Photon & Wave Mode: Equivalent or Not?

[thaiqi]

TL;DR

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 ?

 

[Nugatory]

Where did you hear that? There’s no sensible answer without more context.

 

[thaiqi]

Where did you hear that? There’s no sensible answer without more context.

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.
"

 

[thaiqi]

Where did you hear that? There’s no sensible answer without more context.

See the above. Any suggestions? Thanks.

 

[berkeman]

See the above. Any suggestions? Thanks.

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.
"

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]

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?

Well, let's ignore the link in Chinese for now.
How about that in Loudon's book?
In Loudon's The Quantum Theory of Light 3rd ed. section 1.1 and 1.2(page 9), it said: "
The essence of the quantum theory of the radiation field is thus the association of a quantum harmonic oscillator with each mode of the field. ...
A photon is said to have been created(destroyed) when the electromagnetic energy in the mode is increased(decreased) by a single quantum.
"

 

[vanhees71]

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 $|\vec{p},\lambda \rangle=\hat{a}_{\lambda}^{\dagger}(\vec{p}) |\Omega \rangle$.

 

[thaiqi]

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 $|\vec{p},\lambda \rangle=\hat{a}_{\lambda}^{\dagger}(\vec{p}) |\Omega \rangle$.

Thanks for your reply.

Now that the two are equivalent, cannot we think the classical model of light can explain photo-electric effect?