Pairs of Entangled Photons
Single detection probabilities
Entanglement
Entangled photon pairs are excellent for studying entanglement due to their simple two-dimensional state-space, which simplifies formalism and calculations while still capturing entanglement’s nuances.
They are formally equivalent to spin one-half systems and are particularly relevant in quantum optics. Furthermore, entangled photons are experimentally practical, having been used in many key entanglement experiments and quantum information protocols, unlike spin one-half particles.
A typical setup involves a source emitting two photons, \nu_1 and \nu_2, in opposite directions (-z and +z) with distinct frequencies.
Polarization measurements are performed using polarizers oriented along vectors \mathbf a and \mathbf b at angles \alpha and \beta from the x-axis.
The polarization state of the photon pair is described by:
| \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle = \frac{1}{\sqrt{2}} \left(| \mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle\right)
Let us demonstrate that factorization is impossible. Consider the most general product state form for photons \nu_1 and \nu_2:
|\boldsymbol \Psi \rangle = (\gamma |\mathbf x_1\rangle + \lambda |\mathbf y_1\rangle) \otimes (\mu |\mathbf x_2\rangle + \nu |\mathbf y_2\rangle)
Expanding this expression, we decompose the product state onto the complete basis \{|\mathbf x_1, \mathbf x_2\rangle, |\mathbf x_1, \mathbf y_2\rangle, |\mathbf y_1, \mathbf x_2\rangle, |\mathbf y_1, \mathbf y_2\rangle\}:
|\boldsymbol \Psi \rangle = \gamma \mu |\mathbf x_1, \mathbf x_2\rangle + \gamma \nu |\mathbf x_1, \mathbf y_2\rangle + \lambda \mu |\mathbf y_1, \mathbf x_2\rangle + \lambda \nu |\mathbf y_1, \mathbf y_2\rangle
To represent the state:
|\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle = \frac{1}{\sqrt{2}} \left(| \mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle\right) the coefficients must match.
Specifically, the coefficients of |\mathbf x_1, \mathbf y_2\rangle and |\mathbf y_1, \mathbf x_2\rangle in |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle are zero. This requires \gamma \nu = 0 and \lambda \mu = 0.
However, the terms |\mathbf x_1, \mathbf x_2\rangle and |\mathbf y_1, \mathbf y_2\rangle must be present with non-zero coefficients, implying \gamma \mu \neq 0 and \lambda \nu \neq 0.
This creates a contradiction: if \gamma \nu = 0 and \lambda \mu = 0, then either \gamma=0 or \nu=0, and either \lambda=0 or \mu=0. In either case, either \gamma \mu = 0 or \lambda \nu = 0, contradicting the requirement that both are non-zero.
Therefore, no such factorization exists. This impossibility of factorization defines an entangled state |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle that cannot be expressed as a tensor product of individual photon states.
In quantum mechanics, knowing the state allows us to predict probabilities for any observable measurement. In this experiment, we measure the polarization of each photon.
For polarizer (I) oriented along \mathbf a, we can measure either +1 (polarization along | \boldsymbol \varepsilon_\alpha \rangle):
| \boldsymbol +_{\mathbf a} \rangle = | \boldsymbol \varepsilon_\alpha \rangle = \cos\left(\alpha\right) | \mathbf x \rangle + \sin\left(\alpha\right) | \mathbf x \rangle
or -1 (orthogonal polarization):
| \boldsymbol -_{\mathbf a} \rangle = | \boldsymbol \varepsilon_{\alpha + \pi/2} \rangle = -\sin\left(\alpha\right) | \mathbf x \rangle + \cos\left(\alpha\right) | \mathbf x \rangle
Similarly, for polarizer (II) along \mathbf b, we can measure +1 (polarization along \boldsymbol \varepsilon_\mathbf{\beta}):
| \boldsymbol +_{\mathbf b} \rangle = | \boldsymbol \varepsilon_\beta \rangle = \cos\left(\beta\right) | \mathbf x \rangle + \sin\left(\beta\right) | \mathbf x \rangle
or -1 (orthogonal polarization):
| \boldsymbol -_{\mathbf b} \rangle = | \boldsymbol \varepsilon_{\beta + \pi/2} \rangle = -\sin\left(\beta\right) | \mathbf x \rangle + \cos\left(\beta\right) | \mathbf x \rangle
Joint detection probabilities
Let’s compute first the joint probabilities.
The probability \mathcal P_{++}(\mathbf a, \mathbf b) to obtain +1 for photon \nu_1 and +1 for photon \nu_2 is given by the squared modulus of the projection of the state |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle onto the state |+_{\mathbf a}, +_{\mathbf b}\rangle = |+_{\mathbf a}\rangle_1 \otimes |+_{\mathbf b}\rangle_2:
\mathcal P_{++}(\mathbf a, \mathbf b) = |\langle +_{\mathbf a}, +_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2
We have:
\begin{aligned} | +_{\mathbf a}, +_{\mathbf b} \rangle = & (\cos(\alpha) | \mathbf x_1 \rangle + \sin(\alpha) | \mathbf y_1 \rangle) \otimes (\cos(\beta) | \mathbf x_2 \rangle + \sin(\beta) | \mathbf y_2 \rangle) \\ = & \cos(\alpha)\cos(\beta) | \mathbf x_1, \mathbf x_2 \rangle + \cos(\alpha)\sin(\beta) | \mathbf x_1, \mathbf y_2 \rangle \\ & + \sin(\alpha)\cos(\beta) | \mathbf y_1, \mathbf x_2 \rangle + \sin(\alpha)\sin(\beta) | \mathbf y_1, \mathbf y_2 \rangle \end{aligned}
Projecting onto |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle = \frac{1}{\sqrt{2}} \left(| \mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle\right):
\begin{aligned} \langle +_{\mathbf a}, +_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle & = \frac{1}{\sqrt{2}} \left( \langle +_{\mathbf a}, +_{\mathbf b} | \mathbf x_1, \mathbf x_2 \rangle + \langle +_{\mathbf a}, +_{\mathbf b} | \mathbf y_1, \mathbf y_2 \rangle \right) \\ & = \frac{1}{\sqrt{2}} \left( \cos(\alpha)\cos(\beta) + \sin(\alpha)\sin(\beta) \right) \\ & = \frac{1}{\sqrt{2}} \cos(\alpha - \beta) \end{aligned}
The probability is:
\mathcal P_{++}(\mathbf a, \mathbf b) = \left| \frac{1}{\sqrt{2}} \cos(\alpha - \beta) \right|^2 = \frac{1}{2} \cos^2(\alpha - \beta)
