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I am currently studying the textbook The Quantum Theory of Light, third edition, by R. Loudon. In the introduction, the author says the following:

In the customary photon description of quantum-optical interference experiments, it is never the photons themselves that interfere, one with another, but rather the probability amplitudes that describe their propagation from the input to the output. The two paths of the standard interference experiments provide a sample illustration, but more sophisticated examples occur in higher-order measurements covered in the main text.

The first sentence is a bit unclear. Is the author saying that it is never the photons themselves that interfere with one another, but rather the probability amplitudes (of the photons) that interfere with each other (which sounds weird, since the photons themselves are probability amplitudes, right?)? Or is the author saying that the photons (in the form of probability amplitudes) never interfere with each other at all, and that the photon propagation from input to output is fully described by the probability amplitude (that is, photons do not affect each other at all)? Or is it saying both?

I would greatly appreciate it if people would please take the time to clarify this.

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  • $\begingroup$ A particle itself can not be a probability amplitude! In general you can associate a probability amplitude to a particle only. The author in these lines says, a particle can't interfere with another particle but their probability amplitude (sort of cloud you can imagine around the particle) is interfering. When cloud of one particle comes into the vicinity of another particle's cloud, they interfere. $\endgroup$ – Muthu manimaran Jul 27 at 17:45
  • $\begingroup$ @Muthumanimaran You said "A particle itself can not be a probability amplitude! In general you can associate a probability amplitude to a particle only." These two sentences are contradictory, no? $\endgroup$ – The Pointer Jul 27 at 17:46
  • $\begingroup$ The probability amplitude refers here is probability of finding the particle in some interval $dx$ on the screen, which is given by $|\psi(x)|^2dx$. So in that context the probability amplitudes of two photons lets say |\psi_{1}(x)|^2 and |\psi_{2}(x)|^2 interfere. $\endgroup$ – Muthu manimaran Jul 27 at 17:50
  • $\begingroup$ The "a photon only interferes with itself" is due to Dirac and was taken out of context. The mistakes propagated through textbooks. Ballentine's book books.google.com.br/books/about/… explains this issue in the chapter about the quantum description of the electromagnetic fields. The main point, if I am not mistaken, is that the correlations, such as interference observed in quantum-optical experiments, are field correlations, not particle correlations (even though in some scenarios one can interpret it this way). $\endgroup$ – Karl Pilkington Jul 27 at 17:58
  • $\begingroup$ The probability that two photons hits at particular interval $dx$ is $|\psi_{1}(x)+\psi_{2}(x)|^2dx=|\psi_{1}(x)|^2dx + |\psi_{2}(x)|^2dx+ \psi_{1}(x)^{*}\psi_{2}(x)dx+\psi_{2}(x)^{*}\psi_{1}(x)dx$. So the last two terms is responsible for interference. $\endgroup$ – Muthu manimaran Jul 27 at 17:59
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The confusion seem to be purely one of semantics. Different people associate slightly different concepts with the term "photon."

Take for instance the famous statement of Paul Dirac: "Each photon then interferes only with itself. Interference between different photons never occurs." Compare this with Loudon's statement "...it is never the photons themselves that interfere ..." Then it is clear that these two statement are contradictory. The reason seems to be that they mean different things by the term photon.

For some people, a photon is a dimensionless point traveling on a world line (Eugene Wigner's definition of a particle). Other people associate the wave function with the photon. In view of the fact that one can only establish the existence of a photon as a particle when you measure it, the idea that photons exist as particles even when they are not observed cannot be confirmed experimentally. (There are perhaps some deeper argument that one can present, but I won't go into that unless being asked to do so.) Therefore, it is probably more acceptable to think of the photon in terms of its wave function. The particle nature then only emerges when it is observed.

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  • $\begingroup$ So which of these two interpretations are Dirac and Louden each using? In other words, under which one of these interpretations does the photon interfere only with itself, and under which one do the photons never interfere? $\endgroup$ – The Pointer Jul 31 at 17:26
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    $\begingroup$ Particles cannot show interference because particles don't are not waves. So Loudon considers photons purely as particles and Dirac considers them in terms of their wave functions. $\endgroup$ – flippiefanus Aug 1 at 4:19
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You are confused, and I understand, because the text you are referring to, is a little bit misworded.

