An electronic transition in an atom produces a photon, and the photon's energy is determined by the difference between upper and lower electronic state. For the final photon, its energy is described by its frequency only.

However, the photon can be considered as an electromagnetic wave as well, and then the question is, how many periods does the wave consist of (actually this post might seem to be a duplicate of How many wavelengths does a single photon span?, but it's not!).

I was reading somewhere that the electronic transition is not instantaneous, but takes something in the order of $\Delta t = 10^{-12}$ s (maybe even less, please correct me, as I cannot remember 100%). If the photon wavelength is 500 nm (because the difference of energy levels correspnd to 500 nm, i.e. $\Delta E = E_{upper} - E_{lower} = \frac{h\cdot c}{500 nm}$), then the resulting frequency is $\nu = \frac{c}{500 nm}= 600 $ THz. And thus the pulse train that corresponds to the emitted photon is $N = \Delta t \cdot \nu = 600$ periods long.

But: If the transition time is longer, the pulse train would consist of more periods. And on the other hand, if the transition time is shorter than $1/\nu$, then $N<1$, i.e. there is not even one period...? What is the link between transition time and wavelength?

  • $\begingroup$ Cannot help too much but take a look at Rabi Oscillation for a deeper explanation of interaction of light at atoms. $\endgroup$
    – boyfarrell
    May 15, 2021 at 22:32
  • $\begingroup$ Had a look at Rabi oscillations. But I think the Rabi frequency depends on the lifetime of the upper energy level, whereas the question here is "how long does it take for the electron to fall from the upper into the lower energy level" (only the time for the transition, without the time the electron spent in the upper state). $\endgroup$ May 16, 2021 at 11:55
  • $\begingroup$ I must say I have not studied the QM in depth, but my intuitive explanation of Rabi oscillation was just this: the wave function oscillates between the two states, but it is damped, so it begins to spend longer and longer in one of the states. I think you could say 1/e for this process is the lifetime. This is a really interesting questions! I hope a QM expert gives a proper answer rather than my hand waving explanations! $\endgroup$
    – boyfarrell
    May 16, 2021 at 12:12
  • $\begingroup$ Wikipedia says (en.wikipedia.org/wiki/Atomic_electron_transition): "It appears discontinuous as the electron "jumps" from one energy level to another, typically in a few nanoseconds or less." This would imply that you can calculate the length of the emutted pulse train the photon consists of... $\endgroup$ May 16, 2021 at 12:24
  • $\begingroup$ Following the reference it seems to referring to this nature.com/articles/s41586-019-1287-z (pay wall) $\endgroup$
    – boyfarrell
    May 17, 2021 at 10:25

3 Answers 3


If by the term "transition time" you mean the natural lifetime of the transition, then there is no link to wavelength, except for a general trend that transitions of lower wavelength tend to have shorter natural lifetime. The situation is that if the natural lifetime is $\tau$ then the transition rate $\Gamma = 1/\tau$ typically goes as $$ \Gamma \propto |M_{ij}|^2 \omega^3 $$ where $\omega = 2\pi c/\lambda$ is the transition frequency and $M_{ij}$ is the matrix element describing the coupling between atom and electromagnetic field.

However it seemed to me that your question is also asking something else, namely how can it be that a transition that can take a long time (e.g. natural lifetime could be as long as 1 second) can nevertheless feature a photon whose arrival at a detector may be a sudden event. This touches on the general fact that a wavefunction can be spread over a range of position and time, but a suitably constructed observation of the relevant system can be concentrated in position or time.

In the case of emission of a photon, both aspects are on show. As the light propagates away from the atom, the wavefunction associated with the electromagnetic field goes in all directions, expanding outwards with spherical wavefronts and an amplitude distribution over angle which is typically a dipole emission pattern, but can be other shapes. And yet a position measurement of the photon will find it to be at one particular place.

Similar statements can be made about the timing. Once the atom has been excited, its interaction with the surrounding electromagnetic field (which may typically be in its ground state) begins immediately and the joint evolution of atom and electromagnetic field is that the atomic dipole oscillates and the field picks up an excitation in the form of a wave of many wavelengths, oscillating and decaying in amplitude. The wavefunction of the emitted photon (I am here employing a slightly rough-and-ready concept, in order to avoid the field theory details) has a form roughly given by $$ \psi(r,\theta,\phi,t) = \frac{a(\theta,\phi)}{r} e^{-t/2 \tau} e^{i(kr-\omega t)} $$

For example, if the natural lifetime of the atom is 10 nanoseconds and the frequency of the light is $10^{15}$ Hz then the wave-train will last about 10 ns and will have about $10^6$ wavelengths. However, if you position a detector of electromagnetic energy, such as a photo-multiplier tube, in this wave-train, then it will register an excitation at one particular time, an excitation having the entire energy of the emitted light. That is the marvelous nature of quantum mechanics, and if you wish to look into this process more deeply, you enter into the area known as the quantum measurement problem, which tries to unpack exactly how the many-possible-times-and-places gets resolved into one-particular-time-and-place.

To pick an example to bring this out, the atomic natural lifetime could be much longer, say one second, but the detector can react much faster, say one nanosecond; in this case the detector clicks abruptly at some moment chosen randomly, with a probability distribution over time corresponding to the 1-second exponential decay time of the atom.

In the present example one speaks of "a single photon" and what I called a wave-train can be considered to be a mode of the electromagnetic field. This mode has gained a single degree of excitation; that is why there is just one photon, and why the detector clicks just once. The timescale on which the detector transitions from "no click" to "click" is set by its own internal dynamics.