The probability \mathcal P_{--}(\mathbf a, \mathbf b) to obtain -1 for photon \nu_1 and -1 for photon \nu_2 is given by the squared modulus of the projection of the state |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle onto the state |-_{\mathbf a}, -_{\mathbf b}\rangle = |-_{\mathbf a}\rangle_1 \otimes |-_{\mathbf b}\rangle_2:
\mathcal P_{--}(\mathbf a, \mathbf b) = |\langle -_{\mathbf a}, -_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2
We have:
\begin{aligned} | -_{\mathbf a}, -_{\mathbf b} \rangle = & (-\sin(\alpha) | \mathbf x_1 \rangle + \cos(\alpha) | \mathbf y_1 \rangle) \otimes (-\sin(\beta) | \mathbf x_2 \rangle + \cos(\beta) | \mathbf y_2 \rangle) \\ = & \sin(\alpha)\sin(\beta) | \mathbf x_1, \mathbf x_2 \rangle - \sin(\alpha)\cos(\beta) | \mathbf x_1, \mathbf y_2 \rangle \\ & - \cos(\alpha)\sin(\beta) | \mathbf y_1, \mathbf x_2 \rangle + \cos(\alpha)\cos(\beta) | \mathbf y_1, \mathbf y_2 \rangle \end{aligned}
Projecting onto |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle = \frac{1}{\sqrt{2}} \left(| \mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle\right):
\begin{aligned} \langle -_{\mathbf a}, -_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle & = \frac{1}{\sqrt{2}} \left( \langle -_{\mathbf a}, -_{\mathbf b} | \mathbf x_1, \mathbf x_2 \rangle + \langle -_{\mathbf a}, -_{\mathbf b} | \mathbf y_1, \mathbf y_2 \rangle \right) \\ & = \frac{1}{\sqrt{2}} \left( \sin(\alpha)\sin(\beta) + \cos(\alpha)\cos(\beta) \right) \\ & = \frac{1}{\sqrt{2}} \cos(\alpha - \beta) \end{aligned}
The probability is:
\mathcal P_{--}(\mathbf a, \mathbf b) = \left| \frac{1}{\sqrt{2}} \cos(\alpha - \beta) \right|^2 = \frac{1}{2} \cos^2(\alpha - \beta) = \mathcal P_{++}(\mathbf a, \mathbf b) The probability \mathcal P_{+-}(\mathbf a, \mathbf b) to obtain +1 for photon \nu_1 and -1 for photon \nu_2 is given by the squared modulus of the projection of the state |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle onto the state |+_{\mathbf a}, -_{\mathbf b}\rangle = |+_{\mathbf a}\rangle_1 \otimes |-_{\mathbf b}\rangle_2:
\mathcal P_{+-}(\mathbf a, \mathbf b) = |\langle +_{\mathbf a}, -_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2
We have:
\begin{aligned} | +_{\mathbf a}, -_{\mathbf b} \rangle = & (\cos(\alpha) | \mathbf x_1 \rangle + \sin(\alpha) | \mathbf y_1 \rangle) \otimes (-\sin(\beta) | \mathbf x_2 \rangle + \cos(\beta) | \mathbf y_2 \rangle) \\ = & -\cos(\alpha)\sin(\beta) | \mathbf x_1, \mathbf x_2 \rangle + \cos(\alpha)\cos(\beta) | \mathbf x_1, \mathbf y_2 \rangle \\ & - \sin(\alpha)\sin(\beta) | \mathbf y_1, \mathbf x_2 \rangle + \sin(\alpha)\cos(\beta) | \mathbf y_1, \mathbf y_2 \rangle \end{aligned}
Projecting onto |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle = \frac{1}{\sqrt{2}} \left(| \mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle\right):
\begin{aligned} \langle +_{\mathbf a}, -_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle & = \frac{1}{\sqrt{2}} \left( \langle +_{\mathbf a}, -_{\mathbf b} | \mathbf x_1, \mathbf x_2 \rangle + \langle +_{\mathbf a}, -_{\mathbf b} | \mathbf y_1, \mathbf y_2 \rangle \right) \\ & = \frac{1}{\sqrt{2}} \left( -\cos(\alpha)\sin(\beta) + \sin(\alpha)\cos(\beta) \right) \\ & = \frac{1}{\sqrt{2}} \sin(\alpha - \beta) \end{aligned}
The probability is:
\mathcal P_{+-}(\mathbf a, \mathbf b) = \left| \frac{1}{\sqrt{2}} \sin(\alpha - \beta) \right|^2 = \frac{1}{2} \sin^2(\alpha - \beta) The probability \mathcal P_{-+}(\mathbf a, \mathbf b) to obtain -1 for photon \nu_1 and +1 for photon \nu_2 is given by the squared modulus of the projection of the state |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle onto the state |-_{\mathbf a}, +_{\mathbf b}\rangle = |-_{\mathbf a}\rangle_1 \otimes |+_{\mathbf b}\rangle_2:
\mathcal P_{-+}(\mathbf a, \mathbf b) = |\langle -_{\mathbf a}, +_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle|^2
We have:
\begin{aligned} | -_{\mathbf a}, +_{\mathbf b} \rangle = & (-\sin(\alpha) | \mathbf x_1 \rangle + \cos(\alpha) | \mathbf y_1 \rangle) \otimes (\cos(\beta) | \mathbf x_2 \rangle + \sin(\beta) | \mathbf y_2 \rangle) \\ = & -\sin(\alpha)\cos(\beta) | \mathbf x_1, \mathbf x_2 \rangle - \sin(\alpha)\sin(\beta) | \mathbf x_1, \mathbf y_2 \rangle \\ & + \cos(\alpha)\cos(\beta) | \mathbf y_1, \mathbf x_2 \rangle + \cos(\alpha)\sin(\beta) | \mathbf y_1, \mathbf y_2 \rangle \end{aligned}
Projecting onto |\boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle = \frac{1}{\sqrt{2}} \left(| \mathbf x_1, \mathbf x_2\rangle + |\mathbf y_1, \mathbf y_2\rangle\right):
\begin{aligned} \langle -_{\mathbf a}, +_{\mathbf b} | \boldsymbol \Psi \left( \nu_1, \nu_2 \right) \rangle & = \frac{1}{\sqrt{2}} \left( \langle -_{\mathbf a}, +_{\mathbf b} | \mathbf x_1, \mathbf x_2 \rangle + \langle -_{\mathbf a}, +_{\mathbf b} | \mathbf y_1, \mathbf y_2 \rangle \right) \\ & = \frac{1}{\sqrt{2}} \left( -\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) \right) \\ & = \frac{1}{\sqrt{2}} \sin(\alpha - \beta) \end{aligned}
The probability is:
\mathcal P_{-+}(\mathbf a, \mathbf b) = \left| \frac{1}{\sqrt{2}} \sin(\alpha - \beta) \right|^2 = \frac{1}{2} \sin^2(\alpha - \beta) = \mathcal P_{+-}(\mathbf a, \mathbf b)
Using the calculated joint probabilities, we can now highlight a key characteristic of entangled states: strong measurement correlations.
Single detection probabilities
Let’s now consider polarization measurements on only one photon, say \nu_1.
In quantum mechanics, to determine the probability of an outcome for a partial measurement, we need to consider all possible complete measurements that are compatible with the partial measurement and sum their probabilities. This principle arises from the Born rule and the completeness of the measurement basis. Essentially, when we measure only photon \nu_1’s polarization, we are ignoring the outcome of the measurement on photon \nu_2.
