Why doesn't a photon state have to be infinite in length?

Question

In all discussions I've seen so far (old quantum theory, semiclassical QM, QFT), when we talk about photon states, we seem to say they have a definite momentum. At the same time, we also say a photon is a particle that is localized in space. These two statements seem contradictory to me, given what I know so far.

In de Broglie's proposition, a photon is supposed to have energy $E = \hbar\omega$ and momentum $p = \hbar k$ ($k = 2\pi/\lambda$). Now in this semi-classical / old quantum theory treatment, the photon has a precise momentum, so doesn't that mean by Heisenberg's uncertainty principle that $\Delta x = \infty$? Even classically, a light wave of definite wavelength would have to be an infinite planewave.

In the second quantization formalism, we can think of a photon as a state of the form $\hat{a}_{\vec{k},\vec{\epsilon}}^{\dagger} | 0\rangle$ (where $\vec{k}$ is a definite wavevector and $\vec{\epsilon}$ is a definite polarization). If I'm not mistaken, the electric and magnetic fields would have nonzero amplitude distribution at every point in the entire quantization volume. But the quantization volume is arbitrary, so it seems there is no sense in which the photon is located in any specific point.

So why do we say photons are localized? It seems like they have to be infinite in size, but that doesn't make any sense.

Answer 1

Actually, for both electrons and photons, we are more secure in the understanding of their momentum eigenstates, that are infinite plane waves that cannot be normalised, than their position eigenstates. The nice property that momentum eigenstates hold, is that they are the free particle's Hamiltonian eigenstates.

(Sidenote: And in the mathematical framework of QFT, the S matrix really needs to be used between Hamiltonian eigenstates, because if you do a small mixture of different energy eigenvalues in any of the initial or final states, the accumulation of infinite phase factors will destroy the calculation's sanity. Just work with Hamiltonian eigenstates first, and the use the postulates of quantum theory to figure out what superpositions of different energy eigenstates are supposed to do.)

Every good textbook will tell you that the real thing has thus to be gotten by wavepackets. Only by superposing many different momentum eigenstates can you hope to normalise the wavefunction, and so forth.

Then you should not have as much difficulty with localising the photon too.

We need to be able to localise the photon because when we detect the photon with a detector, the detector is localised, and so we must be able to say that the photon is at the detector. This is just the most convenient and sensible and natural way to think about it.The photon must also be able to pass through slits, which are localised.

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Note that both momentum eigenstates and position eigenstates are not normalisable. That is ok. The correct statements of the postulates of quantum theory are that the quantum state lives in Hilbert space, but the eigenstates of the continuous part of observable operators live in rigged Hilbert space. Those non-normalisable eigenstates cannot be realised, but they can be safely used in superpositions to expand any realisable quantum state.

Answer 2

It's very simple: we say that photons are localized because a real physical photon detector registers the arrival of a photon at a particular place and time. That's fundamental: the math is stories about this physics.