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I have some doubts on the calculations behind Bell's experiment, which goes as follows: a pair of entangled photons is shared between two scientists, Alice and Bob, who possess a bunch of polarizers at $0°, 22.5°$ and $45^°$ with respect to some fixed vertical direction. In particular let us assume that Alice and Bob set up the following combinations of polarizers before measuring their qubit simultaneously:

  1. Alice uses the one at $0°$, Bob uses the one at $90°$;
  2. Alice uses the one at $0°$, Bob uses the one at $45°$;
  3. Alice uses the one at $45^°$. Bob uses the one at $90^°$.

(I will be using the convention to halve the angles when performing an actual rotation, so, for example, $45^°$ corresponds to a rotation by $22.5^°$). Assume the EPR state that describes the two photons is \begin{equation} |\psi\rangle=\frac{1}{\sqrt 2}(|\updownarrow\updownarrow\rangle+|\leftrightarrow\leftrightarrow\rangle) \end{equation}

so that they both have the same polarization, though unknown a priori. In the first setup of the experiment, the global state before the measurement becomes (in the vertical/horizontal basis) \begin{align} (\mathbb{I}_A\otimes R_B(90^°))|\psi\rangle&=\frac{1}{\sqrt 2}\left(|\updownarrow\rangle\otimes \frac{1}{\sqrt 2}(|\updownarrow\rangle-|\leftrightarrow\rangle)+|\leftrightarrow\rangle\otimes\frac{1}{\sqrt 2}(|\updownarrow\rangle+|\leftrightarrow\rangle) \right) \\ &= \frac{1}{2}(|\updownarrow\updownarrow\rangle-|\updownarrow\leftrightarrow\rangle+|\leftrightarrow\updownarrow\rangle+|\leftrightarrow\leftrightarrow\rangle). \end{align} What is the probability for the photons to go through one and only one of the two polarizers? Only the first and the last term contribute, as since both photons are polarized vertically or horizontally, they either go through one or the other, for a total probability of $50\%$.

A similar calculation shows \begin{align} (\mathbb{I}_A\otimes R_B(45^°))|\psi\rangle= \frac{1}{\sqrt 2}(\cos 22.5^°|\updownarrow\updownarrow\rangle-\sin 22.5^°|\updownarrow\leftrightarrow\rangle+\sin 22.5^°|\leftrightarrow\updownarrow\rangle+\cos 22.5^°|\leftrightarrow\leftrightarrow\rangle). \end{align} Again, if the polarization are the same, then they go through either one or the other but not both: therefore $$ p=\frac{1}{2}\cos^2 22.5^°+\frac{1}{2}\cos^2 22.5^°\approx 85\%.$$ I fear that here my reasoning in not correct... in fact, I was expecting to get the complementary $15\%$! Where does it go wrong?

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Short answer: Yes, it is the complementary angle because you have grabbed the wrong probability amplitudes from $|\psi\rangle$.

\begin{align} \text{Pr}(A=\updownarrow, B=\searrow) &= \langle \psi | \left(|\updownarrow\rangle\langle\updownarrow| \otimes |\searrow\rangle\langle\searrow|\right) |\psi \rangle\\ &= \langle \psi | (\mathbb{I}_A \otimes R_B^\dagger(45^\circ) ) |\updownarrow\rangle\langle\updownarrow| \otimes |\leftrightarrow\rangle\langle\leftrightarrow| (\mathbb{I}_A \otimes R_B (45^\circ) ) |\psi \rangle \\ &= |\langle \updownarrow \leftrightarrow | (\mathbb{I}_A \otimes R_B (45^\circ) ) |\psi \rangle|^2 \\ &= \frac{1}{2} \sin^2 (22.5^\circ) \end{align}

where I've used the notation $B=\searrow$ to mean "The detector at $B$ measured a photon perpendicular to the polarization of the detector in the rotated basis", and I will use $B=\nearrow$ for the event that the detector at $B$ measured a photon parallel to the polarization of the rotated detector. Mathematically, $|\nearrow\rangle = R_B^\dagger (45^\circ) |\updownarrow\rangle$ and $|\searrow\rangle = R_B^\dagger (45^\circ) |\leftrightarrow\rangle$ where the $^\dagger$ reflects the fact that you were originally working in a system where you viewed the state as having been rotated instead of the measurement apparatus being rotated.

And so the probability that a photon passes through exactly one polarizer is \begin{align} \text{Pr}([A=\updownarrow, B=\searrow] \cup [A=\leftrightarrow, B=\nearrow]) &= \text{Pr}(A=\updownarrow, B=\searrow) + \text{Pr}(A=\leftrightarrow, B=\nearrow)\\ &= \sin^2 (22.5^\circ) \end{align}

More generally, you can construct measurements to detect the events $[A=\nearrow, B=\searrow]$ and $[A=\searrow, B=\nearrow]$ that indicate that exactly one of the detectors measured a photon with polarization aligned with the detector. Let the system be rotated locally by $\alpha$ in system $A$ and by $\beta$ in system $B$. These angles correspond to the actual directions of polarizations so there's an additional factor of two. Keeping with the convention of rotating the state and not the measurement apparatus, the measurement for the event $[A=\nearrow, B=\searrow]$ is given by:

\begin{align} M_{A\nearrow, B\searrow} &= |\nearrow \rangle \langle \nearrow| \otimes |\searrow \rangle \langle \searrow | \\&= R_A^\dagger (2\alpha) |\updownarrow \rangle \langle \updownarrow| R_A (2\alpha)\otimes R_B^\dagger (2\beta)|\leftrightarrow\rangle \langle \leftrightarrow| R_B (2\beta) \\&\sim \begin{pmatrix} \cos^2\alpha & \cos\alpha\sin\alpha\\ \cos\alpha\sin\alpha & \sin^2\alpha \end{pmatrix} \otimes \begin{pmatrix} \sin^2\beta& -\cos\beta\sin\beta \\ -\cos\beta\sin\beta & \cos^2\beta \end{pmatrix} \end{align}

You can then compute the probability for the event $[A=\nearrow, B=\searrow]$ as

\begin{align} \langle \psi | M_{A\nearrow, B\searrow} | \psi \rangle &= \frac{1}{2} \left( \cos^2\alpha \sin^2\beta- 2 \cos\alpha\sin\alpha \cos\beta\sin\beta +\sin^2\alpha \cos^2\beta\right) \\ &= \frac{1}{2}\left( \cos\alpha\sin\beta - \sin\alpha\cos\beta\right)^2 \\ &= \frac{1}{2} \sin^2 (\alpha - \beta) \end{align}

And you can compute a similar outcome for $[A=\nearrow, B=\searrow]$, and $[A=\searrow, B=\searrow]$, and so on to find the desired event probabilities.

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