Therefore, to find the probability of a specific outcome for photon \nu_1, we must sum the probabilities of all possible outcomes for photon \nu_2, given the specific outcome for photon \nu_1.
The probability of obtaining +1 for photon \nu_1 with polarizer (I) set at angle \alpha, is sum the probabilities of the joint measurements where photon \nu_1 is measured as +1 and photon \nu_2 can be either +1 or -1. This means summing \mathcal P_{++}(\mathbf a, \mathbf b) and \mathcal P_{+-}(\mathbf a, \mathbf b).
Using the probabilities we derived:
\begin{aligned} \mathcal P_{+}(\mathbf a) & = \mathcal P_{++}(\mathbf a, \mathbf b) + \mathcal P_{+-}(\mathbf a, \mathbf b) \\ & = \frac{1}{2} \cos^2(\alpha - \beta) + \frac{1}{2} \sin^2(\alpha - \beta) \\ & = \frac{1}{2} \left( \cos^2(\alpha - \beta) + \sin^2(\alpha - \beta) \right) = \frac{1}{2} \end{aligned}
Similarly, the probability of obtaining -1 for photon \nu_1 with polarizer (I) at angle \alpha is the sum of \mathcal P_{-+}(\mathbf a, \mathbf b) and \mathcal P_{--}(\mathbf a, \mathbf b):
\begin{aligned} \mathcal P_{-}(\mathbf a) & = \mathcal P_{-+}(\mathbf a, \mathbf b) + \mathcal P_{--}(\mathbf a, \mathbf b) \\ & = \frac{1}{2} \sin^2(\alpha - \beta) + \frac{1}{2} \cos^2(\alpha - \beta) \\ & = \frac{1}{2} \left( \sin^2(\alpha - \beta) + \cos^2(\alpha - \beta) \right) = \frac{1}{2} \end{aligned}
Regardless of the orientation \mathbf a of polarizer (I), the probability of measuring polarization along \mathbf a (+1) or orthogonal to \mathbf a (-1) for photon \nu_1 is always \frac{1}{2}. This indicates that photon \nu_1, when considered in isolation, appears to be completely unpolarized.
For photon \nu_2, we want to calculate the probabilities of obtaining +1 or -1 polarization, irrespective of the measurement outcome for photon \nu_1.
The probability \mathcal P_{+}(\mathbf b) to obtain +1 for polarizer (II) aligned along \mathbf b is the sum of the probabilities of joint measurements where photon \nu_2 is +1 and photon \nu_1 can be either +1 or -1:
\mathcal P_{+}(\mathbf b) = \mathcal P_{++}(\mathbf a, \mathbf b) + \mathcal P_{-+}(\mathbf a, \mathbf b) Substituting the previously calculated probabilities:
\begin{aligned} \mathcal P_{+}(\mathbf b) & = \frac{1}{2} \cos^2(\alpha - \beta) + \frac{1}{2} \sin^2(\alpha - \beta) \\ & = \frac{1}{2} \left( \cos^2(\alpha - \beta) + \sin^2(\alpha - \beta) \right) = \frac{1}{2} \end{aligned}
Similarly, the probability \mathcal P_{-}(\mathbf b) to obtain -1 for polarizer (II) aligned along \mathbf b is the sum of the probabilities of joint measurements where photon \nu_2 is -1 and photon \nu_1 can be either +1 or -1:
\mathcal P_{-}(\mathbf b) = \mathcal P_{+-}(\mathbf a, \mathbf b) + \mathcal P_{--}(\mathbf a, \mathbf b) Substituting the probabilities:
\begin{aligned} \mathcal P_{-}(\mathbf b) & = \frac{1}{2} \sin^2(\alpha - \beta) + \frac{1}{2} \cos^2(\alpha - \beta) \\ & = \frac{1}{2} \left( \sin^2(\alpha - \beta) + \cos^2(\alpha - \beta) \right) = \frac{1}{2} \end{aligned}
Therefore, just as with photon \nu_1, the probabilities of finding polarization along \mathbf b (+1) or perpendicular to \mathbf b (-1) for photon \nu_2 are both equal to one-half, regardless of the orientation \mathbf b of polarizer (II). Photon \nu_2 also appears to be fully unpolarized when considered individually.
Correlations
We have observed that individual polarization measurements for photons \nu_1 and \nu_2 yield random outcomes, either +1 or -1 with equal probability of 1/2.
Let’s examine the joint measurement probabilities specifically when the polarizers are parallel, i.e., when the angle between \mathbf a and \mathbf b is zero (\alpha = \beta, so \alpha - \beta = 0). In this case, we have:
\begin{aligned} & \mathcal P_{++}(0) = \frac{1}{2} \cos^2(0) = \frac{1}{2} \\ & \mathcal P_{--}(0) = \frac{1}{2} \cos^2(0) = \frac{1}{2} \\ & \mathcal P_{+-}(0) = \frac{1}{2} \sin^2(0) = 0 \\ & \mathcal P_{-+}(0) = \frac{1}{2} \sin^2(0) = 0 \end{aligned}
Consider the probability of obtaining +1 for photon \nu_1, which we calculated earlier as \mathcal P_{+}(\mathbf a) = 1/2. Now, look at the joint probability \mathcal P_{++}(\mathbf a, \mathbf b) = 1/2.
We can compute the conditional probability of finding +1 for photon \nu_2 given that we have found +1 for photon \nu_1. Using the definition of conditional probability:
\mathcal P(+_{\nu_2} \mid +_{\nu_1}) = \frac{\mathcal P(+_{\nu_1} \cap +_{\nu_1})}{\mathcal P(+_{\nu_1})} = \frac{\mathcal P_{++}(\mathbf a, \mathbf b)}{\mathcal P_{+}(\mathbf a)} = \frac{1/2}{1/2} = 1
These probabilities reveal a total correlation between the measurements on \nu_1 and \nu_2.
The result, 1 (100\%), signifies that if we measure +1 for photon \nu_1, we are absolutely certain to also measure +1 for photon \nu_2, when the polarizers are parallel. Further evidence of total correlation comes from examining:
\mathcal P_{+-}(0) = \mathcal P_{-+}(0) = 0
This zero probability means it is impossible to find +1 for photon \nu_1 and -1 for photon \nu_2 or -1 for \nu_1 and +1 for \nu_2 simultaneously when the polarizers are parallel.
While individual measurements on \nu_1 or \nu_2 appear random, the outcomes are perfectly correlated when measured with parallel polarizers.
If we measure a specific polarization (say, +1 or -1) for photon \nu_1, we will invariably measure the same polarization (+1 or -1 respectively) for photon \nu_2. This perfect correlation is typical for entangled photons.
To quantify the degree of correlation between polarization measurements at polarizers (I) and (II) for arbitrary angles, we use the classical correlation coefficient.