If one wanted to demonstrate experimentally that the electromagnetic field mode here really is spread out in time, as I have said, then one could do it by using a temporal coherence measurement. For example, one would split the wave into two parts at a beam splitter, sending one part on a longer journey via a mirror, and then recombine, and there would be an interference effect. Of course to get convincing evidence such an experiment would have to be repeated many times.

  • $\begingroup$ I saw that I got a down vote, so this comment is to signal that I re-read my answer and am content with it. In case it helps to convince that I know what I am talking about: I have in my own lab run quantum jump experiments on the D-S quadrupole transition at 729nm in Ca+, where the natural lifetime is 1 second. $\endgroup$ Jun 4, 2021 at 8:34
  • $\begingroup$ Just to remark that I am not the one who downvoted it - it would be bad sportsmanship. Although I confess that I find it hard to wrap my mind around your answer... $\endgroup$
    – Roger V.
    Jun 8, 2021 at 15:43
  • 1
    $\begingroup$ @RogerVadim thanks for this friendly comment; it prompted me to edit my answer, which basically agrees with yours but I thought the OP was also asking about some further things which I wanted to comment on too. $\endgroup$ Jun 8, 2021 at 16:44
  • $\begingroup$ @Andrew Steane Thanks for your update! So given your example above, a lifetime of 10 ns leading to $10^6$ periods in the wave-train: In a classical view this would mean the atoms starts to radiate the EM wave (photon) as soon as it is excited (but of course a measurement would lead to a single photon when measured due to the quantum nature)? The longer the lifetime the longer the wave-train? $\endgroup$ Jun 8, 2021 at 21:49
  • $\begingroup$ @CharlesTucker3 "The longer the lifetime the longer the wave-train" YES $\endgroup$ Jun 8, 2021 at 22:28

Because emission of a photon by an atom is a quantum process.

Quantum treatment
Photon is an essentially quantum notion. The fact that it has energy equal to that of the level spacing of a two-level system is a result of energy conservation: e.g., if we consider a two-level atom with energies of the excited and the ground states $E_e$, and $E_g$, and a single-mode photon field of frequency $\omega_0$ that initially contains $n$ photons, then the energy before the transition is: $$ E_i = E_e + n\hbar\omega_0, $$ whereas the energy after the transition is $$ E_f = E_g + (n+1)\hbar\omega_0. $$ The energy conservation requires that $E_i=E_f$, which means that $E_e - E_g=\hbar\omega_0$.

This is not strictly true when a transition happens over a finite time, i.e., when we incorporate the atom-photon coupling, and take the thermodynamic limit. In the simplest case one says that the frequency of the emitted photons have spectrum of width $1/\tau$,w here $\tau$ is the transition time, and the shape of this spectrum is often approximated by a Lorentz curve: $$ \rho(\omega) \propto \frac{1}{(E_e - E_g -\hbar\omega)^2 + \frac{1}{\tau^2}} $$ In other words, the photon energy may deviate from the atomic level spacing, and the transition time characterizes the magnitude of this deviation.

Classical treatment
The calculation above would be impossible, if we considered classical electromagnetic field, and treated the atom as an oscillating dipole exciting the electromagnetic waves. $$ \mathbf{d} = \mathbf{d}_0 e^{-t/\tau}\cos\left(\frac{E_e-E_g}{\hbar}t\right) $$ Substituting $\ddot{d}$ instead of $qa$ in the Larmor formula, and integrating over time, one could calculate the energy passed to the electromagnetic field during the emisson process. This energy willd epend explicitly on the dipole length and the emisison time, and not at all close to $E_e-E_g$.

In fact, a theory, treating electromagnetic field classically, is capable of calculating the transition rates between the atomic states due to stimulated absorption/emission, but it is capable neither of calculating the changes in the field, not predicting the rates of spontaneous emission.

Further reading
For a deeper background it is worthwhile looking into the theory behind the Einstein coefficients, the books on laser physics, such as Quantum electronics by Amnon Yariv, and the books on quantum optics, such as Atim-photon interactions: Basic processes and applications by Claude Cohen-Tannoudji et al.

  • $\begingroup$ Many thanks for your answer! But is $\tau$ in the equation above for $\rho(\omega)$ really the transition time (i.e. the time the electron spends between upper and lower state and radiates the photon)? I learned that $\tau$ leading to natural line broadening (Lorentz profile) is the lifetime of the excited state (e.g. pci.tu-bs.de/aggericke/PC4e_osv/Spectroscopy050119/node7.html). This lifetime is not fixed but only statistically determined (transition will happen sooner or later) and this leads to a Heidenberg uncertainty of the energy and therefore broadening. $\endgroup$ Jun 3, 2021 at 16:07
  • $\begingroup$ If the transitions from the excited to the ground state are only by photon emission, then the transition time and the lifetime are the same. It could be trickier, if there are other ways of relaxation. Both are statistically determined. Also, the link between the quantum description and the dipole moment oscillations is provided by Bloch equations - they describe evolution of a dipole (or magnetic) moment, but can be rigorously derived from the quantum point of view. $\endgroup$
    – Roger V.
    Jun 3, 2021 at 16:51

I don't know if I'm understanding what the question is but... What's wrong with having less than one period? You do not need a full period to determine the wavelenght. It might be harder to measure, but you can know the period with a short part of the oscillation as well. On the other hand, there's nothing wrong with having a transition of 600 periods too.


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