# A question about atomic clocks

I have a rather simple question about atomic clocks. I have read that:

Microwave radiation with a frequency of exactly 9.192.631.770 cycles per second causes the outermost electron of cesium-133 atom to reverse its spin direction. An atomic clock uses this phenomenon to tune moicrowaves to this exact frequency. It then counts 1 second for each 9.192.631.770 cycles.

So does that mean that in an interval of 1 second the outermost electron of the cesium-133 atom changes its spin direction 9.192.631.770 times? And let 1 and 2, be the two possible directions of spin. Does one change mean a change from 1 to 2 , or a change 1 to 2 and back to 1?

If you measure the absorption of microwaves by a gas of caesium atoms you'll get a spectrum looking something like this:

NB not a real spectrum - I drew this as an illustration

When the microwave frequency is 9,192,631,770 Hz the microwave photons have exactly the right energy to flip the outer electron spin, so at this frequency they are more strongly absorbed than at nearby frequencies. Energy is absorbed from the microwaves to flip the electron spin, then the electron either reradiates a photon and relaxes back or it loses the energy and relaxes due to collisions with other caesium atoms.

The caesium clock works by tuning the frequency for maximum absorption i.e. exactly at the peak of the absorption line. Then it's just a matter of counting each cycle of your microwave generator, and after 9,192,631,770 of them one second has passed.

Response to comment:

The electron flips every time a caesium atom absorbs a microwave photon. So if we can work out how many photons per second are absorbed this will tell us how many electrons per second are flipped.

Suppose the microwave power is $P$, and the absorptance (the depth of the trough in the diagram above) is $f$, where $f = 0$ means no microwaves are absorbed and $f = 1$ means all the microwaves are absorbed. The the power absorbed is $Pf$. The energy of a photon is $h\nu$, where $\nu$ is the frequency of the hyperfine absorption 9,192,631,770 Hz. Then the number of photons absorbed per second, which is the same as the number of electrons flipped per second, is:

$$N = \frac{Pf}{h\nu}$$

Alternatively, if $N_C$ is the number of caesium atoms in the clock then the number of times the electron is flipped per second per caesium atom is:

$$N_a = \frac{Pf}{N_Ch\nu}$$

However I have no idea what power microwaves are used in an atomic clock or what fraction of the microwaves are absorbed, so I'm afraid I can't calculate the numbers $N$.

• So in a second, how many times does the electron switch positions? – Bardo Aug 3 '14 at 18:48
• @Bardo: I've updated my answer to respond to your comment – John Rennie Aug 4 '14 at 4:59
• There is something here that I don't understand but I don't know how to express it.. – Bardo Aug 4 '14 at 18:52
• The cesium atoms in a truly accurate atomic clock don't collide, do they? That would introduce way too much noise. Maybe they did in cesium clocks 50 years ago, when they were much less accurate. – Peter Shor Apr 28 '18 at 17:07

It does not mean that it switches the hyperfine state that often. I means that, if you hit the atom with a photon that (by $E = \hbar\omega$) corresponds to exactly that frequency, then the energy of that photon will precisely be the energy to lift the electron from the lower hyperfine state to the upper hyperfine state. If hit, it will then, as with all other orbital excitations, stay a short while in the upper state before dropping into the lower again, sending out a photon of exactly (modulo uncertainty) the same energy again.

All an atomic clock does is now to tune some kind of circuit so that it oscillates exactly with the same frequency as the photons/microwaves associated to this transition. (Not actually that easy)

