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I am wondering whether the light interference is a quantum phenomenon. Or, alternatively, is there any interference in Maxwell's theory understood as a classical field theory?

The reason I am puzzled is as follows. Typically, in the general physics course one presents interference as a classical phenomenon, which is related to the wave nature of light. At the same time, when you look at the derivation, it crucially features complex fields. Namely, one takes two electric fields $E_1$ and $E_2$, each oscillating in time and space with the standard formula

$E=a e^{i(kx+\omega t+\phi)}$

then considers the total electric field by adding them up and computes intensity via

$I = E E^*$.

This discussion does crucially feature complex electric fields. What are these? Can a complex electric field be measured with any instrument? I guess, it cannot. Then, I would say, it cannot be classical, since classical quantities should be physical, in a sense that it should be possible to measure them. I believe that a trick with a complex electric field is intended to mimic a wave function of quantum mechanics and then intensity $I$ is mimicking the probability to observe a particle. Am I missing something? Can one improve the derivation so that fields at any stage are real and get the same effect?

Clarification.

More specifically, I meant the following. There is a classical field theory that describes electromagnetism. From this point of view, solutions to the free equations of motion are provided by plane waves. On the other hand, one can consider a single photon and quantize it in the sense of quantum mechanics. Thus one gets quantum mechanics of a single photon (I do know that the standard quantum mechanics is non-relativistic, while photons do require relativity. Still, one can consider relativistic quantum mechanics of a single photon as in the second section of book by Weinberg on QFT: you can make the wave function transform properly under the Poincare group, you can define the norm and then normalize the wave function so that you have a single particle). Then, a photon becomes a wave, but in a different sense: it is a probability wave.

So, I am confused whether interference refers to interference of waves in the first or in the second sense. Can a single photon interfere with itself? Is there any difference in this type of interference and interference of classical electromagnetic waves?

A related question is how to regard the Klein-Gordon equation. I am used to regard it as an equation for a classical spin-0 field. At the same time Wikipedia says that it is an equation for a wave function of a single spin-0 particle. Which one is correct?

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    $\begingroup$ If you don't see $\hbar$ anywhere in your formulas, it is hard to argue quantum.... $\endgroup$ Commented Feb 7, 2023 at 20:02
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    $\begingroup$ Just an aside: Many phenomena attributed to quantum mechanics arent strictly within its purview. Even a form of tunneling can be demonstrated classically, look up evanescent waves, you can do it with wqves on water (or light) $\endgroup$
    – R. Rankin
    Commented Feb 7, 2023 at 22:17
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    $\begingroup$ A complex number is naturally represented by a 2x2 matrix. Ie. nagwa.com/en/explainers/152196980513 nothing quantum about that $\endgroup$
    – R. Rankin
    Commented Feb 7, 2023 at 22:21
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    $\begingroup$ In the original electron diffraction experiment, you observe effectively a wavelength $\lambda =h/p$, so the link to the electron of momentum p is h. In surveying entangled photons at a distance, you are observing individual photons, which, ipso facto, relies on h in the photodetector construction. Do the math. $\endgroup$ Commented Feb 8, 2023 at 21:22
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    $\begingroup$ The presence of the complex $i$ itself is just a mathematical hack, since it's easier to do calculus and phase calculations with $(e^{it\omega} + e^{-it\omega})/2$ than with $\cos(t\omega)$ even though they're equivalent. By convention you just use the first term (or second, depending on textbook / field) and take the real part of the result when done with the important calculations. Interference follows from the wave equation. Quantum has nothing to do with it. $\endgroup$ Commented Feb 8, 2023 at 22:21

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What are these? Can a complex electric field be measured with any instrument? I guess, it cannot. Then, I would say, it cannot be classical, since classical quantities should be physical, in a sense that it should be possible to measure them.

Complex fields are perfect legitimate and perfectly classical. They are a convenient way to represent a quadrature RF detector such as the kind that we use in magnetic resonance imaging, communications, and many other applications. The imaginary part is simply a 90 degree phase shifted signal, or in other words, a $\sin$ instead of a $\cos$. This stems from the usual representation $ e^{i \theta}=\cos(\theta) + i \ \sin(\theta)$.

So, for example, if we build two detectors, one that is sensitive to fields in the $x$ direction and the other that is sensitive to fields in the $y$ direction, then we would call the one in the $y$ direction the "imaginary" channel and the one in the $x$ direction the "real" channel. Together they would form a complex-valued signal that we could then Fourier transform, demodulate, and otherwise process mathematically as a complex signal.

The complex number is a completely legitimate and powerful representation of this signal. There is no reason to object to complex numbers simply because of the "imaginary" component. That is just a label and does not have anything to do with whether or not it is experimentally measurable.

