense that we can design lasers properly using QM only).

Also I agree with Eugene about the measuring of the same state
repeatedly to see at which time it collapses - How precisely does one
measure the energy of the nucleus (or whatever) with out disturbing it
from the metastable state?

More rigorously, the state consisting of the metastable nucleus evolves
into a mixture of states, one of which is the undecayed nucleus, the
other is the decayed nucleus + a photon. Measurement projects the mixed
state vector onto one or the other. If we _never_ measure the system,
we can never actually say that the decay has taken place! Note that the
evolution of the state into the mixture starts immediately, so the
decay could "take place" immediately. I say "take place", as all we can
really tell is that the decay took place sometime before the
measurement, unless we look for extra data like how far the photon has
travelled.

[ Mod. note: I believe the poster meant "superposition" rather than
"mixture". "Mixture" has a different meaning in quantum statistical
mechanics. [itex]-ik ][/itex]

Just a question - what's the Zeno effect? I've moved off into pure
maths and so have lost that little reality-connecting thread I once
had.

David

Antiphon

Measure the spectral width of the photon (by measuing many of them).

[tex] T \approx 1/(spectral width)[/tex]

Eugene Stefanovich

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nDM Roberts wrote:\n\n> Just a question - what\'s the Zeno effect?\n\nMy quick web search produced the following:\n\n"The so called "Quantum Zeno Effect" was first discussed in detail by\nE. C. George Sudarshan and Baidyanath Misra in 1977, and first\nobserved experimentally by Fischer et al. in 2001. The idea is often\nparaphrased as "a watched pot never boils". Fischer el al. observed the\nescaping time of trapped sodium atoms, and found that repeated\nmeasurements can suppress or enhance the decay of the system, depending\non the frequency of measurements. When the observation frequency is\nsmaller than the characteristic relaxation frequency of the trapped\natoms, the decay is suppressed, we have quantum Zeno effect. When the\nobservation frequency is larger, the decay is enhanced, thus we have\nquantum anti-Zeno effect."\n\nI don\'t know much about this stuff, so I prefer discuss quantum effects\nthat can be observed without involving repetitive measurements,\ni.e., system prepared -> system measured -> system discarded.\n\nEugene.\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>DM Roberts wrote:

> Just a question - what's the Zeno effect?

My quick web search produced the following:

"The so called "Quantum Zeno Effect" was first discussed in detail by
E. C. George Sudarshan and Baidyanath Misra in 1977, and first
observed experimentally by Fischer et al. in 2001. The idea is often
paraphrased as "a watched pot never boils". Fischer el al. observed the
escaping time of trapped sodium atoms, and found that repeated
measurements can suppress or enhance the decay of the system, depending
on the frequency of measurements. When the observation frequency is
smaller than the characteristic relaxation frequency of the trapped
atoms, the decay is suppressed, we have quantum Zeno effect. When the
observation frequency is larger, the decay is enhanced, thus we have
quantum anti-Zeno effect."

I don't know much about this stuff, so I prefer discuss quantum effects
that can be observed without involving repetitive measurements,
i.e., system prepared -> system measured -> system discarded.

Eugene.

Igor Khavkine

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Paul Danaher wrote:\n> We have ("Mossbauer effect and retarded interactions") a fairly subtle\n> debate going on which an outsider with some mathematics but not nearly\n> enough physics feels should be capable of a direct answer. One of the\n> questions posed in this is, how long does it take to emit one photon. I\n> would really welcome a "simple" answer.\n\nTo be really safe, the answer has to be infinity. Of course, we see\nphotons emitted all the time and these events happen rather more\nquickly than that. Of course, this just means that the photon emission\ntime scale is very long compared to some other characteristic length,\nyet still very short compared to the time scales we are used to. To\nidentify what each of these time scales is, we need to look at how\nphoton emission is observed and theoretically analyzed.\n\nFirst lets look at how photons appear theoretically. Take Maxwell\'s\nequations. They form a system of linear translation invariant\nequations. A Fourier transform decouples the fields into independently\nvibrating modes, each with a characteristic frequency. In this form,\nquantization is trivial. Each mode is treated as an independent\nharmonic oscillator.\n\nThe spectrum of stationary states (those with definite energy) of a\nharmonic oscillator is well known. It is discrete and equally spaced,\nwith the spacing given by the frequency. Each state is labeled by an\ninteger n. The energy of the state is basically a multiple of this\nnumber. We like to say that the state n contains n quanta, each with\nenergy proportional to the frequency of the oscillator. These quanta\nare called phonons. Each stationary state of the electromagnetic field\ncan be described by a linear superposition of states with definite\nfinite photon number in each vibration mode.\n\nSo far we have only described stationary states of the EM field and\ngiven them an interpretation in terms of photons. Obviously this does\nnot cover non-stationary states, which describe photon emission for\nexample. If we know all the stationary states and their energies, we\ncan in principle describe all time dependent states of the EM field as\nwell, but *only* of the EM field. To model interesting physical\nsituations, we have to introduce other matter or fields and\ninteractions with them. Unfortunately, for most interacting systems, we\ncan\'t write down a closed form solution for its state spectrum.\nCalculations are usually done in perturbation theory. The crucial\nassumption is that we can approximate the state at the beginning and\nend of the experiment by stationary states of the non-interacting EM\nfield + matter system. The interaction can then be turned on and off in\nbetween. By necessity, for the initial and final states to be well\napproximated by stationary ones, the on/off switching of the\ninteraction must be slow, i.e. adiabatic. That is why we formally place\nthe initial state at time -oo and the final state at time +oo.\n\nSo, with this set up, how do we characterize processes in which photons\nare emitted? It\'s rather simple really. Count the photons in the\ninitial state and count the photons in the final state. If the latter\nnumber is larger than the former, then we say that some photons were\nemitted. Note however, that in the only answer we can give to how long\nit to took for the emission to take place is infinity, which is the\nelapsed time between the initial and final states.\n\nOk, now we know what theorists mean whey they speak of photons. How\nabout experimentalists? If you look at the experimental eveidence that\nis pointed to when speaking of photons, it always comes back to\nquantization of energy, for example the photoelectric effect and\nPlanck\'s radiation spectrum. But to measure energy we need to observe\nthe system for a sufficiently long time. In other words, the\nexperimental setup is such that the system\'s state can be approximated\nby a stationary one. But that\'s exactly where the theoretical\ndescription of photons comes in. Everyone is on the same page here.\n\nFinally, we come to observation of emission. The experimental procedure\nis to prepare the system in some known state (again most likely\napproximated by some stationary state), wait for the emission event and\nfor the detector to register the emission (wait for the detector to\nalso settle close to a stationary state). This matches the theoretical\ndescription as well. But then we face the same uncertainty about the\ntime needed for emission.\n\nHowever, now we have at least some bounds the needed time just by the\namount of time needed to setup the experiment and wait for the resutls.\nWith better technology, the time needed to perform the experiment can\nbe shortened. Can this be done until the least upper bound on the\nemission duration is reached? Here, quantum mechanics says that you can\nget close, but not quite. If you don\'t allow enough time for the system\nto come close to a stationary state, the theoretical approximation and\nthe photon description break down. You will still measure photons\nabsorbed or emitted. But the results of the experiment will become more\nand more erratic. In other words, the uncertainty in the number of\nphotons absorbed or emitted grows the less time you devote to setup of\nthe experiment and to observation. This is nothing but the energy time\nuncertainty relation.\n\nSo, at the end of this perhaps overly verbose explanation, the simple\nanswer is given by the Heisenberg uncertainty principle. If the energy\nof the photon is E, the time needed to emit a photon of this energy is\ngreater than hbar/E. The longer you wait, the more sure you are how\nmany quanta were emitted. The less you wait, the less certainty there\nis about the number of photons emitted, which could also be zero.\n\nHope this helps.\n\nIgor\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Paul Danaher wrote:
> We have ("Mossbauer effect and retarded interactions") a fairly subtle
> debate going on which an outsider with some mathematics but not nearly
> enough physics feels should be capable of a direct answer. One of the
> questions posed in this is, how long does it take to emit one photon. I
> would really welcome a "simple" answer.

