How long does does it take for the probability cloud of an electron to jump?

[Curyos]

Me and a friend at school have been thinking about this question for a while and have come up with a few perspectives, I'm sure there's a 'proper' answer but I haven't managed to find one even after scouring numerous web pages. We are aware that the electron doesn't 'move' in a sense, only the probability of it being in certain positions.

On one hand, I say that upon being 'impacted' by a photon, the electron changes orbital instantly because the photon holds a specific quanta of energy and just as 4+4=8, the adding of extra energy to the system does not take any time.

But on the other hand, my friend argues that it's impossible for no time to have elapsed because the smallest measure of time is the Planck time - which is non zero.

So, exactly, the question is: How long does it take for the probability cloud of an electron to 'jump'?

And from that: Is the time elapsed exactly the same for different consecutive clouds, or different? And why is it not instant (if it isn't)?

Thanks!

- Newbie

Edit: Sorry for the inaccuracy of the use of Planck time, had realized just after about zero being 'a number'!

 

[arkajad]

the smallest measure of time is the Planck time - which is non zero.

- Newbie

I do not think there is an agreement about that. But:

zero is also a number and it is somewhat smaller than Planck's time. So, why not zero?

 

[nismaratwork]

The best that can be said with confidence is the shortest time that can currently be measured... far larger than [tex]t_{p}\ =\ 5.3906(40)\ \times\ 10^{-44}\ s[/tex] or 0.

 

[Curyos]

@ nismaratwork: So what you are saying is that either we don't know because we can't measure it, or that we have tried to measure it and it has always been 'faster than that' so to speak?

 

[arkajad]

I would propose a different question: suppose, by some stretch of imagination, that the beginning and the end of the collapse can be recorded. Then the question is: what is the shortest time that the recording device (necessarily involving some macroscopic elements) needs in order to produce the numbers with some reasonable certainty?

 

[Dr Lots-o'watts]

The state of the art :

http://www.sciencedaily.com/releases/2010/06/100624144109.htm

 

[Curyos]

@ Dr Lots-o'watts: Thanks for that, but it's not quite what i was looking for (although this gives me hope it will be measureable). The article states that the electrons began to 'move' on the order of about 20 attoseconds after being hit, but it doesn't say how long it took for them to be fully ejected, or even to the next consecutive orbital. You don't happen to have any more do you?

@ arkajad: That probably would be more appropriate, but I had hoped it would be possible to work out the exact ( or at least very near ) elapsed time via some sort of rearrangement or solving of an existing set of equations or equation. I suppose that's just not possible?

 

[Dr Lots-o'watts]

Not for the moment, but if I understand your question correctly, I have a strong intuitive feeling that it is a finite time.

1. From basic QM, an energy state is a mode.

2. In the case of a classical rope, it takes a finite time to go from one mode to the other. Experimentally (imagine holding a jump rope fixed to a wall on the other end), this time corresponds to at least about one period of the new wavelength. So if I had to guess using my current level of QM knowledge, I'd confidently say on the order of T = h/E. (E = hf = h/T inverted).

 

[nismaratwork]

@ nismaratwork: So what you are saying is that either we don't know because we can't measure it, or that we have tried to measure it and it has always been 'faster than that' so to speak?

Just so Curyos, and both. We can say that if it happens in a finite time (and I agree with Dr Lots-o'watts that it DOES) the interval is far shorter than anything we can detect. The result is that you can only know with confidence that the interval is shorter than anything we can currently measure.

If you get into something like string theory, then I suppose it happens at or near Plank Time, but if the universe and its constituents are further divisible... who knows? That's where Arkajad is really putting the question in its necessary and proper framework, as much as I understand wanting to know more.

 

[Dr Lots-o'watts]

Interestingly, assuming the e- follows a path from one level to the next, t=d/v, with d=1 Angstrom (plausible distance between orbital shells of an atom), and a v=SQRT(2E/m) (classical KE), E of a visible photon, you get a number of similar order (attoseconds, as with with T=1/f).

 

[Curyos]

Ok, thanks guys - it seems as though the case is one of those 'nobody knows' ones, though it would be plausible to estimate that the time taken is somewhere on the order of a few attoseconds.

@ Dr Lots-o'watts: Although i can understand the nature of the equations you posted, I'm not quite sure what some of the constants/variables are, if you could just explain it would help a lot - cheers!

 

[Dr Lots-o'watts]

Note that I'm doing here is quite elementary and not fully quantum mechanical.

A.

1. Each atomic orbital has an average distance from the nucleus, as can be seen from any diagram of an orbital.

2. One can thus define a distance between two orbitals. Say this distance is d = 1 Angstrom, which is a plausible value.

3. A photon that is absorbed has energy E=hf. (h = Planck's constant, f = frequency of photon). A frequency of f = c/500 nm is typical for a visible (electric transition) photon.

4. I'm assuming the energy is completely absorbed by a lower level electron as classical kinetic energy.

[tex]
E\ =(1/2)mv^{2}\
[/tex]

5. You can isolate v, substitute E for hf, and then use t=d/v to find a time value.

B. The calculation from my other post basically translates to t = 1/f, where f is the frequency of the incident photon.

Both A. and B. give values on the order of attoseconds. Hmmm... makes you wonder why they chose the suffix "atto".

 

[nismaratwork]

Interestingly, assuming the e- follows a path from one level to the next, t=d/v, with d=1 Angstrom (plausible distance between orbital shells of an atom), and a v=SQRT(2E/m) (classical KE), E of a visible photon, you get a number of similar order (attoseconds, as with with T=1/f).

I like this, it's elegant, but I have no way of knowing if this reflects nature. Good idea though.

 

[Dr Lots-o'watts]

Me neither. And as I understand it, Heisenberg argued with Shroedinger whether the electron follows a path within an atom, so the debate is not a new one. (wikipedia, "shroedinger picture" and "heisenberg picture")

 

[Curyos]

Ah the equation makes a lot of sense - but I can see what nismaratwork means, we really have no way of telling if it is right, thanks a bunch any way, has helped quite a bit.

I think i'll need a better understanding of the maths involved in the different 'pictures' before i can begin to talk about them.

 

[Boy@n]

No expert, but wondering... wouldn't the time passed for that electron to change orbit be the distance electron made divided by the speed of light?

And if electron made a distance, say, of 10 electron radius, then the shortest possible time passed would be 10 * 2.8*10^-15m / 3*10^8m/s = 9.3*10^-7s? But should be even longer than that since massive particle cannot travel at the speed of light. (Way more than the Plank's time of 5.4*10^-44s anyway.)

I did notice you saying this 'should' happen instantly, but I guess this isn't the same kind of 'instant' as with the change of spin of two entangled particles?

Anyway, if this indeed happens faster than the time needed for light to travel that distance then I don't see a reason for it to not happen at zerro delay (as with entanglement).