## Monday, January 9, 2017

### Are quanta particles or waves?

The title of this post is an age-old question isn't it? Particle or wave? Wave or particle? Many have rightly argued that the so-called "wave-particle duality" is at the very heart of quantum weirdness, and hence, of all of quantum mechanics. Einstein said it. Bohr said it. Feynman said it. Two out of those three are physics heroes of mine, so that's a majority right there.

Feynman, when talking about what we now call the wave-particle duality, was referring to the famous "double-slit experiment". He wrote (in his famous Feynman Lectures, Chapter 37 of Volume 1, to be precise):
 Richard Feynman (1918-1988) Source: Wikimedia
"We choose to examine a phenomenon which is impossible, absolutely impossible, to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery. We cannot make the mystery go away by “explaining” how it works. We will just tell you how it works. In telling you how it works we will have told you about the basic peculiarities of all quantum mechanics."
So what is Feynman talking about here? Instead of launching on a lengthy exposition of the double-slit experiment, as luck would have it I've already done that, in a blog post about the quantum eraser. That post, incidentally, was No. 6 in the "Quantum measurement" series that starts here. You don't necessarily have to have read all those posts to follow this one, but believe me, it would help a lot. At the minimum, start at No. 6 if you're not already familiar with the double-slit experiment. But you'll get a succinct introduction to the double-slit experiment below anyway.

Alright, back to quantum mechanics. Actually, step back a little bit more, to classical mechanics. In classical physics, there is no duality between waves and particles. Waves are waves, and they would never behave like particles. For example, you can't kick a wave, really, no matter what the surfer types tell you. Particles on the other hand, do not interfere with each other as waves do. You can kick particles (kinda), and you can count them. You can't count waves.

What Bohr, Einstein, and Feynman are trying to tell you is that in quantum mechanics (meaning the real world, because as I have told you before, classical mechanics is an illusion, it does not exist) the same stuff can be either particle OR wave. Not both, mind you. Here's what Einstein said about this, and to tell you the truth, this statement sounds like he's been hanging out with Bohr far too much:
 A. Einstein (1879-1955) Source: Wikimedia

"I
t seems as though we must use sometimes the one theory and sometimes the other, while at times we may use either. We are faced with a new kind of difficulty. We have two contradictory pictures of reality; separately neither of them fully explains the phenomena of light, but together they do".
I've used a picture of Einstein in 1904 here, because you've seen far too many pics of him sticking out his tongue and hair disheveled. He wasn't like that most of the time when he made his most important contributions.

Lest you think that the troubles these 20th century physicists had with quantum mechanics is the stuff of history, think again. In 2012, a mere 5 years ago, experimenters from Germany (in the lab of the very eminent Wolfgang Schleich) claimed that they had collected evidence that a quantum system can be both particle and wave at the same time. Such an observation-if true-would run afoul of Bohr's "duality principle", which declared that a quantum system can only be one or the other, depending on the type of experiment used to examine the system. One or the other, but never both

Rest assured though, analyzing results of the Schleich experiment in a different way reveals that all is well with complementarity after all, as was pointed out by a team at the University of Ottawa, led by the equally eminent Robert Boyd. (You can read an excellent summary of that controversy in Tom Siegfried's piece here.) What all this fighting about duality should teach you is that this is not at all a solved problem. As recently as a few days ago, Steven Weinberg (who, full disclosure, has also been in my pantheon of physicists ever after I read his "First Three Minutes" at a very tender age) wrote about the particle-wave duality in the New York Review of Books. I hope that he reads this post, because it may alleviate some of his troubles.

In this piece, entitled "The Trouble with Quantum Mechanics", Weinberg admits to being as puzzled as his predecessors Einstein, Bohr, and Feynman, about the true nature of quantum physics. How can we understand, he muses, that quantum dynamics is governed by a deterministic equation (the Schrödinger equation), yet when we try to measure something, then all we can muster is probabilities? "So we still have to ask", Weinberg writes, "how do probabilities get into quantum mechanics?"

How indeed. You know of course, from reading my diatribes, that this is a question I am interested in myself. I have obliquely hinted that I think I know where the probabilities are coming from (if you can find the relevant post) and that one day I'll write a detailed account of that idea (it's 3/4 written already, actually). But today is not that day. Having convinced you that the particle-wave duality is still a very hot topic in quantum physics, let me take on that particular subject first.