To be really safe, the answer has to be infinity. Of course, we see
photons emitted all the time and these events happen rather more
quickly than that. Of course, this just means that the photon emission
time scale is very long compared to some other characteristic length,
yet still very short compared to the time scales we are used to. To
identify what each of these time scales is, we need to look at how
photon emission is observed and theoretically analyzed.

First lets look at how photons appear theoretically. Take Maxwell's
equations. They form a system of linear translation invariant
equations. A Fourier transform decouples the fields into independently
vibrating modes, each with a characteristic frequency. In this form,
quantization is trivial. Each mode is treated as an independent
harmonic oscillator.

The spectrum of stationary states (those with definite energy) of a
harmonic oscillator is well known. It is discrete and equally spaced,
with the spacing given by the frequency. Each state is labeled by an
integer n. The energy of the state is basically a multiple of this
number. We like to say that the state n contains n quanta, each with
energy proportional to the frequency of the oscillator. These quanta
are called phonons. Each stationary state of the electromagnetic field
can be described by a linear superposition of states with definite
finite photon number in each vibration mode.

So far we have only described stationary states of the EM field and
given them an interpretation in terms of photons. Obviously this does
not cover non-stationary states, which describe photon emission for
example. If we know all the stationary states and their energies, we
can in principle describe all time dependent states of the EM field as
well, but *only* of the EM field. To model interesting physical
situations, we have to introduce other matter or fields and
interactions with them. Unfortunately, for most interacting systems, we
can't write down a closed form solution for its state spectrum.
Calculations are usually done in perturbation theory. The crucial
assumption is that we can approximate the state at the beginning and
end of the experiment by stationary states of the non-interacting EM
field + matter system. The interaction can then be turned on and off in
between. By necessity, for the initial and final states to be well
approximated by stationary ones, the [itex]on/off[/itex] switching of the
interaction must be slow, i.e. adiabatic. That is why we formally place
the initial state at time -oo and the final state at time +oo.

So, with this set up, how do we characterize processes in which photons
are emitted? It's rather simple really. Count the photons in the
initial state and count the photons in the final state. If the latter
number is larger than the former, then we say that some photons were
emitted. Note however, that in the only answer we can give to how long
it to took for the emission to take place is infinity, which is the
elapsed time between the initial and final states.

Ok, now we know what theorists mean whey they speak of photons. How
about experimentalists? If you look at the experimental eveidence that
is pointed to when speaking of photons, it always comes back to
quantization of energy, for example the photoelectric effect and
Planck's radiation spectrum. But to measure energy we need to observe
the system for a sufficiently long time. In other words, the
experimental setup is such that the system's state can be approximated
by a stationary one. But that's exactly where the theoretical
description of photons comes in. Everyone is on the same page here.

Finally, we come to observation of emission. The experimental procedure
is to prepare the system in some known state (again most likely
approximated by some stationary state), wait for the emission event and
for the detector to register the emission (wait for the detector to
also settle close to a stationary state). This matches the theoretical
description as well. But then we face the same uncertainty about the
time needed for emission.

