I am trying to reconcile, what to me at least, are two slightly different answers to what I think is the same question.

The first answer below to an earlier OP implies to me that there is a definite time period involved in the transition between two bound state energy levels, although the last line of the answer implies that a coordination of times is also required between different atomic clocks in order to keep a worldwide standard.

From Atomic Clocks answer:

"Suppose you generate a microwave signal with a frequency of 9GHz. Then if you want to time 1 second you just count  cycles of the microwave and that gives you one second. Conversely suppose you want to time something. You start counting the microwave cycles when your event starts and stop counting when it finishes. However many cycles you count, you just divide this by  and that gives you the time in seconds.

The trouble is that this only works if you are absolutely sure your microwave is exactly 9GHz. An atomic clock uses the fact that caesium atoms absorb energy at exactly 9.192631770GHz. So what you do is shine your microwave beam through a gas of caesium atoms and adjust the frequency until the caesium atoms start absorbing it. Then you know it's frequency is exactly 9.192631770GHz and you can use it as a reliable timer.

Atomic clocks are synchronised with each other by using GPS and sending each other synchronisation signals. In fact about 200 atomic clocks, all synchronised with each other, are used to define International Atomic Time."

Now the second answer comes from my earlier question regarding the timing of photon emission. I previously asked this question thinking incorrectly in classical terms , which I now, (I hope) have stopped doing.

From my earlier post: time of photon emission and part of the answer I received.

"An individual atom with its electron in an excited state may emit the photon at an arbitrary time t. One has to take a large sample of atoms with the electrons at that energy level and measure the time the photon hits the detector . One then will have a curve characterizing the lifetime of that bound state's collective time behavior"

My assumption is that the second answer, "is more correct" (if you forgive my terminology) because it allows for the uncertainty principle between time and energy. In theory at least, (if I am correct) there is a probability that because any amount of energy could in theory be released in transitions between energy levels, experimentally we need to find the average time, which in the case of caesium atoms is (on average) as stated in the first answer. i.e. the reciprocal of the frequency.

EDIT: the above paragraph is worded very badly. What I mean to ask is, do we need to take a range of measurements to arrive at an average typical transition timescale and that perhaps the uncertainty principle is only useful, if it is at all involved, as a constraint on this process in some way?

Finally my question(s):

  1. Is my assumption that experimentally, we do use the uncertainty principle to establish transition times a correct one, or if this assumption is wrong, where does my reasoning break down? As I say in the edit, the uncertainty principle may not be as important as I originally thought.

  2. My second question is related in that, if we want to induce the "drop" in energy levels for say the caesium atom, we need to initially provide a photon that has an energy level comparable to the photon emitted during the transition? To put this another way, an incoming photon that does not resonate , in wave terms, with the energy of the electron, either by too much or too little, will not produce a transition.

I realise that these are very basic quantum mechanics questions, but as I self study, having these assumptions confirmed, or just as valuable to me, learning how or why I am wrong, is of significant help to me in further study.

I have previously asked similar questions and my apologies if there is repetition here, but I would greatly appreciate as much clarity/certainty as possible with this particular question.


Those two are answers to different questions.

When you talk about the $\nu = $ 9.192631770 GHz, $h \nu $ is the separation between the energy levels involved in the transition. This is precisely defined.

The second concept you talk about is the amount of time the atom remains in the excited state, also known as the excited state lifetime $\tau$. These two concepts are related because the inverse of $\tau$ defines the uncertainty with which the $h \nu$ transition is known. The frequency $\Gamma$ is called line-width of the transition.

For the transition used to define the second, the "excited" state has a lifetime longer than the age of the universe so this uncertainty is not relevant.

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  • $\begingroup$ thanks for that...I may close the question , do some more background and then be in a better position to try again, regards $\endgroup$ – user74893 Apr 13 '15 at 23:42

A few comments, since it seems you're about to rephrase this question.

When dealing with the caesium standard, you can essentially forget about the energy (of the excited state, and of the photon). What matters is the frequency, and the only uncertainty principle that ever gets involved is of the form $\Delta\omega\,\Delta t\gtrsim1$. This is a standard Fourier relation and it has nothing to do with quantum mechanics.

As far as the microwaves are concerned, the caesium atoms are just a system of charges which is capable of oscillating at microwave frequencies, and which has a resonance frequency at which its response is much bigger. It is essentially equivalent to the quartz resonator of electronic watches, or the springs and pendulums in old clocks.

  • $\begingroup$ Hi Emilio thanks...I really need to take what I've learned from the site so far and hit the books again in the light of that, and leave the site alone until then...one thing I have learned is that terminology is physics is much more important than in linguistics :)...so I really want to learn how to ask questions as precisely as possible..as a self study person I did not appreciate that till recently... regards $\endgroup$ – user74893 Apr 14 '15 at 0:10

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