Derivation of $E=h\nu$

Is it possible to derive the relation $E=h\nu$ from Schrodinger equation or the basic principles of quantum mechanics or is it something which is considered to be an axiom with no explanation?

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No. The Schrodinger equation tells you how the state of a quantum system evolves, this is not specific to photons, and cannot be used to derive any facts about them. I'm assuming that by "basic principles" you mean the postulates of quantum mechanics that hold for every quantum system (like Born's rule etc.) in which case the same goes for these.

You need a specific quantum model for the photon to even start talking about photons, and this will involve more specific physical input than just the basic principles of quantum mechanics.

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ok. so can it be derived from any other set of elementary principles or axioms? –  user774025 Mar 28 '13 at 22:56
Actually you can "derive" this result from Schrodinger equation, but that would just be hand-waving and stupid assumptions. The derivation, as you might say, will have no physical meaning. –  Cheeku Mar 29 '13 at 0:37
@Cheeku by "derive" do you mean "it can be found to be compatible"? –  user12345 Mar 29 '13 at 12:18
@user16307 Yep! In other words, mathematically related. Not necessarily physically! On the fun side, that kind of holds true for all of quantum mechanics. –  Cheeku Mar 29 '13 at 14:41

It was originally postulated by Einstein (1905) to explain the photoelectric effect. The claim was that light consists of quanta (corpuscles as he called them, we now know them as photons) with a certain energy that was linearly related to the frequency of the light. This explained why the photoelectric effect was only observed starting from some minimum frequency (thus minimum energy) and independent of the light intensity (the amount of photons).

So this is an experimental relation, not a theoretical one. And it has been proven an extremely useful relation and incredibly important insight, deepened by de Broglie in 1924. (this wikipage is also very much worth mentioning in that respect)

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As I understand the Planck constant, it is defined by $h=E/\nu$; the equation you have asked about is true by definition ($E=(E/\nu)\nu$).

That the Planck constant is, in fact, constant (i.e., that energy is linearly proportional to wavelength) is not similarly tautological; it was established empirically.

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Although historically, the relation $E = h\nu$ was not derived using the Schrödinger equation, you can derive it using the time-independent Schrödinger equation for quantum harmonic oscillator. If you start from the Hamiltonian $\hat{H} = \hat{p}^2/(2m)+m\omega^2\hat{x}^2/2$ and introduce the creation and annihilation operators $\hat{a}^\dagger$, $\hat{a}$, you can cast the Hamiltonian in the form $\hat{H} = h\nu(\hat{a}^\dagger\hat{a}+1/2)$. Then you have the eigenvalue equation $\hat{H}|n\rangle = E_n|n\rangle$ with $E_n = h\nu(n+1/2)$ and you can deduce that the oscillator is quantized with energy quanta of energy $E = h\nu$.

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But what does a harmonic oscillator has to do with a photon? –  user774025 Mar 29 '13 at 13:15
@user774025 That is, of course, another question. But you can quantize the classical vector potential and then you will see that the Hamiltonian of the quantized EM field is the same as the Hamiltonian of the harmonic oscillator. –  Ondřej Černotík Mar 30 '13 at 8:21

Personally, I like the semi-classical approach to the photoelectric effect, which models matter quantum-mechanically and the EM field classically.

After applying perturbation theory, you'll end up with $E=h\nu$ without field quantization, ie the Schrödinger equation is indeed sufficient to derive this relation.

Keep in mind that this model is somewhat simplistic and in particular ignores bulk effects.

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