However, now we have at least some bounds the needed time just by the
amount of time needed to setup the experiment and wait for the resutls.
With better technology, the time needed to perform the experiment can
be shortened. Can this be done until the least upper bound on the
emission duration is reached? Here, quantum mechanics says that you can
get close, but not quite. If you don't allow enough time for the system
to come close to a stationary state, the theoretical approximation and
the photon description break down. You will still measure photons
absorbed or emitted. But the results of the experiment will become more
and more erratic. In other words, the uncertainty in the number of
photons absorbed or emitted grows the less time you devote to setup of
the experiment and to observation. This is nothing but the energy time
uncertainty relation.

So, at the end of this perhaps overly verbose explanation, the simple
answer is given by the Heisenberg uncertainty principle. If the energy
of the photon is E, the time needed to emit a photon of this energy is
greater than [itex]\hbar/E[/itex]. The longer you wait, the more sure you are how
many quanta were emitted. The less you wait, the less certainty there
is about the number of photons emitted, which could also be zero.

Hope this helps.

Igor

Eugene Stefanovich

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nDM Roberts wrote:\n\n> More rigorously, the state consisting of the metastable nucleus evolves\n> into a mixture of states, one of which is the undecayed nucleus, the\n> other is the decayed nucleus + a photon. Measurement projects the mixed\n> state vector onto one or the other. If we _never_ measure the system,\n> we can never actually say that the decay has taken place! Note that the\n> evolution of the state into the mixture starts immediately, so the\n> decay could "take place" immediately. I say "take place", as all we can\n> really tell is that the decay took place sometime before the\n> measurement, unless we look for extra data like how far the photon has\n> travelled.\n>\n> [ Mod. note: I believe the poster meant "superposition" rather than\n> "mixture". "Mixture" has a different meaning in quantum statistical\n> mechanics. -ik ]\n\nThank you -ik. This is an important correction. We are not talking about\nstatistical mixed states described by density matrices (operators).\nWe should use words "superposition" or "linear combination" instead\nof "mixture" everywhere.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>DM Roberts wrote:

> More rigorously, the state consisting of the metastable nucleus evolves
> into a mixture of states, one of which is the undecayed nucleus, the
> other is the decayed nucleus + a photon. Measurement projects the mixed
> state vector onto one or the other. If we _never_ measure the system,
> we can never actually say that the decay has taken place! Note that the
> evolution of the state into the mixture starts immediately, so the
> decay could "take place" immediately. I say "take place", as all we can
> really tell is that the decay took place sometime before the
> measurement, unless we look for extra data like how far the photon has
> travelled.
>
> [ Mod. note: I believe the poster meant "superposition" rather than
> "mixture". "Mixture" has a different meaning in quantum statistical
> mechanics. [itex]-ik ][/itex]

Thank you -ik. This is an important correction[itex]. We[/itex] are not talking about
statistical mixed states described by density matrices (operators).
We should use words "superposition" or "linear combination" instead
of "mixture" everywhere.

Eugene.

David Park

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Paul Danaher" <paul.danaher@watwinc.com> wrote in message\nnews:vuEFe.52883\\$rb6.296@lakeread07...\n> We have ("Mossbauer effect and retarded interactions") a fairly subtle\n> debate going on which an outsider with some mathematics but not nearly\n> enough physics feels should be capable of a direct answer. One of the\n> questions posed in this is, how long does it take to emit one photon. I\n> would really welcome a "simple" answer.\n>\n\nWriting as a student and not as an expert:\n\nI thought that one of the lessons of QM was that there is no such thing as a\ntrajectory? You can calculate the probability that a particle in a certain\nstate at point A later appears at point B in another state. But you can\'t\nsay how it got from A to B. You can\'t say that it took some particular path.\n(In fact in Feynman\'s formulation of QM you would add up all the possible\npaths.)\n\nSo I would suppose that in the same vain it is meaningless to talk about the\n\'process\' of emitting a photon - as if it were some drawn out detailed\nprocess that could be followed in a step by step continuous manner. All you\ncan calculate is the probability that a system in one energy state is later\na system in a lower energy state plus a photon.\n\nSo it is not a question that QM gives an answer to, and probably not even a\nproper question.\n\nDavid Park\ndjmp@earthlink.net\nhttp://home.earthlink.net/~djmp/\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Paul Danaher" <paul.danaher@watwinc.com> wrote in message
news:vuEFe.52883$rb6.296@lakeread07...
> We have ("Mossbauer effect and retarded interactions") a fairly subtle
> debate going on which an outsider with some mathematics but not nearly
> enough physics feels should be capable of a direct answer. One of the
> questions posed in this is, how long does it take to emit one photon. I
> would really welcome a "simple" answer.
>

Writing as a student and not as an expert:

I thought that one of the lessons of QM was that there is no such thing as a
trajectory? You can calculate the probability that a particle in a certain
state at point A later appears at point B in another state. But you can't
say how it got from A to B. You can't say that it took some particular path.
(In fact in Feynman's formulation of QM you would add up all the possible
paths.)

So I would suppose that in the same vain it is meaningless to talk about the
'process' of emitting a photon - as if it were some drawn out detailed
process that could be followed in a step by step continuous manner. All you
can calculate is the probability that a system in one energy state is later
a system in a lower energy state plus a photon.

So it is not a question that QM gives an answer to, and probably not even a
proper question.