For two random variables, A and B, each representing the outcome of a measurement and taking values of +1 or -1, the classical correlation coefficient \mathcal C between A and B is defined as:
\mathcal C = \frac{\overline{AB} - \overline{A}\cdot \overline{B}}{\sqrt{\left(\overline{A^2} - \left(\overline{A}\right)^2\right)\left(\overline{B^2} - \left(\overline{B}\right)^2\right)}} where \overline{X} denotes the statistical average of the random variable X. If A and B are independent or uncorrelated, the average of their product \overline{AB} equals the product of their averages \overline{A} \overline{B}, leading to a correlation coefficient \mathcal C of zero. This is expected for independent random events. However, if the outcomes are dependent, \overline{AB} will differ from \overline{A} \overline{B}, and \mathcal C will be non-zero.
In our specific case of polarization measurements, the outcomes are \pm 1 with equal probabilities (as we found for individual measurements). For such variables, the average value is \overline{A} = \overline{B} = 0, and the average of the square is \overline{A^2} = \overline{B^2} = 1.
Substituting these values into the correlation coefficient formula, we get:
\mathcal C = \frac{\overline{AB} - 0 \cdot 0}{\sqrt{(1 - 0^2)(1 - 0^2)}} = \overline{AB} For our polarization measurements with outcomes \pm 1 and zero average, the correlation coefficient is simply the average value of the product of the outcomes A \cdot B.
In the next steps, we will calculate this average value for the entangled photon state to quantify the polarization correlation.
Considering the four possible outcomes for measurements A and B (\pm 1), the product A \cdot B can take two values: +1 (when A and B have the same sign) and -1 (when A and B have opposite signs).
We can express the average value of the product A \cdot B, which is the correlation coefficient \mathcal C = \overline{AB}, in terms of the joint probabilities:
\begin{aligned} \mathcal C = \overline{AB} = & (+1) \cdot (\mathcal P_{++}(\mathbf a, \mathbf b) + \mathcal P_{--}(\mathbf a, \mathbf b)) \\ & + (-1) \cdot (\mathcal P_{+-}(\mathbf a, \mathbf b) + \mathcal P_{-+}(\mathbf a, \mathbf b)) \\ = & (\mathcal P_{++}(\mathbf a, \mathbf b) + \mathcal P_{--}(\mathbf a, \mathbf b)) \\ & - (\mathcal P_{+-}(\mathbf a, \mathbf b) + \mathcal P_{-+}(\mathbf a, \mathbf b)) \end{aligned}
For correlated photons with parallel polarizers (\alpha - \beta = 0), we have:
\begin{aligned} \mathcal P_{++} & = \mathcal P_{--} = \frac{1}{2} \cos^2(0) = \frac{1}{2} \\ \mathcal P_{+-} & = \mathcal P_{-+} = \frac{1}{2} \sin^2(0) = 0 \end{aligned}
The correlation coefficient for parallel polarizers is:
\mathcal C = \left(\frac{1}{2} + \frac{1}{2}\right) - (0 + 0) = 1
\mathcal C=1 indicates total positive correlation for parallel polarizers.
Now, let’s calculate the correlation coefficient for a general angle difference (\alpha - \beta). Substituting the expressions for the joint probabilities:
\begin{aligned} \mathcal C(\alpha, \beta) = & \left( \frac{1}{2} \cos^2(\alpha - \beta) + \frac{1}{2} \cos^2(\alpha - \beta) \right) \\ & - \left( \frac{1}{2} \sin^2(\alpha - \beta) + \frac{1}{2} \sin^2(\alpha - \beta) \right) \\ = & \cos^2(\alpha - \beta) - \sin^2(\alpha - \beta) \\ = & \cos(2(\alpha - \beta)) \end{aligned}
When the polarizers are parallel, \alpha - \beta = 0, \mathcal C = \cos(0) = 1, there is total positive correlation. The correlation is total when |\mathcal C| = 1, which occurs when \cos(2(\alpha - \beta)) = \pm 1. This happens when (\alpha - \beta) = n\pi/2.
Hidden variable theories
Let’s consider a scenario where a measurement at polarizer (I) is performed first, oriented along \mathbf a, followed by a measurement at polarizer (II). Quantum calculations tell us that at polarizer (I), oriented along \mathbf a, we can obtain either +1 (polarization along | \boldsymbol +_{\mathbf a} \rangle) or -1 (polarization along | \boldsymbol -_{\mathbf a} \rangle) with equal probabilities, whatever the orientation \mathbf a of the polarizer.
We take polarizer (I) along \mathbf x, (\alpha = 0) and the measurement yields +1.
After that measurement at polarizer (I), we use the “projection postulate”, which states that upon measurement, the initial state vector is projected onto the eigenspace associated with the result of the measurement.
Just after the measurement yielding +1 for \nu_1, associated with | \boldsymbol +_{\mathbf a, 1} \rangle = |\mathbf x_1 \rangle polarization for photon \nu_1, the initial state vector must be projected on the subspace associated with | \boldsymbol +_{\mathbf a, 1} \rangle = |\mathbf x_1 \rangle.
The initial state is entangled and the projection operator for measuring +1 for photon 1 is:
|P_{+}^{(1)} = |\boldsymbol +_{\mathbf a, 1}\rangle \langle \boldsymbol +_{\mathbf a, 1}| \otimes \mathbb{I}_2 = |\mathbf x_1\rangle \langle \mathbf x_1| \otimes \mathbb{I}_2
The projected, unnormalized state is:
\begin{aligned} |\Psi^\prime\rangle &= P_{+}^{(1)} |\Psi\rangle \\ &= (|\mathbf x_1\rangle \langle \mathbf x_1| \otimes \mathbb{I}_2) \frac{1}{\sqrt{2}} (|\mathbf x_1\rangle |\mathbf x_2\rangle + |\mathbf y_1\rangle |\mathbf y_2\rangle) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle \langle \mathbf x_1| |\mathbf x_1\rangle |\mathbf x_2\rangle + |\mathbf x_1\rangle \langle \mathbf x_1| |\mathbf y_1\rangle |\mathbf y_2\rangle \right) \\ &= \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle (\langle \mathbf x_1| |\mathbf x_1\rangle) |\mathbf x_2\rangle + |\mathbf x_1\rangle (\langle \mathbf x_1| |\mathbf y_1\rangle) |\mathbf y_2\rangle \right) \end{aligned}
Since \langle \mathbf x_1| |\mathbf x_1\rangle = 1 and \langle \mathbf x_1| |\mathbf y_1\rangle = 0:
|\Psi^\prime\rangle = \frac{1}{\sqrt{2}} \left( |\mathbf x_1\rangle |\mathbf x_2\rangle + 0 \right) = \frac{1}{\sqrt{2}} |\mathbf x_1\rangle |\mathbf x_2\rangle
After normalization, the projected state vector is:
|\Psi_{\mathrm{proj}}\rangle = \frac{|\Psi^\prime\rangle}{|||\Psi^\prime\rangle||} = \frac{\frac{1}{\sqrt{2}} |\mathbf x_1\rangle |\mathbf x_2\rangle}{\frac{1}{\sqrt{2}}} = |\mathbf x_1\rangle |\mathbf x_2\rangle
with \alpha = \beta = 0.
