Bell’s inequality, which was formulated by John Bell in 1964, highlights one of the strangest features of quantum mechanics: entanglement, a phenomena which connects the physical attributes of two particles that can be infinitely distant from one another. In this article we will describe the experiment conceived by John Bell, the predictions of this experiment as derived from quantum mechanics, and why they prove the phenomena of entanglement is real.
Electron spin and photon polarization
The Bell experiment can be performed either on photons or electrons. In the case of photons, the property of the particle that is measured is the polarization of the photon, which corresponds to the direction of the electric field to which the photon belongs. In the case of the electron, the property that is being measured is the spin of the electron, which determines how a magnetic field would affect it. A detailed description of these properties is beyond the scope of this article, however we will describe how they are measured and what experiments actually show.
In the case of a photon the polarization can be measured by a polarizer, in which case the polarization will either turn out to be horizontal if the photon passed through the polarizer, or vertical if it got blocked by the polarizer. If a photon passed the polarizer, then it is guaranteed to pass any subsequent polarizer with the same orientation, and get blocked by any subsequent polarizer with a perpendicular orientation. If the subsequent polarizer is oriented at 45o relative to the first polarizer, then the photon will pass it with 50% probability. For other angles the probability for passing the polarizer will change depending on the exact angle.
In the case of an electron, the spin can be measured using a Stern-Gerlach device. This device emits an inhomogeneous magnetic field perpendicular to the motion of the electron along a segment of the electrons trajectory, causing a change in the electrons direction of motion. Due to the magnetic field the electron can either move up, in which case we say it has a spin up, or down, in which case we say it has a spin down. As before, if an electron was measured to have spin up, it is guaranteed that in all subsequent measurements it will also measure with spin up, and vice versa for spin down. If however a subsequent measurement will be taken such that the magnetic field is oriented at 90o degrees relative to the first magnetic field, then with 50% probability the electron will be detected either as having spin up or down (relative to the rotated measurement device). Again, for other angles of rotation the probability to be detected in either spin depends on the exact angle.
For both photons and electrons, it is possible to generate pairs of particle such that their total polarization (for photons) or spin (for electrons) is zero. Taking photons as an example, this means that each photon on its own will display the same behavior as described above, but if it was measured to have horizontal polarization then it is guaranteed that the other photon will measure to have vertical polarization, and vice versa. Furthermore, this behavior is not affected by the absolute orientation of the polarizers: if both polarizers are rotated by some angle the behavior of both photon will not change. It is possible of course to measure each photon with a polarizer at an arbitrary angle, such that the polarizers are not aligned. In this case it is possible that both photons will get blocked or pass through their polarizers, and the rate at which this happens depends only on the relative angle between the polarizers. At the extreme example, the polarizers can be oriented with a 90o angle between them, in which case it is guaranteed that both photon will measure the same.
All this applies also for electrons and measurement of their spin. Both photon polarization and electron spin behaviors are modeled in the same way by quantum mechanics, as we will later describe in more detail. Two particles demonstrating the above behavior are referred to as entangled particles.
It seems natural to assume that when dealing with entangled particles the state of each particle is randomly determined once the pair is created, and the measurements simply reveal what the state was all along. This world view is known as a ‘hidden variable’ theory, alluding to some kind of unknown physical quantity that is not directly observed in these experiments. However, this is not how quantum mechanism describes it. According to quantum mechanism, before the measurement the state of the particles is in a superposition of both ‘vertical’ and ‘horizontal’ states, and the measurement of one particle instantly changes the state of both particles. This phenomena is the famous ‘spooky action at a distance’ that bothered Einstein so much. At first glance, the two world views seem indistinguishable: in either case we get the same measurement results. However, in 1964 during a sabbatical year, the physicist John Bell came up with a brilliant experiment which enables us to discern which world view is true.
The Bell experiment
The Bell experiment consists of a source that emits two entangled particles at a time, each traveling in a different direction. From now on we will assume the particles we use are photons measured by polarizers. Unlike the previous experiments we’ve described where each photon is measured by a polarizer at some fixed angle, Bell proposed a twist: what if each scientist measuring one of the photons decides at random how to orient his polarizer before making the measurement?