David Park
djmp@earthlink.net
http://home.earthlink.net/~djmp/

Eugene Stefanovich

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz wrote:\n\n>>In QM we describe the photon emission as a decay process\n>>A -> B + C\n\n\n> I am not happy with the assumption that there are two states, existed\n> and emitted and nothing in between.\n>\n> We can have a superposition of \'emitted\' and \'excited\' which seems to me\n> to be as good a description of \'partly emitted\' as one could wish to\n> hope for.\n\nIn QM the two states A and B+C are described by orthogonal subspaces\nin the Hilbert space. Of course, the state vector of the system may\nwander somewhere in between these subspaces: it may have non-zero\nprojections on both A and B+C. However, our measuring devices cannot\nreach into that territory. All we can measure is either A or B+C.\nNot every projection in the Hilbert space correspond to a measurement.\n\nIf you ever saw tracks of particles in bubble chambers, you probably\nnoticed that at each point in time there is a well-defined collection\nof particles. E.g., at some points you definitely see a muon, at later\npoints you definitely see an electron (plus two neutrinos). You never\nsee a "mixture" or "linear combination" of these two states.\n\n\n\n>\n> Now how about the following scenario.\n> Here we measure 100,000 changes of A -> (B+C) ensuring A is always in\n> the same state before we start. This apparatus can force the system to\n> be in either A or (B+C) in 1/10ns. We get:\n>\n> %A %(B+C)\n> 2.0ns 100 0\n> 2.1 99 1\n> 2.2 97 3\n> 2.3 85 15\n> 2.4 60 40\n> 2.5 50 50 (OK its convenient...)\n> 2.6 40 60\n> 2.7 15 85\n> 2.8 3 97\n> 2.9 1 99\n> 3.0 0 100\n>\n> Now you can decide what you like, but I would say that this system takes\n> about 0.6 ns to emit a photon to first order. Mind you IMHO it would be\n> a slightly odd system that behaved like the above.\n\nThis is a completely hypothetical system. Real unstable systems do not\nbehave in this way. They have exponential decay laws. My point was to\nshow that the decay rate of the exponent has nothing to do with\n"the time in which the system decays" or "the time it takes to\nemit one photon". This time is always very short,\nbut the decay rate could be in millions of years in some radioactive\nnuclei.\n\n\n>>I think that your question boils down to the following:\n>>"how fast is the collapse of the wave function?" My best guess is that\n>>it happens instantaneously.\n>\n>\n> My guess is that it happens typically rather slowly and isn\'t a collapse\n> at all but a gradual change via complex superposition of states.\n\nIf it were so, then we would be able to describe this gradual collapse\nby some sort of evolution equation (like Schroedinger equation). I\'ve\nnever heard of such a thing. The Schroedinger equation describes the\ndecay law, i.e., the gradual change of probabilities in time, but it has\nnothing to do with the random (and instantaneous)\nquantum jump (or collapse) between two states.\n\nSo, even if quantum collapse takes a measurable time, quantum mechanics\nis silent about that. From the point of view of QM, this collapse is\ninstantaneous. Maybe there is a theory better than QM? I doubt it.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz wrote:

>>In QM we describe the photon emission as a decay process
>>A [itex]-> B + C[/itex]

> I am not happy with the assumption that there are two states, existed
> and emitted and nothing in between.
>
> We can have a superposition of 'emitted' and 'excited' which seems to me
> to be as good a description of 'partly emitted' as one could wish to
> hope for.

In QM the two states A and B+C are described by orthogonal subspaces
in the Hilbert space. Of course, the state vector of the system may
wander somewhere in between these subspaces: it may have non-zero
projections on both A and B+C. However, our measuring devices cannot
reach into that territory. All we can measure is either A or B+C.
Not every projection in the Hilbert space correspond to a measurement.

If you ever saw tracks of particles in bubble chambers, you probably
noticed that at each point in time there is a well-defined collection
of particles. E.g., at some points you definitely see a muon, at later
points you definitely see an electron (plus two neutrinos). You never
see a "mixture" or "linear combination" of these two states.

>
> Now how about the following scenario.
> Here we measure 100,000 changes of [itex]A -> (B+C)[/itex] ensuring A is always in
> the same state before we start. This apparatus can force the system to
> be in either A or [itex](B+C)[/itex] in [itex]1/10ns[/itex]. We get:
>
> %A [itex]%(B+C)[/itex]
> 2.0ns 100
> 2.1 99 1
> 2.2 97 3
> 2.3 85 15
> 2.4 60 40
> 2.5 50 50 (OK its convenient...)
> 2.6 40 60
> 2.7 15 85
> 2.8 3 97
> 2.9 1 99
> 3. 100
>
> Now you can decide what you like, but I would say that this system takes
> about .6 ns to emit a photon to first order. Mind you IMHO it would be
> a slightly odd system that behaved like the above.

This is a completely hypothetical system. Real unstable systems do not
behave in this way. They have exponential decay laws. My point was to
show that the decay rate of the exponent has nothing to do with
"the time in which the system decays" or "the time it takes to
emit one photon". This time is always very short,
but the decay rate could be in millions of years in some radioactive
nuclei.

>>I think that your question boils down to the following:
>>"how fast is the collapse of the wave function?" My best guess is that
>>it happens instantaneously.
>
>
> My guess is that it happens typically rather slowly and isn't a collapse
> at all but a gradual change via complex superposition of states.

If it were so, then we would be able to describe this gradual collapse
by some sort of evolution equation (like Schroedinger equation). I've
never heard of such a thing. The Schroedinger equation describes the
decay law, i.e., the gradual change of probabilities in time, but it has
nothing to do with the random (and instantaneous)
quantum jump (or collapse) between two states.

So, even if quantum collapse takes a measurable time, quantum mechanics
is silent about that. From the point of view of QM, this collapse is
instantaneous. Maybe there is a theory better than QM? I doubt it.

Eugene.