What I want to do in this blog post is to make you think differently about the complementarity principle. What I'm going to tell you is that you should stop thinking in terms of "particle or wave". It is a false dichotomy. It is a false dilemma because quantum systems are neither particle nor wave. Those two are classical concepts, after all. Strictly speaking, quantum systems are quantum fields. But this is not the time to delve into quantum field theory, so instead I will try to marshal the tools of quantum information theory to tell you what is really complementary in quantum measurement, what it is that you can have "only one of", and what it is that is being "traded-off". You don't exchange a bit of particle for a bit of wave, this much I can tell you right here.

To do this, I have to introduce you to some very counter-intuitive quantum stuff. Now, you might argue: "All quantum stuff is counter-intuitive", and I'd have to agree with you if all your intuition is classical. What I am going to tell you is stuff that even baffles seasoned quantum physicists. I'm going to tell you about quantum experiments where the "nature" of the quantum experiment that you perform can be changed after you've already completed the experiment!

Let me remind you right here, that the--also very eminent--Niels Bohr tried to teach us that whether a quantum system appears as a particle or as a wave depends on the type of experiment you subject it to. Here I'm telling you that this is a bunch of hogwash, because I'll show you that when you do an experiment, you can change whether it is a "particle"- or a "wave"-experiment long after the data have been collected!

I know you're not shocked at my dissing Bohr as I have a habit of doing so. But I'm in good company, by the way, if you read what Feynman wrote about Bohr in his "Surely You're Joking" series.

"Alright I bite", one of you readers exclaimed just now, "how do you retroactively change the type of experiment you make?"

Glad you asked. Because now I can talk about John Archibald Wheeler. Wheeler was not a conventional physicist: Even though his early career as a nuclear physicist led to several important contributions to the Manhattan project, he was also interested in many other areas of physics. Indeed, he was a central figure in the "revival" of general relativity theory. (That theory had gone a bit out of fashion when people realized that many predictions of the theory were difficult to measure.) Wheeler co-authored what many (including myself) think is the best book on the topic: "Gravitation" (with Charles Misner and Kip Thorne). That book is often just referred to as "MTW".

 John Archibald Wheeler (1911-2008). Source: University of Texas
I never got to meet Wheeler, perhaps because I entered the field of quantum gravity too late. While Wheeler has been influential in the field of quantum information, it really was his gravity work that had the most lasting impact. He invented the terms "black hole" and "wormhole", after all. His most influential contribution to quantum information science is, undoubtedly, the "delayed choice" gedankenexperiment. Let me explain that to you.

Wheeler's thought experiment examines the question of whether a photon, say, takes on wave or particle nature before it interacts with the experiment, sensing (in a way) what kind of experiment is going to be performed on it. In the simplest version of the delayed choice experiment, the nature of the experiment would be changed "after the photon had made up its mind" whether it was going to play the role of particle, or whether it would make an appearance as a wave. Needless to say, this is of course not how quantum mechanics works, and Wheeler was fully aware of it. His interpretation was that a photon is neither wave nor particle, and that it takes on one of the two "coats" only when it is being observed. I'm going to tell you that I agree with the first part (the photon is neither wave nor particle), but I disagree with the second part: it does not in fact take on either particle or wave nature after it is observed. It never ever takes on such a role.

If you think about it, the idea that a system only "comes into being by being observed" is preposterous (however, such a thought was quite in line with some other of Wheeler's philosophies). Measurements are interactions with other systems just as much as any other interactions are: there is nothing special about measurement. This is, in essence, what I'm going to try to convince you of.

Even though the reasoning behind the delayed-choice experiment is preposterous, it has generated an enormous amount of work. Let's first look at how we may set up such an experiment. Below is an illustration of a double-slit experiment from Feynman's famous lecture, where he replaced photons by electrons shot out of an electron gun (such devices are perfectly reasonable and feasible). Note that Caltech, where Feynman spent the majority of his career, has made these lectures freely available. The particular chapter can be accessed here

 Fig. 1: An interference experiment with electrons. (Source: Feynman Lectures on Physics)
Later on, we're going to be using photons instead of electrons for the quantum system, because experiments are much easier with photon beams as opposed to electron beams.  In that case, we are going to assume that any light is going to be so faint that it can't be thought of as the classical light waves that give rise to Young's interference fringes. Then, at any point in time, there will be at most one photon between the double-slit and the detector, so you have to think about single photons either taking one or the other, or both paths, through the double-slit experiment.

