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Jan 26, 2023 at 16:48 comment added DanielSank @RyderRude That may be true, but the original post also reveals a misunderstanding, the resolution of which I think will lead its author to better understanding. This happens a lot here.
Jan 26, 2023 at 7:20 comment added Ryder Rude But OP is asking what mathematically causes the electromagnetic field to be quantised. Your answer is assuming the particle behavior of the field, so this answer just assumes what the OP asked to be justified.
Jan 25, 2023 at 20:10 comment added DanielSank This is much easier to understand in second quatization language, for whatever that's worth.
Jan 25, 2023 at 20:09 history edited DanielSank CC BY-SA 4.0
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Jan 25, 2023 at 20:07 comment added DanielSank @RyderRude you are correct: what I wrote about $\hat a$ being discrete is imprecise. However, the states in e.g. the particle-in-a-well problem being discrete is just like classical normal modes being discrete. And as soon as we remember that we can have more than one excitation (i.e. particle) there's a clear correspondence between $\hat a$ and the classical Fourier amplitudes.
Jan 24, 2023 at 5:50 comment added Ryder Rude I don't think this answer is correct because : 1. Quantum discretization is a consequence of the boundary conditions of the TISE, like OP says. Discretization happens in the case of bound states because we want the wavefunction to exponentially fall at infinity (not necessarily the spatial infinity. It can be an infinity in an abstract space like that of field strength) . 2. The amplitudes of waves, $a_k$ and $b_k$, do not become discrete integers in Quantum Field Theory. They become operators. It's the eigenvalues of $a^{\dagger}a$ and $b^{\dagger}b$ that become discrete.
Jan 23, 2023 at 4:47 vote accept Mikayla Eckel Cifrese
Jan 23, 2023 at 4:47
Jan 23, 2023 at 4:36 history answered DanielSank CC BY-SA 4.0