Re: The time it takes to emit one photon

Eugene Stefanovich <eugenev@xxxxxxxxxxxx> writes
>
>
>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
>"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 _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 0
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
3.0 0 100

Now you can decide what you like, but I would say that this system takes
about 0.6 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@xxxxxxxxxxxxxxxxxxxx [ozacoohdb@xxxxxxxxxxxxx functions].
BTOPENWORLD address has ceased. DEMON address has ceased.

.