The projected state |\mathbf x_1\rangle |\mathbf x_2\rangle is a factorable state, indicating that each photon possesses a definite polarization. Specifically, photon \nu_1 is polarized along \mathbf x, as expected from the measurement outcome, and remarkably, photon \nu_2 is also polarized along \mathbf x. Consequently, if a subsequent measurement is performed on \nu_2 with polarizer (II) also oriented along \mathbf x (\beta = 0), the outcome will invariably be +1.
Similarly, consider the case where the measurement on \nu_1 yields -1. This corresponds to finding photon \nu_1 to be polarized along \mathbf y.
Applying the projection postulate in this scenario would project the initial entangled state onto the eigenspace associated with the -1 outcome for polarizer (I) along \mathbf x, onto the state |\mathbf y_1\rangle. The projected state would then become |\mathbf y_1\rangle |\mathbf y_2\rangle. This indicates that photon \nu_2 is now also polarized along \mathbf y.
Therefore, a measurement on \nu_2 with polarizer (II) along \mathbf x (\beta = 0) would now yield -1.
These results derived from the projection postulate are consistent with the total correlation we initially found using global probability calculations for parallel polarizers. When polarizers are parallel, measuring +1 (or -1) for \nu_1 guarantees that we will measure +1 (or -1) for \nu_2 as well.
Now, let’s consider the case where polarizer (II) is oriented at an angle \beta relative to polarizer (I), which remains aligned along \mathbf x (\alpha = 0).
If we have measured +1 for photon \nu_1, the projected state is |\mathbf x_1\rangle |\mathbf x_2\rangle. To determine the probability of measuring +1 or -1 for photon \nu_2 with polarizer (II) at angle \beta, we need to project |\mathbf x_2\rangle onto the eigenstates of polarizer (II), which are |\boldsymbol +_{\mathbf b, 2}\rangle = \cos(\beta) |\mathbf x_2\rangle + \sin(\beta) |\mathbf y_2\rangle and |\boldsymbol -_{\mathbf b, 2}\rangle = -\sin(\beta) |\mathbf x_2\rangle + \cos(\beta) |\mathbf y_2\rangle.
The probability of measuring +1 for photon \nu_2 is given by the squared magnitude of the projection of |\mathbf x_2\rangle onto |\boldsymbol +_{\mathbf b, 2}\rangle:
\begin{aligned} \mathcal P(+_{\nu_2} \mid +_{\nu_1}) & = |\langle \boldsymbol +_{\mathbf b, 2} | \mathbf x_2 \rangle|^2 \\ & = |\langle (\cos(\beta) \mathbf x_2 + \sin(\beta) \mathbf y_2) | \mathbf x_2 \rangle|^2 \\ & = |\cos(\beta) \langle \mathbf x_2 | \mathbf x_2 \rangle + \sin(\beta) \langle \mathbf y_2 | \mathbf x_2 \rangle|^2 \end{aligned} Since \langle \mathbf x_2 | \mathbf x_2 \rangle = 1 and \langle \mathbf y_2 | \mathbf x_2 \rangle = 0, we have:
\mathcal P(+_{\nu_2} \mid +_{\nu_1}) = |\cos(\beta)|^2 = \cos^2(\beta) Similarly, the probability of measuring -1 for photon \nu_2 is:
\begin{aligned} \mathcal P(-_{\nu_2} \mid +_{\nu_1}) & = |\langle \boldsymbol -_{\mathbf b, 2} | \mathbf x_2 \rangle|^2 \\ & = |\langle (-\sin(\beta) \mathbf x_2 + \cos(\beta) \mathbf y_2) | \mathbf x_2 \rangle|^2 \\ & = |-\sin(\beta) \langle \mathbf x_2 | \mathbf x_2 \rangle + \cos(\beta) \langle \mathbf y_2 | \mathbf x_2 \rangle|^2 = |-\sin(\beta)|^2 \\ & = \sin^2(\beta) \end{aligned}
These conditional probabilities, derived from the projection postulate, are proportional to \cos^2(\beta) and \sin^2(\beta), respectively. This cosine squared dependence is exactly what we found in our initial global calculations for the joint probabilities, confirming the consistency between the two approaches and providing a clear image of how the quantum correlations manifest in ordinary space through the projection postulate.
Einstein was famously uncomfortable with the implications of quantum entanglement and the apparent instantaneous connection between entangled particles, especially as revealed by the projection postulate. His discomfort stemmed from the principle of locality: relativity dictates that no information or physical influence can travel faster than the speed of light. The idea that measuring the polarization of photon \nu_1 could instantaneously determine the polarization state of photon \nu_2, regardless of the distance separating them, seemed to violate this fundamental principle.
Einstein believed that quantum mechanics, while successful in its predictions, was an incomplete description of reality. He argued that the probabilistic nature of quantum mechanics and the instantaneous collapse of the wave function upon measurement suggested that there must be a more fundamental, deterministic theory underlying it. He proposed that there might be “hidden variables” – parameters not accounted for in the standard quantum mechanical description – that, if known, would fully determine the outcomes of measurements and eliminate the need for probabilistic interpretations and instantaneous influences.
In this view, the apparent randomness and correlations in quantum measurements would not be fundamental features of nature, but rather a consequence of our incomplete knowledge. Just as statistical mechanics describes the macroscopic behavior of gases based on the deterministic laws governing individual atoms, Einstein hoped that a more complete theory with hidden variables would provide a deterministic and local explanation for quantum phenomena, including entanglement. This would restore a picture of the world where influences are always local and propagate at or below the speed of light, aligning with the principles of relativity.
This is precisely the core idea behind hidden variable theories. Let’s imagine that at the moment of their creation, each pair of entangled photons is endowed with a set of hidden properties, represented by \lambda. This \lambda is a variable (or possibly a set of variables) that is not described by standard quantum mechanics but is nonetheless real and predetermined at the source. This hidden variable \lambda carries all the necessary information to determine the outcomes of polarization measurements for both photons.
For photon \nu_1, when measured with polarizer (I) oriented along \mathbf a, the outcome (+1 or -1) is not fundamentally random, but is entirely dictated by the value of \lambda and the orientation \mathbf a. We can represent this deterministic outcome by a function A(\lambda, \mathbf a). This function A is dichotomic:
A(\lambda, \mathbf a) = \pm 1
Similarly, for photon \nu_2, measured with polarizer (II) oriented along \mathbf b, the outcome is given by another dichotomic function B(\lambda, \mathbf b), also determined by \lambda and \mathbf b:
B(\lambda, \mathbf b) = \pm 1
To account for the observed randomness in individual measurements, even with these deterministic functions A and B, we introduce a probability distribution \rho(\lambda). This distribution describes the statistical ensemble of photon pairs emitted from the source. Different pairs are emitted with different values of \lambda, distributed according to \rho(\lambda). When we perform a measurement on a single photon, we don’t know the specific \lambda for that pair. We only know that \lambda is drawn from the distribution \rho(\lambda). This distribution is properly normalized, meaning:
\int \rho(\lambda) \mathrm d\lambda = 1
for continuous \lambda (or \sum_\lambda \rho(\lambda) = 1 for discrete \lambda), and is always positive, \rho(\lambda) \ge 0, as required for a probability distribution.
