energy of light ( wave or particle)

jd12345

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

Chronos

E = hc/f is the accepted formula.

jd12345

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?

Jano L.

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.

Bill_K

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)e^ik·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.

ibysaiyan

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