Specifically, two orientations for each polarizer are used. If we label one of the photons as $A$ and the other as $B$, and their respective polarizer angles as $a$ and $b$, then for photon $A$ we will use the angles $a_1 = 0^{\circ}$ and $a_2 = 45^{\circ}$, and for photon B the angles $b_1 = 22.5^{\circ}$ and $b_2 = 67.5^{\circ}$:
Each scientist at his measuring device chooses one of these orientations with probability 50% and records the result:
- If the photon passed through the polarizer (horizontally polarized) we denote this as 1
- If the photon got blocked by the polarizer (vertically polarized) we denote this as 0
After many such trials we would get two logs, one from each scientist, that will look something like this:
| $n$ | $a$ | A | $b$ | B |
|---|---|---|---|---|
| 1 | $0^{\circ}$ | 1 | $67.5^{\circ}$ | 1 |
| 2 | $45^{\circ}$ | 1 | $22.5^{\circ}$ | 1 |
| 3 | $45^{\circ}$ | 0 | $22.5^{\circ}$ | 0 |
| 4 | $45^{\circ}$ | 1 | $22.5^{\circ}$ | 1 |
| 5 | $0^{\circ}$ | 0 | $67.5^{\circ}$ | 1 |
| 6 | $45^{\circ}$ | 0 | $67.5^{\circ}$ | 1 |
| 7 | $0^{\circ}$ | 0 | $67.5^{\circ}$ | 0 |
| 8 | $0^{\circ}$ | 1 | $22.5^{\circ}$ | 0 |
| … | … | … | … | … |
For each one of the four possible polarizer orientations we can measure the correlation between $A$ and $B$, which in this case is given by:
$$
E(a,b) = \Pr(A=B) – \Pr(A \neq B)\tag{1}
$$
We can further combine these correlations for all four angle pairs into one metric $S$ which is defined as:
$$
S \triangleq -E(a_1,b_1) + E(a_1,b_2) – E(a_2,b_1) – E(a_2,b_2)\tag{2}
$$
The value of $S$ can be estimated from the log file recorded during the experiment: let $N$ be the total number of trials and let $x_n = 1$ if $A = B$ in experiment $n$ and $0$ otherwise, then the empirical value of $S$ comes out to be:
$$
\hat{S} = \frac{1}{N} \sum_{n=1}^N (2x_n – 1) \cdot (-1)^{I_n} \tag{3}
$$
Where $I_n = 0$ if $(a,b) = (0^{\circ},67.5^{\circ})$ and $1$ otherwise. We can of course increase $N$ as much as we’d like to get better and better estimations of $S$.
As it turns out, in this specific setup the value of $S$ as predicted by quantum mechanics is higher than any value allowed by a hidden value theory, as we will now show.
What quantum mechanics predicts
A single particle
In both cases of the photon polarization and the electron spin, the measurement has only two possible outcomes. Quantum mechanics describes such systems by constructing a two-dimensional vector space over the complex numbers. Each of the two possible outcomes is assigned a basis vector, which we can denote using the ket notation as $|0\rangle$ and $|1\rangle$. A general vector in this space takes the form:
$$
|\Psi\rangle = \alpha_1 |0\rangle + \alpha_2 |1\rangle
$$
Where $\alpha_1,\alpha_2 \in \mathbb{C}$. In addition, we require that the coefficients are normalized as follows:
$$
|\alpha_1|^2 + |\alpha_2|^2 = 1
$$
To this space we attach an additional two-dimensional dual vector space, whose elements are denoted using the bra notation: $\langle |$. The dual space is spanned by the basis vectors $\langle 0|$ and $\langle 1|$, which are related to the original space by the following inner product identities:
$$
\langle 0|0 \rangle = \langle 1|1 \rangle = 1 \;\;\; ; \;\;\; \langle 0|1 \rangle = \langle 1|0 \rangle = 1
$$
Physically, the dual space is used to calculate probabilities of measurement outcomes. If we want to know the probability of finding the system at some state $|\Phi\rangle$:
$$
|\Phi\rangle = \beta_1 |0\rangle + \beta_2 |1\rangle
$$
Then first we define the dual vector $\langle\Phi|$ which is the Hermitian conjugate of $|\Phi\rangle$:
$$
\langle\Phi| = \beta_1^* \langle 0 | + \beta_2^*\langle 1 |
$$
And the probability to observe the state $|\Phi\rangle$ is then given by:
$$
\Pr(|\Phi\rangle) = | \langle \Phi|\Psi\rangle |^2 = |\beta_1^*\alpha_1 + \beta_2^*\alpha_2|^2
$$
Once the measurement is performed there are two possible outcomes: either the system will be found in state $|\Phi\rangle$ or not. If it was found to be in state $|\Phi\rangle$, which happens with the probability we’ve just calculated, then the state of the system will change from $|\Psi\rangle$ to $|\Phi\rangle$. If not, it will change to $|\Phi\rangle$’s orthogonal state:
$$
|\Phi_{\perp}\rangle = -\beta_2^* |0\rangle+\beta_1^*|1\rangle
$$
This abrupt change in the state of the system following a measurement is what’s commonly known as the ‘wave function collapse’. In the experiments we’ve described at the beginning it was not specified what is the state of the photon before measuring it. Nevertheless, let’s suppose its state was:
$$
|\Psi\rangle = \frac{1}{\sqrt{2}} \big|0\rangle + \frac{i}{\sqrt{2}} \big|1\rangle \tag{4}
$$
This state represents a superposition of horizontal and vertical polarizations, with equal weight for each one. When measuring this photon’s polarization the probability for detecting it for example in horizontal state is given by:
$$
\Pr(1) = |\langle 1|\Psi \rangle|^2 = \frac{1}{2}
$$
And the probability for vertical detection is similarly also 50%. If the polarizer is rotated by an angle $\theta$, then the possible outcomes of the measurement are still ‘horizontal’ or ‘vertical’, only now with respect to the rotated polarizer. We can denote the rotated horizontal and vertical states as $|\tilde{1}\rangle$ and $|\tilde{0}\rangle$, which are related to the original states via a rotation matrix:
$$
\left[ \begin{matrix} |\tilde{1}\rangle \\ |\tilde{0}\rangle \end{matrix} \right] =
\left[ \begin{matrix} \cos(\theta) & \sin(\theta) \\ -\sin(\theta) & \cos(\theta) \end{matrix} \right]
\left[ \begin{matrix} |1\rangle \\ |0\rangle \end{matrix} \right] \tag{5}
$$
The inner product of state (4) and state $|\tilde{1}\rangle$ is given by:
$$
\langle \tilde{1}|\Psi \rangle = \frac{\cos(\theta) + i\sin(\theta)}{\sqrt{2}}
$$
And therefore the probability that a particle in state (4) will measure to be in state $|\tilde{1}\rangle$ by our rotated polarizer is given by:
$$
\Pr(\tilde{1}) = |\langle \tilde{1}|\Psi \rangle|^2 = \frac{\cos^2(\theta) + \sin^2(\theta)}{2} = \frac{1}{2}
$$
And similarly $\Pr(\tilde{0}) = 50\%$.