Quantum mechanics predicts that a single electron takes both paths to create the interference pattern in the figure above at (c). Thus, it must somehow interfere with itself, which is difficult to imagine if you think of the electron as a particle. (Which of course it is not). Can we force it to behave as a particle? Suppose you put a particle detector between the wall and the backstop: one behind slit 1, and one behind slit 2. If you get a "hit" on either detector, then you know which path the electron travelled. (You can do this experiment without actually removing the electron, so that you can still get patterns on the screen.) When you obtain this "which-path" information, the interference pattern disappears: you've forced the electron to behave as a particle.

Wheeler's idea was this: Suppose the distance between the wall and the backstop is very, very large. If you do not put the contraption that will measure which path the electron took (the "which-path detector") into the experiment, the electron would have no choice but to go along both paths, ready to interfere with itself and create the interference pattern on the screen. But suppose you bring in the "which-path" apparatus after the electron has passed the slit, but before it is going to hit the screen. Is the electron wave function that is on the "other path" going to "change its mind", or go backwards? What would happen? The thought experiment very nicely illustrates how preposterous the idea is that the experiment itself determines "what the quantum system is", as changing the experiment mid-flight cannot possibly change the nature of the electron.

The experiment I'm going to describe to you (the delayed-choice quantum eraser experiment) has in fact been carried out several times now, and drives Wheeler's idea to the extreme. The choice of experiment (insert the "which-path" detector or not), can be made after the electron has hit the screen! If you are a reader for whom this is immediately obvious, then congratulations (and consider a career in quantum physics, if this is not already your career). It is indeed completely obvious if you understand quantum mechanics, but let me walk you through it anyway.

First, if it was the experiment that determines the nature of the quantum system (particle or wave), how can you change the experiment after it already has occurred? That this is possible is also due to the peculiarities of quantum physics, and it is also the hardest to explain. I'll do it with photons rather than electrons, as this is the experiment that was carried out, and it is also the description I used in the paper that I'm really writing about. You knew this was coming, didn't you?

We can do double-slit experiments with photons just as with electrons: we just have to turn down the intensity of light such that individual photons can be registered on a phosphorescent screen. When you see the screen light up at a particular spot (or, in more modern times, a pixel on a CCD detector lights up), you interpret it that a photon has hit there. Often, the double-slit is replaced by a Mach-Zehnder interferometer, but you shouldn't worry about such technicalities: you can in fact use either.

To pull off this feat of changing the experiment after the fact, you have to create an entangled pair of photons first. You already know what an entangled pair (a "Bell-state") is, because I wrote about it several times: for example in the context of black holes here, and in the context of quantum teleportation and superdense coding here. This pair of photons is also sometimes called an Einstein-Podolsky-Rosen (EPR) pair, because that trio first described a similar entangled state in a very famous paper in 1935.

Let's create such a pair by entangling the "polarization" degree of freedom of the photon. This is the part that is a bit more complicated: to understand it, you have to understand polarization.

Every photon can come in two different polarization states, but what these states are depends on how you decide to measure them. This will be crucial, because this is in fact how you change the measurement after the fact. The thing to know about an entangled pair is that it is in a superposition of those two states. Suppose we use as basis for the photon polarization the "horizontal/vertical" basis. That means that if a photon is polarized horizontally, and you put a filter in front of it that only allows vertical polarization to go through, then out comes nothing. Polarization is, if you will, a photon's way of wiggling. Below is a picture which shows the photon wiggling in the "vertical" and in the horizontal way. But they can also wiggle in the "circular-left" and "circular-right" way. In fact, it can wiggle in an infinite number of "opposing ways", and these are related to each other by a unitary transformation.

 Fig. 2: One way of depicting photon polarization.
The way a photon is polarized can be changed by an optical element (a "wave plate"), and this ability will be key in the experiment. Suppose we begin with a pair of photons A and B in a Bell-state, written in terms of the horizontal $|h\rangle$ and vertical $|v\rangle$ polarization eigenstates:

$|\Psi\rangle_{AB}=\frac1{\sqrt2}(|h\rangle_A|v\rangle_B+|v\rangle_A|h\rangle_B)$          (1)

You notice that neither of the photons has a defined state, but if I measure one of them (say A) and find that my detector says it is in an $|h\rangle$ state, then I can be sure that measuring B will give you "v", no matter whether you do the measurement now, or a year later with a detector placed a light year away. This is precisely what Einstein could not stomach, calling this mysterious bond "spooky action at a distance", but a careful analysis reveals that there is no "action" at all: signals cannot be sent using this bond.

