# What is the meaning of the complex valued field in diffraction?

This is a philosophical physics question which came to me when I was calculating diffraction-limited focal spots during my PhD.

When discussing diffraction, most textbooks (such as Hecht's Optics or Wikipedia) begin with Maxwell's equations, for example:

$$\nabla \times\vec{E}=-\frac{\partial \vec{B}}{\partial t} \tag{1}$$

They then derive the wave equation and go on through the Huygens principle and the Kirchhoff integral and may finally end up at Fraunhofer diffraction arriving at something like the following formula:

$$\tilde{E}(\tilde{x})=\int_{-\infty}^{\infty}E(x)\exp(ikx\tilde{x}/D)dx \tag{2}$$

My question is basically: what is the meaning of the complex variable $E$ in Equation (2)? Let me explain. Hecht is pretty clear that this is the same as the electric field in Equation (1) (it stayed this way all the way through the derivation) and that the observed intensity is the modulus-squared of some rapidly oscillating electric field. The fact that the electric field $q\vec{E}=\vec{F}$ is complex is thought of merely as a mathematical tool - though it is not clear what the mathematical relation is to the good old real-valued DC field.

Edit2: Note that $E$ in Equation (2) is truly complex valued. If I had given the same equation in the Fresnel limit, for example, a perfect thin lens at the slit would not change the amplitude, but would introduce a phase of the form $\exp(ikx^2/f)$. So $E$ is not in general real.

Maxwell's equations are certainly the last word in classical electrodynamics. And everyone has done single- and double-slit experiments to verify the eventual result of Equation (2) as far back as highschool, perhaps by shining a laser through some slits. The fact that a continuous electric field interferes to create minima and maxima seems sensible.

However, we also know that this idea of a continuous electric field is misleading: the diffracting light consists of billions of photons. We could just as easily perform the experiment sending photons through slits at a rate of 1/minute. Then, there is no way that two successive photons could interfere with each other. In that case, if we had a CCD for example, rather than seeing a smooth intensity profile increasing in intensity with each new photon, we would see a grainy picture emerge as each photon stikes a particular pixel on the CCD. Only after a statistically significant number of photons would we begin to approximate the smooth e.g. $\mathrm{sinc}(\tilde{x})$ function.

One explanation is clear: the complex-valued variable in Equation (2) whose modulus-squared we must take is not the electric field, but something like the wavefunction (or a relativistic equivalent). This would explain the paragraph above and allow the electric field to be real. But then, why does it emerge directly from Maxwell's equation, like Hecht insists the electric field has? Clearly Equations (1) and (2) make very accurate experimental predictions, but how are they reconciled with quantum field theory?

Edit: How is an electric field reconciled with e.g. electron diffraction, which can form an identically "grainy" distribution?

By the way, after many debates with research physicists, the best answer I've heard so far is: "Well, I'm a physicist, so ultimately everything is a field theory"

• In classical physics you're supposed to take the real part, because the complex numbers are a mathematical convenience and actually physically significant. The electric field is not a wavefunction. – Javier Mar 7 '17 at 19:01
• My original comment was edited - title included - because it was assumed that the answer to my question was "the electric field" and that I was confused about why it should be complex. That is not the case. My question is how quantum field theory and the diffraction of particles - photons, electrons, etc - can be reconciled philosophically with Maxwell's equations of classical electrodynamics. – Valentin Aslanyan Mar 7 '17 at 19:21

What is the meaning of the complex variable $E$?

The electric field $E(\mathbf r, t)$ is a real function of position and time. Because of this it's Fourier transform (in space or time) is Hermitian, i.e. $\tilde E^*(\omega) = \tilde E(-\omega)$, where $\tilde E(\omega) = \mathcal F[E(t)]$ is the Fourier transform of the electric field. This property means that the negative frequencies carry the same information as the positive frequencies.

Since the negative frequency components do not carry any information, we do not need to compute the evolution of both the positive and negative frequency components of the field to represent all of the physics of a problem. Therefore, we can define the function $$\tilde E_a(\omega) = \begin{cases} \tilde E(\omega) & \text{if \omega \ge 0}\\ 0 & \text{otherwise} \end{cases},$$ which we can use to represent our field without any loss of any information. Taking the inverse Fourier transform of this function gives us $$E_a(t) \equiv \mathcal F^{-1}[\tilde E_a(\omega)].$$ The Fourier Transform preserves all information, so therefore acting on the complex function $E_a(t)$ fully represents all the information in any physical problem. This complex function $E_a(t)$ is known as the analytic representation of the electric field, and represents exclusively the positive frequency components of $E(t)$ (which however contain all information of the problem).

A good detailed introduction to the analytic representation of fields can be found in chapter 3 of the Book "Coherence and Quantum Optics" by Mandel and Wolf

However, we also know that this idea of a continuous electric field is misleading

This is not exactly true. The field solutions from Maxwell's equation are perfectly valid, even in the quantum regime. The difference comes from the fact that the field amplitudes for a given field configuration (e.g. a contiuous solution of Maxwell's equations) become operator-valued.

