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Typically a traveling photon is described as being in a superposition of frequency modes $\hat{E} = \int g(\omega) a^\dagger_\omega d\omega + h.c. $ where often the $g(\omega)$ is some kind of pulse. Can $g(\omega)$ really be a pulse of any bandwidth? Putting it another way: Is the spectral profile of the pulse connected to the frequencies in some kind of way? For instance if these two things ($g(\omega)$ and $a_\omega^\dagger$) are independent, what stops me from having tiny attosecond pulses with small frequences associated with huge wavelengths (which won't physically span even a fraction of their wavelength across the distance of the pulse)?

Isn't that sort of contradictory? I guess if we're only paying attention to the energy of a photon and not really saying that the "wavelength" is physical then it makes sense to have a tiny pulse associated with small frequencies.

Steven Sagona
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1 Answers1

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This is a comment:

The photon is a quantum mechanical particle. The wave function describing it mathematically and thus the frequency associated with the wave,$Ψ$ gives the probability as $Ψ^*Ψ$ , a real number, of finding the photon at (x,y,z,t). Te frequency in the wavefunction is the energy of the photon = $h*ν$

When talking of quantum mechanical particles free in vacuum, the plane wave solutions of the quantum mechanical equation are not a good model for them. One uses the wavepacket formalism, but it is a wave packet in the probability space, not in space and time.

which won't physically span even a fraction of their wavelength across the distance of the pulse

There is no distance of the pulse in space in the way you are discussing it. Only the probability of finding the particle at a specific (x,y,z,t).

anna v
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    I am aware that $g(\omega)$ is a probability amplitude. I use the word "wavefunction" to refer to it in the question, so I'm not sure how this is relevant. – Steven Sagona Sep 17 '20 at 07:39
  • Really nice answer. Do you think this explanation is correct: https://physics.stackexchange.com/questions/580182/pair-production-in-vacuum/580291#580291 – Árpád Szendrei Sep 17 '20 at 16:24
  • @StevenSagona That is why I call it a comment. How do you propose on a light beam of infrared frequency(huge wavelengths ) to modulate photons' probabilities ? Look how individual photons behave probabilistically and then build the classical em wave .https://www.google.gr/search?q=swiss+double+slit+single+photon&ie=utf-8&oe=utf-8&client=firefox-b&gfe_rd=cr&ei=sGFmWYrFCO_v8AfrtLGADQ – anna v Sep 17 '20 at 17:21
  • the wave packet complex conjugate squared is the probability distribution for theindividual photon. – anna v Sep 17 '20 at 17:57
  • This answer is incomplete and disregards many aspects of what is possible in modern quantum optics. $\Psi^* \Psi$ is not the only relevant quantity when characterizing quantum light. One can easily see that this does not even characterize classical light fields completely, since you loose the phase through the quadrature. Besides, I don't see an answer to the OP's question. I am voting -1 for now. – Wolpertinger Sep 18 '20 at 10:13
  • @Wolpertinger Photons are the basic building blocks of light and quantum optics. They are elementarpy point particles with no spatial extent, of energy h*nu, zero mass and spin +/-1 https://en.wikipedia.org/wiki/Elementary_particles basic building blocks in the current mainstream physics. and their confluence builds up classical EM. https://motls.blogspot.com/2011/11/how-classical-fields-particles-emerge.html . I have said again that this is mainly a comment, to clarify the question, as the question is confused. – anna v Sep 18 '20 at 12:15