But here's the thing: I can measure photon B either in the h,v coordinate system, or in another one. This will become crucial, so keep this in mind. But for the moment let's forget that a "copy" of photon A (the entangled partner) is flying out there, possibly to a measurement device a light-year away. Actually, there is nothing a light year away from us, so let's say we are far in the future and the detector is on Proxima Centauri, about 4 and a quarter light years away. It'll just be a longer experiment.

Photon A now goes through a double-slit, just as the electrons in Figure 1. Now we'll do the "are you a particle or a wave" measurement. We do this by putting so-called "quarter-wave plates" in the path of the photons. When you do this, you entangle the polarization of the photon with the spatial degree of freedom (namely "left slit" or "right slit"). Once you've done this, you only have to measure the polarization of photon A to know whether it went through the left or right slit. In a way, you've tagged the photon's path by the polarization. After doing this, you will lose the interference pattern. You can either have an interference pattern (and we say that the photon wavefunction is "coherent"), or you can have "which-path" information, which makes the wavefunction incoherent. Or so people thought for a long time. It turns out that you can also have a a little bit of both, but you can't have both full which-path information, and full coherence: there is a tradeoff. And that tradeoff depends on the angle by which you rotate the polarization basis. In the description above, we used "quarter-wave" plates, which give you full information, and zero coherence. Choose something other than 45 degrees (that's the quarter wave), and you can get a little bit of both.

It turns out that there is a simple relationship that quantifies this tradeoff in terms of the angle you choose to do the tagging with. Let's call this angle $\phi$. We can then define the "distinguishability" D and the "visibility" V, where $D^2$ measures how well you can distinguish the photon paths (a measure of which-path information), while $V^2$ quantifies the visibility of the interference fringes (a measure of the coherence of the wavefunction). A celebrated inequality (due to Greenberger and Yasin [1]) states that
$D^2+V^2\leq1$     (2)

Now, according to what I just wrote, choosing the angle of the wave plate when performing the which-path entangling operation chooses the experiment for you: Set it at 0 degree and you do not entangle at all, so that no which-path information is obtained (then $D^2=0$ and $V^2=1$). Set it at $\phi=\pi/4$, and you get perfect which-path information, and no visibility. How can you choose the experiment after the fact, when you have to choose the angle when setting up the experiment? How?

So the following is what makes quantum mechanics so beautiful. You can actually do this because when I described the experiment to you, I did not (it turns out) use an entangled EPR pair as the input, I used a photon in a defined polarization state, such as $|h\rangle$. I did not tell you about this because it would have confused you. I needed you to understand how to extract which-path information first, and how doing it gradually will gradually destroy coherence.

Now take a deep breath, and read very slowly.

If the input to the two-slits (and therefore to the "which-path" detector that entangles polarization and path) is the EPR state Eq. (1), you actually do not get any which-path information using the quarter-wave plate. This is because when the photon "comes in", it is not in a defined polarization state. If it was not in a defined state, you extract nothing. So for that setup, $V^2=1$ even though $\phi=\pi/4$.

Now one more deep breath after you digested this bit. Maybe take two, just to be safe.

Whether the state that comes in to the two slits is indeed Eq. (1) is up to the person at Proxima Centauri, a year after that data was recorded on the CCD screen on Earth.  This is because of what is $|h\rangle$ and what is $|v\rangle$ is determined by how you measure it. A quantum system does not have a state until you say how you measure it. It will be in the h,v basis if that is the basis of your measurement device. It will be in the R,L  (right-circular, left-circular) basis, if that is instead what you will choose to examine it with. Or it could be anything in between.

I wrote about this at length in the blog post about the collapse of the wavefunction, within the "On quantum measurement" series. (Rightfully, the present post really should be "On quantum measurement. Part 8, but I decided to make it stand alone). Please go back to that if the two breaths did not help. There is also an intriguing parallel to how Shannon entropy is not defined until you determine how you will be measuring it, as I wrote about in "What is Information-Part 1".  The deeper reason for this is that all of physics is about the relative state of measurement devices. Mark my words.