"In the customary photon description of quantum-optical interference experiments, it is never the photons themselves that interfere, one with another, but rather the probability amplitudes that describe their propagation from the input to the output.", please read it very carefully again, "one with another".

The author is referring to, the fact, that this experiment is done shooting one photon at a time. Thus, the photons coming after each other, temporally separated, cannot interfere with each other physically.

Rather, you need to understand what is causing the interference pattern to appear. "rather the probability amplitudes that describe their propagation from the input to the output.", is referring to the setup itself, the boundary conditions, and the entanglement of the slits and the photons.

Since the photons are coming from the same laser pump, the setup is the same for all photons, the quantum mechanical properties of the photons are the same, and the boundary conditions are the same for all photons coming from the pump, and the photons are all entangled with the slits. Contrary to popular belief, this is what causes the pattern.

So when the author says "probability amplitudes that describe their propagation from the input to the output", this is referring to the setup itself, and the boundary conditions, which is the same for all photons coming from the pump. Saying that these interfere, is a little bit confusing, that is why you are confused. A better notion is, that these, the setup, and the boundary conditions are all the same, unchanged, and this causes the interference pattern.

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  • $\begingroup$ Why the downvote? $\endgroup$ – Árpád Szendrei Aug 1 at 16:34
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The term photon applies to an electromagnetic wave packet of finite size and a total energy determined by the frequency of the wave. The strength of the two fields at any point determines the probability that all of the energy and momentum of the packet will be absorbed by some other entity (often an electron) at that point. Since this “collapse of the wave” is difficult to conceive, the common assumption is that, instead of being distributed as energy density in the fields, the energy (and momentum) of the packet is carried by a “point-like particle” which apparently wanders at random throughout the packet. In the quote you referred to, the author was using the term photon to denote the point-like particle while leaving any interference effects to the wave.

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  • $\begingroup$ You state that "The term photon applies to an electromagnetic wave packet of finite size and a total energy determined by the frequency of the wave." This statement does not describe a photon but a classical wave packet, itself giving the probability distribution of photons. The size of a photon is unknown, small and not related to its energy or momentum. $\endgroup$ – my2cts Jul 27 at 19:07
  • $\begingroup$ my2cts; See the answer from flippiefanus. $\endgroup$ – R.W. Bird Jul 29 at 13:30
  • $\begingroup$ My point is that the electromagnetic wave equation is to photons what the Schrodinger equation is to non- relativistic massive particles. From its solution one can find the probability of detecting a photon. The field should not be equated with photons, as the statement I quote above appears to do. $\endgroup$ – my2cts Jul 29 at 16:41
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If you send a single photon state $|10\rangle$ through a $50:50$ beam-splitter, the output is the state $$|\psi_m\rangle=\frac{1}{\sqrt{2}}(|10\rangle+|01\rangle)$$ This is a two-mode state in the mode picture, which will be responsible for interference, e.g. in a Mach-Zender interferometer.

In the particle picture this state is given by $$|\psi_p\rangle=\frac{1}{\sqrt{2}}(|a\rangle+|b\rangle)$$ since it is the state of one particle (or you can work in coordinate representation); the $50:50$ beam-splitter merely rotated the input single-photon particle state $|a\rangle$ into $|\psi_p\rangle$ above. This particle can only interfere with itself in this sense.

A state $|20\rangle$ through a $50:50$ beam-splitter will produce the two-mode state $$|\phi_m\rangle=\frac{1}{2}(|20\rangle+\sqrt{2}|11\rangle+|02\rangle)$$ which corresponds to the two-particle state $$|\phi_p\rangle=(\frac{|a\rangle+|b\rangle}{\sqrt{2}})\otimes(\frac{|a\rangle+|b\rangle}{\sqrt{2}})$$. The beam-splitter then acts as a collective rotation $U\otimes U$ in the particle picture and each particle can only interferes with itself in this sense.