Two particles
When the system consists of two particles, we can measure each one of them independently and detect either $|0\rangle$ or $|1\rangle$. We therefore have in total four measurement outcomes of the system: two for the first particle times two for the second one. The way to describe such a system is by constructing a four-dimensional vector space over the complex numbers, and assign each pair of outcomes a basis vector. We can denote these basis vectors as: $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$, where the first digit describes the state of the first particle (photon $A$) and the second digit describes the state of the second particle (photon $B$). A general vector in this space takes the form:
$$
|\Psi\rangle = \alpha_1 |00\rangle + \alpha_2 |01\rangle + \alpha_3 |10\rangle + \alpha_4 |11\rangle
$$
Where again $\alpha_i \in \mathbb{C}$ and the coefficients are normalized as follows:
$$
|\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 + |\alpha_4|^2 = 1
$$
The state change when measuring only one particle is a bit technical and we will not go over the details in this article. For our purposes it is enough do describe a measurement of both particles, which is essentially the same as for the case of a single particle. Again, we have a four-dimensional dual space over the complex numbers which is spanned by the basis vectors: $\langle 00|$, $\langle 01|$, $\langle 10|$, $\langle 11|$, and the probability for finding the system in some state:
$$
|\Phi\rangle = \beta_1 |00\rangle + \beta_2 |01\rangle + \beta_3 |10\rangle + \beta_4 |11\rangle
$$
Is given by:
$$
\Pr(\Phi) = |\langle\Phi|\Psi\rangle|^2 = \left|\sum_{i=1}^4\alpha_i\beta_i^*\right|^2
$$
In our experiment with two photons exhibiting opposite polarization to one another, we describe their state before measurement with a *maximally entangled* state vector given by:
$$
\Psi = \frac{1}{\sqrt{2}} \big|10\rangle – \frac{1}{\sqrt{2}} \big|01\rangle \tag{6}
$$
Let’s say we measure both photons with a $0^{\circ}$ oriented polarizer. The probability for $|11\rangle$ (both photons passed through the polarizers), is given by:
$$
\Pr(11) = |\langle 11|\Psi \rangle|^2 = 0
$$
Similarly we get $\Pr(00) = 0$, and $\Pr(10) = \Pr(01) = 50\%$. So far this agrees with what we observe in real life. What happens if we rotate both polarizers by an angle $\theta$? In this case the possible measurement outcomes are $|\tilde{0}\tilde{0}\rangle$, $|\tilde{0}\tilde{1}\rangle$, $|\tilde{1}\tilde{0}\rangle$, or $|\tilde{1}\tilde{1}\rangle$. To calculate the probability for each outcome for a particle pair in state (6) we need to express them using $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$. This can again be done independently for each particle using the rotation matrix (5). For example, the state $|\tilde{1}\tilde{1}\rangle$ can be written as:
$$
\begin{align}
|\tilde{1}\tilde{1}\rangle
&= (\cos(\theta)|1\rangle +\sin(\theta)|0\rangle)(\cos(\theta)|1\rangle +\sin(\theta)|0\rangle) \\
&= \cos^2(\theta)|11\rangle + \sin^2(\theta)|00\rangle + \cos(\theta)\sin(\theta)(|10\rangle + |01\rangle)
\end{align}
$$
Taking the inner product with $\Psi$ gives:
$$
\langle\tilde{1}\tilde{1}|\Psi\rangle = 0
$$
Which implies that $\Pr(\tilde{1}\tilde{1}) = 0$. A similar calculation will show that we also get $\Pr(\tilde{0}\tilde{0}) = 0$, and $\Pr(\tilde{1}\tilde{0}) = \Pr(\tilde{0}\tilde{1})= 0.5$. So we see that rotating both polarizers together doesn’t change the outcome: the photons will always turn out to have opposite polarization, and each polarization is observed 50% of the times, which is again consistent with what we see in practice.
But we of course did not have to rotate both polarizers by the same angle. We can easily generalize this result to a case where the polarizer measuring photon A is rotated by some angle $a$ and the polarizer measuring photon B is rotated by some angle $b$. Using $C_a, S_a, C_b, S_b$ as abbreviations for $\cos(a), \sin(a), \cos(b), \sin(b)$ respectively, we get that the state $|\tilde{1}\tilde{1}\rangle$ is now given by:
$$
\begin{align}
|\tilde{1}\tilde{1}\rangle
&= (C_a|1\rangle + S_a|0\rangle)(C_b|1\rangle + S_b|0\rangle) \\
&= C_aC_b|11\rangle + C_aS_b|10\rangle + S_aC_b|01\rangle + S_aS_b|00\rangle
\end{align}
$$
The complete dependency of the rotated states on the original states can be summarized in a matrix form:
$$
\left[ \begin{matrix} |\tilde{1}\tilde{1}\rangle \\ |\tilde{1}\tilde{0}\rangle \\ |\tilde{0}\tilde{1}\rangle \\ |\tilde{0}\tilde{0}\rangle \end{matrix} \right] =
\left[ \begin{matrix}
C_aC_b & C_aS_b & S_aC_b & S_aS_b \\
-C_aS_b & C_aC_b & -S_aS_b & S_aC_b \\
-S_aC_b & -S_aS_b & C_aC_b & C_aS_b \\
S_aS_b & -S_aC_b & -C_aS_b & C_aC_b \\
\end{matrix} \right]
\left[ \begin{matrix} |11\rangle \\ |10\rangle \\ |01\rangle \\ |00\rangle \end{matrix} \right]
$$
We can now use this to calculate the probability of each outcome for a particle pair in state (6):
$$
\begin{align}
\Pr(\tilde{1}\tilde{1}) &= \frac{1}{2}(\cos(a)\sin(b)-\sin(a)\cos(b))^2 \\
\Pr(\tilde{1}\tilde{0}) &= \frac{1}{2}(\cos(a)\cos(b)+ \sin(a)\sin(b))^2 \\
\Pr(\tilde{0}\tilde{1}) &= \frac{1}{2}(-\sin(a)\sin(b)-\cos(a)\cos(b))^2 \\
\Pr(\tilde{0}\tilde{0}) &= \frac{1}{2}(-\sin(a)\cos(b)+ \cos(a)\sin(b))^2 \\
\end{align}
$$
Using the trigonometric angle addition identity simplifies this into:
$$
\begin{align}
\Pr(\tilde{1}\tilde{1}) &= \Pr(\tilde{0}\tilde{0}) = \frac{1}{2}\sin^2(a-b) \\
\Pr(\tilde{1}\tilde{0}) &= \Pr(\tilde{0}\tilde{1}) = \frac{1}{2}\cos^2(a-b) \\
\end{align}
$$
And from this we can calculate the correlation between $A$ and $B$ according to (1):
$$
\begin{align}
E(a,b) &= \Pr(A = B) – \Pr(A \neq B) \\
&= \Pr(\tilde{1}\tilde{1}) + \Pr(\tilde{0}\tilde{0}) – \Pr(\tilde{1}\tilde{0}) – \Pr(\tilde{0}\tilde{1})\\
&= \sin^2(a-b) – \cos^2(a-b)\\
&= -\cos(2(a-b))
\end{align} \tag{7}
$$
Prediction for the Bell experiment
Returning to the Bell experiment, we recall that after a photon pair is created each scientist chooses at random one of two angles with which to orient the polarizer measuring his photon. If we denote these angles as $(a_1,a_2)$ and $(b_1,b_2)$, then by combining the expression for $S$ in (2) with the correlation predicted by quantum mechanics in (7) we get that $S$ should be:
$$
S = \cos(2(a_1-b_1)) – \cos(2(a_1-b_2)) + \cos(2(a_2-b_1)) + \cos(2(a_2-b_2))
$$
This is a four valued function. However, without loss of generality we can choose $a_1 = 0$, making $S$ a three valued function. We can further simplify matters by requiring that for a given choice of $a_2 = \theta$, the angles for polarizer $B$ will be:
$$
b_1 = \theta/2 \;\; ; \;\; b_2 = 3\theta/2
$$
Under this constraint, the value of $S$ reduces to:
$$
S = 3\cos(\theta) – \cos(3\theta)
$$
Which looks like this:
The maximal value of this function is found at $\theta = 45^{\circ}$ where it is equal to:
$$
S_{\text{max}} = 2\sqrt{2}
$$
And this leads us to the choice of angles in the Bell experiment which we described at the beginning:
$$
(a_1,a_2) = (0^{\circ},45^{\circ}) \;\; ; \;\; (b_1,b_2) = (22.5^{\circ},67.5^{\circ})
$$
As a reference, the probabilities related to each choice of polarizer angles are summarized in the following table:
| $a$ | $b$ | $a-b$ | $\Pr(A = B)$ | $\Pr(A \neq B)$ | $E(a,b)$ |
|---|---|---|---|---|---|
| $0^{\circ}$ | $22.5^{\circ}$ | $-22.5^{\circ}$ | 0.146 | 0.854 | -0.708 |
| $0^{\circ}$ | $67.5^{\circ}$ | $-67.5^{\circ}$ | 0.854 | 0.146 | 0.708 |
| $45^{\circ}$ | $22.5^{\circ}$ | $22.5^{\circ}$ | 0.146 | 0.854 | -0.708 |
| $45^{\circ}$ | $67.5^{\circ}$ | $-22.5^{\circ}$ | 0.146 | 0.854 | -0.708 |
So there we have it. Quantum mechanics predicts that if we perform the Bell experiment as described at the beginning and calculate the value of $S$ using equation (3), the result will be approximately 2.82. Let’s see now why this value is not possible if nature uses hidden variables.
