# A Photon's wavelength is Color, but an Electron's wavelength is a Probability? [closed]

If a photon of 475nm strikes my retina, my brain registers it as "blue," whereas a photon of 650nm is "red." If I ask the question "what is oscillating, and therefore causing the change in wavelength (and thus frequency) that I observe?" I'll receive an answer invoking alternating orthogonal electric and magnetic fields. Since a single photon is itself a quanta of the EM field, there is an easily understood relationship between the wavelength of light that I access directly when my eyes register the color "Blue," and the QED entity described by "the photon has a wavelength of 475nm."

My question is this:

Why is the wavelength of a photon discussed and treated as a physically real component of the world, while the de Broglie wavelengths of other (massive) particles are dismissed as "waves of probability."

The color blue registered by my retina is the product of very real wavelength in an EM field, which is itself composed of (built up of) field quanta of the same wavelength (within uncertainty distribution). Why do we alter the way we conceptualize de Broglie wavelengths when we speak about particles/atoms/molecules, when diffraction and other characteristics remain, and the maths remain the same as well?

## closed as unclear what you're asking by Emilio Pisanty, Jon Custer, Yashas, ZeroTheHero, David HammenJun 30 '17 at 12:37

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• de Broglie wavelength is just as physical, it's the inverse momentum – Kosm Jun 28 '17 at 18:58
• Possible duplicate of Treating matter waves as light waves? – Kosm Jun 28 '17 at 19:14
• The signals registered by your cone cells in your retina are individual photons of a given energy, not some sort of wave. The statistics of the signal recall the wavefunction waviness, similarly to electron signals on a screen. Your cone cells are not electrometers. – Cosmas Zachos Jun 28 '17 at 22:33
• @CosmasZachos The difference between a photon of energy x and a photon of energy y IS the wavelength by E=hc/λ. – JPattarini Jun 29 '17 at 13:05
• Of course it is. Same for momenta of electrons. My point, however, is that the conceit that our eyes/experience, etc... detects waves directly for light is unsound. Classical optics is merely a practical limit of quantum optics, and should not be misconstrued as something more fundamental. As you indicate, wavelength talk is shorthand for the more fundamental energy (momentum) entities. – Cosmas Zachos Jun 29 '17 at 13:44

That's a complete misunderstanding of the statement of how matter waves work, and in particular it is a confusion regarding the role of wavelength and amplitude in waves. If you take a standard sinusoidal wave with waveform $$f(x,t) = A \sin(kx-\omega t), \tag 1$$ then the wavelength is $\lambda=2\pi/k$, and the amplitude is $A$.

• For visible light, as detected by your eyes, the colour that you will perceive is determined by the frequency $\omega$, which is tied to the wavelength via $\omega=2\pi c/\lambda$. The amplitude $A$ determines, through its square $|A|^2$, the intensity of the light.
• For matter waves, the probability aspect is described via the squared amplitude, $|A|^2$, which describes the peak probability density for finding the particle in that region. The wavelength, on the other hand, acts separately, through the de Broglie relation $p=h/\lambda=\hbar k$, and it is directly tied with the momentum of the particle. And, as far as the wave characteristics of matter go, such as its ability to diffract of of apertures, the de Broglie wavelength acts exactly in the same role that the wavelength of diffracting light does.

Thus, the role of the wavelength in the wave dynamics remains completely unchanged. The only change in going from light to matter waves is that the quantity that is actually doing the oscillations (along $f$ and along $A$) changes to a probability amplitude with all the conceptual problems that brings with it, but those have been treated to death elsewhere and there's no point in rehashing them here.

• "The only change in going from light to matter waves is what is actually doing the oscillations changes to a probability amplitude" Photons are quantum objects just as much as massive particles, so there's no theoretical justification for why photon wavelengths should be comprised of oscillations in anything fundamentally different than an electron's, or anything else for that matter. That's the point. – JPattarini Jun 28 '17 at 19:08
• You can't treat a photon wavelength as something real and an electron wavelength as something happening only in probability. You can, but it's not based in any theoretical backing I'm aware of. The intensity = likelihood of detection relationship is well understood, and amplitude is well treated (and understood) as relating the square of the wavefunction to probability for both massless and massive particles. There's no confusion there. – JPattarini Jun 28 '17 at 19:10
• @JamesPattarini You can indeed, and you have heard of the theoretical backing ─ it's called quantum mechanics. If you want to understand the nuts and bolts of why the wavelength of light (but not photons! their wavefunction is rather more complicated) happens over the electric field, while the wavelength of a massive particle describes oscillations in its probability amplitude, there's no substitute for a full rigorous quantum mechanical treatment. Take it on faith, or learn the maths. – Emilio Pisanty Jun 28 '17 at 19:14
• Again, I'm not saying that an ensemble of a few billion photons aren't best described by Maxwell's equations for an EM field, I'm saying that in the quantum mechanical description of an individual photon, the λ in λ=h/p cannot be treated as somehow being a real oscillation in an electric field while the λ in λ=h/p for an electron is treated as an oscillation in probability only. Either they're both probabilities, or neither are. – JPattarini Jun 29 '17 at 13:23
• @JamesPattarini As noted in the linked question, the quantum-mechanical description of a single photon does not even involve a wavefunction for the photon, so you're mixing apples and oranges. (To be precise: the quantum state of a photon plays out in the counting statistics of the energy content of a given mode, which is apples compared to the oranges of a massive particle's wavefunction.) Your mangled version of QM is indeed problematic, but that's because you mangled it. – Emilio Pisanty Jun 29 '17 at 13:49