Oz

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Eugene Stefanovich <eugenev@synopsys.com> writes\n>Oz wrote:\n>\n>>>In QM we describe the photon emission as a decay process\n>>>A -> B + C\n>\n>\n>> I am not happy with the assumption that there are two states, existed\n>> and emitted and nothing in between.\n>>\n>> We can have a superposition of \'emitted\' and \'excited\' which seems to me\n>> to be as good a description of \'partly emitted\' as one could wish to\n>> hope for.\n>\n>In QM the two states A and B+C are described by orthogonal subspaces\n>in the Hilbert space. Of course, the state vector of the system may\n>wander somewhere in between these subspaces: it may have non-zero\n>projections on both A and B+C.\n\nPrecisely.\n\n>However, our measuring devices cannot\n>reach into that territory.\n\nExactly.\nIts a very limited quantised detector that cannot detect superpositions.\n\n>All we can measure is either A or B+C.\n\nExactly.\n\n>If you ever saw tracks of particles in bubble chambers, you probably\n>noticed that at each point in time there is a well-defined collection\n>of particles. E.g., at some points you definitely see a muon, at later\n>points you definitely see an electron (plus two neutrinos). You never\n>see a "mixture" or "linear combination" of these two states.\n\nOf course not. The energy of the interaction is such that the chamber\ncould not resolve it even if it could. By definition the chamber can\nonly detect either a muon or an electron (see above).\n\n>> Now how about the following scenario.\n>> Here we measure 100,000 changes of A -> (B+C) ensuring A is always in\n>> the same state before we start. This apparatus can force the system to\n>> be in either A or (B+C) in 1/10ns. We get:\n>>\n>> %A %(B+C)\n>> 2.0ns 100 0\n>> 2.1 99 1\n>> 2.2 97 3\n>> 2.3 85 15\n>> 2.4 60 40\n>> 2.5 50 50 (OK its convenient...)\n>> 2.6 40 60\n>> 2.7 15 85\n>> 2.8 3 97\n>> 2.9 1 99\n>> 3.0 0 100\n>>\n>> Now you can decide what you like, but I would say that this system takes\n>> about 0.6 ns to emit a photon to first order. Mind you IMHO it would be\n>> a slightly odd system that behaved like the above.\n>\n>This is a completely hypothetical system.\n\nYes.\n\n>Real unstable systems do not\n>behave in this way. They have exponential decay laws.\n\nSimple ones, yes.\n\nMore complex (as in requiring many steps, in effect) ones, such as the\ndecay of an unstable nucleus, might well have this sort of shape. A very\nlong decay one measured in years followed by the ejection of a gamma in\na very short time. These are characterised by (typically) being\nirreversible, that is you cannot reverse the process by hitting the\nstable atom by the right gamma.\n\nYou can, of course, readily do this for simpler systems because they are\ntypically reversible.\n\n>My point was to\n>show that the decay rate of the exponent has nothing to do with\n>"the time in which the system decays" or "the time it takes to\n>emit one photon". This time is always very short,\n\nWell, its not for most atomic (as against nuclear) transitions.\nTed has quoted \'forbidden\' transitions measured in megayears.\n\nKinsler has told me that exact (or close to it) time-evolution of an\nexcited H atom has been done and is modelled by a superposition of\nstates with the two ends (ie excited H to e- + h+ + v) oscillating at a\nparticular frequency. To me this provides a theoretical expression of\n\'time of emission of a photon\'.\n\n>but the decay rate could be in millions of years in some radioactive\n>nuclei.\n\nThat\'s actually the decay of a complex nuclear reaction.\nThe emission of the gamma is merely the final step of the decay of some\nmetastable internal state.\n\n>>>I think that your question boils down to the following:\n>>>"how fast is the collapse of the wave function?" My best guess is that\n>>>it happens instantaneously.\n>>\n>> My guess is that it happens typically rather slowly and isn\'t a collapse\n>> at all but a gradual change via complex superposition of states.\n>\n>If it were so, then we would be able to describe this gradual collapse\n>by some sort of evolution equation (like Schroedinger equation). I\'ve\n>never heard of such a thing.\n\nKinsler assures me it has been done for the isolated excited H atom.\nI believe anything more complex is beyond the reach of QM algorithms.\nOf course reasonable models can probably be had that are somewhat less\nexact.\n\n>The Schroedinger equation describes the\n>decay law, i.e., the gradual change of probabilities in time, but it has\n>nothing to do with the random (and instantaneous)\n>quantum jump (or collapse) between two states.\n\nWe may have to agree to differ here.\nIMHO it is the measuring device which enforces this apparent effect.\n\n>So, even if quantum collapse takes a measurable time, quantum mechanics\n>is silent about that.\n\nIn essence yes. This is because detectors are hugely too complex for the\ntransition to be modelled. The solution is to concentrate on those\nthings we can say about in the interaction. This is absolutely normal\nand usually gives excellent results. After all no scientist attempts to\nanalyse the precise atomic/crystal/mechanical properties when\nconsidering the clash of two billiard balls.\n\n>From the point of view of QM, this collapse is\n>instantaneous.\n\nI\'m not sure that\'s actually true.\nSchroedinger does not need to assume such a thing and nor do many\noptical analyses.\n\n>Maybe there is a theory better than QM? I doubt it.\n\nI\'m absolutely certain there is a better theory.\nI am astonished that you are not hoping to find such a thing.\n\n--\nOz\nThis post is worth absolutely nothing and is probably fallacious.\n\nUse oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].\nBTOPENWORLD address has ceased. DEMON address has ceased.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Eugene Stefanovich <eugenev@synopsys.com> writes
>Oz wrote:
>
>>>In QM we describe the photon emission as a decay process
>>>A [itex]-> B + C[/itex]
>
>
>> I am not happy with the assumption that there are two states, existed
>> and emitted and nothing in between.
>>
>> We can have a superposition of 'emitted' and 'excited' which seems to me
>> to be as good a description of 'partly emitted' as one could wish to
>> hope for.
>
>In QM the two states A and B+C are described by orthogonal subspaces
>in the Hilbert space. Of course, the state vector of the system may
>wander somewhere in between these subspaces: it may have non-zero
>projections on both A and B+C.

Precisely.

>However, our measuring devices cannot
>reach into that territory.