What a hidden variables theory predicts
Upper bound on the value of $S$
According to the hidden variables world view, the outcome of the photons measurement is determined once the photon pair is created. Under this framework, the choice of each scientists polarizer orientation does not affect the outcome of the measurement of the other photon. We can describe this mathematically by assuming that the measurement outcome of each photon is determined by some function of the form:
$$
A(a,\lambda) \in \{0,1\} \;\; ; \;\; B(b,\lambda) \in \{0,1\}
$$
The parameter $\lambda$ encapsulates all the unknowns, or ‘hidden variables’, about the photon which determine the measurement results for each choice of polarizer angle. We can imagine that each time we create a photon pair, the value of $\lambda$ is chosen from some unknown distribution $p(\lambda)$. Under this very general model, the correlation $E$ is given by:
$$
\begin{align}
E(a,b) &= \Pr(A(a) = B(b)) – \Pr(A(a) \neq B(b)) \\
&= \int_{\lambda: A(a,\lambda)=B(b,\lambda)} {p(\lambda) d\lambda} – \int_{\lambda: A(a,\lambda)\neq B(b,\lambda)} {p(\lambda) d\lambda}\\
&= \int_{\lambda} {(-1)^{A(a,\lambda)+B(b,\lambda)}p(\lambda) d\lambda}
\end{align}
$$
And the value of $S$ from (2) is given by:
$$
S = \int_{\lambda} {\left[-(-1)^{A_1+B_1} + (-1)^{A_1+B_2} – (-1)^{A_2+B_1} – (-1)^{A_2+B_2} \right]p(\lambda) d\lambda}
$$
Where we abbreviated $A(a_i,\lambda), B(b_i,\lambda)$ with $A_i, B_i$. We can further write $S$ as:
$$
S = \int_{\lambda} {\left[(-1)^{A_1}\left[(-1)^{B_2}-(-1)^{B_2}\right] – (-1)^{A_2}\left[(-1)^{B_2}+(-1)^{B_1}\right]\right]p(\lambda) d\lambda}
$$
For a given value of $\lambda$, either $B_1 \neq B_2$ in which case we get:
$$
\left|(-1)^{B_2}-(-1)^{B_1}\right| = 2 \;\; ; \;\; \left|(-1)^{B_2}+(-1)^{B_1}\right| = 0
$$
Or $B_1 = B_2$ in which case we get:
$$
\left|(-1)^{B_2}-(-1)^{B_1}\right| = 0 \;\; ; \;\; \left|(-1)^{B_2}+(-1)^{B_1}\right| = 2
$$
So we have:
$$
\begin{align}
|S| &\leq \int_{\lambda} {\left|(-1)^{A_1}\left[(-1)^{B_2}-(-1)^{B_1}\right] – (-1)^{A_2}\left[(-1)^{B_2}+(-1)^{B_1}\right]\right|p(\lambda) d\lambda} \\
&\leq \int_{\lambda} {\left|(-1)^{A_1}\left[(-1)^{B_2}-(-1)^{B_1}\right]\right| + \left|(-1)^{A_2}\left[(-1)^{B_2}+(-1)^{B_1}\right]\right|p(\lambda) d\lambda} \\
&= \int_{\lambda} {\left|(-1)^{B_2}-(-1)^{B_1}\right| + \left|(-1)^{B_2}+(-1)^{B_1}\right|p(\lambda) d\lambda} \\
&= \int_{\lambda} {2p(\lambda) d\lambda} \\
&= 2
\end{align}
$$
We find that for a hidden variable theory the value of $S$ can not exceed 2.
