Number of photon in the electromagnetic wave

In summary, the energy of an electromagnetic wave in a unit volume is given by \varepsilon_0 E^2. This does not necessarily mean that the number of photons is \varepsilon_0 E^2/\hbar\omega, as this is only an average calculation for a coherent state. The concept of a single-photon state is well-defined in relativistic quantum field theory and is not easily prepared. It is important to note that creating one-photon states with a dimmed laser is incorrect, as lasers produce classical em. waves and not single-photon states. The historical approach to introducing photons as "light particles" is outdated, but the concept remains valid in quantum theory.
  • #1
jackychenp
28
0
From EM, the energy of electromagnetic wave in unit volume is [itex] \varepsilon_0 E^2[/itex]. Does that mean the number of photon is [itex]\varepsilon_0 E^2/\hbar\omega [/itex]( [itex]\omega [/itex] is frequency of wave)? In 1-D, [itex]E=E_0cos(\omega t+kx)[/itex], then the number of photon in average is [itex]\frac{1}{(2\pi/\omega)}\int_{0}^{2\pi /\omega}\varepsilon_0 E_0^2cos^2(\omega t+kx)/\hbar\omega dt = \varepsilon_0 E_0^2/\hbar\omega[/itex]. This result looks very weird. Please correct me if something is wrong!
 
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  • #2
Photons are mostly introduced in the most unfortunate way to beginners in quantum physics. The reason is that usually people prefer a historical approach to introduce quantum theory, and the photon has been the first quantum phenomenon ever since Planck discovered the quantum behavior of nature through the quantum nature of thermal electromagnetic radiation. This lead Einstein to his famous 1905 article on a "heuristic point of view" of "light particles", nowadays called photons. One can not emphasize enough that this is an outdated picture of affaires, although it has been a genius invention at the time.

Nowadays a one-photon state is a well-defined concept in relativistic quantum field theory. It's given by

[tex]|\psi \rangle=\int \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} \hat{a}^{\dagger}(\vec{p},\lambda) A(\vec{p}) |\Omega \rangle.[/tex]

Here, [itex]\hat{a}^{\dagger}(\vec{p},\lambda)[/itex] denotes the creation operator for a photon of momentum, [itex]\vec{p}[/itex] and a polarization indicated by [itex]\lambda[/itex] (e.g., the helicity, i.e., left or right circular polarized states). Further [itex]A[/itex] is some square integrable function, and [itex]|\Omega \rangle[/itex] is the vacuum state.

Such a single-photon state is not so easy to prepare, although nowadays it's common in quantum-optics labs: One uses some crystal to create an entangled pair of photons and measuring one of them to make sure that one is left with a pure one-photon state.

Often, it's stated, one creates one-photon states by a dimmed laser. This is wrong! Lasers produce something close to a classical em. wave with a small spectral width, i .e., with high spatial and temporatl coherence but never single-photon states (nor any state of determined photon number, or Fock state). Quantum mechanically such a thing is described by a coherent state, which is the superposition of many states of different photon number. For a dimmed laser, where the intensity is such that the energy divided by [itex]\hbar \omega[/itex] is small, e.g., close to 1, the dominating state is in fact the vacuum state superposed by the one-photon state. The photon number in such a state is Poisson distributed with an expectation value (and thus also variance!) of 1. So, what you calculate with [itex]\mathcal{E}/(\hbar \omega)=\epsilon_0 \int \mathrm{d}^3 x E^2/(\hbar \omega)[/itex] is an average photon number for the given (classical approximation of a) coherent state. This is the only physical sense you can make out of this number.

Of course, a plane wave can never be realized, since this implies and infinite energy. Plane waves are modes, used to provide a Fourier representation of the electromagnetic field, it's not somthing physical!
 
  • #3
jackychenp said:
From EM, the energy of electromagnetic wave in unit volume is [itex] \varepsilon_0 E^2[/itex]. Does that mean the number of photon is [itex]\varepsilon_0 E^2/\hbar\omega [/itex]( [itex]\omega [/itex] is frequency of wave)? In 1-D, [itex]E=E_0cos(\omega t+kx)[/itex], then the number of photon in average is [itex]\frac{1}{(2\pi/\omega)}\int_{0}^{2\pi /\omega}\varepsilon_0 E_0^2cos^2(\omega t+kx)/\hbar\omega dt = \varepsilon_0 E_0^2/\hbar\omega[/itex]. This result looks very weird. Please correct me if something is wrong!
Your equations are correct. The time average of E^2 just gives a factor of 1/2.
 
  • #4
jackychenp said:
This result looks very weird.

Why?
 
