
#1
Feb1212, 09:55 PM

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 physicsrelated studies 



#3
Feb1212, 11:42 PM

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? 



#4
Feb1412, 11:58 PM

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 [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.




#5
Feb1512, 01:03 PM

Sci Advisor
Thanks
P: 3,851

Of course they are the same. They come from different theories, but the theories must agree.
The classical Hamiltonian is H = ½∫(E^{2} + B^{2}) d^{3}x. In the radiation gauge Φ = 0 and ∇·A = 0 this can be written as H = ½∫(A^{·2} + (∇ x A)^{2}) d^{3}x. 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}] d^{3}k 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) d^{3}k. 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) d^{3}k, which is basically ħω times the number of photons present in each mode. 



#6
Feb1512, 01:10 PM

P: 437




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