Single probabilities in entangled photon pairs
In my previous posts (here and here), I discussed entangled photon pairs and calculated the joint probabilities for polarization measurements. Now, I turn my attention to the probabilities of measuring polarization on just one photon of the pair, considered in isolation.
Let’s recall the entangled 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)
To find the probability of measuring a specific polarization for photon \nu_1, say +1 along direction \mathbf a, we must consider all possible outcomes for photon \nu_2. In quantum mechanics, this means summing the joint probabilities where photon \nu_1 is +1 and photon \nu_2 can be either +1 or -1.
Thus, the probability \mathcal P_{+}(\mathbf a) of measuring +1 for photon \nu_1 is:
\mathcal P_{+}(\mathbf a) = \mathcal P_{++}(\mathbf a, \mathbf b) + \mathcal P_{+-}(\mathbf a, \mathbf b)
Using the joint probabilities I derived earlier: \mathcal P_{++}(\mathbf a, \mathbf b) = \frac{1}{2} \cos^2(\alpha - \beta) and \mathcal P_{+-}(\mathbf a, \mathbf b) = \frac{1}{2} \sin^2(\alpha - \beta), we get:
\begin{aligned} \mathcal P_{+}(\mathbf a) & = \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 a) of measuring -1 for photon \nu_1 is:
\mathcal P_{-}(\mathbf a) = \mathcal P_{-+}(\mathbf a, \mathbf b) + \mathcal P_{--}(\mathbf a, \mathbf b) = \frac{1}{2}
Both probabilities are \frac{1}{2}, and importantly, they are independent of the polarizer angles \alpha and \beta. This means that no matter how I orient the polarizer for photon \nu_1, the probability of measuring either polarization is always 50%. Photon \nu_1, when considered alone, appears to be unpolarized.
The same holds true for photon \nu_2. The probability \mathcal P_{+}(\mathbf b) of measuring +1 for photon \nu_2 is:
\mathcal P_{+}(\mathbf b) = \mathcal P_{++}(\mathbf a, \mathbf b) + \mathcal P_{-+}(\mathbf a, \mathbf b) = \frac{1}{2}
And the probability \mathcal P_{-}(\mathbf b) of measuring -1 for photon \nu_2 is:
\mathcal P_{-}(\mathbf b) = \mathcal P_{+-}(\mathbf a, \mathbf b) + \mathcal P_{--}(\mathbf a, \mathbf b) = \frac{1}{2}
Again, both probabilities are \frac{1}{2} and independent of the polarizer angles. Thus, photon \nu_2 also appears unpolarized when measured individually.
This result is a key aspect of entanglement. While individual entangled photons seem random and unpolarized, their polarizations are strongly correlated when measured jointly. This contrast highlights the non-classical nature of entanglement and one of the ways quantum correlations manifest.
For more insights into this topic, you can find the details here.