Using the functions:
\frac{1}{2}\left[ A(\lambda, \mathbf a) + 1\right]
which assume +1 for |\boldsymbol + \rangle and:
\frac{1}{2}\left[ 1 - A(\lambda, \mathbf a)\right]
which assume +1 for |\boldsymbol - \rangle, the probabilities for the single and joint detection can be written as:
\begin{aligned} & \mathcal P_{+}(\mathbf a) = \int \rho(\lambda)\left\{\frac{1}{2}\left[ A(\lambda, \mathbf a) + 1\right] \right\}\mathrm d \lambda \\ & \mathcal P_{+-}(\mathbf a, \mathbf b) = \int \rho(\lambda)\left\{\frac{1}{2}\left[ A(\lambda, \mathbf a) + 1\right]\right\} \left\{\frac{1}{2}\left[ 1 - B(\lambda, \mathbf b)\right]\right\} \mathrm d \lambda \end{aligned}
and the correlation assumes the form:
\begin{aligned} \mathcal C = & \overline{AB} = (\mathcal P_{++}(\mathbf a, \mathbf b) + \mathcal P_{--}(\mathbf a, \mathbf b)) \\ & - (\mathcal P_{+-}(\mathbf a, \mathbf b) + \mathcal P_{-+}(\mathbf a, \mathbf b)) \\ = & \int \rho(\lambda) \left\{ \frac{1}{2}\left[ A(\lambda, \mathbf a) + 1\right] \frac{1}{2}\left[ B(\lambda, \mathbf b) + 1\right] \right. \\ & + \frac{1}{2}\left[ 1 - A(\lambda, \mathbf a)\right] \frac{1}{2}\left[ 1 - B(\lambda, \mathbf b)\right]\\ & - \frac{1}{2}\left[ A(\lambda, \mathbf a) + 1\right] \frac{1}{2}\left[ 1 - B(\lambda, \mathbf b)\right] \\ & \left. - \frac{1}{2}\left[ 1 - A(\lambda, \mathbf a)\right] \frac{1}{2}\left[ B(\lambda, \mathbf b) + 1\right] \right] \mathrm d \lambda \\ = & \frac{1}{4} \int \rho(\lambda) \left\{ [A(\lambda, \mathbf a) + 1][B(\lambda, \mathbf b) + 1]\right. \\ & + [1 - A(\lambda, \mathbf a)][1 - B(\lambda, \mathbf b)] \\ & - [A(\lambda, \mathbf a) + 1][1 - B(\lambda, \mathbf b)] \\ & \left. - [1 - A(\lambda, \mathbf a)][B(\lambda, \mathbf b) + 1] \right\} \mathrm d \lambda \\ = & \frac{1}{4} \int \rho(\lambda) \left[ A(\lambda, \mathbf a)B(\lambda, \mathbf b) + A(\lambda, \mathbf a) + B(\lambda, \mathbf b) + 1 + 1 \right. \\ & - A(\lambda, \mathbf a) - B(\lambda, \mathbf b) + A(\lambda, \mathbf a)B(\lambda, \mathbf b) \\ & - (A(\lambda, \mathbf a) - A(\lambda, \mathbf a)B(\lambda, \mathbf b) + 1 - B(\lambda, \mathbf b)) - (B(\lambda, \mathbf b) \\ & \left. + 1 - A(\lambda, \mathbf a)B(\lambda, \mathbf b) - A(\lambda, \mathbf a)) \right] \mathrm d \lambda \\ = & \frac{1}{4} \int \rho(\lambda) \left[ 2 A(\lambda, \mathbf a)B(\lambda, \mathbf b) + 2 \right.\\ & - (A(\lambda, \mathbf a) - A(\lambda, \mathbf a)B(\lambda, \mathbf b) + 1 - B(\lambda, \mathbf b)) - (B(\lambda, \mathbf b) \\ & \left. + 1 - A(\lambda, \mathbf a)B(\lambda, \mathbf b) - A(\lambda, \mathbf a)) \right] \mathrm d \lambda \\ = & \frac{1}{4} \int \rho(\lambda) \left[ 2 A(\lambda, \mathbf a)B(\lambda, \mathbf b) + 2 - A(\lambda, \mathbf a) \right. \\ & + A(\lambda, \mathbf a)B(\lambda, \mathbf b) - 1 + B(\lambda, \mathbf b) - B(\lambda, \mathbf b) - 1 \\ & \left. + A(\lambda, \mathbf a)B(\lambda, \mathbf b) + A(\lambda, \mathbf a) \right] \mathrm d \lambda \\ = & \frac{1}{4} \int \rho(\lambda) \left[ 4 A(\lambda, \mathbf a)B(\lambda, \mathbf b) \right] \mathrm d \lambda \\ = & \int \rho(\lambda) A(\lambda, \mathbf a) B(\lambda, \mathbf b) \mathrm d \lambda \end{aligned}
With this framework, the observed randomness of individual measurement outcomes reflects our ignorance of the specific value of \lambda for each photon pair.
The correlations between measurements on entangled photons then arise because both measurement outcomes are determined by the same shared hidden variable \lambda.
It’s important to note that models of this kind, based on local hidden variables, can indeed be constructed to reproduce certain types of correlations predicted by quantum mechanics.
Let’s consider a model where each emitted photon pair possesses a predetermined linear polarization angle, denoted as \lambda, relative to the vertical axis.
We postulate that the outcome of a polarization measurement is not intrinsically random, but rather determined by this hidden variable \lambda and the orientation of the polarizer. For polarizer one, oriented at an angle \mathbf a, the measurement result is given by a function A(\lambda, \mathbf a). We propose the functional form:
A(\lambda, \mathbf a) = \operatorname{sgn}\left[ \cos(2(\mathbf a - \boldsymbol \lambda))\right]
Similarly, for polarizer two, oriented at an angle \mathbf b, the outcome is given by:
B(\lambda, \mathbf b) = \operatorname{sgn}\left[ \cos(2(\mathbf b - \boldsymbol \lambda))\right]
This function is dichotomic, yielding +1 if the photon’s polarization angle \lambda is within \pi/4 of the polarizer axis \mathbf a, and -1 otherwise.
