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From this question :Why is the light reflected at the same angle from mirror?

and this part of the answer: The starting point it that a single silver atom is far smaller than the wavelength of light, so any scattering from it will be isotropic i.e. it will scatter the light equally in all directions.

In a case where only a single photon is fired at the mirror surface and is scattered by the silver atom, would it be scattered isotropically?

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Nope, you can't split a photon but its direction will be random with an isotropic distribution. As you get more and more photons they will trace the expected distribution on your detector.

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  • $\begingroup$ any specific laws predicting those properties? $\endgroup$
    – Loogai
    Commented Dec 14, 2015 at 0:39
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    $\begingroup$ The lazy way is to assume the field is classical, and use the Raleigh approximation to get the differential cross section. A normalized version of the differential cross section will give you the probability of finding the photon at that particular point. $\endgroup$
    – Mikhail
    Commented Dec 14, 2015 at 0:45
  • $\begingroup$ "You can't split a photon" is a bit misleading (at best and wrong at worst). The single photon scatters into the same field distribution as a classical field and so is delocalized in space. If you have a detector at some point in space, you have some probability of localizing/collapsing the state at the detector (with probability proportional to the classical intensity of the scattered field by Born's rule). If you want to describe this as a photon being emitted into a random direction, you need to explain that it is a coherent superposition of directions (not a classic/incoherent "random"). $\endgroup$ Commented Dec 14, 2015 at 4:31
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When talking about photons one is in the quantum mechanical framework. A photon is an elementary particle in the standard model of particle physics.

Let us keep it simple.

In first quantization, one has a photon impinging on an atom . This atom even if neutral has a spill over electric field and one can calculate a quantum mechanical probability for the scattering of the photon from the field. This probability distribution may or may not be isotropic , depending on the spin orientation of the photon and the silver atom.

So the angle of scatter will add up to a distribution of many photons scattering in the same boundary conditions with a silver atom. The angle of the individual photon will seem random.

In second quantization one uses simple solutions of the quantum mechanical equations (dirac or klein gordon or a quantized version of maxwell's equations) as ground states on which creation and annihilation operators act to define the scatter of the photon (which is the answer by garyp). The Feynman diagrams which give the integrals for the scattering cross section are expressed in this system. In this formulation one can show how the classical wave emerges consistently from the confluence of innumerable photons. The mathematics is not simple.

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The scattered wave function fills all of space. It doesn't pick out a preferred direction for the scattering. The electric field mode, and hence any excitation of the field (a quantum) fills all of space. The destruction of the quantum can occur anywhere the wave function is not zero and there is some means to absorb the energy and momentum of the excitation: some kind of detector. That can occur just about anywhere in space, but the spot at which it happens is entirely unpredictable. The field loses one quantum of excitation at a particular place, and transfers a particular amount of energy and momentum to the detector. It behaves for all the world as if a particle collided with the detector at that point, and then vanished.

So each excitation is detected at some particular, unpredictable place. If you repeat this experiment many times, you find that detection events can occur almost everywhere.

In my opinion the picture of a photon as a particle is seriously problematic. You get questions like yours whose answer goes against intuition, and you have to accept the fact that the particle can simply vanish at the detection event. In our intuitive notion of a particle, it doesn't simply disappear. With care, the picture is useful, but safe use of the metaphor requires some understanding of the fuller picture so that the limits of the particle picture are recognized.

Pedantic note: should you be able to launch a single "photon" toward a silver atom, the scattering would not be isotropic. The scattering distribution depends on the polarization. (That's why I haven't used the word isotropic.) But that point has no bearing on the question.

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Atoms emit photons isotropically in random directions. If only one photon is emitted it will be in a random direction. I believe that is what experiment shows.

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  • $\begingroup$ Which experiment? $\endgroup$
    – Loogai
    Commented Dec 20, 2015 at 11:23
  • $\begingroup$ Any experiment will show it to be isotropic. As John was saying in his answer the photons are emitted isotropically but if The atoms are packed close together as in a mirror surface then they will effect each other but there still being emitted from the electrons isotropically. If you see something in a mirror you can move around to any place so that you can see that same object at every other point on that Mirrors surface because photons are being emitted from that object to every point on that mirror. $\endgroup$ Commented Dec 20, 2015 at 17:35

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