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Energy of light ( wave or particle)

by jd12345
Tags: energy, light, particle, wave
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jd12345
#1
Feb12-12, 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 physics-related studies
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Chronos
#2
Feb12-12, 10:20 PM
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E = hc/f is the accepted formula.
jd12345
#3
Feb12-12, 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?

Jano L.
#4
Feb14-12, 11:58 PM
PF Gold
P: 1,165
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.
Bill_K
#5
Feb15-12, 01:03 PM
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P: 4,160
Of course they are the same. They come from different theories, but the theories must agree.

The classical Hamiltonian is H = ∫(E2 + B2) d3x. In the radiation gauge Φ = 0 and ∇A = 0 this can be written as H = ∫(A2 + (∇ x A)2) d3x. Transform to momentum space using A(x,t) = ∫(2ω)-ε(k,λ)[a(k,λ)e-ikx + a*(k,λ)eikx] 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
#6
Feb15-12, 01:10 PM
P: 442
Quote Quote by Jano L. View Post
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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