  • #5
vanhees71 said:
Often, it's stated, one creates one-photon states by a dimmed laser. This is wrong! Lasers produce something close to a classical em. wave with a small spectral width, i .e., with high spatial and temporatl coherence but never single-photon states (nor any state of determined photon number, or Fock state). Quantum mechanically such a thing is described by a coherent state, which is the superposition of many states of different photon number. For a dimmed laser, where the intensity is such that the energy divided by [itex]\hbar \omega[/itex] is small, e.g., close to 1, the dominating state is in fact the vacuum state superposed by the one-photon state. The photon number in such a state is Poisson distributed with an expectation value (and thus also variance!) of 1. So, what you calculate with [itex]\mathcal{E}/(\hbar \omega)=\epsilon_0 \int \mathrm{d}^3 x E^2/(\hbar \omega)[/itex] is an average photon number for the given (classical approximation of a) coherent state. This is the only physical sense you can make out of this number.
I just wanted to "second" this excellent description of a coherent state.
 
  • #6
Vanhees, thanks for your fabulous comments.
clem, I forgot a factor 1/2. Thanks for pointing it out.

vanhees71 said:
Photons are mostly introduced in the most unfortunate way to beginners in quantum physics. The reason is that usually people prefer a historical approach to introduce quantum theory, and the photon has been the first quantum phenomenon ever since Planck discovered the quantum behavior of nature through the quantum nature of thermal electromagnetic radiation. This lead Einstein to his famous 1905 article on a "heuristic point of view" of "light particles", nowadays called photons. One can not emphasize enough that this is an outdated picture of affaires, although it has been a genius invention at the time.

Nowadays a one-photon state is a well-defined concept in relativistic quantum field theory. It's given by

[tex]|\psi \rangle=\int \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 \omega_{\vec{p}}}} \hat{a}^{\dagger}(\vec{p},\lambda) A(\vec{p}) |\Omega \rangle.[/tex]

Here, [itex]\hat{a}^{\dagger}(\vec{p},\lambda)[/itex] denotes the creation operator for a photon of momentum, [itex]\vec{p}[/itex] and a polarization indicated by [itex]\lambda[/itex] (e.g., the helicity, i.e., left or right circular polarized states). Further [itex]A[/itex] is some square integrable function, and [itex]|\Omega \rangle[/itex] is the vacuum state.

Such a single-photon state is not so easy to prepare, although nowadays it's common in quantum-optics labs: One uses some crystal to create an entangled pair of photons and measuring one of them to make sure that one is left with a pure one-photon state.

Often, it's stated, one creates one-photon states by a dimmed laser. This is wrong! Lasers produce something close to a classical em. wave with a small spectral width, i .e., with high spatial and temporatl coherence but never single-photon states (nor any state of determined photon number, or Fock state). Quantum mechanically such a thing is described by a coherent state, which is the superposition of many states of different photon number. For a dimmed laser, where the intensity is such that the energy divided by [itex]\hbar \omega[/itex] is small, e.g., close to 1, the dominating state is in fact the vacuum state superposed by the one-photon state. The photon number in such a state is Poisson distributed with an expectation value (and thus also variance!) of 1. So, what you calculate with [itex]\mathcal{E}/(\hbar \omega)=\epsilon_0 \int \mathrm{d}^3 x E^2/(\hbar \omega)[/itex] is an average photon number for the given (classical approximation of a) coherent state. This is the only physical sense you can make out of this number.

Of course, a plane wave can never be realized, since this implies and infinite energy. Plane waves are modes, used to provide a Fourier representation of the electromagnetic field, it's not somthing physical!

clem said:
Your equations are correct. The time average of E^2 just gives a factor of 1/2.

jtbell said:
Why?
 

1. What is the relationship between the number of photons and the electromagnetic wave?

The number of photons in an electromagnetic wave is directly proportional to the energy of the wave. This means that as the number of photons increases, the energy of the wave also increases.

2. How is the number of photons in an electromagnetic wave measured?

The number of photons in an electromagnetic wave is typically measured using a device called a photodetector. This device can detect and count individual photons, allowing for an accurate measurement of the number of photons present in a wave.

3. Can the number of photons in an electromagnetic wave be changed?

Yes, the number of photons in an electromagnetic wave can be changed by altering the energy of the wave. This can be done by adjusting the frequency or amplitude of the wave.

4. Is there a minimum or maximum number of photons in an electromagnetic wave?

No, there is no minimum or maximum number of photons in an electromagnetic wave. The number of photons present in a wave can vary greatly, depending on the energy of the wave.

5. How does the number of photons in an electromagnetic wave affect its properties?

The number of photons in an electromagnetic wave is directly related to its properties, such as energy and intensity. As the number of photons increases, the energy and intensity of the wave also increase, leading to a more powerful and brighter wave.

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