# energy of light ( wave or particle)

by jd12345
Tags: energy, light, particle, wave
 P: 260 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. Lets 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 dont know - i'm still a beginner in quantum physics-related studies
 Sci Advisor PF Gold P: 9,183 E = hc/f is the accepted formula.
 P: 260 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?
P: 1,027

## energy of light ( wave or particle)

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.
 Sci Advisor Thanks P: 3,856 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)-½ ∑ 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.
P: 437
 Quote by Jano L. 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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