# Change in Wavelength of a Photon Relation to Energy (specifically Compton Effect)

Given a photon dropping from $\lambda_1$ to $\lambda_2$, its energy will drop from $\frac{hc}{(\lambda_1)}$ to $\frac{hc}{(\lambda_2)}$. However, I was wondering if there is any significance in the energy of the change in wavelength itself. For $\Delta \lambda = \lambda_2 - \lambda_1$, this change in wavelength has an energy $\frac{hc}{(\Delta \lambda)}$, yet this value does not correspond at all (as far as I can see) with $\Delta E = \frac{hc}{(\lambda_1)} -\frac{hc}{(\lambda_2)}$. Is this just a subtlety in the math, or is there actually meaning behind the value $\frac{hc}{(\Delta \lambda)}$?

One common way that this happens is through spontaneous parametric down-conversion.

From Wiki: an important process in quantum optics, used especially as a source of entangled photon pairs, and of single photons. A nonlinear crystal is used to split photons into pairs of photons that, in accordance with the law of conservation of energy, have combined energies and momenta equal to the energy and momentum of the original photon, are phase-matched in the frequency domain, and have correlated polarizations.

Say an example of a photon dropping from λ1 to λ2: * a green photon λ1=630 nanometers, hc/λ1 = 1.97 evolts * splits in two and one photon is λ2=830 nanometers, hc/λ2 = 1.49 evolts

The difference in energy, hc/λ1-hc/λ2 = 0.48 evolts and would end up with the other photon. the other photons wave length would be 2583 nanometers based on its 0.48 evolts energy.

The difference you reference Δλ = λ2−λ1 of 200 nanometers converted to hc/Δλ is quite high at 6.2 evolts. There is not much sense to the quantity hc/Δλ and I dont think there is any particular meaning to it.

• OK, that works for me. It was a mistake I made on a question, and I just wanted to know if there was any meaning at all to it. It doesn't seem like there should be, so thanks! – A4Treok Dec 9 '13 at 6:41

If $\Delta \lambda$ is much smaller compared to either $\lambda_1$ or $\lambda_2$(it doesn't really matter, it should be much smaller than both), then we can make the following approximation:

$$|\Delta E|= \left|\Delta \left( \frac {hc}{\lambda}\right)\right|\approx\left|\frac {hc\Delta\lambda}{\lambda^2}\right|$$

• You're missing a negative sign on the right hand side. Also, how exactly does this give a physical interpretation of $hc/\Delta\lambda$? – joshphysics Dec 9 '13 at 2:04
• @joshphysics Nope I'm not missing a minus sign; I intentionally didn't put there, because normally it doesn't have to be there unless you are super-strict with your definitions. Anyway, about the second question, it simply doesn't. I think the question is wrong! – Ali Dec 9 '13 at 2:31
• Ok well I'm relatively certain that without the minus sign, you're using notation that's sufficiently unconventional that it's just plain confusing. I think it's a rather large exaggeration to assert that using the convention $\Delta E = E(\lambda + \Delta\lambda) - E(\lambda)$ is "super-strict;" I feel compelled to downvote answer (v1). – joshphysics Dec 9 '13 at 3:04
• @joshphysics I'm correcting it, in a way that you will be satisfied. – Ali Dec 9 '13 at 3:09