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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?

3 Answers3

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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 $\delta p \cdot \delta x \ge \hbar$.

ohneVal
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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).

flippiefanus
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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.

WillO
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