# Does a 9,192,631,770 Hz hyperfine transition create 9,192,631,770 photons per second, or fewer?

where Emilio Pisanty says:

Transitions which change spin direction are electric-dipole-forbidden, and they can only take place from magnetic-dipole onwards, which means that they're suppressed (in likelihood, not in energy) compared to transitions which don't flip any spins. In both of the cases above, the photons will have a low photon energy, and they will also be dipole-forbidden, which means that they will be relatively unlikely.

where John Rennie says:

Caesium has a single electron in the outermost 6s orbital, and this electron can be aligned with or against the nuclear spin. These two configurations differ in energy by about 0.000038 eV, and transitions between them produce/absorb light with a frequency of 9,192,631,770 Hz. This is the frequency used to measure time.

Now based on this, the frequency of light is exactly 9,192,631,770 Hz, that is, each and every time the transition happens, the energy of the atom/electron system changes by 0.000038eV, and that energy should go somewhere, thus a photon should be emitted. Now this could mean that with every single transition, a single photon is emitted, meaning in one second, exactly 9,192,631,770 number of photons need to be produced. The second is defined by the number of transitions, but it does not say anything about photon emission.

But if this transition is dipole forbidden, and the emission of the photon is relatively unlikely but the frequency of light is exactly 9,192,631,770 Hz, then this could mean that some transitions emit photons, others do not (or that the transition itself is relatively unlikely), and either there are exactly 9,192,631,770 number of photons produced per second, or there are less photons produced, but the photons that are actually produced have a frequency of 9,192,631,770 Hz and energy of 0.000038eV.

Just to clarify, I am trying to ask, whether a single atom, whenever making a hyperfine transition should always emit a photon?

Question:

1. Does a hyperfine transition always cause a photon emission?
• in one second, exactly 9,192,631,770 number of photons need to be produced You seem to be confusing “a transition” with “an oscillation”. May 18, 2020 at 17:56
• @G.Smith would you please elaborate on that? "The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transition between the two hyperfine levels of the ground state of the caesium-133 atom" Do they not mean by radiation the emission of photons? May 18, 2020 at 18:25
• @G.Smith what I am trying to ask is whether the hyperfine transition (of a single atom) always causes a photon emission or not. May 18, 2020 at 18:43
• Yes, they mean by emission of photons. But you don’t get a photon every period. May 18, 2020 at 19:02
• @G.Smith thank you, can you please elaborate on why "you don’t get a photon every period."? May 18, 2020 at 19:03

Now this could mean that with every single transition, a single photon is emitted, meaning in one second, exactly 9,192,631,770 number of photons need to be produced

You're confusing the frequency of light and frequency of transition events.

Frequency $$\nu$$ of light, which is $$9\,192\,631\,770\;\mathrm{Hz}$$ here, is the number of periods of EM field at a given point per second. It defines the energy of photons generated by the transition in question: $$E=h\nu$$.

Frequency of transition events, OTOH, is number of transitions per second, i.e. number of times an atom's state changes from state $$A$$ to state $$B$$ or vice versa. This quantity, when called "frequency", is not well-defined because photons are emitted at random times instead of periodically. It would make more sense to measure number of such random events per second in becquerels instead of hertz, although this unit is not really used for anything other than radioactive decays.

• Thank you so much. I get now that the frequency of the EM signal (the field) that the atoms are in, is exactly 9192631770Hz, and the atom reacts on that external field (absorbing the photons) by making these transitions (then emitting), and around 0K, the more atoms transition, the more precise the clock is. May 18, 2020 at 20:57

Perhaps it may help to note that if the transition frequency is $$f$$, and the total energy which has been emitted is $$E$$, then the number of photons which have been emitted is $$E/(h f)$$ where $$h$$ is Planck's constant. A single excited atom will emit just one photon on its journey to the lower-energy state of any given transition.

(There do exist much rarer processes where two photons, each of half the frequency, are emitted, but the question was not about that.)

If the atom is then left alone, it will not make any more transitions. Not $$1$$ per second, not $$1$$ per hour, not $$1$$ per year, just none at all. To get it to make another transition you have to excite it again.

The time it takes an atom to emit an electromagnetic wave on any given transition can itself be quite long, such as a second or more if the transition has a very accurately defined frequency. This can happen when the upper state lifetime is long, and it is what you want for the kind of transitions involved in atomic clocks. The photon picture can be confusing here. If the frequency is precise to within one hertz, then the duration of the process is certainly at least one second, because only a pulse of that duration can have such a well-defined frequency. But the process of photon detection can, and often does, have a much faster timescale associated with it. For this reason atomic clocks do not usually involve spontaneously emitted photons, but rather they drive the atom using a microwave source (or an optical source for some modern clocks) and thus control the duration of the interaction between the atom and other things.

• thank you, "A single excited atom will emit just one photon on its journey to the lower-energy state of any given transition." do you think that in one second, the single atom makes exactly 9,192,631,770 transitions or just fewer because that (the transition) is a random process? May 19, 2020 at 0:40
• @ÁrpádSzendrei As Andrew said, the oscillation is being driven by a microwave source. So it's like pushing a child on a swing: you get optimum energy transfer when the frequency of the pushes equals the natural oscillation frequency of the swing. See en.wikipedia.org/wiki/Atomic_fountain & en.wikipedia.org/wiki/Ramsey_interferometry May 20, 2020 at 7:19
• @ÁrpádSzendrei (cont) "In broad strokes, the Ramsey method involves exposing a cloud of atoms to a brief radiofrequency (rf) electromagnetic field; waiting a time T; briefly exposing the cloud to the rf field again; and then measuring what fraction of the atoms in the cloud have transitioned. If the frequency of the rf field is identical to the atomic transition frequency, 100% of the atoms will have transitioned; if the frequency of the field differs slightly from the transition frequency, some of the atoms will not have transitioned" May 20, 2020 at 7:20
• @PM2Ring thank you so much, this is the correct description of the clock. May 20, 2020 at 15:39

Árpád, sometimes it is helpful to use analogies to better understand what is happening. The next one I am talking about is only for the better imagination and is not the only correct explanation.

