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I have recently studied optics. But I feel having missed something important: how can amplitudes of light waves be complex numbers?

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3 Answers 3

With any simple harmonic oscillator there are two quantities we are interested in, the phase and the amplitude. Complex numbers are an easy way to represent both of these in a single value, especially as a complex number can be written in the form $Ae^{i\theta}$ where $A$ is the amplitude and $\theta$ is the phase. This doesn't mean light has some "imaginary" component. The complex number is just a mathematical model for the light.

There are lots of ways in which complex numbers make it easier to construct mathematical models. If you have access to a copy of Roger Penrose's book "The Road to Reality" have a look at chapter 4 where he discusses this aspect of complex numbers.

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So, if I reiterate all those calculations with real amplitudes, I am to get exactly the same results? I still feel a bit skeptical: as QM has been built upon optics, I could ask the same question, though in that case complex amplitudes make part of an axiom... I couldt not formulate a precise question, but I am still not satisfied ... –  Isaac Dec 23 '11 at 17:16
    
So, if I reiterate all those calculations with real amplitudes Yes, but only if you keep track of the phases as well. Remember that waves with opposite phases will cancel so you can't just add up amplitudes and ignore the phase. You can do this without using complex numbers, but it's a lot more work. –  John Rennie Dec 23 '11 at 17:25

Isaac, let me say you are not the only one who feels this way. I had recently been tutoring an undergrad course (on nonlinear optics) and was almost shocked to find most of the students getting muddled in the usage of complex no.s. In fact, in the process, they had a hard time in also understanding/appreciating the beautiful physics.

In my opinion, it's disappointing that the lecturers/professors do not emphasize enough that the complex representation of electric field amplitude contains an additional term c.c. (or sometimes H.c.)

c.c. stands for complex conjugate; H.c. means Hermetian conjugate and you can see that the addition of this term would make the overall quantity (on LHS) as real.

So, an electric field of the form $ E(z,t) = E_0 e^{i(kz - \omega t + \phi_0 )} + c.c.$ = $ 2\cdot E_0 \cos(kz - \omega t + \phi_0)$ indeed describes a real/physical wave.

Of course, while doing the maths it may become cumbersome to carry around the c.c. term through a series of equations, and so it is dropped (but implicitly, it is still there).

Perhaps the simplest reason for justifying the usage of this representation is that multiplication of two or more light waves - which can be encountered in several phenomena such as interference - can be simply understood by the addition or subtraction of the terms in the exponent. As in, two waves $e^{i\omega_1 t}$ and $e^{i\omega_2 t}$ will produce terms $\propto$ $e^{i(\omega_1+\omega_2) t}$, $e^{i(\omega_1-\omega_2) t}$ etc.

Compare this with having to use trigonometric identities and you'll understand the beauty of using complex no.s in optics.

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John Rennie's Answer, that complex numbers simplify calculations with sinusoidally varying quantities by letting you do linear operations with complex exponentials and then reverting back to sinusoids at the end of your calculation, is altogether correct and a summary of what is called the phasor method for dealing with any quantity that varies sinusoidally with time (usually - occasionally a co-ordinate replaces time).

But there is also a very special relationship between complex numbers and Maxwell's equations (and thus with optics) that holds regardless of whether the variation is time-harmonic or not and is unique to Maxwell's equations. This is the notion of diagonalising or decoupling Maxwell's equations. Let's look at the Faraday and Ampère laws:

$$\partial_t \vec{B} = -\nabla\times \vec{E} \tag{1}$$ $$\partial_t \vec{E} = c^2\,\nabla\times \vec{B} \tag{2}$$

two coupled first order equations. You can solve them by brute force elimination, thus ending up with second order equations. Alternatively, to simplify things, we can uncouple them by diagonalising them. We seek a field $\alpha\vec{E} + \beta\vec{B}$ which does this: it so happens that if we put $\vec{F}_{\pm} = \vec{E} \pm i\,c\,\vec{B}$ then we get two decoupled, first order equations by adding $\pm i\,c$ times (1) to (2)

$$\partial_t \vec{F}_\pm = \mp\,i\,c\,\nabla\times \vec{F}_\pm \tag{3}$$

which simplifies the solution of Maxwell's equations (often the $i$ is written on the LHS to emphasise that Maxwell's equations are the first quantised Schrödinger equation for the photon). Try this: there are no other superposition weights that work: $\pm i\,c$ are the unique ones that do diagonalise Maxwell's equations.

So we could choose either $\vec{F}_\pm$, assume $\vec{E},\, \vec{B}$ are real-valued (which of course they are in the laboratory), solve Maxwell's equations with the complex $\vec{F}_+$ (or $\vec{F}_-$: it doesn't matter which) then split our complex number solution into real and imaginary parts to get the real valued $\vec{E}$ and $c\,\vec{B}$ at the end. This trick, and the vectors $\vec{F}_\pm$, are named the Riemann-Silberstein method and vectors, respectively (after Bernhard Riemann and Ludwik Silberstein).

However, it turns out that if instead we keep only the positive frequency parts of both $\vec{F}_\pm$ (i.e. we replace $\cos(\omega\,t+\delta)$ by $\exp(-i\,\omega\,t-\delta)$) exactly as in the phasor method, then it can be shown that a wholly left circular polarised field has zero positive frequency component in $\vec{F}_+$ - only $\vec{F}_-$ is nonzero. Likewise, a wholly right circular polarised has only a zero $\vec{F}_-$ part and the $\vec{F}_+$ is nonzero.

So, not only are Maxwell's equations diagonalised by the eigenvalues $\pm i\,c$, their diagonalisation by this unique, complex number splits the fields precisely into their circularly polarised components.

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