Why the Quantum | A Response to Wheeler’s 1986 Paper

Wheeler’s opening statement in his 1986 paper, “How Come the Quantum?” holds as true today as it did then [1]

The necessity of the quantum in the construction of existence: out of what deeper requirement does it arise? Behind it all is surely an idea so simple, so beautiful, so compelling that when — in a decade, a century, or a millennium — we grasp it, we will all say to each other, how could it have been otherwise? How could we have been so stupid for so long?

In this Insight, I will answer Wheeler’s question per its counterpart in quantum information theory (QIT), “How come the Tsirelson bound?” Let me start by explaining the Tsirelson bound and its relationship to the Bell inequality, then it will be obvious what that has to do with Wheeler’s question, “How Come the Quantum?” The answer (the Tsirelson bound is a consequence of conservation per no preferred reference frame (NPRF)) may surprise you with its apparent simplicity, but that simplicity belies a profound mystery, as we will see.

The Tsirelson bound is the spread in the Clauser-Horne-Shimony-Holt (CHSH) quantity

\begin{equation}\langle a,b \rangle + \langle a,b^\prime \rangle + \langle a^\prime,b \rangle – \langle a^\prime,b^\prime \rangle \label{CHSH1}\end{equation}

created by quantum correlations. Here, we consider a pair of entangled particles (or “quantum systems” or “quantum exchanges of momentum”). Alice makes measurements on one of the two particles with her measuring device set to ##a## or ##a^\prime## while Bob makes measurements on the other of the two particles with his measuring device set to ##b## or ##b^\prime##. There are two possible outcomes for either Bob or Alice in either of their two possible settings given by ##i## and ##j##. For measurements at ##a## and ##b## we have for the average of Alice’s results multiplied by Bob’s results on a trial-by-trial basis

\begin{equation}\langle a,b \rangle = \sum (i \cdot j) \cdot P(i,j \mid a,b) \label{average}\end{equation}

That’s a bit vague, so let me supply some actual physics. The two entangled states I will use are those which uniquely give rise to the Tsirelson bound [2-4] , i.e., the spin singlet state and the ‘Mermin photon state’ [5]. The spin singlet state is ##\frac{1}{\sqrt{2}} \left(\mid ud \rangle – \mid du \rangle \right)## where ##u##/##d## means the outcome is displaced upwards/downwards relative to the north-south pole alignment of the Stern-Gerlach (SG) magnets (Figure 1).

Figure 1. A Stern-Gerlach (SG) spin measurement showing the two possible outcomes, up and down, represented numerically by +1 and -1, respectively. Figure 42-16 on page 1315 of Physics for Scientists and Engineers with Modern Physics, 9th ed, by Raymond A. Serway and John W. Jewett, Jr.

This state obtains due to conservation of angular momentum at the source as represented by momentum exchange in the spatial plane P orthogonal to the source collimation (“up or down” transverse). This state might be produced by the dissociation of a spin-zero diatomic molecule [6] or the decay of a neutral pi meson into an electron-positron pair [7], processes which conserve spin angular momentum. For more information about the spin singlet state and the spin triplet states, see this Insight.

The Mermin state for photons is ##\frac{1}{\sqrt{2}} \left(\mid VV \rangle + \mid HH \rangle \right)## where ##V## means the there is an outcome (photon detection) behind one of the coaligned polarizers and ##H## means there is no outcome behind one of the co-aligned polarizers. This state obtains due to conservation of angular momentum at the source as represented by momentum exchange along the source collimation (“yes” or “no” longitudinal). Dehlinger and Mitchell created this state by laser inducing spontaneous parametric downconversion in beta barium borate crystals [8], a process that conserves spin angular momentum as represented by the polarization of the emitted photons. At this point we will focus the discussion on the spin single state for total anti-correlation, since everything said of that state can be easily transferred to the Mermin photon state.

Let us investigate what Alice and Bob discover about these entangled states in the various contexts of their measurements (Figure 2). Alice’s detector responds up and down with equal frequency regardless of the orientation ##\alpha## of her SG magnet. This is in agreement with the relativity principle, aka “no preferred reference frame” (NPRF), where different SG magnet orientations relative to the source constitute different “reference frames” in quantum mechanics just as different velocities relative to the source constitute different “reference frames” in special relativity (see this Insight).

Figure 2. Alice and Bob making spin measurements in the xz plane on a pair of spin-entangled particles with their Stern-Gerlach (SG) magnets and detectors.