Quantum mechanics uses complex numbers for the same reason as classical mechanics does: to represent phase. Phase is more important in quantum mechanics than in classical mechanics, but that does not imply that every use of phase is quantum.

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    $\begingroup$ Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on Physics Meta, or in Physics Chat. Comments continuing discussion may be removed. $\endgroup$ Commented Feb 13, 2023 at 0:15
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Interference is a wave phenomenon. E.g., sound waves interfere, waves on water surface interfere.

Interference is quantum to the extent that quantum objects exhibit both wave an particle properties. While for classical particles, like electrons, quantum interference appears surprizing, there is nothing special about it for light - which is classically already a wave, exhibiting interference, diffraction, etc. What is surprizing for quantum light is when it exhibits particle-like properties - like, e.g., photon countability.

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  • $\begingroup$ ok, I see. I studied this stuff quite a while ago and things got blurred. I guess, the key thing is that for light electric and magnetic fields add up, while what a detector measures is intensity which is (E1+E2)^2 +(B_1+B_2)^2. So, due to cross terms intensities do not add up. As a result, each source separately may have equal intensity at all points of the screen, while when you switch them on together an interference pattern occurs. So, the use of complex electric and magnetic fields is just a misleading trick? Is that what you mean? $\endgroup$
    – Dr.Yoma
    Commented Feb 8, 2023 at 20:58
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    $\begingroup$ @Dr.Yoma uding complex fields is a useful trick (more generally, one uses Fourier transform) - one could do the same calculations with the real fields (in terms of sines and cosines.) $\endgroup$
    – Roger V.
    Commented Feb 9, 2023 at 6:04
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Others already answered the question, but I wanted to add a simple example of how interference shows up mathematically without using complex numbers. Consider two electric fields $E_1 = A_1 \cos(\omega t)$ and $E_2 = A_2 \cos(\omega t + \phi)$. If both fields are present at a certain location, the combined field amplitude at that point will be $E = E_1 + E_2$, and the intensity will then be $$I = (E_1 + E_2)^2 = A_1^2 \cos^2(\omega t) + A_2 \cos^2(\omega t + \phi) + 2A_1 A_2 \cos(\omega t) \cos(\omega t + \phi).$$

The last term $2 A_1 A_2 \cos \omega t \cos(\omega t + \phi)$ gives precisely the interference between the two waves!

Note that the typical way complex numbers are introduced in classical electromagnetics is by defining the field phasor as a complex field $\mathcal{E}$ such that the (real) electric field at a time $t$ is given by $Re(\mathcal{E} e^{i \omega t})$. Note that with this definition the actual electric field is still a real-valued function, but by representing it by a complex phasor, we make the math of working with it easier.

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  • $\begingroup$ This is true for electric fields for example in an antenna .... 2 photons would excite 2 free electrons ... the net E being dependent upon phase. But the statement is NOT true for say 2 photons hitting a laser power meter for example .... in this case the energy of both photons is measured .... i.e. there's no "interference" in this case even though the photons can have differing phase. $\endgroup$ Commented Feb 8, 2023 at 16:37
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Can a complex electric field be measured with any instrument?

What you are asking for is a "vector voltmeter". They exist. They aren't common as stand-alone instruments, but usually part of a vector network analyzer.

Remember, both real and complex numbers are mathematical abstractions, products of human imagination. We use them as models of physics, but they don't exist in the physical world. When we're smart, we use the abstraction that best fits the physical phenomenon we're modeling. Sometimes it's complex electric fields.

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  • $\begingroup$ It's only a "complex electric field" when there's a well-defined frequency, and even then you're not measuring a complex value - you're measuring a phase difference between input and output and representing that with a complex number. Calling it a "complex field" suggests that it might exist as a static phenomenon or be measurable without knowing the input. $\endgroup$ Commented Feb 10, 2023 at 16:51
  • $\begingroup$ @SarahMesser Yes, every model has its domain. $\endgroup$
    – John Doty
    Commented Feb 10, 2023 at 17:21
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Interference is the ability of waves of certain nature to combine in a way that makes the resultant oscillation have a larger or smaller amplitude. Waves of different nature, such as acoustic or surface waves on water, can interfere. There is little special about the interference of intensive light waves in this sense.

There is a special quantum property of light though. As its intensity decreases, it becomes more and more pronounced that the oscillations constituting the light can exist only in individual chunks. We know these chunks as photons and the existence of these chunks is one of the major distinctions of quantum physics (compared to classical).

The photons constituting the light waves also have wave structures associated with them. These structures are known as wave-functions and correspond to probability amplitudes for photon location. As wave functions corresponding, say, to different paths taken by a single-photon can interfere, the probability of this single-photon to take different paths can exhibit interference patterns.