The really interesting case is given by the input state $|11\rangle$, which a $50:50$ beam-splitter converts to $$|\xi_m\rangle=\frac{1}{\sqrt{2}}(|20\rangle-|02\rangle)$$
and in the particle view is given by $$|\xi_p\rangle=\frac{1}{\sqrt{2}}(|aa\rangle-|bb\rangle)$$ which is an entangled two-particle state which displays an interference between the two individual particles. This state is present in the Hong-Ou-Mandel effect https://arxiv.org/abs/2005.08239, displaying second-order optical quantum correlations. For an explanation of mode picture and particle picture: https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.112.150501.

Hence, I would say that a single-photon state (any number of modes) indeed only interferes with itself, but multi-photon states can interfere with each other.

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Photons are one of the elementary particles in the standard model of particle physics, on par with electrons, quarks etc.

Photons wave functions are given by solutions of a quantized version of Maxwell's equations.

It can be shown mathematically that classical electromagnetic light emerges as a comfluence of the wavefunctions of zillions of individual photons with the photon $energy =hν$, where $ν$ is the frequency of the classical wave. The photon is a point particle as seen here with single photons at a time.

singlphot

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The above illustrates the statement you quote "it is never the photons themselves that interfere, one with another, but rather the probability amplitudes that describe their propagation from the input to the output".

Each photon follows the path of its interaction with the slits individually, the probability is shown in the many photon frames, which show the probability of the photon hitting the screen at a particular x,y.

At first order there are no photon-photon interactions, that is why two light beams go through each other without scattering off each other.

Or is the author saying that the photons (in the form of probability amplitudes) never interfere with each other at all,

It is the statement that photons do not interact with each other to first order. To see the higher very improbable orders see here.

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    $\begingroup$ Classical light is not the superposition of zillions of individual photons. The superposition of a zillion single photons is still a single photon. $\endgroup$ – flippiefanus Jul 28 at 9:06
  • $\begingroup$ @flippiefanus maybe I should say the superpositions of the wavefunctions , although I do not understand your statement. $\endgroup$ – anna v Jul 28 at 10:57
  • $\begingroup$ If $|\phi_1\rangle$ and $|\phi_2\rangle$ are both single photon states, then the superposition $|\phi\rangle=|\phi_1\rangle\alpha+|\phi_2\rangle\beta$ is still a single photon state. One can generalize that to any number of terms in the superposition, the result remains a single photon state. $\endgroup$ – flippiefanus Jul 28 at 13:06
  • $\begingroup$ The same applies to the superposition of single photon wave functions. The get a multi-photon state, one needs the tensor product of single photon states (or wave functions). $\endgroup$ – flippiefanus Jul 28 at 13:08
  • $\begingroup$ @flippiefanus You are talking of a denstiy matrix type of formalism? The link deals with it with quantum field theory. $\endgroup$ – anna v Jul 28 at 19:28
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There is no difference between the two interpretations that you list. Photons do not interfere with themselves or other photons. The wave function should not be identified with photons. It gives the probability of detecting photons. It can be seen as the average of a poisson distribution describing the number of photons that can be detected.

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  • $\begingroup$ So the equation $|\psi_{1}(x)+\psi_{2}(x)|^2dx=|\psi_{1}(x)|^2dx + |\psi_{2}(x)|^2dx+ \psi_{1}(x)^{*}\psi_{2}(x)dx+\psi_{2}(x)^{*}\psi_{1}(x)dx$, as presented in the comments, means that the amplitudes of photons interfere, but that does not mean that the photons themselves interfere (that is, the photons themselves never interfere)? $\endgroup$ – The Pointer Jul 27 at 19:19
  • $\begingroup$ What you are saying is actually contradictory. In the quantum context, the fields in Maxwell's equations are probability amplitudes (their squared modules give the probability to detect a photon.) So if these equation are wave equations for photons, then they are probability amplitudes. $\endgroup$ – flippiefanus Jul 28 at 9:04
  • $\begingroup$ @flippiefanus I repeat that photons are not probability amplitudes. Again etc. $\endgroup$ – my2cts Jul 28 at 19:33

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