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The total energy density $(U)$ of an electromagnetic wave is given by the equation $$U=\frac{1}{2}\epsilon_0{E_0}^2+\frac{1}{2}\frac{{B_0}^2}{\mu_0}\tag{1}$$

Also quantum energy $(E)$ of a photon in the electromagnetic wave is given as $$E=hf\tag{2}$$

This might sound absurd, but can we arrive at Equation $(1)$ by means of any mathematical methods (For example: Integration) performed on Equation $(2)$. Since the first equation, gives the total energy density of the wave and the second gives the energy of the particle (photon) which is a constituent of the wave. I asked whether if we could find out by integration(I know it's not as simple as performing some standard integration) because the energy of an electromagnetic wave could be a summation of all the tiny energy packets of the electromagnetic wave right?

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    $\begingroup$ No, it is not simple . see my answer here physics.stackexchange.com/questions/113574/… unfortunately the link to Motl is not accessible that gives the mathematics. try arxiv.org/abs/1201.5536 $\endgroup$
    – anna v
    Commented Jan 10 at 8:50
  • $\begingroup$ After reviewing your answer, I have questions, not claiming complete comprehension but having gained some understanding. 1. Where do these photons flow in the electromagnetic wave, in the changing Electric field or the Magnetic field? 2. Can we exactly find out the position of each photon in an electromagnetic wave which travels at the speed of light? 3. Do these photons exist in a static Electric field? 4. How do these photons form all of a sudden when there is a changing Electric field (in an Electromagnetic wave), which in turn constitutes the energy for the same wave. $\endgroup$ Commented Jan 10 at 10:16
  • $\begingroup$ See my question here physics.stackexchange.com/questions/455321/… $\endgroup$ Commented Jan 10 at 10:27
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    $\begingroup$ @JeffyJames 1. The photons build up the classical wave. They have velocity c, and different angles between them , which means that lorentz vectors enter in the description.2, No photons are quantum entities, one needs QED. 3 the electric and magnetic fields that describe the classical em wave are mathematically built up by photons in a complicated way. 4. The photons form the wave , the classsical wave is a quantum mechanical confluence of photons. $\endgroup$
    – anna v
    Commented Jan 10 at 10:35
  • $\begingroup$ @JeffyJames Quantum electrodynamis, the quantum field theory of the electromagnetic interactions, answers all your questions. You find this in the relevant text books. $\endgroup$
    – Hyperon
    Commented Jan 10 at 10:36

2 Answers 2

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No, because the equation (2) refers to something much more special, an energy difference of subsequent states of an harmonic oscillator, and one can't recover the concepts of electric and magnetic fields and expressions involving these fields giving energy density just from (2) - this is a general expression describing any harmonic oscillator.

The equation (2) can be actually arrived at by starting with the equation (1), then performing quantization of the classical theory of free EM field (radiation in a box). This quantization is a formal procedure that modifies meaning of symbols in classical theory, and adds the assumption that the new concepts obey commutation relations involving the Planck constant $h$, to get the corresponding quantum theory. This is why $h$ appears in (2) and is not present in (1).

In the resulting quantum theory of radiation, one arrives at (2) as an expression of difference of subsequent eigenvalues of quantum EM field Hamiltonian in the same mode.

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I give a brief answer. More details, in particular on the conjugated variables used for the electromagnetic field see my post The origin of quantization (Note that I use the convention $\hbar=1$ and the angular frequency $\omega = 2\pi f$ instead of simple frequency $f$ in that post).

The formula

$$ U =\frac{1}{2}\left(\epsilon_0 E_0^2 + \frac{B_0^2}{\mu_0^2}\right)\quad\text{respectively}\quad H(\equiv E) = \frac{1}{2}\int dV \left(\epsilon_0 E_0^2 + \frac{B_0^2}{\mu_0^2}\right)$$

is the energy density (respectively energy $E$ or Hamiltonian $H$ ) in classical electrodynamics. In order to get $$E=hf$$ the electromagnetic field has to be quantized.

Actually, not every electromagnetic field can be quantized, in particular static fields cannot be quantized. The field should have some oscillatory behaviour, or in other words be a (superposition of) electromagnetic wave(s).

In order to achieve that one first has to find the conjugated variables the Hamiltonian depends on. Once one has the Hamiltonian written in conjugated variables $\mathbf{Q_k}$ and $\mathbf{P_k}$ we can recognize that the form of the Hamiltonian looks like a summation of Hamiltonians of the harmonic oscillator which distinguish by different wave vectors $\mathbf{k}$. Up to now it is still classical physics.

Now Quantum physics: We can study harmonic oscillators as a quantum system, if we do this --- well documented in every textbook on Quantum Mechanics -- we learn that the energy spectrum of a quantum harmonic oscillator is no longer continous, but can take on only discrete values ($N_\mathbf{k}$ is an integer number which can be 0 (no excitation) or 1 (first excitation), 2, .... 100 ("high" excitation),.... etc.)

$$ E_\mathbf{k} = hf_\mathbf{k} \left(N_\mathbf{k}+\frac{1}{2}\right)$$

$\mathbf{k}$ is the parameter which distinguishes the different harmonic oscillators since we have many of them in the electromagnetic field. Therefore the total energy of quantum electromagnetic field can written as

$$ E = \sum_{\mathbf{k}} h f_\mathbf{k} \left(N_k+\frac{1}{2}\right) \quad\tag{1}$$

The last step is a conceptual one. For study just select a single harmonic oscillator characterized by the wave vector $\mathbf{k}$ from the ensemble of oscillators which represent the electromagnetic field. (you can also choose another oscillator with wave vector $\mathbf{h}$ -- it does not change anything) The energy excitations of a single quantum harmonic oscillator (and this course also valid for the one chosen in the selection) with a fixed frequency $f_\mathbf{k}$ are like a ladder which constant energy steps. One single step in the excitation of the energy of the electromagnetic field is called a photon with the energy (step):

$$ E_k = h f_k $$

Of course one can go all steps backwards from (1). For getting something which looks like the integral over the energy density one would need to define field strength operators of $\mathbf{E}$ and $\mathbf{B}$. They can actually be constructed from $\mathbf{Q_k}$ and $\mathbf{P_k}$, but right now I don't know the exact formula since in modern descriptions of the quantum electromagnetic field they are very rarely used. The discrete harmonic oscillator summation over different wave vectors $\mathbf{k}$ can always be written as a continous integral with corresponding math tools.

At some point one would to deprive $\mathbf{E}$ and $\mathbf{B}$ of the operator status in order to get the classical formula back. But it is a bit weird, because once you are quantum one would stay quantum. But if the quantum numbers $N_\mathbf{k}$ are of high value one can kind of neglect that the concerned operators fulfill (non-zero) commutator relations.

EDIT

This transition from the quantum world back to classical physics is based on the Correspondence Principle which says that in case of high quantum numbers there is no sensible difference between classical physics and quantum physics anymore. In case of high quantum numbers, i.e. in case one deals with a high number of photons one would be able to describe such a situation with classical electrodynamics, i.e. also with a classical energy density to be computed from the field strengths.

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