Energy of light ( wave or particle)

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SUMMARY

The energy of light can be calculated using two distinct approaches: as a wave and as a particle. When treated as a wave, the energy density is expressed as 1/2εE² dV, while as a particle, the energy is represented by nhf, where n is the number of photons in volume dV. Although both formulas arise from different theoretical frameworks, they ultimately describe the same physical phenomenon, as established by the Hamiltonian of free radiation. The relationship between these formulations is crucial for understanding quantum mechanics and electromagnetic theory.

PREREQUISITES
  • Understanding of electromagnetic theory, specifically Maxwell's equations.
  • Familiarity with quantum mechanics concepts, including the photoelectric effect.
  • Knowledge of Hamiltonian mechanics and its application in classical and quantum physics.
  • Basic grasp of wave-particle duality and its implications in physics.
NEXT STEPS
  • Study the derivation of the Hamiltonian for electromagnetic fields in classical physics.
  • Explore the implications of wave-particle duality in quantum mechanics.
  • Investigate the photoelectric effect and its relationship to E = hf.
  • Learn about the De Broglie wavelength and its connection to E = hc/f.
USEFUL FOR

Students and professionals in physics, particularly those focused on quantum mechanics, electromagnetic theory, and anyone interested in the fundamental properties of light.

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 E = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. 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 E = \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2. 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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