To account for the apparent randomness observed in experiments, we assume that the hidden variable \lambda varies from photon pair to photon pair, following a uniform probability distribution \rho(\lambda):
\rho(\lambda) = \frac{1}{2\pi}
We can now compute the probabilities. For single detection:
\begin{aligned} \mathcal P_{+}(\mathbf a) & = \int \rho(\lambda)\left\{\frac{1}{2}\left[ A(\lambda, \mathbf a) + 1\right] \right\}\mathrm d \lambda \\ & = \int_0^{2\pi} \frac{1}{2\pi} \left\{\frac{1}{2}\left[ \operatorname{sgn}\left[ \cos(2(\mathbf a - \boldsymbol \lambda))\right] + 1\right] \right\}\mathrm d \lambda \\ & = \frac{1}{4\pi} \int_0^{2\pi} \operatorname{sgn}\left[ \cos(2(\mathbf a - \boldsymbol \lambda))\right] \mathrm d \lambda + \frac{1}{4\pi} \int_0^{2\pi} 1 \mathrm d \lambda \\ & = 0 + \frac{1}{4\pi} 2\pi = \frac{1}{2} \end{aligned}
The first integral is zero because the function \operatorname{sgn}[\cos(2(\mathbf a - \boldsymbol \lambda))] is periodic and symmetric with respect to zero over its period, and we are integrating over an integer number of periods (in fact two periods since the period of \operatorname{sgn}[\cos(2(\mathbf a - \boldsymbol \lambda))] with respect to \boldsymbol \lambda is \pi). The positive and negative contributions cancel each other out exactly over each period.
A similar results is true for the others single detections and therefore:
\mathcal P_{+}(\mathbf a) = \mathcal P_{-}(\mathbf a) = \mathcal P_{+}(\mathbf b) = \mathcal P_{-}(\mathbf b) =\frac{1}{2} These results are consistent with the results given by quantum mechanics.
For the correlation:
\begin{aligned} \mathcal C = & \int \rho(\lambda) A(\lambda, \mathbf a) B(\lambda, \mathbf b) \mathrm d \lambda \\ = & \int_0^{2\pi} \frac{1}{2\pi} \operatorname{sgn}\left[ \cos(2(\mathbf a - \boldsymbol \lambda))\right]\operatorname{sgn}\left[ \cos(2(\mathbf b - \boldsymbol \lambda))\right] \mathrm d \lambda \end{aligned}
We define \alpha=\mathbf a-\mathbf b and note the function \operatorname{sgn}[\cos(2x)] has period \pi:
\begin{aligned} \mathcal C(\alpha) & =\frac1{2\pi}\int_0^{2\pi} \operatorname{sgn}\left[\cos(2x)\right] \operatorname{sgn}\left[ \cos(2(x+\alpha))\right]\mathrm d x \\ & =\frac1{\pi}\int_0^{\pi}\operatorname{sgn}\left[\cos(2x)\right] \operatorname{sgn}\left[\cos(2(x+\alpha))\right]\mathrm d x \end{aligned}
On the interval x\in[0,\pi], the quantity \cos(2x) is positive on the two subintervals:
\biggl[0,\frac{\pi}{4}\biggr], \quad \biggl[\frac{3\pi}{4},\pi\biggr]
and it is negative on the interval:
\left[ \frac{\pi}{4},\frac{3\pi}{4} \right]
\operatorname{sgn}[\cos(2x)] is a “square-wave” taking values +1 on the first and last subintervals, and -1 in between.
for \operatorname{sgn}[\cos(2(x+\alpha))], the boundaries \tfrac{\pi}{4} and \tfrac{3\pi}{4} shift to x=\tfrac{\pi}{4}-\alpha and x=\tfrac{3\pi}{4}-\alpha (mod \pi), so is positive on the interval:
\biggl[0,\frac{\pi}{4} - \alpha \biggr], \quad \biggl[\frac{3\pi}{4} - \alpha,\pi\biggr]
and it is negative on the interval:
\left[ \frac{\pi}{4} - \alpha,\frac{3\pi}{4} - \alpha \right]
We can then find how much of [0,\pi] has both factors with the same sign versus the different signs, which depends linearly on \alpha, so it varies linearly with \mathbf a - \mathbf b. Putting all together:
\mathcal C(\alpha) =\begin{cases} 1-\frac{4\alpha}{\pi},&0\le\alpha\le\frac{\pi}{2}\\ \frac{4\alpha}{\pi}-3,&\frac{\pi}{2}\le\alpha\le\pi \end{cases}
where we set \alpha=|\mathbf a-\mathbf b| taken mod \pi, so 0\le\alpha\le\pi.
As in the quantum mechanical case, the correlation \mathcal C depends only on the relative angle (\mathbf a, \mathbf b) of the polarized, the discrepancy with the quantum mechanical prediction and the difference is zero for the angles that corresponds to the cases when the correlation is zero or total (0, \pi/4, \pi/2, etc.).
Bell’s inequalities
I have been already presented the Bell’s theorem here. In this section, I will now provide a mathematical analysis of Bell’s inequalities.
Bell formalized Einstein’s local realism by introducing hidden variables \lambda that determine photon measurement outcomes via functions A(\mathbf a, \lambda) and B(\mathbf b, \lambda), where \mathbf a and \mathbf b are polarizer orientations.
Locality implies A depends only on \mathbf a (not \mathbf b) and B depends only on \mathbf b (not \mathbf a). The hidden variable distribution \rho(\lambda) is independent of \mathbf a and \mathbf b. The correlation coefficient is the average of A B over \lambda:
\mathcal C(\mathbf a, \mathbf b) = \int \rho(\lambda) A(\lambda, \mathbf a) B(\lambda, \mathbf b) \mathrm d \lambda
Let’s define the quantity S as:
\begin{aligned} S = & A(\lambda, \mathbf a) B(\lambda, \mathbf b) - A(\lambda, \mathbf a) B(\lambda, \mathbf b^\prime) \\ & + A(\lambda, \mathbf a^\prime) B(\lambda, \mathbf b) + A(\lambda, \mathbf a^\prime) B(\lambda, \mathbf b^\prime) \end{aligned}
Since A(\lambda, \mathbf a), A(\lambda, \mathbf a^\prime), B(\lambda, \mathbf b), B(\lambda, \mathbf b^\prime) can only take the value \pm 1, this quantity is bounded and can be factorized as:
\begin{aligned} S = & A(\lambda, \mathbf a) B(\lambda, \mathbf b) - A(\lambda, \mathbf a) B(\lambda, \mathbf b^\prime) \\ & + A(\lambda, \mathbf a^\prime) B(\lambda, \mathbf b) + A(\lambda, \mathbf a^\prime) B(\lambda, \mathbf b^\prime) \\ = & A(\lambda, \mathbf a) \left[ B(\lambda, \mathbf b) - B(\lambda, \mathbf b^\prime) \right] \\ & + A(\lambda, \mathbf a^\prime) \left[ B(\lambda, \mathbf b) + B(\lambda, \mathbf b^\prime) \right] \end{aligned}
Let’s examine the terms B(\lambda, \mathbf b) - B(\lambda, \mathbf b^\prime) and B(\lambda, \mathbf b) + B(\lambda, \mathbf b^\prime).