Why we can not exceed this bound
At this point you’re hopefully convinced by the math, but you might be left wondering what is it exactly in the hidden variable model that prohibits $S$ to be larger than 2. To understand this better, let’s try to construct functions $A(a,\lambda)$ and $B(b,\lambda)$ and a distribution $p(\lambda)$ that will give the same result as predicted by quantum mechanics. We can start with the correlation between two photons only for the choice of angles $a = 0^{\circ}, b = 22.5^{\circ}$. There are four possible measurement outcomes, and listing them with the variable $\lambda$ directly gives us the desired functions:
| $\lambda$ | $A(0^{\circ},\lambda)$ | $B(22.5^{\circ},\lambda)$ | $p(\lambda$) |
|---|---|---|---|
| 1 | 0 | 0 | 0.073 |
| 2 | 0 | 1 | 0.427 |
| 3 | 1 | 0 | 0.427 |
| 4 | 1 | 1 | 0.073 |
The specific values of $\lambda$ are of course not important, we just need them to distinguish between the different options. Also, we did not need to specify $A(a,\lambda)$ and $B(b,\lambda)$ for any angles other than $a = 0^{\circ}, b = 22.5^{\circ}$, since these are the only angles we use in this case. So we see that when using static polarizer angles, the experimental result can be described by a hidden variable theory. But what happens if we have two possible angle pairs? Let’s say we keep $a$ at $0^{\circ}$ but choose $b$ to be $22.5^{\circ}$ or $67.5^{\circ}$ with 50% probability. In this setup a hidden variable theory must specify for each selection of polarizer angles the measurement result for both photons. There are 8 options in total which we can summarize in a table:
| $\lambda$ | $A(0^{\circ},\lambda)$ | $B(22.5^{\circ},\lambda)$ | $B(67.5^{\circ},\lambda)$ | $p(\lambda$) |
|---|---|---|---|---|
| 1 | 0 | 0 | 0 | 0.0365 |
| 2 | 0 | 0 | 1 | 0.0365 |
| 3 | 0 | 1 | 0 | 0.3905 |
| 4 | 0 | 1 | 1 | 0.0365 |
| 5 | 1 | 0 | 0 | 0.0365 |
| 6 | 1 | 0 | 1 | 0.3905 |
| 7 | 1 | 1 | 0 | 0.0365 |
| 8 | 1 | 1 | 1 | 0.0365 |
The probability for each $\lambda$ is chosen to match the predictions of quantum mechanics. For example, we can see that:
$$
\Pr(A(0^{\circ}) =0, B(22.5^{\circ})=0) = 0.073 = p(1) + p(2)
$$
So we see that also in this case it is possible to find a hidden variables model that explains all observations. But what happens if we try to find such a model for the case of 4 angle pairs like we have in the Bell experiment? in this case there are 16 possible scenarios which we need to account for, for all the possible values of $A(0^{\circ}), A(45^{\circ}), B(22.5^{\circ}), B(67.5^{\circ})$. We need to assign a probability for each scenario, and we need these probabilities to sum up to the probabilities predicted by quantum mechanics. But this is not possible in this case. We will not prove this carefully, but the problem here is that each measurement outcome is the union of 4 events from this list, and it’s probability is therefore the sum of the 4 corresponding probabilities. Some of these outcomes have probability 0.073, which forces the corresponding constituent probabilities to be small. But each one of them also appears in a combination that should sum to 0.427, however since $0.427 > 4\times0.073$ this is not possible.
This conclusion shouldn’t be surprising, as we’ve already proved in the previous part that this setup is not possible, but hopefuly it gives some insight as to what exactly breaks down for the hidden variables theory specifically in the Bell experiment.
Closing remarks
When the Bell experiment was actually performed the value of $S$ was indeed found to be $2\sqrt{2}$, as predicted by quantum mechanics, thus disproving any local variable theory. The philosophical implications of this result are still debated, and will probably continue to be debated for a long time, but the bottom line is that quantum mechanics, which was already well established at the time, became even more widely accepted as the best theory describing the particles in our universe.
The metric $S$ we used in the experiment is known as the Bell-CHSH parameter, and was derived in 1969 by physicists John Clauser, Michael Horne, Abner Simony and Richard Holt following John Bells original publication in 1964. In his original paper, Bell used a different setup which requires three angles instead of two per each detector. Due to this and other experimental challenges, the CHSH version is the one commonly used today in Bell experiments. John Clauser himself, together with Alain Aspect and Anton Zeilinger, have won the 2022 Nobel physics prize for conducting the Bell experiments based on the CHSH parameter and proving that the quantum mechanical prediction is indeed correct. For more details see here.