• This is getting clearer to me, but I have asked @John the same question I am asking you now: So in a second, how many times does the electron switch positions? – Bardo Aug 3 '14 at 18:49
• @Bardo: It doesn't switch back and forth, unless the same atom/electron is repeatedly hit by multiple photon. For a single photon, the process is just: "lower state -(photon hit)-> upper state -(photon emission)-> lower state." You could make them switch rapidly by turning up the intensity of the radiation (so that they are hit very rapidly), but that has nothing to do with the transition as such, and has no relation to the frequency of the photon, but would be purely a property of your particular setup. – ACuriousMind Aug 3 '14 at 18:54
• Ok, but here timeanddate.com/time/international-atomic-time.html it says that "The International System of Units (SI) defines one second as the time it takes a Cesium-133 atom at the ground state to oscillate exactly 9,192,631,770 times. ". So, for the Cesum-133 atom to oscillate, that outer electron needs to switch states, right? So it needs to switch states 9.192.631.770 time .. ? – Bardo Aug 4 '14 at 18:52
• So, let me se if I get it: we somehow, I honestly don't care how yet, we somehow manage to apply a radiation of a given frequency $f$ that makes the Cesium-133 atom to change state. But it doesn't change its state 9 billion times in a second. The frequency we used, has that value. So an atomic clock works like this: we have Cesium-133 atoms. We hit them with different frequencies that cause them to behave differently. We have a device that allows us to see when an atom has gone from its lower energy state, to its highest. We save the frequency we used to attain this state, and that's it – Bardo Aug 4 '14 at 19:04
• Then we come up with a clever way to make something oscillate with the exact same frequency that produced that phenomenon, and at the same time to be able to count the oscillations. And we have a clock! Is that right? – Bardo Aug 4 '14 at 19:07

edit: Actually I don't think the picture below for the different hyperfine states is quite correct. I'll try to update it with a more correct picture at a later time.

Consider the Cs atom to have two states. A ground state and an excited state. The ground state corresponds to the nuclear spin pointing down (along the $-z$ axis) and the excited state corresponds to the nuclear spin point up*. These starts are separated in energy by $E = h \times 9,192,631,770 \text{ Hz}$ where $h$ is Plank's constant. In both of these states we can also imagine electron to have a total spin and orbital angular momentum in the $+z$ direction as well. To imagine this angular momentum, imagine the electron is orbiting the nucleus in the $x-y$ plane. This means it is creating a current loop around the nucleus that creates a magnetic field at the location of the nucleus. Since the nucleus has a magnetic moment and it is in a magnetic field this means there will be an energy difference between the nuclear spin pointing up along the magnetic field of the electron and the nuclear spin pointing down against the magnetic field created by the electron*.

It is known that light composed of photons with the same energy (that is microwave radiation at $9.192... \text{ GHz}$) can cause a transition from the ground to the excited state. However, additionally it is possible to use a special pulse of light of that same frequency to put the atom into a superposition of ground and excited states. See Rabi Cycle.

One quick point on how the radiation is able to affect the internal quantum state of the atom. Recall that the nuclear spin has a magnetic moment. Good. Also recall the light is composed in a part of a magnetic field. Good. The reason the light is able to change the state of the nucleus is because the magnetic field of the light is able to put a force on the magnetic moment of the nucleus. That is, the light is literally tugging on the nucleus a little bit.

If a spin is in a superposition of point down and pointing up then that means it can be thought of as pointing sideways. This is a little trick of quantum spin physics.

Which direction sideways is the spin pointing? Well that depends the direction (and phase) of the magnetic field vector of light when it interacted with the atom. This is something which can be controlled experimentally. Let's say the experimentalists choose to have the atom point in the $+x$ direction after the first pulse of light.

Now the spin of the nucleus is pointing along the $+x$ direction. However, recall that the nuclear spin is sitting in the magnetic field of the electron (which is still whizzing about in the $x-y$ plane.) If a magnetic moment is transverse to a magnetic field then it will under go Larmor precession. Like a gyroscope, the magnetic moment will rotate around the magnetic field. At what speed will the magnetic moment rotate? It will rotate at the timescale set by the energy difference between the two states. The period of oscillation will be

$$T = \frac{h}{E} = \frac{1}{9,192,631,770 \text{ Hz}}$$

It has occasionally been asked 'what is ticking in an atomic clock?', trying to make a comparison to a grandfather clock, for example, in which case it is obvious that the pendulum's motion under gravity is the source of the ticking. The answer is that in microwave atomic clocks it is the nuclear spin precessing about the magnetic field of the electron which 'ticks'.

To answer your question briefly: $9,192,631,770 \text{ Hz}$ is the timescale corresponding to the precession of the nuclear spin about the magnetic field of the electron. In addition, this also corresponds to the frequency of light which will cause a transition of the atom from the ground to the excited state. It is easy to see how these two concepts might get confused. Because these two frequencies are the same, if we are able to measure the transition frequency we have also measured the precession or ticking frequency.

*Note that I am giving an intuitive description of the hyperfine splitting which gives rise to the energy difference between the two states we are discussing here. Having a microscopic picture for the two different states greatly helps in understanding the functioning of an atomic clock.