Energy of light ( wave or particle)

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Discussion Overview

The discussion revolves around the calculation of the energy of light, considering both wave and particle perspectives. Participants explore the implications of these two models in terms of energy expressions and their potential equivalence.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant suggests that the energy of light as a wave can be expressed as 1/2εE^2 dV, while as a particle, it is nhf, raising the question of whether these energies are the same.
  • Another participant states that E = hc/f is the accepted formula for energy, which is challenged by a subsequent post that insists E = hf is correct.
  • A participant asserts that while the two energy expressions arise from different theories, they should ultimately agree, citing classical and quantum mechanical formulations of the Hamiltonian.
  • Further elaboration is provided on the connection between the classical Hamiltonian and quantum mechanics, indicating that the Hamiltonian in quantum mechanics relates to the number of photons present.
  • Another participant reiterates the connection between the two energy expressions, emphasizing that they originate from different theoretical frameworks.

Areas of Agreement / Disagreement

Participants express differing views on whether the energy expressions for light as a wave and as a particle are equivalent. While some argue for their equivalence, others highlight that they stem from distinct theories, indicating an unresolved debate.

Contextual Notes

Participants reference various theoretical frameworks, including classical electromagnetism and quantum mechanics, without reaching consensus on the equivalence of energy expressions. The discussion includes specific mathematical formulations that may depend on interpretations and assumptions within those theories.

jd12345
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How can we calculate the energy of light?
IF we consider it as a wave then energy is 1/2εE^2 or rather it is the energy density
IF we multiply it by small volume dV we get energy in that volume as 1/2εE^2 dV

But if we consider as a particle its energy is hf. Let's say there are n photons in volume dV
So energy will be nhf

So both the energies will be different or same? Should be same but i don't know - I'm still a beginner in quantum physics-related studies
 
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E = hc/f is the accepted formula.
 
I'm pretty sure E = hf not hc/f ( f is frequency)

Well my question was - energy of light considering it as a wave is 1/2εE^2 dV
And energy of light considering it as a particle is nhf ( n is number of photons in dV volume)
So both of them are equal or not?
 
E = nhf is proper value of the Hamiltonian of free radiation, which is based on the formula [itex]E = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2[/itex]. So there is a connection. But strictly, they are formulae from different theories and are not the same.
 
Of course they are the same. They come from different theories, but the theories must agree.

The classical Hamiltonian is H = ½∫(E2 + B2) d3x. In the radiation gauge Φ = 0 and ∇·A = 0 this can be written as H = ½∫(A·2 + (∇ x A)2) d3x. Transform to momentum space using A(x,t) = ∫(2ω)ε(k,λ)[a(k,λ)e-ik·x + a*(k,λ)eik·x] d3k where ε(k,λ) are the two transverse polarization vectors. Put this expression in for A and you get H = ∫ω ∑a*(k,λ)a(k,λ) d3k. This is still classical.

But in quantum mechanics, a*(k,λ)a(k,λ) is ħ times the number operator N(k,λ) that tells you how many photons you have. So H = ∫ħω ∑ N(k,λ) d3k, which is basically ħω times the number of photons present in each mode.
 
Jano L. said:
E = nhf is proper value of the Hamiltonian of free radiation, which is based on the formula [itex]E = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2[/itex]. So there is a connection. But strictly, they are formulae from different theories and are not the same.

E = hc/f comes from the De-Broglie's relation while E=hf is of photoelectric effect.
 

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