The complex-valued variable in Equation (2) whose modulus-squared we must take is not the electric field, but something like the wavefunction.

I actually find this to be a useful way of thinking about this, and I'm not alone (see e.g. this paper). However, you have to be careful with this picture. Unlike massive particles (such as electrons), photons are massless and thus there is no proper non-relativistic theory for photons. Some people take this to mean that you can't have a first-quantized theory of light (only a second-quantized/quantum field theory description). However, technically the first quantization picture of electrons is also an approximation of a more complete fully relativistic quantum field theory, so I think it's hypocritical to say one first-quantized approximate theory is okay but another isn't. The biggest problem for photons that you don't have for electrons, is that photon number is never conserved (whereas electron number is conserved at low energy). So as long as you are only using the "single-photon wavefunction" description for situations where photon-number is conserved (such as linear propagation and even up to, but not after the measurement), this picture works fine.

• The other thing that sets photons apart is that there is no position representation, but then position representations are only valid for nonrelativistic QM anyway, so this lack could be construed as being part of your general lack of nonrelativistic QM anyway. You can get a probability amplitude to destructively detect a photon at position $\vec{r}$, which is proportional to $\vec{E} = \langle 0 | \hat{E}(\vec{r},\,t)|\psi\rangle$, and this is analogous to Bialynicki-Birula's field representation of the one-photon state. In any case, I find the interpretation of the nonrelativistic .... – WetSavannaAnimal Mar 8 '17 at 0:59
• ... position representation of the electron state unclear and incomprehensible from an experimental standpoint aside from just being a particular, abstract co-ordinate representation of the electron's state- does one really seriously imagine there to be a measurement that would reach into an orbital and "find" the electron? – WetSavannaAnimal Mar 8 '17 at 1:02
• You might also be interested in this answer of mine, given your interest in Bialynicki-Birula's work. – WetSavannaAnimal Mar 8 '17 at 1:04
• I like the final paragraph and thanks for the arXiv link. A couple of points, however: (1) As I have now clarified, what you call $E(\vec{r},t)$ is not purely real. And the FT is only used as an example (far-field Fraunhofer diffraction). (2) As with every other discussion here, you have ignored the words "something like" (now changed to bold) in front of "wavefunction". I'm not asserting that it is the sort of wavefunction from non-relativistic QM (which is an asymptotic approximation, as Newtonian mechanics is in classical mechanics). – Valentin Aslanyan Mar 8 '17 at 11:49
• @ValentinAslanyan $E(r,t)$ is real, but the analytic representation $E_a(r,t)$ is complex. People are usually sloppy (in language/notation) and just assume the field and the analytic representation are equivalent due to generally being interchangeable. Diffraction formulas involving Fourier transforms are simply (approximate) solutions to the wave equation in terms of Green's functions that only include the positive frequency solutions, and thus are formulas to compute the propagation of $E_a$, not $E$. This is why even a real field appears to becomes complex upon propagation. – Punk_Physicist Mar 16 '17 at 22:05

The meaning of the complex $E$ is that it indicates a phase difference. The electric field is continuous, photons are excitations of this field (don't think of them as particles).

You can only talk about wavefunctions in quantum mechanics, which is a single particle approximation of the full quantum field theory of the electromagnetic field.

• Photons are excitations of $A^\mu$ in quantum field theory, not $\vec E$. – JamalS Mar 7 '17 at 19:06
• Yes, you are right, it doesn't change my point – JgL Mar 7 '17 at 22:02
• So you acknowledge there is a factual error in your answer (which albeit doesn't affect the overall point), and you won't correct it? -1. – JamalS Mar 7 '17 at 22:08
• I won't correct it because it is not a factual error but a matter of semantics. You could quantize $E_i$ instead of $A_\mu$, which would be more tedious (as the Lorentz invariance of its dynamics would not be directly obvious). Furthermore, the Maxwell equations would afterward give rise to what we already know if you write it in $A_m$. But whether you call the 'photon' the quantum of the 'electric' or of the 'electromagnetic' field is as meaningless as debating, whether a wavefunction is a function of 'position' or 'position and momentum'. – JgL Mar 7 '17 at 22:19

The sinusoidal E for light has a phase component which allows the summation of many EM waves (photons) so that a net E field is calculable. Classically everyone assumed the double slit pattern was a result of net zero summation and the bright spots at 2 times E, which I think now is generally assumed incorrect. The multi-single photon observations still result in the "pattern" but it should not be called an interference pattern. The pattern is a result of the available photon pathways the photons take, bright spots are well populated and dark spots are not.

The equations above allow one to calculate an interference pattern if one assumes the photons spread out and interfere and are then summed. But, my opinion, what's needed is a formula that derives the actual pathways. For example we have cavity modes in lasers, they are a result of the cavity dimensions. Similarity the slit, source, screen all have dimensions. Maybe the calculated modes will show the correct pattern in 2 slit "interference" (but its not really interference).