As far as I understand the definition of a second, the Cs-133 atom has two hyperfine ground states (which I don't really understand what they are but it's not really important), with a specific energy difference between them. Whenever the atom transitions from the higher-energy to the lower energy state, the difference in energy is released as a photon. A photon with that energy is equivalent to EM radiation of a specific frequency. A second is then defined as 9192631770 divided by this frequency.

In many places I see people claiming that the Cesium atom oscillates between the two states, transitioning from one to the next 9192631770 times per second, and that this is what the definition is based on. This makes no sense to me, and seems incompatible with the interpretation above - which is based on the energy of a single transition, not to rapid transitions. So I usually just dismiss it and/or correct the person claiming this.

When I saw the "oscillations" interpretation repeated in a video by the hugely popular Vsauce, I started to think maybe I got it all wrong. Maybe the second is defined by oscillations after all? Or maybe the two interpretations are somehow equivalent?

So, is there any truth to Vsauce's description? And if not, why is the misconception of oscillations so popular?

  • 2
    $\begingroup$ Although it is probably irrelevant to the working of an atomic clock, it could be noted that a Cs atom prepared in the state (|↑⟩+|↓⟩)/√2 would oscillate at the reference frequency between that state and (|↑⟩–|↓⟩)/√2. $\endgroup$ – Edgar Bonet May 17 '16 at 8:07
  • 1
    $\begingroup$ @Edgar's comment is precisely on the money, with the bonus that it is directly relevant to the workings of atomic clocks. In many implementations you do prepare the state (|↑⟩+|↓⟩)/√2 and watch for oscillations into (|↑⟩-|↓⟩)/√2 and back. $\endgroup$ – Emilio Pisanty May 18 '16 at 0:25
  • $\begingroup$ In addition to that, the actual definition of the SI second (shown explicitly in the video) makes it clear that it is periods the radiation itself, which does oscillate (and does so resonantly when it's at exactly at the right frequency). $\endgroup$ – Emilio Pisanty May 18 '16 at 0:28
  • $\begingroup$ This question seems to be essentially a duplicate of this question. $\endgroup$ – ACuriousMind May 18 '16 at 17:56

You're correct and the video is mistaken. In fact, if cesium atoms were constantly oscillating between the two hyperfine states, cesium beam clocks wouldn't work at all!

In its simplest form, a cesium beam clock uses a magnet to separate a stream of atoms into two streams based on their hyperfine state; one state is selected to continue down the tube to be exposed to an oscillating magnetic field in the microwave range, and the others are wasted. After the microwave chamber, the stream is magnetically separated again, with one state (differing from the state that was selected the first time by a certain energy) hitting a target that responds to cesium atoms by producing an electrical signal.

The effect is something like the crossed polarizers of an LCD display. Since one state is selected before the microwave chamber, and a different state is selected afterwards, there is no signal unless atoms changed state in between. "Ordinarily", this doesn't happen, but if the microwave tube is bombarding the atoms with energy that corresponds to the desired hyperfine transition, then some of the atoms will absorb energy, make the transition, and be detected at the far end. By incorporating the beam and detector into the control loop of a variable oscillator, the microwave frequency can be maintained at the frequency that causes the hyperfine transition, independent of outside conditions.

The part of this that's crucial to your question is the statement that the cesium atoms don't change state between the A and B selectors unless something causes them to. If they were changing states at >9GHz, then small variations in the travel times for the atoms (which move at hundreds or thousands of m/s, but nowhere near the speed of light) would result in a completely random signal at the detector. Instead, we get a coherent signal because the rate of spontaneous hyperfine transitions is small compared to the time the atoms spend in the tube. Any type of interaction that can scramble the hyperfine state of an atom reduces the sensitivity of the clock, and eliminating these interactions is a big part of maximizing accuracy.


The definition for the cesium clock is:

9192631770 cycles per second is frequency of the radio waves which cause maximum resonance, a physically measurable condition, in the cesium atoms.

This corresponds to a particular tuning of the radio. Keeping it tuned provides the reference frequency cited.

  • 2
    $\begingroup$ +1: That's the shortest explanation of an atomic clock I have ever seen... and it works for me! :-) $\endgroup$ – CuriousOne May 16 '16 at 23:10
  • 24
    $\begingroup$ Brevity is important when teac $\endgroup$ – Peter Diehr May 16 '16 at 23:20
  • $\begingroup$ Make everything as simple as impossible, but never simpler. $\endgroup$ – a CVn May 17 '16 at 14:52
  • $\begingroup$ @PeterDiehr SQUIRREL!! $\endgroup$ – hBy2Py May 17 '16 at 16:39

Yes, they really are oscillating between two different states (not simply driven in one direction), but as you suspect they are not oscillating at the reference frequency.

Rather than "just" sending radiation at the atoms to absorb, they also interact with an oscillating magnetic field (which is at the reference frequency). This field spurs some of the atoms to absorb or emit radiation. So both states are driven to the other state by the conditions inside.

A rather technical description of a cesium clock can be found at leapsecond.com

Some of the relevant items from it:

An applied radiation field of frequency appropriate to the transition. under discussion induces atoms in the F = 4 state to emit a quantum of energy, and those in the F = 3 state to absorb a quantum of energy; they are reversed in their magnetic moments following such transition, and are said to have flopped.


The magnetic component of the injected microwave field interacts with the atoms. If the frequency is at the transition frequency, 9,192,631,774.3133 Hz then atoms flop to the other transition energy state. Since their effective magnetic moment is 1hereby reversed in its direction, a second state selector magnet, the "B" magnet, can selectively deflect flopped atoms to the detector.

  • $\begingroup$ So, in the end Vsauce was wrong when he said "After 9192631770 oscillations between those two levels, one second has passed." $\endgroup$ – M. Enns May 17 '16 at 0:16
  • 2
    $\begingroup$ @M.Enns I didn't watch the video, but I agree that the quote you have above would not be correct. $\endgroup$ – BowlOfRed May 17 '16 at 0:18
  • $\begingroup$ @M.Enns yes. That's a mistake on a par with thinking that the electron in a hydrogen atom is changing energy levels 1.4 billion times a second because the "21 cm line" transition has an energy that corresponds to 1.4GHz photons. In reality that transition happens spontaneously about once per 70 million years per atom. $\endgroup$ – hobbs May 17 '16 at 4:23
  • 1
    $\begingroup$ @BowlOfRed: 1. Note that I've linked to the time marker in the video where he starts talking about this, so you only need to watch 30 seconds to see what we mean. 2. It's clear to me that the atom can also absorb energy and go up a level, that the radiation can also be stimulated, and that this can happen several times a second - I was just contesting the idea that this happens at a predefined, and extremely high, frequency. $\endgroup$ – Meni Rosenfeld May 17 '16 at 7:22
  • $\begingroup$ @MeniRosenfeld Yes, I'm agreeing with you. I wouldn't object to someone merely saying the electrons are "oscillating" between the states, but it doesn't happen near the frequency of microwave radiation. To equate it with the resonance frequency is incorrect. Only a fraction of the population will "flop" on any passage. $\endgroup$ – BowlOfRed May 17 '16 at 7:35

protected by Qmechanic May 17 '16 at 10:27

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.