There are similar questions to mine on this site, but not quite what I am asking (I think). The de Broglie relations for energy and momentum

$$ \lambda = \frac{h}{p}, \\ \nu = E/h .$$

equate a specific frequency and wavelength to a particle, yet we know that a wave packet is a linear combination of an infinite range of frequencies and wavelengths. How is it that we (or nature) choose one frequency and wavelength out of the range? Does this have to do with the collapse of the wave packet when measured? And if so, is the resulting measured frequency a random outcome? Similarly, when an electron jumps from one energy level to another in an atom, it emits a photon of frequency

$$ \ \nu = \Delta E/h .$$

Since the photon is not a pure sinusoidal wave, how can a single frequency be ascribed to the photon?


Let's say an isolated atom emits a photon. The excited state in the atom has some lifetime $\tau$. Through the energy-time uncertainty relation, that gives the excited state some uncertainty in energy $\delta E\sim h/\tau$ (not the same as $\Delta E$, which is a difference in energy between atomic states). The photon then has the same uncertainty $\delta E$ in its energy, which corresponds to an uncertainty in frequency. The photon isn't in an eigenstate of energy.

For many real-life examples such as a visible photon emitted by a hydrogen atom, or gamma-rays emitted by beta-decay daughters, $\tau$ is very long compared to $h/\Delta E$, so we have $\delta E \ll \Delta E$. The uncertainty $\delta E$ is also often very small compared to the limitations imposed by, e.g., Doppler shifts or the resolution of the detector.

Yes, when you measure the energy of the photon, you get a random outcome. However, there is a quantum-mechanical correlation between this energy and the energy of the atom, so that energy is exactly conserved (not just statistically, on an average basis).

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  • $\begingroup$ Despite your answer is excellent, it simply answers none of the question of PMay. You could have just started from a historical point of view, saying that energy (exchange) is quantised according to the postulate of quantum mechanics, or Bohr's principle. Then you could have elaborated about the statistical nature of the decay, which is not unitary, and thus much more complicated to understand I believe. $\endgroup$ – FraSchelle Jun 19 '13 at 3:55
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    $\begingroup$ @Oaoa: I don't understand your comment. Maybe you could expand it into an answer...? It's not true that decay is not unitary. All processes in quantum mechanics are unitary. $\endgroup$ – user4552 Jun 19 '13 at 3:57
  • $\begingroup$ My bad, you're right. I confound time symmetry breaking and unitarity. Nevertheless, the lifetime of an atom is explained via the coupling to an infinite reservoir modes. The simplest picture is (as I believe) to first discuss the isolated atoms, for which the quantisation of the energy level is a postulate. This perfectly quantised level gets broader by statistical effects. You just described the second part. I'm feeling your answer unlikely to be understood by someone stuck at the Broglie's relationship. $\endgroup$ – FraSchelle Jun 19 '13 at 4:22

This is a very good question.

As Ben Crowell says, the photon life-time is finite and so its energy is spread by a range of frequencies, i.e. it's not a delta-function over a specific frequency.

The photon's energy can be obtained integrating over all the spectrum and might be the one corresponding to the peak, E= h•f where f is the frequency where there is the peak.

Anyway, in this case should we assign the frequency "f" to this photon and expect that when it's absorbed it will transfer h•f eV to the absorber electorn?

Another question is how the wave packet is in the space since it has to respect Maxwell equations and therefore you cannot place a square window with N periods and left it happily... Gaussian beams may be good candidates...

I am sorry if I cannot be more accurate about this very interesting matter and I also apologize by my English.

Best regards, Sergio

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  1. according to de Broglie, the wave packet is moving with the group velocity vg and the phase wave is moving with vp=f*lamda. the de Broglie relation lamda=h/p where p=m*vg but lamda=vp/f. this equation involve both wave-particle duality.(p=m*vg for matter and vp=f*lamda). therefore, in this relation, the frequency is refer to phase wave but not the the particle(wave packet) itself

  2. each emitted photon consists energy E=hf. getting the energy, we can simply calculate the frequency with f=E/h. hope this will help and sorry for my broken english

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