Exactly.
Its a very limited quantised detector that cannot detect superpositions.

>All we can measure is either A or B+C.

Exactly.

>If you ever saw tracks of particles in bubble chambers, you probably
>noticed that at each point in time there is a well-defined collection
>of particles. E.g., at some points you definitely see a muon, at later
>points you definitely see an electron (plus two neutrinos). You never
>see a "mixture" or "linear combination" of these two states.

Of course not. The energy of the interaction is such that the chamber
could not resolve it even if it could. By definition the chamber can
only detect either a muon or an electron (see above).

>> Now how about the following scenario.
>> Here we measure 100,000 changes of [itex]A -> (B+C)[/itex] ensuring A is always in
>> the same state before we start. This apparatus can force the system to
>> be in either A or [itex](B+C)[/itex] in [itex]1/10ns[/itex]. We get:
>>
>> %A [itex]%(B+C)>> 2[/itex].0ns 100
>> 2.1 99 1
>> 2.2 97 3
>> 2.3 85 15
>> 2.4 60 40
>> 2.5 50 50 (OK its convenient...)
>> 2.6 40 60
>> 2.7 15 85
>> 2.8 3 97
>> 2.9 1 99
>> 3. 100
>>
>> Now you can decide what you like, but I would say that this system takes
>> about .6 ns to emit a photon to first order. Mind you IMHO it would be
>> a slightly odd system that behaved like the above.
>
>This is a completely hypothetical system.

Yes.

>Real unstable systems do not
>behave in this way. They have exponential decay laws.

Simple ones, yes.

More complex (as in requiring many steps, in effect) ones, such as the
decay of an unstable nucleus, might well have this sort of shape. A very
long decay one measured in years followed by the ejection of [itex]a \gamma[/itex] in
a very short time. These are characterised by (typically) being
irreversible, that is you cannot reverse the process by hitting the
stable atom by the right [itex]\gamma[/itex].

You can, of course, readily do this for simpler systems because they are
typically reversible.

>My point was to
>show that the decay rate of the exponent has nothing to do with
>"the time in which the system decays" or "the time it takes to
>emit one photon". This time is always very short,

Well, its not for most atomic (as against nuclear) transitions.
Ted has quoted 'forbidden' transitions measured in megayears.

Kinsler has told me that exact (or close to it) time-evolution of an
excited H atom has been done and is modelled by a superposition of
states with the two ends (ie excited [itex]H to e- + h+ + v)[/itex] oscillating at a
particular frequency. To me this provides a theoretical expression of
'time of emission of a photon'.

>but the decay rate could be in millions of years in some radioactive
>nuclei.

That's actually the decay of a complex nuclear reaction.
The emission of the [itex]\gamma[/itex] is merely the final step of the decay of some
metastable internal state.

>>>I think that your question boils down to the following:
>>>"how fast is the collapse of the wave function?" My best guess is that
>>>it happens instantaneously.
>>
>> My guess is that it happens typically rather slowly and isn't a collapse
>> at all but a gradual change via complex superposition of states.
>
>If it were so, then we would be able to describe this gradual collapse
>by some sort of evolution equation (like Schroedinger equation). I've
>never heard of such a thing.

Kinsler assures me it has been done for the isolated excited H atom.
I believe anything more complex is beyond the reach of QM algorithms.
Of course reasonable models can probably be had that are somewhat less
exact.

>The Schroedinger equation describes the
>decay law, i.e., the gradual change of probabilities in time, but it has
>nothing to do with the random (and instantaneous)
>quantum jump (or collapse) between two states.

We may have to agree to differ here.
IMHO it is the measuring device which enforces this apparent effect.

>So, even if quantum collapse takes a measurable time, quantum mechanics
>is silent about that.

In essence yes. This is because detectors are hugely too complex for the
transition to be modelled. The solution is to concentrate on those
things we can say about in the interaction. This is absolutely normal
and usually gives excellent results. After all no scientist attempts to
analyse the precise atomic/crystal/mechanical properties when
considering the clash of two billiard balls.

>From the point of view of QM, this collapse is
>instantaneous.

I'm not sure that's actually true.
Schroedinger does not need to assume such a thing and nor do many
optical analyses.

>Maybe there is a theory better than QM? I doubt it.

I'm absolutely certain there is a better theory.
I am astonished that you are not hoping to find such a thing.

--
Oz
This post is worth absolutely nothing and is probably fallacious.

Use oz@farmeroz.port995.com [ozacoohdb@despammed.com functions].
BTOPENWORLD address has ceased. DEMON address has ceased.

Eugene Stefanovich

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nDavid Park wrote:\n\n> So I would suppose that in the same vain it is meaningless to talk about the\n> \'process\' of emitting a photon - as if it were some drawn out detailed\n> process that could be followed in a step by step continuous manner. All you\n> can calculate is the probability that a system in one energy state is later\n> a system in a lower energy state plus a photon.\n>\n> So it is not a question that QM gives an answer to, and probably not even a\n> proper question.\n\nI tend to agree with you. At least the time of photon emission cannot\nbe understood by following the time evolution of the wave function of\nthe emitting system (e.g., atom).\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>David Park wrote:

> So I would suppose that in the same vain it is meaningless to talk about the
> 'process' of emitting a photon - as if it were some drawn out detailed
> process that could be followed in a step by step continuous manner. All you
> can calculate is the probability that a system in one energy state is later
> a system in a lower energy state plus a photon.
>
> So it is not a question that QM gives an answer to, and probably not even a
> proper question.

I tend to agree with you. At least the time of photon emission cannot
be understood by following the time evolution of the wave function of
the emitting system (e.g., atom).

Eugene.