1. The excited state in an atom is given to an electron by a pulse. It can come from another electron that wiggles around, or from an incident photon. The wiggling electron looses energy (photon) and the excited get it. To be in the excitation, some part of the incoming pulse (from another electron or an photon) gets swallowed and the residual value spat out.
The electron is now in a metastable state and falls back into the more stable state, emitting in most cases exactly one photon. This photon has the same energy in the case, the temperature of the material is the same. Otherwise, the whole system has a higher (hotter) or lower (colder) energy content and the transitions take place through slightly different energy packets (photons).
Take it like this: The electrical interaction between the nucleus and the electron is reduced. All emitted photons on their way to the stable state of each electron around the nucleus take some of the electrical field interaction away and the discrete (the only possible) emission of photons ends at some distance and the electron is stable.

2. The interaction between the electrons around a nucleus is determined by the magnetic dipoles of the electrons. The best stability have the noble gases. Other elements have pairs of electrons and thus their electron (pairs) are more stable then for elements with unpaired electrons. Take 5 electrons in an outer shell; the fifth electron is indecisive about his spin orientation.

3. To tilt the spin state of an electron, energy again is needed. The incoming pulse from a photon gets absorbed by the electron (plus releasing back the not needed energy), the spin flips. Now it is clear, why the hyperfine transition (back to the more stable state) is accompanied by the emission of a photon of exactly the observed frequency.

Compare my pictorial explanations - which I have drawn only for your better understanding - with the correct four answers. Is there a common intersection?

• thank you so much. May 20, 2020 at 15:40
• Árpád, so has my pictorial explanation an intersecting set with the other answers? From my point of education and research yes, but I’m not sure that I explained all details, which are needed for a consistent understanding. May 20, 2020 at 16:02
• yes this is the part what the others explain too "To tilt the spin state of an electron, energy again is needed. The incoming pulse from a photon gets absorbed by the electron (plus releasing back the not needed energy), the spin flips. Now it is clear, why the hyperfine transition (back to the more stable state) is accompanied by the emission of a photon of exactly the observed frequency." May 20, 2020 at 16:03
• @ÁrpádSzendrei Want to read this physics.stackexchange.com/questions/553309/… please? May 21, 2020 at 6:34
• Really nice answer. Here is an interesting one:physics.stackexchange.com/questions/572605/… Aug 20, 2020 at 16:34

Two things should be clear. Photons are produced or absorbed in the change in energy levels, and all photons are spin 1 and their energy is given by $$E=hν$$ where $$E$$ is the difference in the energy levels.

Each atom has its own energy levels , even though labeled by the same quantum numbers, each s p d f level characterizes the atom. The only thing that can be carried by the photon is the energy, which is different for each atom, and the spin orientation, + or - its direction of motion.

This is how an atomic clock works

Since 1967, the official definition of a second is 9,192,631,770 cycles of the radiation that gets an atom of the element called cesium to vibrate between two energy states.

Inside a cesium atomic clock, cesium atoms are funneled down a tube where they pass through radio waves . If this frequency is just right 9,192,631,770 cycles per second then the cesium atoms "resonate" and change their energy state.

A detector at the end of the tube keeps track of the number of cesium atoms reaching it that have changed their energy states. The more finely tuned the radio wave frequency is to 9,192,631,770 cycles per second, the more cesium atoms reach the detector.

The detector feeds information back into the radio wave generator. It synchronizes the frequency of the radio waves with the peak number of cesium atoms striking it. Other electronics in the atomic clock count this frequency. As with a single swing of the pendulum, a second is ticked off when the frequency count is met.

So it is a complicated combination of atomic physics transitions under zillion of photons impinging on the cesium atoms at the correct energy . Photons are emitted and absorbed as there is a contiuous radio wave field flooding everything. It is a resonance type of effect that decides the second, as far as I can see. A better description is here.

But it is to be understood that people would not choose forbidden transitions to make the clokcs!

The fact that one measures a hyperfine structure means that the difference in energy levels has been measured. If the transition is not forbidden there will be photons with small energy and some probability coming out.

• thank you so much, do you think that whenever a single atom does a hyperfine transition, it must emit a photon? May 18, 2020 at 18:35
• all transtions in materials are electromagnetic transitions, so the energy has to be released as a photon of the appropriate energy, (or absorbed). It could be that more complicated exchanges of very low frequecies are better modeled in new ways, in which case the photons would be virtual exchanges simplified in new forms, as with phonons. May 19, 2020 at 3:25

The frequency,9,192,631,770 Hz. does not refer at all to the number of photons. The number of photons is given by the number of atoms that undergo that transition. You need to determine the intensity of emitted light to determine the number of photons. So your question about the number of photons is not answerable. The hyperfine transition most likely emits one single photon of frequency 9,192,631,770 Hz. A 133 gram sample of Caesium contains around $$10^{23}$$ atoms. So how many of them that decay in a second is roughly the number of photons of frequency 9,192,631,770 Hz that are produced per second. One transition most often produces one photon. 9,192,631,770 transitions produce about 9,192,631,770 photons.