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?


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:

Radio wave

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.


The photon's field of influence in this case really is as big as the antenna! If you think of gathering photons one by one from distant stars, then the quantum state of each photon propagates through space and time precisely following Maxwell's equations. I talk about this idea in great detail in my answers here, here and here.

So, for one photon at a time reception, or many unentangled photons at a time, classical optics / electromagnetics is indeed exact for the design of the antenna systems. If you image with a smaller aperture, a photon's field of influence is going to lie more and more outside the antenna and the probability that it will be absorbed by matter outside the antenna increases relative to the probability that it will be registered by the antenna. You calculate the probability density from Maxwell's equations as a normalised intensity at all times: this gives the probability density that the photon will be absorbed at a single point (i.e. a lone absorber atom) at different places. Your design goal is to maximise the probability that this absorber will belong your detection hardware.


The electromagnetic fields of a photon are confined in the cylinder with length of half the wavelength, whose radius is proportional to square root of the wavelength. Here, it has the definite frequency and is different from the well-known wavepacket which consist of the plane waves of different frequencies, as described in the paper: Shan-Liang Liu. Electromagnetic fields, size, and copy of a single photon. ArXiv: 2016, 1604, 03869.

  • $\begingroup$ What you you mean bu the radious of a photon? Are you thinking in terms of wave packages here? Could you explain more withour refering to your own newly published ariv paper. $\endgroup$ – Mikael Fremling Oct 22 '16 at 11:22

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