Thus, interference of single photons is established mathematically by the same effect, as interference of intensive light beams (which is very much classical). It is the existence of individual single photons which is non-classical.

Edit: a paradigmatic example of single particles interfering with themselves is the double-slit experiment.

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    $\begingroup$ If we consider the Feynman path integral we can bridge how individual photons actions or paths develop in into the classical behaviour of many photons .... the fact that the math works out classically is just a product of the fact that photons like to travel path lengths that are multiples of their wavelength .... as shown by the path integral. $\endgroup$ Commented Feb 8, 2023 at 16:31
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    $\begingroup$ This is very close to what I actually meant to ask. What I was wondering is the following. In the classical description of light you have waves. At the same time, you can consider a single photon quantum mechanics, which involves probability waves. So, I was wondering whether interference happens due to electromagnetism being waves in the classical field theory, or due to a single photon being a wave when it is quantized. $\endgroup$
    – Dr.Yoma
    Commented Feb 8, 2023 at 21:36
  • $\begingroup$ @Dr.Yoma Many photons are just acting like single photons .... their paths are all governed by the EM field. Dirac said every photon interferes with itself ..... but the modern interpretation is to say every photon determines its own path .... but this is even better said as the EM field (all the electrons, all the excited electrons, protons, etc are important) .... but Feynman integral shows that EM field likes to resonate (like in a laser cavity or interference filter) so in the DSE we see photon paths that are lambda multiples. $\endgroup$ Commented Feb 9, 2023 at 22:22
  • $\begingroup$ @Dr.Yoma I am not sure if this is a big study area in QM. Feynman used the path integral to advance nuclear science and QFT. The DSE is very old (and maybe not too important ). $\endgroup$ Commented Feb 9, 2023 at 22:25
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    $\begingroup$ @Dr.Yoma in the picture I'm most comfortable with, the EM field is first expanded over a basis of the spatial modes that are geometric solutions to Maxwell's equations. These modes are plane/radial waves in many practical cases, and each is a harmonic oscillator distributed in space, so to speak. These distributed oscillators can exhibit interference fully compatible with the classical description. If needed, the harmonic oscillators are quantized, and in the relevant intensity ranges, they exhibit interferences incompatible with classics, but still in geometry determined by classical eqns. $\endgroup$
    – And R
    Commented Feb 10, 2023 at 9:57
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You can think of imaginary numbers in real physical systems as being a trick to make the math easier.

Imaginary numbers can be represented in a vector form with an inner product, so there's nothing "imginary" about them and can be simply thought of as linear algebra.

The essence of "quantum" is that things are waves of "probability amplitudes" that slosh around and form probabilities. Classical waves are just normal things that slosh around.

But to technically answer your question:

Is interference of light a quantum phenomenon?

The answer is yes. While interference can be explained with classical physics, the reality is that light is made of photons and the interference of such a system is a "quantum phonemenon." Sometimes people define things as "quantum" as being systems that can uniquely be defined as quantum, and cannot be explained by classical physics. In that definition then this is not "quantum" in the general case.

Interference of single photons of light is certainly a quantum phenomenon in all definitions though though.

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    $\begingroup$ "Interference of single photons of light is certainly a quantum phenomenon". The interference is entirely a wave phenomenon, accurately computable with classical models. It's the photon detection that's quantum. $\endgroup$
    – John Doty
    Commented Feb 7, 2023 at 22:15
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    $\begingroup$ see HOM intereference and a quantum mach-zender interferometer $\endgroup$ Commented Feb 8, 2023 at 11:31
  • $\begingroup$ Is an excited laser cavity quantum? Yes and maybe no. Certainly the bandgap and hence the laser wavelength is quantum ... a direct result of Schrodinger's equation and that fact that electrons only occupy/jump defined energy levels. The fact that the optical cavity needs to be tuned (fixed length) and photons only exist when the cavity is multiples of the wavelength is a wave property .... not something we typically call "quantum" ... for example many antennas and microwave waveguides exploit the wave phenomenon but we do not use the word quantum. $\endgroup$ Commented Feb 8, 2023 at 16:46
  • $\begingroup$ Feynman's path integral gives insight into why there is no energy (photons) in the dark areas of the typical "interference" pattern ..... and why all the energy (photons) end up in the bright areas. Historically we call this "interference". $\endgroup$ Commented Feb 8, 2023 at 16:48
  • $\begingroup$ @PhysicsDave, point of my response is to try to break down the many types of meaning of if something is "quantum." There are many different ways of defining something as quantum, but insisting something /isn't/ quantum because it has "wavelike" properities is incorrect. The key to QM is that discrete things act like "waves of probability" and it is that wave-like behavior that makes it quantum. $\endgroup$ Commented Feb 9, 2023 at 0:30

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