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When trying to visualize a photon, I imagine it as an electromagnetic wave of very short length. Is this accurate?

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  • $\begingroup$ No. In that case it would be a classical thing. BTW, what do you mean by short length? A wave packet containing just a few wavelengths? $\endgroup$ – G. Smith Nov 30 '19 at 1:28
  • $\begingroup$ yes, a few wavelengths (less than 10, for example) $\endgroup$ – set5 Nov 30 '19 at 1:30
  • $\begingroup$ That visualization wouldn’t explain how photons manage to go through both slits in the double-slit experiment. You shouldn’t expect any visualization of quantum particles. $\endgroup$ – G. Smith Nov 30 '19 at 1:36
  • $\begingroup$ @G.Smith: That visualization wouldn’t explain how photons manage to go through both slits in the double-slit experiment. Well, it would if the length of the wave train is $\gtrsim$ the distance between the slits. $\endgroup$ – user4552 Nov 30 '19 at 1:37
  • $\begingroup$ Right. I was assuming a distance between the slits significantly larger than the packet size. I should have said so. $\endgroup$ – G. Smith Nov 30 '19 at 1:39
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The visualization of a photon is not really part of physics, because physics is a science and as such it only make statements about things that can be tested by observations. The existence of photons are inferred from the interactions of light with matter. We can therefore say that photons are involved when such interactions take place, i.e., when light is emitted or absorbed.

If we entertain the question: how does light produce these quantized phenomena?, then we can think about light in terms of photons even when it does not interact with matter. (But we cannot confirm this view scientifically.) Quantum mechanics does not tell us how the photons behave individually. All it tells us is the probability to observe a photon given certain experimental conditions. One can derive the probability distribution for the observation of a photon from the electromagnetic field. However, the photon itself would then have to be a dimensionless point particle. (Let me stress again, this view is strictly non-scientific.)

Just to remove a misconception that is being spread on PhysicsSE: an optical field is NOT the coherent superposition of lots of photons. In quantum mechanics the state of a photon can be represented by a Fock state $|\psi_n^{(1)}\rangle$, where $\psi_n$ represents the wave function of the photon and the superscript indicates that it is a single excitation (one photon) of the wave function. One can form the coherent superposition of many different single photons states like that $$ |\phi\rangle = \sum_n |\psi_n^{(1)}\rangle C_n , $$ such that $$ \sum_n |C_n|^2 = 1 . $$ What is $|\phi\rangle$ then? Is it the state of a multi-photon optical field? No! It is still a single photon state. All that happened is that its wave function is now the coherent superposition of all the wave functions of those single photon states.

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When trying to visualize a photon, I imagine it as an electromagnetic wave of very short length. Is this accurate?

The question would then be, short compared to what? Quantum mechanics only has one built-in scale, which is Planck's constant. Since Planck's constant doesn't have units of length, there is nothing intrinsic in quantum mechanics that we can compare with in order to say whether an electromagnetic wave has a short wavelength.

What makes a classical electromagnetic wave different from a single photon isn't the wavelength, it's that the classical wave consists of a coherent superposition of a large number of photons.

There is a more subtle issue, which is that although we can talk about the wavefunction of an electron as a function $\Psi(x)$, we can't really talk about the wavefunction of a photon as a function $\Psi(x)$. This doesn't seem to be what you're asking about, but we could go into it if you like. This has to do with the fact that photons are inherently relativistic.

G. Smith asked in a comment:

Do you think there is a good way to visualize photons?

I do visualize them as electromagnetic waves, with the amplitude quantized so as to satisfy $E=h\nu$. However, I'm conscious of the limitations of this visualization (because there is no $\Psi(x)$ for a photon).

G. Smith also asked in a comment:

Based on the OP’s response to another comment of mine, they didn’t mean a short wavelength but rather a short wavepacket that is only a few wavelengths in size, regardless of the wavelength. Could you address this visualization as well?

Hmnm...so you're going to make me work at this, are you!? :-)

So say we take "a few" to mean $n$, with the implication that if $n$ is small, then there will be errors in the classical description which are of relative size $1/n$. This implies that if $n$ is large, the classical description is a good approximation. But quantum mechanics doesn't predict that this is so, and experiments show that it is false. A nice example is K.C. Lee et al., "Generation of room-temperature entanglement in diamond with broadband pulses," Oxford, 2012. (This is a thesis, and it can be accessed by googling. See also https://arxiv.org/abs/1601.07927 .) In this experiment, Lee demonstrated entanglement between diamond crystals separated by a distance of 15 cm. This is a bajillion times greater than any of the wavelengths involved, showing that quantum effects exist even for very large $n$.

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  • $\begingroup$ Based on the OP’s response to another comment of mine, they didn’t mean a short wavelength but rather a short wavepacket that is only a few wavelengths in size, regardless of the wavelength. Could you address this visualization as well? $\endgroup$ – G. Smith Nov 30 '19 at 1:52
  • $\begingroup$ Can you show your work on how you arrived at "bajillion times greater"? $\endgroup$ – BioPhysicist Nov 30 '19 at 2:07
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    $\begingroup$ @AaronStevens: That's left as an exercise for the reader. $\endgroup$ – user4552 Nov 30 '19 at 2:18
  • $\begingroup$ The Planck scale is not really part of quantum mechanics, because quantum mechanics does not include gravity, which is a requirement for the definition of the Planck scale. Moreover, the Planck scale is just a hypothetical scale since we don't really know if it exists. $\endgroup$ – flippiefanus Nov 30 '19 at 4:37
  • $\begingroup$ @flippiefanus This answer didn’t say anything about the Planck scale. $\endgroup$ – G. Smith Nov 30 '19 at 4:55

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