# How light splits up into different colours when passed through prism using QED?

I want to know how light gets splits up into different colours when it is passed through prism? How light interacts with atoms and electrons of the prism? Can someone explain this to me using Quantum Electrodynamics?

## 2 Answers

I assume that you mean you want the general idea to be presented "Oppan Feynman-style" rather than for me to start talking about a QED Lagrangian that describes the sort of prism which would generate refraction.

Photons which go into the prism will in general have the same frequency $f$, because wave-fronts and troughs cannot enter the material any faster than they enter it: but they will effectively change their wavelength due to effectively traveling at a different velocity through the medium. Now it would be nice if we could say that there was just one velocity $v$ such that every new wavelength was $\lambda = v / f$, but for the materials which make good prisms, there is instead dispersion, which is a fancy way of saying that there is some complicated $\lambda(f)$ function rather than such a simple one.

Now suppose we have an emitter E at some point $(0, Y)$ and a detector $D$ at some point $(X, -Y)$ inside the prism, and we want to sum up all of the amplitudes seen for light going through the surface at point $(x, 0)$. We'll assume that the light shines towards the surface with some wavefunction $\gamma(\phi)$ where $\phi = 0$ goes towards $x = +\infty$ and $\phi=\pi$ goes towards $x=-\infty$. Together, this gives us an integral:$$\psi = \int_{-\infty}^\infty dx~ \frac{\gamma(~\tan^{-1}(x/Y)~)}{\sqrt{(x^2 + Y^2)[(X - x)^2 + Y^2]}} \exp\left({2\pi i \sqrt{x^2 + Y^2} \over \lambda_0} + {2\pi i \sqrt{(X - x)^2 + Y^2} \over \lambda_1}\right)$$The term on the right is a phasor that rotates wildly in general; assuming $\gamma$ and the other term vary slowly relative to this, the main contribution to this amplitude will be come from the narrow band around the point $x$ where the parenthesized stuff changes least:$$\frac{d}{dx}\left(\sqrt{x^2 + Y^2} + {\lambda_0\over\lambda_1} \sqrt{(X - x)^2 + Y^2} \right) = 0$$ Assuming that $\lambda_{0,1}$ are independent of $x$ this gives just:$$\left(\frac{x}{\sqrt{x^2 + Y^2}} - {\lambda_0\over\lambda_1} \frac{X - x} {\sqrt{(X - x)^2 + Y^2}} \right) = 0$$which if you like trigonometry is actually just Snell's law in disguise:$$\sin\theta_0 - \frac{\lambda_0}{\lambda_1} ~ \sin\theta_1 = 0.$$This now has the meaning, "the main contribution comes from two angles $\theta_0, \theta_1$ which have the above relationship." You can hypothetically solve for both of them by first solving for $x$ and then using the $X$ and $Y$ that you presumably already know. But of course we are also free to choose $\gamma$ to be very localized at some angle $\phi_0$, kind of like a laser beam. The jargon term is that such light is "collimated". Then the combination of these two effects means that we won't see a large amplitude unless we happen to place our detector at the exact $\theta_1$ such that $\theta_0 = \phi_0$, so you can also interpret this the "conventional way", that a beam of light gets "bent" on entering the medium.

If the first medium is sufficiently vacuum-like, $\lambda_0 \approx c/f,$ and for a nondispersive medium $\lambda_1 \approx v/f$ and the factor relating the two is $c/v$, having nothing to do with frequency or wavelength at all.

However, supposing that $\lambda_1$ has some dependency on $f$, and $\gamma$ is collimated white light centered at some particular $\theta_0$, then different colors (frequencies) will have different $\theta_1$s which optimally refract them, and the incoming white light will "split" into different colors. That's all that's really happening.

Now if you want to know why the sum of all of the paths from $x$ through the glass to the detector might have this dispersive character, there are some models of that, too, but the basic thing to ask yourself is, well, why wouldn't it have this dispersive character? Surely there is some complicated relationship which has some Taylor expansion of $1/\lambda_1$ at roughly optical wavelengths, which are such a tiny fraction of the electromagnetic spectrum: you can certainly expect some $\alpha + \beta ~f$ expansion or so for these frequencies.

As you know white light is made of many different wavelengths. Each wavelength of light takes a slightly different path through the prism. In QED we have a spinning clock which turns at different rates for different wavelengths of light. We calculate the probability of a certain path the light can take using these clocks which have associated probability amplitudes. It turns out that because blue light has a clock which turns at a different rate than red that it bends more or rather is more likely to travel through the prism along a more bent path.