My question is really a request for an intuitive explanation as to why only harmonic frequencies of photons allowed between two conduction plates. Why do the plates have to be conductive? And can real photons with unmatching frequencies exist between the two plates in the casimir experiment?
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$\begingroup$ 1. try strumming a guitar. 2 lightly touch the strings half way down with one finger, while strumming. 3. etc. $\endgroup$– Andrew SteaneSep 15, 2022 at 10:00
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$\begingroup$ just for interested readers , a good review of the effect and experiments arxiv.org/abs/quant-ph/0609145 $\endgroup$– anna vSep 15, 2022 at 11:41
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$\begingroup$ @AndrewSteane interesting though, a guitar string (not being touched) has the appearance of a half wave between its two anchors. There's obviously a lot more going on with a vibrating string, but presumably a photon can't have half the wave-length of the plate distance. $\endgroup$– rghomeSep 15, 2022 at 13:37
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1 Answer
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Plates do not have to be conducting. Various dielectric materials can be used, and the Casimir force on the plates will depend on the characteristics of those materials such as dielectric constants.
But perfectly conducting plates are the simplest toy model, because perfect conductor means electric field is zero in and at the plates.
For two infinitely long perfectly conducting plates separated by distance $a$, relevant frequencies used in the Fourier expansion have to obey the relation (due to Maxwell equations and the mentioned boundary condition)
$$ \frac{\omega_n^2}{c^2} = \frac{n^2\pi^2}{a^2} + k_x^2 + k_y^2. $$
So we can see allowed frequencies are harmonics only in 1D world, where there is no $k_x,k_y$. In 3D space, any frequency is allowed, due to the fact $k_x,k_y$ can have any real value.
answered Sep 15, 2022 at 14:52
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$\begingroup$ Ok, so if i had a hollow ball of dielectric metirial, the point in the middle would be limited to harmonics in every direction in 3d space? $\endgroup$– NazuidSep 16, 2022 at 8:30
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1$\begingroup$ I am not sure, but I don't think eigenfrequencies have to be limited to harmonics. Harmonics are multiples of some basic frequency. But eigenfrequencies depend on the shape of the cavity, and need not be multiples of anything. It's like in QT; eigenvalues of Hamiltonian are multiples of some basic value (and thus form an arithmetic sequence) only in special case, such as 1D harmonic oscillator. In general, they are not. $\endgroup$ Sep 16, 2022 at 14:25
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$\begingroup$ I see. Then what about photons that are not allined with the normal of the plates? Would they still be limited to harmonics depending on the angle of the approch? $\endgroup$– NazuidSep 16, 2022 at 15:12
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$\begingroup$ Not harmonics, but frequencies $\omega_n$ obeying the relation above. $\endgroup$ Sep 16, 2022 at 16:19