## [SOLVED] The time it takes to emit one photon

<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">&nbsp;&nbsp;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.

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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 $A -> B + C$ 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" $\omega_A(t)$ 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 $\omega_A(t)$. 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 $_the same_$ metastable system. E.g., prepare system in A at time $t=0,$ measure the state of the system at $t=1ns,$ then measure the same system $at t=2ns,$ 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 $B+C,$ and not somewhere in between): 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. 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.



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? $*-The$ 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

## [SOLVED] The time it takes to emit one photon

>
>
>In QM we describe the photon emission as a decay process

$>A -> B + C$
>
>Quantum mechanics allows us to calculate the function

$>\omega_A(t)$. In most cases this is an
>exponential function of t with the decay parameter called
>
>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 $_the same_$ metastable system. E.g.,
>prepare system in A at time $t=0,$ measure the state of the system at

$>t=1ns,$ then measure the same system $at t=2ns,$ 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 $B+C,$ and not somewhere in between):

Note carefully the above sentence. In essence one has FORCED the system
into state A or $(B+C)$. Not unsurprisingly one sees ONLY A or $(B+C).$ This
says nothing about the state of the system before you enforced a
measurement$. It$ 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
$A -> (A+B)$ happens in the absence of the measuring apparatus.

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

$%A %(B+C)$
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].



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. $-ik ]$ 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

 Measure the spectral width of the photon (by measuing many of them). $$T \approx 1/(spectral width)$$


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.



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 $on/off$ 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 $\hbar/E$. 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



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. $-ik ]$ Thank you -ik. This is an important correction$. We$ 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.



"Paul Danaher" 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/



Oz wrote: >>In QM we describe the photon emission as a decay process >>A $-> B + C$ > 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 $A -> (B+C)$ ensuring A is always in > the same state before we start. This apparatus can force the system to > be in either A or $(B+C)$ in $1/10ns$. We get: > > %A $%(B+C)$ > 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.



Eugene Stefanovich writes >Oz wrote: > >>>In QM we describe the photon emission as a decay process >>>A $-> B + C$ > > >> 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 $A -> (B+C)$ ensuring A is always in >> the same state before we start. This apparatus can force the system to >> be in either A or $(B+C)$ in $1/10ns$. We get: >> >> %A $%(B+C)>> 2$.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. 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 $a \gamma$ 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 $\gamma$. 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 $H to e- + h+ + v)$ 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 $\gamma$ 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.



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.



"Igor Khavkine" 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 $\hbar/E$. 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. $~ \hbar/\delta E)$ 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 $10e-7$ 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



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 $B + C$ 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 $at t=0$ 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 $at t=0$ to a non-zero (but very small) value after time T. A typical lifetime of Mossbauer nuclei is about $10^{-7} s$. So, we can say that the entire process takes about $10^{-7} s.$ 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 $10^{-7} s$. 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 $10^{-7}s$ (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 $of 10^{-11}s$ 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.



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 $\alpha-particle,$ 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 $\alpha-particle$. So, the actual time of emission (i.e., the time during which the $\alpha-particle$ leaves the nucleus) is very short$. It$ 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.



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. $$1/freq$$ Richard Perry