If B(\lambda, \mathbf b) = B(\lambda, \mathbf b^\prime), B(\lambda, \mathbf b) - B(\lambda, \mathbf b^\prime) = 0 and B(\lambda, \mathbf b) + B(\lambda, \mathbf b^\prime) = 2 B(\lambda, \mathbf b) which can be \pm 2. Then S = A(\lambda, \mathbf a) \times 0 + A(\lambda, \mathbf a^\prime) \times 2 B(\lambda, \mathbf b) = 2 A(\lambda, \mathbf a^\prime) B(\lambda, \mathbf b). Since A(\lambda, \mathbf a^\prime) \in \{\pm 1\} and B(\lambda, \mathbf b) \in \{\pm 1\}, S \in \{\pm 2\}.
If B(\lambda, \mathbf b) \neq B(\lambda, \mathbf b^\prime), B(\lambda, \mathbf b) - B(\lambda, \mathbf b^\prime) = \pm 2 and B(\lambda, \mathbf b) + B(\lambda, \mathbf b^\prime) = 0. Then S = A(\lambda, \mathbf a) \times [B(\lambda, \mathbf b) - B(\lambda, \mathbf b^\prime)] + A(\lambda, \mathbf a^\prime) \times 0 = A(\lambda, \mathbf a) [B(\lambda, \mathbf b) - B(\lambda, \mathbf b^\prime)]. Since A(\lambda, \mathbf a) \in \{\pm 1\} and B(\lambda, \mathbf b) - B(\lambda, \mathbf b^\prime) \in \{\pm 2\}, S \in \{\pm 2\}.
In both cases, S is either 2 or -2, so S is bounded by \pm 2.
We can then take the average of S over \lambda which is going to be bounded:
\left|\int \rho(\lambda) S(\lambda) \mathrm d \lambda \right| \le 2
Using the definition of correlation, we can rewrite this inequality as:
\begin{aligned} \left|\int \rho(\lambda) S(\lambda) \mathrm d \lambda \right| = & \left|\int \rho(\lambda) \left[ A(\lambda, \mathbf a) B(\lambda, \mathbf b) - A(\lambda, \mathbf a) B(\lambda, \mathbf b^\prime) \right.\right.\\ & \left.\left. \vphantom{\int} + A(\lambda, \mathbf a^\prime) B(\lambda, \mathbf b) + A(\lambda, \mathbf a^\prime) B(\lambda, \mathbf b^\prime)\right]\mathrm d \lambda \right|\\ & \left| \mathcal C(\mathbf a, \mathbf b) - \mathcal C(\mathbf a, \mathbf b^\prime) + \mathcal C(\mathbf a^\prime, \mathbf b) + \mathcal C(\mathbf a^\prime, \mathbf b^\prime) \right|\le 2 \end{aligned}
This mathematical inequality need to be satisfied for all the combinations of orientations.
We can consider the orientations:
\begin{aligned} &(\mathbf a, \mathbf b) = \frac{\pi}{8}\\ &(\mathbf a, \mathbf b^\prime) = \frac{3\pi}{8}\\ &(\mathbf a^\prime, \mathbf b) = \frac{\pi}{8}\\ &(\mathbf a^\prime, \mathbf b^\prime) = \frac{\pi}{8} \end{aligned}
Using the quantum mechanics result:
\mathcal C(\mathbf a, \mathbf b) = \cos(2(\mathbf a, \mathbf b))
The correlations are:
\begin{aligned} C(\mathbf a, \mathbf b) &= \cos\left(2\frac{\pi}{8}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\\ C(\mathbf a, \mathbf b^\prime) &= \cos\left(2\frac{3\pi}{8}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}\\ C(\mathbf a^\prime, \mathbf b) &= \cos\left(2\frac{\pi}{8}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\\ C(\mathbf a^\prime, \mathbf b^\prime) &= \cos\left(2\frac{\pi}{8}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \end{aligned}
And the Bell’s inequality gives:
\begin{aligned} \left| \mathcal C(\mathbf a, \mathbf b) - \mathcal C(\mathbf a, \mathbf b^\prime) + \mathcal C(\mathbf a^\prime, \mathbf b) + \mathcal C(\mathbf a^\prime, \mathbf b^\prime) \right| &= \left| \frac{\sqrt{2}}{2} - \left(-\frac{\sqrt{2}}{2}\right) + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right| \\ &= \left| \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} \right| \\ &= \left| 4 \frac{\sqrt{2}}{2} \right| \\ &= 2\sqrt{2} > 2 \end{aligned}
Bell’s inequality, with a limit of 2, is violated by the quantum mechanical prediction of 2\sqrt{2} for the considered correlations. This violation demonstrates that local hidden variable models cannot reproduce all quantum predictions, specifically those for entangled particles.
This result challenges interpretations of quantum mechanics based on supplementary parameters or hidden variables, which were proposed as a way to understand the statistical nature of quantum predictions similar to how statistical mechanics describes gases.
Furthermore, it questions Einstein’s local realist view where systems possess inherent properties determined by local parameters, even when considering probabilistic outcomes rather than strictly deterministic ones.
Bell’s theorem thus highlights a fundamental conflict between quantum mechanics and local realism, indicating that the correlations predicted by quantum mechanics cannot be explained by any local hidden variable theory.
Bell’s inequalities are usually satisfied in observed correlations, making violations exceptional and specific to systems like the correlations we described at particular angles.
Initially, it was uncertain whether these violations indicated a failure of quantum mechanics. Therefore, experiments were essential to test quantum mechanics in these specific scenarios where Bell’s inequalities are predicted to be violated.
Early experiments by Clauser, Freedman, Fry, and Thompson showed agreement with quantum mechanics and violated Bell’s inequality, but they didn’t enforce locality by switching polarizers during photon flight.
Static polarizer setups could allow for unwanted interactions violating Bell’s assumptions, specifically that measurement settings on one side should not influence the other, and that the hidden variable distribution is independent of measurement settings.
Bell proposed switching polarizers faster than the photon travel time to enforce locality through relativistic causality. The Orsay experiment in 1982 successfully implemented this fast switching, demonstrating a violation of Bell’s inequality by six standard deviations. This confirmed entanglement even under locality conditions measured in a laboratory.
These experiments confirm violations of Bell’s inequalities where quantum mechanics predicts them. Bohr’s view of quantum mechanics as complete is supported by these violations, while Einstein’s local realism is contradicted.
Einstein himself considered the implications of rejecting his view as “unacceptable,” presenting two options: instantaneous “telepathic” influence between separated systems, or denying independent reality to spatially separated systems.
Modern physics, after experimental verification, accepts that either quantum non-locality (instantaneous influence) or quantum holism (entangled systems are inseparable wholes) must be true, despite Einstein’s reservations.
I described an apparatus that could be used for testing the violation of these inequalities here, and created a program to simulate experiments for entangled particles, which is accessible at the GitHub repository here.
References
BELL, John S., 1964. On the Einstein Podolsky Rosen paradox. Physics 1, 195 (1964).
ASPECT, Alain, 2004. Bell's theorem: the naive view of an experimentalist. arXiv: Quantum Physics, pp. 119-153.