Bob observes the same regarding his SG magnet orientation ##\beta##. Thus, the source is rotationally invariant in the spatial plane P orthogonal to the source collimation. When Bob and Alice compare their outcomes, they find that their outcomes are perfectly anti-correlated (##ud## and ##du## with equal frequency) when ##\alpha – \beta = \theta = 0## (Figure 3). This is consistent with conservation of angular momentum per classical mechanics between the pair of detection events (again, this fact defines the state). The degree of that anti-correlation diminishes as ##\theta \rightarrow \frac{\pi}{2}## until it is equal to the degree of correlation (##uu## and ##dd##) when their SG magnets are at right angles to each other. In other words, whenever the SG magnets are orthogonal to each other anti-correlated and correlated outcomes occur with equal frequency, i.e., conservation of angular momentum in one direction is independent of the angular momentum changes in any orthogonal direction. Thus, we wouldn’t expect to see more correlation or more anti-correlation based on conservation of angular momentum for transverse results in the plane P when the SG magnets are orthogonal to each other. As we continue to increase the angle ##\theta## beyond ##\frac{\pi}{2}## the anti-correlations continue to diminish until we have totally correlated outcomes when the SG magnets are anti-aligned. This is also consistent with conservation of angular momentum, since the totally correlated results when the SG magnets are anti-aligned represent momentum exchanges in opposite directions in the plane P just as when the SG magnets are aligned, it is now simply the case that what Alice calls up, Bob calls down and vice-versa.

The counterpart for the Mermin photon state is simply that angular momentum conservation is evidenced by ##VV## or ##HH## outcomes for coaligned polarizers. When the polarizers are at right angles you have only ##VH## and ##HV## outcomes, which is still totally consistent with conservation of angular momentum as ‘not ##H##’ implies ##V## and vice-versa [8]. In other words, a polarizer does not have a ‘north-south’ distinction (longitudinal rather than transverse momentum exchange). In particular, having rotated either or both polarizers by ##\pi## one should obtain precisely ##VV## or ##HH## outcomes again.

Nothing is particularly mysterious about the entangled states for electron spin or photon polarization described here so far because we have been thinking as if conservation of angular momentum holds for each experimental trial, as in classical mechanics. Truth is, since Alice and Bob can only measure +1 or -1 (quantum exchange of momentum per NPRF), we can only get conservation of angular momentum in any particular trial when their SG magnets/polarizers are co-aligned. And, we cannot use classical probability theory to account for the conservation of angular momentum on average.

In particular, the probability that Alice and Bob will measure ##uu## or ##dd## at angles ##\alpha## and ##\beta## for the spin singlet state is
\begin{equation}P_{uu} = P_{dd} = \frac{1}{2} \mbox{sin}^2 \left(\frac{\alpha – \beta}{2}\right) \label{probabilityuu}\end{equation}
And, the probability that Alice and Bob will measure ##ud## or ##du## at angles ##\alpha## and ##\beta## for the spin singlet state is
\begin{equation}P_{ud} = P_{du} = \frac{1}{2} \mbox{cos}^2 \left(\frac{\alpha – \beta}{2}\right) \label{probabilityud}\end{equation}
Using these in Eq. (\ref{average}) where the outcomes are +1 (##u##) and -1 (##d##) gives Eq. (\ref{CHSH1}) of
\begin{equation}-\cos(a – b) -\cos(a – b^\prime) -\cos(a^\prime – b) +\cos(a^\prime – b^\prime) \label{CHSHspin}\end{equation}
Choosing ##a = \pi/4##, ##a^\prime = -\pi/4##, ##b = 0##, and ##b^\prime = \pi/2## minimizes Eq. (\ref{CHSHspin}) at ##-2\sqrt{2}## (the Tsirelson bound).

Likewise, for the Mermin photon state we have
\begin{equation}P_{VV} = P_{HH} = \frac{1}{2} \mbox{cos}^2 \left(\alpha – \beta \right) \label{probabilityVV}\end{equation}
and
\begin{equation}P_{VH} = P_{HV} = \frac{1}{2} \mbox{sin}^2 \left(\alpha – \beta \right) \label{probabilityVH}\end{equation}
Using these in Eq. (\ref{average}) where the outcomes are +1 (##V##) and -1 (##H##) gives Eq. (\ref{CHSH1}) of
\begin{equation}\cos2(a – b) +\cos2(a – b^\prime) +\cos2(a^\prime – b) -\cos2(a^\prime – b^\prime) \label{CHSHmermin}\end{equation}
Using ##a = \pi/8##, ##a^\prime = -\pi/8##, ##b = 0##, and ##b^\prime = \pi/4## maximizes Eq. (\ref{CHSHmermin}) at ##2\sqrt{2}## (the Tsirelson bound). So, we have two mysteries.