Color is a property of light, classical electromagnetic wave, as perceived by the receptors in our eye. So the perceived color is not always the color of the spectral frequency. So let us narrow it down to the colors of the spectrum that have one to one correspondence with the frequency.

A photon is an elementary particle,the quantum of electromagnetic radiation, and has only energy , momentum and spin. The formula E=h*nu assigns a frequency to a photon which is the frequency that the classical electromagnetic wave will have when composed by an enormous number of superposed photons of this energy.

With this in mind:

Why is the wavelength of a photon discussed and treated as a physically real component of the world, while the de Broglie wavelengths of other (massive) particles are dismissed as "waves of probability."

the phrase "wavelength of a photon" refers to its energy in the one to one correspondence of E=hc/λ . There is no electric and magnetic field associated with a real photon. There exists a wave function of a photon, which is the solution of a quantized maxwell equation, which carries information of the electric and magnetic fields , and its complex conjugate square will give the probability of finding the photon at (x,y,z,t).

In this sense, the photon also has a debroglie wavelength which will describe its probability locus, given by λ=h/p, the same as for the electron or another particle.

It just happens that the ensemble of superposed zero mass photons builds up a classical electromagnetic wave that has a one to one correspondence with the optical spectrum, but an ensemble of ( for example) protons make a beam of protons at the LHC. The superposition is additive for massive particles.

So both an electron and a photon have a probability associated with them, it so happens that the photon supeposed ensembles have a macroscopic collective behavior that is wavelength dependent and is perceived as color.

• I'll try to frame my difficulty a little differently: We talk of the physically real energy of photons described by E=hc/λ . If E is lower because λ is larger, my eye will tell me it's a "more red" photon. It's real. Why λ=h/p should be treated as a probability, while E=hc/λ is treated as the energy of a physically real object, is maybe the best way to get at my conceptual issue here. Because it seems for photons we treat them as interchangeable and rarely speak of probability, but for massive particles probability reigns. – JPattarini Jun 29 '17 at 13:15
• Both photons and electrons (and the other particles) have a debroglie wavelenghth and a probability locus in spacetime ( a wavefunctions complex conjugate squared). Because classical electromagnetism as a wave theory is so successful people for most electromagnetic problems use classical electromagnetism, because it is accurate if not of dimensions comensurate to h_bar. For partilces such a macroscopic setup does not exist. Macroscopically they are treated with classical mechanics and any waves in classical mechanics are a meta-level on the classical fluid motions. – anna v Jun 29 '17 at 13:33
• The basic difference is that the Hamiktonian of classical mechanics is not quantized in the way that the electromagnetic one is quantized. The same differential equations for classical electromagnetism are use as quantizetion operators and thus the solutions blend into each other, what for classical electromagnetism are sinusoidal electric and magnetic field expressions giving waves, for the quantized form the same solutions give wavefunctions, probability distributions . Superposition of wave functions which contain the E and B fields in their complex form, add up to light. – anna v Jun 29 '17 at 13:37
• Your eye does not really see an individual photon but the interaction of the photon with a molecule which will send a signal of the energy it has that we identify as color. That is why there is not a one to one correspondence of color with frequency as far as the eye is concerned. There is a biological transformation. See wiki link above. It just happens that the spectrum colors raise in the eye the red or blue sense. In the spectrum it is just the energy. – anna v Jun 29 '17 at 13:41
• I'm with you so far (I think), but it seems like you're saying the smooth, 1:1 transition from photon probability locus to classical E/B waves in EM is treated as somewhat of a happy coincidence/special case then, since we don't treat the de broglie wavelengths of massive particles as something real despite their ability to predict diffraction, etc. – JPattarini Jun 29 '17 at 13:47