$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 4.20}$

Let $O_{\theta_1}$ be the single-qubit observable with $+1$-eigenvector

$\ket{v_1} = \cos\theta_1\ket{0} + \sin\theta_1\ket{1}$

and $-1$-eigenvector

$\ket{v_1^\perp} = -\sin_1\theta\ket{0} + \cos\theta_1\ket{1}.$

Similarly let $O_{\theta_2}$ be the single-qubit observable

with $+1$-eigenvector

$\ket{v_2} = \cos\theta_2\ket{0} + \sin\theta_2\ket{1}$

and $-1$-eigenvector

$\ket{v_2^\perp} = -\sin\theta_2\ket{0} + \cos\theta_2\ket{1}.$

Let $O$ be the two-qubit observable $O_{\theta_1}\otimes O_{\theta_2}$. We consider various measurements on the EPR state $\ket{\psi} = \frac{1}{\sqrt{2}}(\ket{00} + \ket{11})$. We are interested in the probability that the measurements $O_{\theta_1}\otimes I$ and $I\otimes O_{\theta_2}$, if they were performed on the state $\ket\psi$, would {\it agree} on the two qubits in that either both qubits are measured in the $1$-eigenspace or both are measured in $-1$-eigenspace of their respective single-qubit observables.

As in Example \ref{bit-equality}, we are not interested in the specific outcome of the two measurements, just whether or not they would agree. The observable $O = O_{\theta_1}\otimes O_{\theta_2}$ gives exactly this information.

a) Find the probability that the measurements $O_{\theta_1}\otimes I$ and $I \otimes O_{\theta_2}$, when performed on $\ket\psi$, would agree in the sense of both resulting in a $+1$ eigenvector or both resulting in a $-1$ eigenvector. (Hint: Use the trigonometric identities $\cos(\theta_1 - \theta_2) = \cos(\theta_1)\cos(\theta_2) + \sin(\theta_1)\sin(\theta_2)$ and $\sin(\theta_1 - \theta_2) = \sin(\theta_1)\cos(\theta_2) - \cos(\theta_1)\sin(\theta_2)$ to obtain a simple form for your answer.)

b) For what values of $\theta_1$ and $\theta_2$ do the results always agree?

c) For what values of $\theta_1$ and $\theta_2$ do the results never agree?

d) For what values of $\theta_1$ and $\theta_2$ do the results agree half the time?

e) Show that whenever $\theta_1 \ne \theta_2$ and $\theta_1$ and $\theta_2$ are chosen from $\{-60^\circ, 0^\circ, 60^\circ \}$, then the results agree $1/4$ of the time and disagree $3/4$ of the time.

a) Let be the two pairs of othogonal vectors $|v_{\theta_{i}}\rangle=\cos({\theta_{i}})|0\rangle+\sin({\theta_{i})}|1\rangle$ and $|v^{\bot}_{\theta_{i}}\rangle=-\sin({\theta_{i}})|0\rangle+\cos({\theta_{i}})|1\rangle$ , $i\in\{1,2\}$ corresponding to photon 1 and photon 2. An observable for only one photon $i$ is:

$O_{\theta_{i}}=|v_{\theta_{i}}\rangle \langle v_{\theta_{i}}|-|v^{\bot}_{\theta_{i}}\rangle \langle v^{\bot}_{\theta_{i}}|$

and an observable for observing photon 1 without observing photon 2 is:

$O_{\theta_{1}}\otimes I=|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\otimes|I\rangle \langle I|-|v^{\bot}_{\theta_{1}}\rangle \langle v^{\bot}_{\theta_{1}}|\otimes|I\rangle \langle I|$

with $|I\rangle=(1/\sqrt{2})(|0\rangle +|1\rangle)$ for photon 2.

The first term of $O_{\theta_{1}}\otimes I$ corresponds to the eigenvalue +1, the second one to the eigenvalue -1 (see Exercice 4.19).

The projector for photon 1 for the eigenvalue +1 is then:

$P_{1}^{+1}=|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\otimes|I\rangle \langle I|$

$P_{1}^{+1}=|v_{\theta_{1}}\rangle |I\rangle (\cos({\theta_{1}})\langle 0|+\sin({\theta_{1})}\langle 1|)(1/\sqrt{2})(\langle 0|+\langle 1|)$

$P_{1}^{+1}=|v_{\theta_{1}}\rangle |I\rangle (1/\sqrt{2})(\cos({\theta_{1}})\langle 00|+\cos({\theta_{1}})\langle 01|+\sin({\theta_{1})}\langle 10|+\sin({\theta_{1})}\langle 11|)$

When this projector is applied to the entangled state $|\psi\rangle =(1/\sqrt{2})(|00\rangle+|11\rangle)$ the result is:

$P_{1}^{+1}|\psi\rangle=|v_{\theta_{1}}\rangle |I\rangle (1/2) (\cos(\theta_{1})+\sin(\theta_{1}))$

with the probability (for photon 1 to have the outcome corresponding to the +1 eigenvalue):

$\Pi_{1}^{+1}=|P_{1}^{+1}|\psi\rangle|^2=(1/4)(\cos(\theta_{1})+\sin(\theta_{1}))^2$

In the same way, the probability for photon 1 to have the outcome corresponding to the -1 eigenvalue is:

$\Pi_{1}^{-1}=(1/4)(-\sin(\theta_{1})+\cos(\theta_{1}))^2$

An observable for photon 2 without observing photon 1 is:

$I\otimes O_{\theta_{2}}=|I\rangle \langle I|\otimes|v_{\theta_{2}}\rangle \langle v_{\theta_{2}}|-|I\rangle \langle I|\otimes|v^{\bot}_{\theta_{2}}\rangle \langle v^{\bot}_{\theta_{2}}|$

In the same way as for photon 1, the probability for photon 2 to have the outcome corresponding to the +1 eigenvalue is:

$\Pi_{2}^{+1}=(1/4)(\cos(\theta_{2})+\sin(\theta_{2}))^2$

and for the outcome corresponding to the -1 eigenvalue:

$\Pi_{2}^{-1}=(1/4)(-\sin(\theta_{2})+\cos(\theta_{2}))^2$

The sum of all these probabilities is 1:

$\Pi_{1}^{+1}+\Pi_{1}^{-1}+\Pi_{2}^{+1}+\Pi_{2}^{-1}=1$

An observable to observe both photon 1 and photon 2 is:

$O_{\theta_{1}}\otimes O_{\theta_{2}}=|v_{\theta_{1}}\rangle|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\langle v_{\theta_{2}}|-|v_{\theta_{1}}\rangle |v^{\bot}_{\theta_{2}}\rangle \langle v_{\theta_{1}}| \langle v^{\bot}_{\theta_{2}}|-|v^{\bot}_{\theta_{1}}\rangle |v_{\theta_{2}}\rangle \langle v^{\bot}_{\theta_{1}}| \langle v_{\theta_{2}}|+|v^{\bot}_{\theta_{1}}\rangle |v^{\bot}_{\theta_{2}}\rangle \langle v^{\bot}_{\theta_{1}}| \langle v^{\bot}_{\theta_{2}}|$

Using the projectors this becomes:

$O_{\theta_{1}}\otimes O_{\theta_{2}}=P_{12}^{+1}-P_{12}^{+1-1}-P_{12}^{-1+1}+P_{12}^{-1}$

where $P_{12}^{+1}$ is the projector for both photons having an outcome corresponding to eigenvalues +1, $P_{12}^{+1-1}$ the projector for photon 1 with eigenvalue +1 and photon 2 with -1, and so on. One has for the projector $P_{12}^{+1}$:

$P_{12}^{+1}=|v_{\theta_{1}}\rangle|v_{\theta_{1}}\rangle \langle v_{\theta_{1}}|\langle v_{\theta_{2}}|$

$P_{12}^{+1}=|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (\cos(\theta_{1})\langle 0|+\sin(\theta_{1})\langle 1|)(\cos(\theta_{2})\langle 0|+\sin(\theta_{2})\langle 1|)$

$P_{12}^{+1}=|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (\cos(\theta_{1})\cos(\theta_{2})\langle 00|+\cos(\theta_{1})\sin(\theta_{2})\langle 01|+\sin(\theta_{1})\cos(\theta_{2})\langle 10|+\sin(\theta_{1})\sin(\theta_{2})\langle 11|)$

Applied to the entangled state $|\psi\rangle =(1/\sqrt{2})(|00\rangle +|11\rangle)$ this gives:

$P_{12}^{+1}|\psi\rangle =|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (1/\sqrt{2}) (\cos(\theta_{1}) \cos(\theta_{2})+\sin(\theta_{1}) \sin(\theta_{2}))$

$P_{12}^{+1}|\psi\rangle =|v_{\theta_{1}}\rangle|v_{\theta_{2}}\rangle (1/\sqrt{2})\cos(\theta_{1}-\theta_{2})$

The probability for both photons to have the outcome corresponding to the +1 eigenvalue is:

$\Pi_{12}^{+1}=|P_{12}^{+1}|\psi\rangle|^2=(1/2)\cos(\theta_{1}-\theta_{2})^2$

The same way, the probability for both photons to have the outcome corresponding to the -1 eigenvalue is:

$\Pi_{12}^{-1}=|P_{12}^{-1}|\psi\rangle|^2=(1/2)\cos(\theta_{1}-\theta_{2})^2$

So, the probability for both photons to have the same outcome is:

$\Pi_{12}^{same}=\Pi_{12}^{+1}+\Pi_{12}^{-1}=\cos(\theta_{1}-\theta_{2})^2$

The same way, the probability for both photons to have different outcomes is:

$\Pi_{12}^{different}=\Pi_{12}^{+1-1}+\Pi_{12}^{-1+1}=\sin(\theta_{1}-\theta_{2})^2$

and, of course:

$\Pi_{12}^{same}+\Pi_{12}^{different}=1$

b) The results always agree for $\theta_{1}=\theta_{2}$ (modulo $180\,^{\circ}$).

c) The results never agree for $\theta_{1}-\theta_{2}=90\,^{\circ}$ (modulo $180\,^{\circ}$).

d) The results agree half the time for $\theta_{1}-\theta_{2}=45\,^{\circ}$ (modulo $180\,^{\circ}$).

e) When $\theta_{1}=0$, which happens 1/3 of the time, then $\theta_{2}=\pm60\,^{\circ}\neq\theta_{1}$ and $\Pi_{12}^{same}=1/4$.

When $\theta_{1}=60\,^{\circ}$, which happens 1/3 of the time, then $\theta_{2}=0$ or $-60\,^{\circ}$ and $\Pi_{12}^{same}=1/4$.

The same when $\theta_{1}=-60\,^{\circ}$.

So, the probability that the results agree (with $\theta_{1}\neq\theta_{2}$) is 3x(1/3)x(1/4)=1/4 of the time.

(If $\theta_{1}$ is allowed to be equal to $\theta_{2}$, the results agree 1/2 the time).

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