First, as explained by Mermin [5], suppose you restrict Alice and Bob’s measurement angles ##\alpha## and ##\beta## to three possibilities, setting 1 is ##0^o##, setting two is ##120^o##, and setting three is ##-120^o##. Eq. (\ref{probabilityud}) says the probability of getting opposite results is 1 when ##\alpha = \beta## (1/2 ##ud## and 1/2 ##du##) and 1/4 otherwise (1/8 ##ud## and 1/8 ##du##). Now, if the source emits particles with definite properties that account for their outcomes in the three possible measurement settings, and we have to get total anti-correlation for like settings, then the particles’ so-called “instruction sets” must be opposite for each of the three settings. For example, suppose we have 1(##u##)2(##u##)3(##d##) for Alice and 1(##d##)2(##d##)3(##u##) for Bob. That guarantees the total anti-correlation for like settings, i.e., 11 gives ##ud##, 22 gives ##ud##, and 33 gives ##du##. And, for unlike settings we get anti-correlation in two combinations, i.e., 12 gives ##ud## and 21 gives ##ud##. In fact, for any instruction set with two ##u## and one ##d## we get anti-correlation for unlike settings in two of the six possible unlike combinations (12,13,21,23,31,32). The only other way to make a pair of instruction sets is to have one with all ##u## and the other with all ##d##. In that case, we get anti-correlation for all six unlike combinations. That means the instruction sets necessary to guarantee anti-correlation for like settings lead to an overall anti-correlation greater than 2/6 for unlike settings, which is greater than the quantum probability for anti-correlation in unlike settings of 1/4. This is Mermin’s version of the Bell inequality [9] (fraction of anti-correlated outcomes for unlike settings must be greater than 2/6) and the manner by which it is violated by quantum correlations (1/4 is less than 2/6). Thus, instruction sets (“counterfactual definiteness”) assumed by classical probability theory cannot account for quantum correlations in this case.

The counterpart to this for the CHSH quantity is that classical correlations give a range of -2 to 2 for the CHSH quantity (“CHSH-Bell inequality”). And, as we saw above, the Tsirelson bound violates the CHSH-Bell inequality. Experiments show that the quantum results can be achieved (violating the Bell inequality), ruling out an explanation of these correlated momentum exchanges via instruction sets per classical probability theory.

The second mystery is that even in cases where we don’t violate the Bell inequality, e.g., ##a = b = 0## and ##a^\prime = b^\prime = \pi/2## which give a CHSH value of 0, we still have conservation of angular momentum. Why is that mysterious? Well, it’s not when the SG magnets are co-aligned, since in those cases we always get a +1 outcome and a -1 outcome for a total of zero. But, in trials where ##\alpha – \beta = \theta## does not equal zero, we need either Alice or Bob, at minimum, to measure something less than 1 to conserve angular momentum. For example, if Alice measures +1, then Bob must measure ##-\cos{\theta}## to conserve angular momentum for that trial. But, again, Alice and Bob only measure +1 or -1 (quantum exchange of momentum per NPRF, which uniquely distinguishes the quantum joint distribution from its classical counterpart [10]), so that can’t happen (Figure 4). What does happen? We conserve angular momentum on average in those trials.

It is easy to see how this follows by starting with total angular momentum of zero for binary (quantum) outcomes +1 and -1 (I am suppressing the factor of ##\hbar/2## and I’m referring to the spin singlet state here [11], Figure 3).

Figure 3. Outcomes (yellow dots) in the same reference frame, i.e., outcomes for the same measurement (blue arrows represent SG magnet orientations), for the spin singlet state explicitly conserve angular momentum.

Alice and Bob both measure +1 and -1 results with equal frequency for any SG magnet angle (NPRF) and when their angles are equal they obtain different outcomes giving total angular momentum of zero. The case (a) result is not difficult to understand via conservation of angular momentum, because Alice and Bob’s measured values of spin angular momentum cancel directly when ##\alpha = \beta##, that defines the spin singlet state. But, when Bob’s SG magnet is rotated by ##\alpha – \beta = \theta## relative to Alice’s, the situation is not as clear (Figure 6).

In classical physics, one would say the projection of the angular momentum vector of Alice’s particle ##\vec{S}_A = +1\hat{a}## along ##\hat{b}## is ##\vec{S}_A\cdot\hat{b} = +\cos{(\theta)}## where again ##\theta## is the angle between the unit vectors ##\hat{a}## and ##\hat{b}## (Figure 2). From Alice’s perspective, had Bob measured at the same angle, i.e., ##\beta = \alpha##, he would have found the angular momentum vector of his particle was ##\vec{S}_B = -1\hat{a}##, so that ##\vec{S}_A + \vec{S}_B = \vec{S}_{Total} = 0##. Since he did not measure the angular momentum of his particle at the same angle, he should have obtained a fraction of the length of ##\vec{S}_B##, i.e., ##\vec{S}_B\cdot\hat{b} = -1\hat{a}\cdot\hat{b} = -\cos{(\theta)}## (Figure 4).

Figure 4. The projection of the angular momentum of Bob’s particle ##\vec{S}_B## along his measurement direction ##\hat{b}##. This does not happen with spin angular momentum due to NPRF.

Of course, Bob only ever obtains +1 or -1 per NPRF, so Bob’s outcomes can only average the required ##-\cos{(\theta)}##. Thus, NPRF dictates