by Jed Brody
Alice and Bob decide it’s festive to paint their buttons, so they paint the A and B buttons green, and they paint the A′ and B′ buttons red (figure 11).
Now, Alice and Bob decide to award themselves points according to the following rules:
Figure 11 Alice and Bob install RED and GREEN lights on their analyzers, and red and green buttons.
•If Alice pushes her green button and Bob pushes his red button, they get a point if their lights flash different colors (one green, one red).
•For all other combinations of button presses, they get a point if their lights flash the same color (both green or both red).
After many measurements, over all combinations of button presses, they find that 85 percent of the time, they get a point.
We want to figure out how the photons manage to award points 85 percent of the time. We assume that the photons don’t have any advance notice about which buttons will be pressed; perhaps the buttons are pressed while they are in flight. We also assume one photon can’t send messages to the other about which button was pushed at its end. So, each photon is isolated from the other when it’s compelled to cause either the green light or red light to flash.
So, the only information available to each photon is the direction of the analyzer that it encounters. The press of a red or green button sets the direction. What strategies are available to each photon? We can think of only four:
1.It can make the green light flash no matter which button was pressed.
2.It can make the red light flash no matter which button was pressed.
3.It can cause a flash of the light with the same color as the button that was pressed.
4.It can cause a flash of the light with a different color than the button that was pressed.
The photon could decide to randomly flash either the green light or the red light, but this is equivalent to randomly choosing between Strategy 1 and Strategy 2. So, the four strategies, including any kind of random selection among them, form a comprehensive set of options.
It’s not necessary for both photons in a pair to be encoded with the same strategy; for example, one photon may follow Strategy 1, while the other follows Strategy 2. So, there are sixteen possible combinations of strategies for each photon pair: for each of the four choices available to one photon, the same four choices are available to the other photon. I’ll go through just four of the sixteen combinations, the cases in which both photons follow the same strategy.
Suppose both photons follow Strategy 1, so that both lights flash green no matter which buttons are pressed. How often do Alice and Bob get a point? If both lights flash the same color, Alice and Bob get a point unless Alice presses her green button and Bob presses the his red button—and this happens 25 percent of the time because each of the four combinations of button presses is equally likely. So, Alice and Bob get a point 75 percent of the time if both photons employ Strategy 1. Yet, in fact, Alice and Bob get a point 85 percent of the time. So, Strategy 1, when employed by both photons, is inconsistent with experiment.
If both photons follow Strategy 2, both lights flash red no matter what, and again Alice and Bob get a point 75 percent of the time (in all cases except when Alice presses her green button and Bob presses his red button). So Strategy 2 isn’t any more successful.
If both photons follow Strategy 3, the light that flashes is the same color as the button pressed. So let’s examine all four combinations of equally likely button presses: When Alice and Bob both press their green button, both lights flash green, and they get a point. When both press their red button, both lights flash red, and they get a point. When Alice presses green and Bob presses red, the lights flash different colors, but they get a point because this is the one combination of button presses in which a point is awarded for flashing two different colors. But when Alice presses red and Bob presses green, the lights flash different colors, and they don’t get a point. So a point is awarded 75 percent of the time, and again the strategy is unsuccessful at reproducing experimental results.
Finally, if both photons follow Strategy 4, the light that flashes is different from the color of the button that was pressed. When Alice and Bob both press green, both lights flash red, and they get a point. When they both press red, both lights flash green, and they both get a point. When Alice and Bob press buttons of different colors, the lights flash different colors. This results in a point when Alice presses green and Bob presses red, but not when Alice presses red and Bob presses green. Once again, they get a point during only three of the four combinations of button presses.
We can also look at cases in which the two photons follow different strategies. But we’ll never find a combination of strategies that awards a point more than 75 percent of the time. So no combination of strategies can produce the experimental result that a point is awarded 85 percent of the time.
This means we made at least one false assumption. We assumed locality: each photon is uninfluenced by the measurement made on the other photon. We also assumed realism: each photon has properties (a “strategy”) that predetermine that outcome of any possible measurement; in other words, we assumed the measurement reveals a property that the photon already had.
Identical Twins: An Analogy
We’ll establish another bona fide Bell inequality, without really using any math at all. All we need is basic logic. This example comes from Anton Zeilinger, who drew on earlier examples from Bernard d’Espagnat and Eugene Wigner.4
First, let’s imagine a number of people in an auditorium. Some of the people, perhaps, have brown hair, and some, perhaps, have brown eyes. Let’s compare the number of people with brown hair and brown eyes, to the number of people with brown hair and any color eyes:
(# with brown hair and brown eyes) ≤ (# with brown hair).
The left side indicates the number of people with brown hair and brown eyes. The right side indicates the number of people with brown hair and any eye color. The left side is more restrictive. The right side is larger than the left side by the number of people with brown hair and non-brown eyes. The number of people with brown hair and non-brown eyes may be 0, which is why we use ≤ instead of <. We’re about to use this kind of logic to derive a Bell inequality.
Now we want to think about identical twins, subject to the following conditions:
•Both twins in a pair have the same height: tall or short.
