3
$\begingroup$

It is common to see photons depicted as a sine wave with multiple peaks and troughs. Is that correct? I am familiar with the energy-wavelength and frequency-wavelength equations for a photon. But my understanding is that "wavelength" is a spacial period, not the length of the entire excitation, so you could have an excitation, spanning an arbitrary number of wavelengths. Knowing that a photon is the least energetic possible excitation with that frequency in the electromagnetic field: how many wavelengths does a single photon span? And if it doesn't make sense to ask this, please explain why.

If possible I'd like to ignore the de-Broglie matter-wave complication.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
4
$\begingroup$

This answer elaborates on the good answer by Alexander, which was posted just before I finished writing this one.

For a moment, forget about photons and think about the classical electromagnetic field instead. How many wavelengths does a classical wavepacket span? Once this question is answered, answering the corresponding question for photons will be easy.

What is the "wavelength" of a classical wavepacket? Strictly speaking, wavelength is a property of an infinitely-extended sinusoid, not a property of a localized wavepacket. To get something localized, we need to superimpose lots of different wavelengths. We can still define the average wavelength of the resulting wavepacket, but not "the" wavelength. (This is an average in the same sense that the center of mass of a bunch of objects is an "average," not an "average" in the statistical sense.)

Clearly we can construct a classical wavepacket whose length is much greater than its average wavelength. (An infinitely-extended sinusoid is a limiting case of this.) Can we construct a classical wavepacket whose length is less than its average wavelength? Yes, we can, but we need to be careful what we ask for! We can construct a classical wavepacket whose average wavelength is arbitrarily large and whose length is arbitrarily small... but it won't stay that way. As time passes, it will spread out very quickly. The smaller the wavepacket is initially, the more quickly it will spread. After some time, its length will end up being much greater than its average wavelength.

Whatever the answer is for a classical wavepacket, that answer carries over to the case of an individual photon, as explained in mathematical detail here:

Relation between radio waves and photons generated by a classical current

The word "photon" refers to something that can be counted, whether or not it happens to be localized. Roughly, the possible configurations of a photon are in one-to-one correspondence with the possible configurations of a classical wavepacket ignoring its overall amplitude. The overall amplitude of a classical wavepacket can be varied continuously, but photons are discrete: they can be counted. As far as the present question is concerned, that's the only important difference. The question about wavepacket-length versus wavelength is exactly the same for a photon as it is for a classical wavepacket.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ Why would a wavepacket whose length is less than its average wavelength spread out over time until its length becomes longer than its average wavelength? $\endgroup$
    – Dunois
    Commented Aug 1, 2020 at 16:10
  • $\begingroup$ @Dunois Here's some intuition: If a wavepacket is localized within a region of size $\sigma$, then its Fourier transform must be spread over a range of wavenumbers of size $\sim 1/\sigma$. Wavenumber is proportional to momentum (for a free particle), so the spread of momenta must be $\sim h/\sigma$ where $h$ is Planck's constant. The smaller $\sigma$ is, the larger the spread of momenta must be, so the more rapidly the wavepacket will spread out (because different momentum-components are traveling with different velocities). $\endgroup$
    – Chiral Anomaly
    Commented Aug 1, 2020 at 19:30
4
$\begingroup$

The essential quantum numbers to specify a unique "quantized electromagnetic field configuration" are -

  • Number of excitation (photons) $N$
  • Energy $E$
  • Momentum $\vec{k}$

To build a general state you can superimpose any $|N,E,\vec{k}\rangle$ states, which not necessarily all have the same $N$ or $E$ or $\vec{k}$. Energy $E$ constraints the magnitude of momenta, by $E=\hbar |\vec{k}|c$ but not its direction, thus the momenta degree of freedom.

If you ask about a single photon, it means you're asking what is the general state for the subspace $N=1$. So you see, you can still superimpose different energy and momenta at this subspace. You can have a single narrow wavepacket propagating at specific direction or 100 wave-packets, each propagates in different direction.

The measurements are to be considered separately, since most of them are projective.

For references see any book on quantization of electromagnetic field. "Photons and Atoms: Introduction to Quantum Electrodynamics", book by Claude Cohen-Tannoudji, Gilbert Grynberg, and Jacques Dupont-Roc is recommended.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ There is also the polarization (spin) quantum number. Since energy is already determined by the magnitude of the momentum vector, it is strictly speaking not a separate independent quantum number. The general situation is a bit more complicated in that the $N$ excitations can each carry a different momentum and spin. $\endgroup$
    – flippiefanus
    Commented Apr 15, 2019 at 8:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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