When detecting radio waves in space, we use very large telescopes or arrays of telescopes. But according to QM, aren't photons point particles when measured? Does a photon with a large wavelength have a wider probability distribution with regard to where it can be expected to be detected? Is that why photons with long wavelengths require large telescopes to be detected relative to shorter wavelength photons?
Quantum Mechanics – Is the Size of a Photon Dependent on Its Wavelength?
astronomyphotonsquantum mechanicswave-particle-duality
Related Solutions
The wavelength of light, and for any wave in general, is measured along the direction of propagation. It has every bit of the intuitive meaning that the wavelength of a water surface wave does.
One of the most meaningful ways to visualize light is as an oscillation of the electric and magnetic fields over space and time:
The electric field $\vec E$, in particular, is orthogonal to the direction of propagation, along $\vec k$. This explains why antennas often have metal rods orthogonal to the direction they point in: the electric field accelerates charges along those rods and creates a voltage between their tips, which can be measured. This is not unusual for waves either: water surface waves are also transverse waves, and wave power stations also use this transverse motion.
The double slit experiment is also consistent with this fact. For one, it is perfectly easy to observe in water waves, as is done, for example, in this Veritasium video:
The reason that the wavelength gets to play a role in a direction that you perceive to be orthogonal to the propagation direction is that the double-slit interference depends on having point sources which radiate waves in all directions, so that each wave will also have a different phase at different points along the screen; the differences in these phases then make up the interference pattern.
The photon is an elementary particle in the standard model of particle physics. It does not have a wavelength. It is characterized in the table as a point particle with mass zero and spin one. Its energy is given by $E=h\nu$, where $\nu$ is the frequency of the classical electromagnetic wave which can be built up by photons of the same energy.
This is where the confusion comes. The wavelength and frequency characterize the emergent electromagnetic wave from very many photons. How the classical wave emerges can be seen here although it needs a quantum field theory background to understand it. The photon, as a quantum mechanical entity, has a quantum mechanical wavefunction. This wavefunction complex conjugate squared gives the probability density for the specific photon to be at $(x,y,z,t)$. The frequency in the wavefunction is the frequency of the possible emergent classical wave, but for the individual photon it is only connected with probability of manifestation, as for example in the single photon double slit experiments.
single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames
you ask:
"How can a photon have a wavelength of a few kilometres and yet still be thought of as a particle?
It does not. It takes zillions of photons to build up the classical electromagnetic wave. In the photos above each individual photon gives a little dot. The build up gives the probability density distribution for photons, and lo, there is a frequency associated with the interference pattern, even though the photon manifests individually as a dot at the $(x,y)$ of the screen.
That is why we need quantum mechanics.
Edit after this question became the main duplicate of another one, where I have a long answer/comment that might be of interest to readers.
Best Answer
We can describe light as either a wave or a particle, and we normally choose whichever is the best description of the situation we are modelling. If you're trying to understand diffraction then a wave description is simplest to work with, but if you're dead set on using photons then it can be done.
The radio waves coming from space are delocalised in the sense that they have no precise position. If you insist on modelling these as photons you would have to treat the wave as a superposition of all possible photon positions. When we detect the photon with our radio dish we have pinned the photon position down to somewhere in our dish. We know the photon interacted with our dish somewhere, but we don't know precisely where. If our dish has a diameter $d$ let's say the uncertainty in the photon position, $\Delta x$ is $d$. Then we appeal to the uncertainty principle to get $\Delta p = h/d$.
Because there is an uncertainty in the momentum there is an uncertainty in the angle from which the photon came. To see this look at the following diagram:
The uncertainty in momentum produces an uncertainty in the angle of $\alpha$ given by:
$$ \tan\alpha = \frac{\Delta p}{p} $$
If we substitute our expression above for $\Delta p$ and use the expression for the momentum of a photon, $p = h/\lambda$, we get:
$$ \tan\alpha = \frac{\lambda}{d} \tag{1} $$
And that's why if you want good resolution you need a large dish, and why large wavelengths give poorer resolution than short wavelengths.
That equation (1) ought to ring a bell, because it's very similar to the classical expression for the angle of the first minimum in the Airy disk diffraction pattern:
$$ \sin\theta \approx 1.22 \frac{\lambda}{d} $$
By choosing a particle description, invoking the uncertainty principle and waving our arms around a bit we have ended up at essentially the same result as doing a classical analysis using a wave.