•Both twins in a pair have the same hair color: brown or blond.
•Both twins in a pair have the same eye color: brown or blue.
We’ll restrict our attention to twins with traits listed above. So, for example, we’ll exclude twins with red, gray, or white hair, or no hair. We’ll also assume that the twins don’t dye their hair (or if they do, they dye it identically, either brown or blond).
According to these conditions, tall, brown-haired twins have either brown or blue eyes:
(# pairs tall, brown hair) = (# pairs tall, brown hair, brown eyes) + (# pairs tall, brown hair, blue eyes).
Let’s establish the fact that
(# pairs tall, brown hair, brown eyes) ≤ (# pairs with brown hair, brown eyes).
This is true because the right side is less restrictive. The right side, but not the left, includes any pairs of brown-haired, brown-eyed twins who are short. Let’s also use the fact that
(# pairs tall, brown hair, blue eyes) ≤ (# pairs tall, blue eyes).
This is true for a similar reason: the right side is less restrictive and includes any pairs of tall, blue-eyed twins who have blond hair. We can combine these two facts with the bold-faced equation above. Each term in the bottom line of the bold-faced equation will be replaced by a term that is at least as large:
(# pairs tall, brown hair) ≤ (# pairs with brown hair, brown eyes) + (# pairs tall, blue eyes).
The left side of this inequality indicates the number of pairs of twins who are tall and brown-haired, regardless of eye color. The tall, brown-haired twins with brown eyes are included
in the first condition on the right side of the inequality, and the tall, brown-haired twins with blue eyes are included in the final condition. Thus, every pair of twins counted on the left side is counted once on the right side. The right side includes two additional sets of twins: any brown-haired, brown-eyed twins who are short, and any tall, blue-eyed twins who have blond hair. This is why the right side may be larger than the left.
We recognize that both twins are tall if one twin is tall, and both twins have brown hair if one twin has brown hair. When I write, “one twin is tall,” I don’t mean that only one twin is tall. Instead, I mean we might observe that one twin is tall, and we immediately infer that the other twin must also be tall. Now we can write
(# pairs tall, brown hair) = (# pairs in which one twin is tall and the other twin is brown-haired),
or, more succinctly,
(# pairs tall, brown hair) = #(one tall, one brown-haired).
Similarly,
(# pairs with brown hair, brown eyes) = #(one brown-haired, one brown-eyed),
and
(# pairs tall, blue eyes) = #(one tall, one blue-eyed).
Rewriting our bold-faced inequality this way, we obtain
#(one tall, one brown-haired) ≤ #(one brown-haired, one brown-eyed) + #(one tall, one blue-eyed).
This is a Bell inequality for pairs of twins. Its validity seems incontrovertible. But how does it apply to entangled photons? We saw in chapter 3 that we can create entangled photon pairs such that:
•If one polarizer is horizontal and one polarizer is vertical, there’s no possibility of a coincidence. (If one photon goes through a vertical polarizer, there’s no possibility that the other photon goes through a horizontal polarizer.)
Vertical polarization and horizontal polarization are mutually exclusive, like brown hair and blond hair (in our simple dichotomy). We can think of mutually exclusive hair colors as analogues of mutually exclusive polarization directions. Let’s associate vertical polarization with brown hair, and horizontal polarization with blond hair. What about eye color, and height? We’ll associate them with other mutually exclusive polarizer angles that we saw in the previous chapter:
•There’s no possibility of a coincidence when one polarizer is set to 30° and the other is set to 120°.
•There’s no possibility of a coincidence when one polarizer is set to −30° and the other is set to 60°.
We’ll associate 30° with tall, and 120° with short. Similarly, we’ll associate −30° with blue eyes, and 60° with brown eyes. Here’s the summary of polarizer angles and their associated traits:
0° (vertical): brown hair
90° (horizontal): blond hair
30°: tall
120°: short
−30°: blue eyes
60°: brown eyes
Now we simply translate our Bell inequality for twins,
#(one tall, one brown-haired) ≤ #(one brown-haired, one brown-eyed) + #(one tall, one blue-eyed),
into a Bell inequality for entangled photons:
#(one 30°, one 0°) ≤ #(one 0°, one 60°) + #(one 30°, one −30°).
Rewriting this more compactly,
N(30°,0°) ≤ N(0°,60°) + N(30°, −30°),
where N(30°,0°) is the number of coincidences measured in some time interval when one polarizer is set to 30°, and the other is set to 0° (vertical). N(0°,60°) and N(30°,−30°) are defined similarly. This inequality is violated by measured data!
The Bell inequality, when applied to twins, is absolutely irrefutable. If you reread the steps leading to the result, I hope you’ll agree that there are no dubious assumptions at all. If you actually do a survey of twins (who are either tall or short, either brown-haired or blond-haired, and either brown-eyed or blue-eyed), the inequality will certainly be satisfied. So why does the inequality fail for entangled photons? We have to think carefully about the (very reasonable) assumptions that we made about twins.