Kit Adams

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n"Igor Khavkine" <igor.kh@gmail.com> wrote in message\nnews:1122574188.254883.275030@g14g2000cwa.googlegroups.com...\ n\nsnip...\n\n> So, at the end of this perhaps overly verbose explanation, the simple\n> answer is given by the Heisenberg uncertainty principle. If the energy\n> of the photon is E, the time needed to emit a photon of this energy is\n> greater than hbar/E. The longer you wait, the more sure you are how\n> many quanta were emitted. The less you wait, the less certainty there\n> is about the number of photons emitted, which could also be zero.\n\nI like your point about photon number only being a meaningful description\nfor the steady state of the field - after all the creation and annihilation\noperators are the Fourier coefficients of monochromatic field modes.\n\nHowever I would say that the time needed to create a field excitation (emit\na photon) is proportional to the inverse of the uncertainty in the energy\n(i.e. ~ hbar/delta E) of the excitation rather the inverse of the absolute\nenergy. This follows directly from Fourier analysis and ties in nicely with\nthe lifetime of excited states of atoms and nuclei being the inverse of the\nlinewidth i.e. the idea that the excited state radiates throughout its\nlifetime.\n\nFor the Mossbauer effect using iron57 that gives an emission (and\nabsorbtion) time of 10e-7 secs, giving plenty of time for the recoil to\npropagate through the 200000 nuclei required to allow resonance absorbtion\n(see http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/mossfe.html#c1\n)\n\nRegards,\nKit\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:1122574188.254883.275030@g14g20...egroups.com...

snip...

> So, at the end of this perhaps overly verbose explanation, the simple
> answer is given by the Heisenberg uncertainty principle. If the energy
> of the photon is E, the time needed to emit a photon of this energy is
> greater than [itex]\hbar/E[/itex]. The longer you wait, the more sure you are how
> many quanta were emitted. The less you wait, the less certainty there
> is about the number of photons emitted, which could also be zero.

I like your point about photon number only being a meaningful description
for the steady state of the field - after all the creation and annihilation
operators are the Fourier coefficients of monochromatic field modes.

However I would say that the time needed to create a field excitation (emit
a photon) is proportional to the inverse of the uncertainty in the energy
(i.e. [itex]~ \hbar/\delta E)[/itex] of the excitation rather the inverse of the absolute
energy. This follows directly from Fourier analysis and ties in nicely with
the lifetime of excited states of atoms and nuclei being the inverse of the
linewidth i.e. the idea that the excited state radiates throughout its
lifetime.

For the Mossbauer effect using iron57 that gives an emission (and
absorbtion) time of [itex]10e-7[/itex] secs, giving plenty of time for the recoil to
propagate through the 200000 nuclei required to allow resonance absorbtion
(see http://hyperphysics.phy-astr.gsu.edu...mossfe.html#c1
)

Regards,
Kit

Eugene Stefanovich

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Paul Danaher wrote:\n> We have ("Mossbauer effect and retarded interactions") a fairly subtle\n> debate going on which an outsider with some mathematics but not nearly\n> enough physics feels should be capable of a direct answer. One of the\n> questions posed in this is, how long does it take to emit one photon. I\n> would really welcome a "simple" answer.\n\nI did a little bit of thinking on this subject (which is always a good\nthing to do), and now, I believe, I am able to formulate a coherent\nsolution.\n\nFirst, I still think that the questions like "how long does it take\nto emit\none photon" or "how much time the collapse of the wavefunction takes"\nare beyond quantum mechanics. They cannot be answered from solutions of\nthe Schroedinger equation.\n\nHowever, in the spirit of quantum mechanics, we can formulate a\ndescription of the emission-recoil process averaged over ensemble,\ni.e., provided by the time-dependent wave function of the system.\n\nThe initial state of the system is A="crystal + radioactive nucleus in the\nexcited state at rest". The final state of the system is B + C\nwhere B="crystal + radioactive nucleus in the ground state moving at a\nconstant speed" and C="photon". Since we are considering the Mossbauer\neffect, the state B does not involve any phonons.\nThe state of the system develops in time from A at t=0 to something\nvery close to B+C after T="the lifetime of the radioactive nucleus".\nAccordingly, the expectation value of the velocity of the\ncrystal changes from zero at t=0 to a non-zero (but very small) value\nafter time\nT. A typical lifetime of Mossbauer nuclei is about 10^{-7} s. So, we can\nsay that the entire process takes about 10^{-7} s. During this time\nthe photon is emitted with high probability, and the crystal picks up\nthe recoil speed.\n\nIf we accept this description, then the recoil is certainly superluminal\nif the size of the crystal is greated than the distance passed by light\nduring 10^{-7} s. This distance is about 30m. I haven\'t seen such big\ncrystals, so the whole idea looks worthless.\n\nI think, one can find radioactive nuclei whose decay time is much\nshorter than 10^{-7}s (these nuclei are not useful for the Mossbauer\nspectroscopy,\nbut they could be useful for the superluminal argument I am developing;\nfor example, the lifetime of 10^{-11}s corresponds to the reasonable\ndistance of only 3mm). Although, I believe that in such cases the recoil\nmomentum\ntransfers superluminally, its value divided by all atoms in the crystal\nis so small that there is no chance it can be observed.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Paul Danaher wrote:
> We have ("Mossbauer effect and retarded interactions") a fairly subtle
> debate going on which an outsider with some mathematics but not nearly
> enough physics feels should be capable of a direct answer. One of the
> questions posed in this is, how long does it take to emit one photon. I
> would really welcome a "simple" answer.

I did a little bit of thinking on this subject (which is always a good
thing to do), and now, I believe, I am able to formulate a coherent
solution.

First, I still think that the questions like "how long does it take
to emit
one photon" or "how much time the collapse of the wavefunction takes"
are beyond quantum mechanics. They cannot be answered from solutions of
the Schroedinger equation.

