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It seems possible to detect a single photon.[1, 2]

But the photon is a free particle. Its momentum is decided precisely and it means that the position of the photon is uncertain. The photon can exist everywhere and the probability of detecting a photon at the finite region would be zero. To detect the photon, we would have to put our detector infinite time.

So my knowledge of a free photon and the possibility of detecting it seems contradictory. How can we detect a single photon?

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  • $\begingroup$ This is a reductio ad absurdum argument. By this logic you should not be able to see anything since your eyes will never detect a photon. $\endgroup$
    – Jon Custer
    Jul 16, 2020 at 14:32
  • $\begingroup$ Why link to photons at the edge of detectability (infrared)? Why not think of gamma radiation or x-rays? $\endgroup$
    – user137289
    Jul 16, 2020 at 14:34
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    $\begingroup$ You are assuming that the wavefunction of the photon is an infinite plane wave, and it's true that this is zero everywhere. But no real photon is an infinite plane wave because every photon got created somewhere and will be absorbed somewhere i.e. all photons are localised to some extent. And for a localised photon $|\psi|^2$ is not zero everywhere. $\endgroup$ Jul 16, 2020 at 15:18
  • $\begingroup$ Nice one John I was writing the same argument simultaneously hehehe $\endgroup$
    – ohneVal
    Jul 16, 2020 at 15:22
  • $\begingroup$ why do you think its momentum is decided precisely ? you could claim th same for other particles $\endgroup$
    – trula
    Jul 16, 2020 at 16:36

3 Answers 3

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As usual one has to be very careful when invoking the uncertainty principle. A free photon means that it is not interacting. So a free photon has a theoretical energy proportional to its momentum, $p$, however we have to measure that $p$ or at least produce such photons by some mechanism. The production mechanism has a given precision $\delta p$, so you could say your local photon manufacturer assures you, his photons come with $p \pm \delta p$.

Once produced he/she sends them to you for detection. Let us imagine you put some detector, some screen of some sort... The impact of the photon on the screen will produce some interaction (e.g. chemical in a film) from which we can read of the position, again up to some precision $\delta x$.

The uncertainty principle just relates $\delta p \cdot \delta x \ge \hbar$.

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  • $\begingroup$ But even if you could read the screen with arbitrarily great precision, the uncertainty principle would still apply. $\endgroup$
    – WillO
    Jul 16, 2020 at 16:55
  • $\begingroup$ Yes that is true, it is a property of nature, so even with idealized measurement devices the relation should hold, in practice the measurement devices are nowhere close to such an idealization however $\endgroup$
    – ohneVal
    Jul 16, 2020 at 18:14
  • $\begingroup$ I am confused by your statement. What you say is that the photon produced is the superposition of photons with similar but not equal momentums for the uncertainty of the momentum. So if we represent the photon state in Fock space, the state would be the multiple photon state. But what they claim is that they detect the single photon. $\endgroup$ Jul 17, 2020 at 3:16
  • $\begingroup$ Yes, but then your issue has more to do with the measuring problem, under the "standard" view of wave-function collapse, it makes sense that this superposition just ends up taking one of the eigenstates. $\endgroup$
    – ohneVal
    Jul 17, 2020 at 6:57
  • $\begingroup$ Where does the uncertainty of momentum come from? If a photon is emitted from an atom by decaying of electron state, the difference between energy levels is well-defined. And the emitted photon should have the energy corresponding to the quantum jump to satisfy the conservation of energy. $\endgroup$ Jul 19, 2020 at 9:25
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A single photon can have any wave function. In general, it would not be just a single plane wave with a fixed momentum. A general wave function can be expressed as a spectrum of plane waves (also called an angular spectrum). The expectation value for the momentum of the photon is then given by a simple calculation involving the spectrum. When you transform the spectrum to the spatial domain you'll get the probability amplitude, from which you can get the detection probability for the photon at a specific location (integrated over the area of the detector).

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  • $\begingroup$ You are right. I was thinking something non-sense again. Thanks!! $\endgroup$ Jul 17, 2020 at 4:51
  • $\begingroup$ @Kevin And of course the "specific location" isn't a mathematical point, it's a small region. We integrate over that region to calculate the probability of detecting the photon in that region. $\endgroup$
    – PM 2Ring
    Jul 17, 2020 at 5:13
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A photon cannot have a well-defined momentum because the momentum operator has no square-integrable eigenstates. So it is never correct to say of a photon that "its momentum is decided precisely", and this is exactly where your argument goes wrong.

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  • $\begingroup$ But in the quantum field theory, the field is quantized with the plane wave. Are these plane waves just an artificial convenient tool that don't exist in reality? $\endgroup$ Jul 17, 2020 at 4:54
  • $\begingroup$ I don't know what "quantized with the plane wave" means. If you're referring to the fact that we sometimes pretend the eigenstates of momentum and position belong to the state space, then yes, it's well understood that this is a convenient and essentially harmless lie. If it really bothers you, you can deal with it through the formalism of rigged vector spaces, but a saner approach might be to accept it just as we accept other convenient and harmeless lies like one-dimensional wires and frictionless planes. $\endgroup$
    – WillO
    Jul 17, 2020 at 5:23

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