The reason our person at Proxima Centauri handling photon B actually prepares the state is because photon A is not "projected" at any point of the experiment. This could be done, of course, but that is a different experiment. So now we can see how the delayed-choice experiment works: If Proxima Centauri person (PCP, for short) measures at an angle $\theta=0$ with respect to the preparation Eq. (2), then the photon is in a defined state (no matter whether the outcome is h or v) and only then do you actually extract which-path information. In that case, visibility $V^2=0$. If PCP measures at $\theta=\pi/4$ on the other hand, the entanglement operation (the "tagging") does not work: it is as if the measurement by PCP "erased" the tagging, and $V^2=1$ instead. So indeed, a measurement far in the future (well, here more than four years in the future) will determine what kind of an experiment is done on the photon. The event far in the future will determine whether the photon appeared as a particle, or a wave. Weird, right?

What is that you ask? How can an event far in the future affect the data that are stored on a device far in the past?

I didn't say it did, did I? Of course it does not. The truth is much more magical. Without going into all the details here (but which you can read about in any paper about the Bell-state quantum eraser, or indeed my own paper referenced below), the result of the measurement by PCP in the future contains crucial information about how to decode the data in the past, information that is akin to the key in a cryptography procedure.

Yes, cryptographic. That is indeed what I wrote. You will only be able to decipher $D^2$ and $V^2$ when the measurement in the future (which is really a state preparation in the past) is available to you. That is the true magic of quantum mechanics. Without it, you won't be able to see any fringes in the data. But with it, you may be able to reconstruct them to full visibility, if that is how the photon was measured at Proxima Centauri.

How do I know any of this is true? Because we (my student Jennifer Glick and I) analyzed the entire experiment in terms of quantum information theory, and ultimately were able to write down the equations that describe discrimination and visibility (coherence) entirely in terms of entropies and information, in [2] (Jennifer did all the calculations and wrote the first draft of the manuscript). Clearly, "which-path information" should have an obvious information-theoretic rendering, but it turns out that this is actually a little bit tricky because it really is a "conditional information". But it turns out that "coherence" (or "visibility") can also be measured information-theoretically. And lo and behold, the two are related. In our description, they are related by a common information-theoretic identity: the chain rule for entropies. According to that identity, information I and coherence C (as a function of the PCP angle $\theta$) are related so that
$I(\theta)+C(\theta)=1$        (3) .
In a simple qubit model, the information and coherence take on extremely simple forms, namely $I(\theta)=H[\sin^2(\theta+\pi/4)]$ with $C(\theta)=1-H[\sin^2(\theta+\pi/4)]$, where $H[p]$ is the standard Shannon entropy function $H[p]=-p\log(p)-(1-p)\log(1-p)$. And take a look at how our information-theoretic quantities compare to the quantum optical measures of discrimination and visibility in Fig. 3 below. It almost looks like that discrimination and visibility (coherence) should have been defined information-theoretically from the outset, doesn't it?
 Fig.3: Top: Which-path information (solid line) and coherence (dashed line) in terms of quantum information theory. Bottom: Discrimination (solid) and visibility (dashed)  in quantum optics. Q refers to the quantum state at the beam-splitter, and $D_A$  and $D_B$ refer to polarization detectors. From [2].
So what does all this teach us about quantum mechanics in the end (besides, of course, that quantum mechanics is awesome)? We have learned at least two things. Quantum systems are not either particle or wave. They are in fact neither because both concepts are classical in nature. This, to some extent, I stipulate we knew already. Wheeler knew it.  (Bohr, I contend, not so much). But what I've shown you is that quantum systems don't "change their colors" after measurement either, as Wheeler had advocated. They remain "neither", even when we think we pinned them down, because what I've shown you is that you can have them take on this coat or that, or any in between, years after the ink has dried (I mean, after the data were recorded). They (the photons, electrons, etc.) are not one or the other. They appear to you the way you choose you want to see them, when you interrogate a quantum state with classical devices.

Those devices cannot reveal to you the reality of the quantum state, because the devices are classical. Don't hate them because of their limitations. Instead, use them wisely, because what I just showed you is that, if used in a clever manner, they enable you to learn something about the true nature of quantum physics after all. As, for example, the experiment in [3] does.

References

[1] D.M. Greenberger and A. Yasin, "Simultaneous wave and particle knowledge in a neutron interferometer. Physics Letters A 128 (1988) 391-394.
[2] J.R. Glick and C. Adami, "Quantum information theory of the Bell-state quantum eraser". Phys. Rev. A 95 (2017) 012105. Full text also on arXiv
Note: Jennifer Glick is first author on this paper because she performed all calculations in it and wrote the first draft.
[3] Y.H. Kim, R. Yu, S.P. Kulik, Y.H. Shih, and M.O. Scully, “Delayed “choice” quantum eraser,” Phys Rev Lett 84 (2000) 1-5.