However, in the spirit of quantum mechanics, we can formulate a
description of the emission-recoil process averaged over ensemble,
i.e., provided by the time-dependent wave function of the system.

The initial state of the system is A="crystal + radioactive nucleus in the
excited state at rest". The final state of the system is [itex]B + C[/itex]
where B="crystal + radioactive nucleus in the ground state moving at a
constant speed" and C="photon". Since we are considering the Mossbauer
effect, the state B does not involve any phonons.
The state of the system develops in time from A [itex]at t=0[/itex] to something
very close to B+C after T="the lifetime of the radioactive nucleus".
Accordingly, the expectation value of the velocity of the
crystal changes from zero [itex]at t=0[/itex] to a non-zero (but very small) value
after time
T. A typical lifetime of Mossbauer nuclei is about [itex]10^{-7} s[/itex]. So, we can
say that the entire process takes about [itex]10^{-7} s.[/itex] During this time
the photon is emitted with high probability, and the crystal picks up
the recoil speed.

If we accept this description, then the recoil is certainly superluminal
if the size of the crystal is greated than the distance passed by light
during [itex]10^{-7} s[/itex]. This distance is about 30m. I haven't seen such big
crystals, so the whole idea looks worthless.

I think, one can find radioactive nuclei whose decay time is much
shorter than [itex]10^{-7}s[/itex] (these nuclei are not useful for the Mossbauer
spectroscopy,
but they could be useful for the superluminal argument I am developing;
for example, the lifetime [itex]of 10^{-11}s[/itex] corresponds to the reasonable
distance of only 3mm). Although, I believe that in such cases the recoil
momentum
transfers superluminally, its value divided by all atoms in the crystal
is so small that there is no chance it can be observed.

Eugene.

Eugene Stefanovich

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>Oz wrote:\n\n> Well, its not for most atomic (as against nuclear) transitions.\n> Ted has quoted \'forbidden\' transitions measured in megayears.\n\nThere are two different times associated with decays. One is\nthe lifetime of the metastable system. Another is the time during\nwhich each particular system disintegrates. Let\'s take for example\na nucleus which emits an alpha-particle, and whose lifetime is\n1 million years.\n\nIf you prepare such a nucleus, it will sit there for hundreds\nof thousands or millions of years doing absolutely nothing.\nThen one day it will emit so long-awaited alpha-particle.\nSo, the actual time of emission (i.e., the time during which the\nalpha-particle leaves the nucleus) is very short. It could be\neven instantaneous. However the time of wait (the lifetime) is very\nlong.\n\n>>Maybe there is a theory better than QM? I doubt it.\n>\n>\n> I\'m absolutely certain there is a better theory.\n> I am astonished that you are not hoping to find such a thing.\n\nThis is a matter of personal belief. I think that\nquantum mechanics is safe. I am hoping to find interesting new\nphysics in other places. In the web of physics, there are links much\nweaker than QM.\n\nEugene.\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Oz wrote:

> Well, its not for most atomic (as against nuclear) transitions.
> Ted has quoted 'forbidden' transitions measured in megayears.

There are two different times associated with decays. One is
the lifetime of the metastable system. Another is the time during
which each particular system disintegrates. Let's take for example
a nucleus which emits an [itex]\alpha-particle,[/itex] and whose lifetime is
1 million years.

If you prepare such a nucleus, it will sit there for hundreds
of thousands or millions of years doing absolutely nothing.
Then one day it will emit so long-awaited [itex]\alpha-particle[/itex].
So, the actual time of emission (i.e., the time during which the
[itex]\alpha-particle[/itex] leaves the nucleus) is very short[itex]. It[/itex] could be
even instantaneous. However the time of wait (the lifetime) is very
long.

>>Maybe there is a theory better than QM? I doubt it.
>
>
> I'm absolutely certain there is a better theory.
> I am astonished that you are not hoping to find such a thing.

This is a matter of personal belief. I think that
quantum mechanics is safe. I am hoping to find interesting new
physics in other places. In the web of physics, there are links much
weaker than QM.

Eugene.

RP

<jabberwocky><div class="vbmenu_control"><a href="jabberwocky:;" onClick="newWindow=window.open('','usenetCode','toolbar=no,location=no, scrollbars=yes,resizable=yes,status=no,width=650,height=400'); newWindow.document.write('<HTML><HEAD><TITLE>Usenet ASCII</TITLE></HEAD><BODY topmargin=0 leftmargin=0 BGCOLOR=#F1F1F1><table border=0 width=625><td bgcolor=midnightblue><font color=#F1F1F1>This Usenet message\'s original ASCII form: </font></td></tr><tr><td width=449><br><br><font face=courier><UL><PRE>\n\nPaul Danaher wrote:\n\n> We have ("Mossbauer effect and retarded interactions") a fairly subtle\n> debate going on which an outsider with some mathematics but not nearly\n> enough physics feels should be capable of a direct answer. One of the\n> questions posed in this is, how long does it take to emit one photon. I\n> would really welcome a "simple" answer.\n\n1/freq\n\nRichard Perry\n\n\n\n</UL></PRE></font></td></tr></table></BODY><HTML>');"> <IMG SRC=/images/buttons/ip.gif BORDER=0 ALIGN=CENTER ALT="View this Usenet post in original ASCII form">  View this Usenet post in original ASCII form </a></div><P></jabberwocky>Paul Danaher wrote:

> We have ("Mossbauer effect and retarded interactions") a fairly subtle
> debate going on which an outsider with some mathematics but not nearly
> enough physics feels should be capable of a direct answer. One of the
> questions posed in this is, how long does it take to emit one photon. I
> would really welcome a "simple" answer.

[tex]1/freq[/tex]

Richard Perry

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