We assumed realism: eye color, for example, is completely independent of whether anyone is observing it. (The role of observation is not explicit in our derivation of the inequality. But before counting up pairs of twins with certain combinations of traits, someone has to observe those traits.) Prior to observation, the twin’s eye color is not some indeterminate mixture of brown and blue, which instantly coalesces to one color or the other at the moment of observation. Indeed, the DNA, shared by both twins, is the hidden variable that causes identical twins to have identical traits regardless of whether anyone is observing them. (DNA is the hidden variable as long as the twins don’t dye their hair or wear colored contact lenses.)
We also assumed locality: one twin’s height does not depend on whether the other twin’s height, hair, or eyes are being observed! This assumption is so obviously true that we’re not even aware that we’re making it. Yet this assumption, combined with the assumption of realism, leads to the Bell inequality that definitely does not apply to entangled photons.
When we rewrote the inequality to apply to entangled photons, we implicitly assumed that every photon all along has properties that predetermine whether it will pass through a polarizer at any chosen angle (just as the twins all along have definite eye color, hair color, and height). Indeed, the photons would have to satisfy the inequality if they had fixed polarization properties all along (realism), and if each photon was uninfluenced by the other photon’s polarizer (locality). But in fact, the entangled photons violate the inequality.
So, we’ve done it again! Assuming local realism, we derived a constraint that is violated by measurements of entangled photons. We used practically no math at all, just pure logic. Why, then, do physicists usually use Bell inequalities based on calculus? Are physicists being deliberately difficult to confuse outsiders and make themselves seem smart?
I think there’s a legitimate reason why other Bell inequalities are valuable. In the derivations of other Bell inequalities, the assumptions of realism and locality appear very explicitly in specific steps. The derivation above, though as simple as we can hope for, somewhat obscures the exact role of the assumptions (which we’re not even aware we’re making when thinking about twins).
Indeed, the photons would have to satisfy the inequality if they had fixed polarization properties all along (realism), and if each photon was uninfluenced by the other photon’s polarizer (locality).
Let’s consider the source of entangled photons described in chapter 3: half the photon pairs are horizontally polarized, and half are vertically polarized. Let’s think a little about the quantum prediction for the three terms in the inequality,
N(30°,0°) ≤ N(0°,60°) + N(30°, −30°).
For our entangled photons, the quantum prediction (confirmed by measurement) is that the term on the left side is 50 percent larger than the sum on the right. We can actually prove this result using just a few ideas:
•As stated in chapter 3, a photon has a 75 percent chance of passing through a polarizer tilted 30° from the photon’s polarization. And, a photon has a 25 percent chance of passing through a polarizer tilted 60° from the photon’s polarization.
•Assume one photon is measured slightly (or much) earlier than the other. The first photon has a 50 percent chance of passing through the polarizer (regardless of its orientation). This is a fact that we haven’t previously stated, but it’s true.
•Let’s suppose, just for convenience, that the measurement of one photon immediately creates a definite polarization for both photons. This is spooky action at a distance: when one photon passes through a polarizer, the distant photon immediately acquires the same polarization (for all practical purposes). Some physicists strenuously reject spooky action at a distance, which is why I include the disclaimer “for all practical purposes.” (We will discuss interpretations of quantum physics in chapter 6.)
To predict N(30°,0°), N(0°,60°), and N(30°, −30°), we need to think about the efficiency of the polarizers and detectors. It’s possible to buy very good polarizers th
at behave almost ideally: they let through all the photons they’re supposed to, and they block all the photons they’re supposed to. Single-photon detectors, though, usually have efficiencies much less than 100 percent: they fail to respond to some of the photons that reach them. But detector efficiency affects every coincidence count by the same factor. If all the terms in our Bell inequality are reduced by the same factor, there’s no change in which side is larger. So we’ll ignore detector efficiency and briefly revisit it in chapter 6.
Let’s calculate N(30°,0°), the number of coincidences detected when one polarizer is set to 30°, and the other is set to 0°. Suppose the first photon reaches the 30° polarizer. The photon has a 50 percent chance of passing through because the first photon in a pair to reach a polarizer (at any angle) has a 50 percent chance of passing through. If it passes through, the other photon acquires the 30° polarization and thus has a 75 percent chance of passing through the 0° polarizer. Consequently, the probability that a photon pair passes through both polarizers is 50% × 75% = 37.5% = . In other words, when one polarizer is set to 30° and the other to 0°, of the photon pairs pass through both polarizers. If the total number of photon pairs encountering the polarizers is Ntotal, then N(30°,0°) = .
Let’s do the same for N(0°,60°). If the first photon encounters the 0° polarizer, it has a 50 percent chance of passing through. If it passes through, the other photon acquires the 0° polarization and thus has a 25 percent chance of passing through the 60° polarizer. (The angular difference between the polarizers is now 60°; it was 30° in the previous case.) Thus, the probability that a photon pair passes through both polarizers is 50% × 25% = 12.5% = . In other words, when one polarizer is set to 0° and the other to 60°, of the photon pairs pass through both polarizers, and N(0°,60°) = . So N(0°,60°) is of N(30°,0°) because there are fewer coincidences when there’s a greater difference between polarizer angles.5
