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thaiqi
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- Are photon and wave mode description equivalent?
I heard there is a saying that photon and light in a certain wave mode are equivalent. Is it so ?
For example,Nugatory said:Where did you hear that? There’s no sensible answer without more context.
See the above. Any suggestions? Thanks.Nugatory said:Where did you hear that? There’s no sensible answer without more context.
thaiqi said:See the above. Any suggestions? Thanks.
And exactly how is a link that is in Chinese going to help us reply to you? Is that the only link you can find? If so, what do you think that might mean?thaiqi said:For example,
https://zhuanlan.zhihu.com/p/37072280
It is in Chinese.
The main idea is that electromagnetic wave forms stationary waves in an empty cavity. The wave mode is some way equivalent to the photon.
In Loudon's The Quantum Theory of Light 3rd ed. section 1.1 and 1.2(page 9), it said: "
A photon is said to have been created(destroyed) when the electromagnetic energy in the mode is increased(decreased) by a single quantum.
"
berkeman said:And exactly how is a link that is in Chinese going to help us reply to you? Is that the only link you can find? If so, what do you think that might mean?
Thanks for your reply.vanhees71 said:That's a bit vague but essentially correct. I'd reformulate it somewhat:
First you show that the free electromagnetic field operator in the Heisenberg picture can be written in terms of annihilation and creation operators referring to a complete set of single-photon energy eigenstates (the most simple ones are momentum-helicity eigenstates) . The annihilation and creation operators fulfill the canonical commutation relations, describing bosonic fields, as it must be according to the spin-statistics theorem and appropriate for photons.
The annihilation and creation operators and thus the field operators themselves act in Fock space. A complete basis of this Fock space are the eigenstates of the occupation numbers of the modes ##\hat{N}_{\lambda}(\vec{p})## with ##\hat{N}_{\lambda}(\vec{p}) = \hat{a}_{\lambda}^{\dagger}(\vec{p}) \hat{a}_{\lambda}(\vec{p})##. It's also convenient to first use a finite (large) volume with periodic boundary conditions for the fields. Then the single-photon momenta a discretized (for a cube of length ##L##, ##\vec{p} \in \frac{2 \pi \hbar}{L} \mathbb{Z}^{3}##). A photon is a Fock state with a total photon number. In the finite-volume case you can have a proper (i.e., normalizable) single-mode single-photon state with ##N_{\lambda}(\vec{p})=1## for exactly one momentum and one helicity ##(\vec{p},\lambda)##. In the infinite-volume limit you need to construct a wave packet with a finite width in momentum in order to get a proper normalizable single-photon state.
The ground state of the system is the vacuum state ##|\Omega \rangle##, where all occupation numbers of photons are 0. It's characterized uniquely by ##\hat{a}_{\lambda}(\vec{p}) |\Omega \rangle=0##. The single-photon state is then given by ##|\vec{p},\lambda \rangle=\hat{a}_{\lambda}^{\dagger}(\vec{p}) |\Omega \rangle##.
The main difference between photon and wave mode is the way in which energy is propagated. In photon mode, energy is carried by individual particles called photons, while in wave mode, energy is carried by a continuous wave.
No, photon and wave mode are not equivalent. While they both describe the movement of energy, they do so in different ways. In certain situations, such as in the quantum world, photon mode is a more accurate description of energy propagation.
Yes, energy can exist in both photon and wave mode simultaneously. This is known as wave-particle duality and is a fundamental concept in quantum mechanics.
Photon and wave mode are both ways of describing electromagnetic radiation, which includes all types of light. The electromagnetic spectrum is a range of frequencies of electromagnetic radiation, with photon mode being associated with higher frequencies and wave mode with lower frequencies.
Photon mode is more useful in situations involving very small particles, such as atoms and subatomic particles, where the discrete nature of energy is more apparent. Wave mode is more useful in situations involving larger objects, such as sound waves or ocean waves, where the continuous nature of energy is more evident.