Problem:
Calculate
\(\swich{\psi}{P_{v_1} \otimes P_{v_2}}{\psi}\)
for \(\ksi\) is the Bell vector
\(\isqt{1} \parenl{ \kt{00} + \kt{11} }\).
Solution:
First observe that
$$\eqb{
P_{v_1} \otimes P_{v_2}
& \eql &
\otr{v_1}{v_1} \otimes \otr{v_2}{v_2} \\
& \eql &
\kt{v_1 \otimes v_2} \br{v_1 \otimes v_2}
}$$
Next,
$$\eqb{
\kt{v_1}\otimes \kt{v_2}
& \eql &
\parenl{ \cos\theta_1\kt{0} + \sin\theta_1\kt{1} }
\otimes
\parenl{ \cos\theta_2\kt{0} + \sin\theta_2\kt{1} } \\
& \eql &
\cos\theta_1 \cos\theta_2 \kt{00}
+ \sin\theta_1 \sin\theta_2 \kt{11}
+ \cos\theta_1 \sin\theta_2 \kt{01}
+ \sin\theta_1 \cos\theta_2 \kt{10}
}$$
Thus,
$$\eqb{
& \; &
P_{v_1} \otimes P_{v_2} \\
& \eql &
\kt{v_1 \otimes v_2} \br{v_1 \otimes v_2} \\
& \eql &
\parenl{ \cos\theta_1 \cos\theta_2 \kt{00}
+ \sin\theta_1 \sin\theta_2 \kt{11}
+ \cos\theta_1 \sin\theta_2 \kt{01}
+ \sin\theta_1 \cos\theta_2 \kt{10} } \\
& \; &
\parenl{ \cos\theta_1 \cos\theta_2 \br{00}
+ \sin\theta_1 \sin\theta_2 \br{11}
+ \cos\theta_1 \sin\theta_2 \br{01}
+ \sin\theta_1 \cos\theta_2 \br{10} } \\
& \eql &
\cos^2\theta_1 \cos^2\theta_2 \otr{00}{00} \\
& \; &
+ \sin\theta_1 \cos\theta_1 \sin\theta_2 \cos\theta_2
\otr{00}{11} \\
& \; &
+ \sin^2\theta_1 \sin^2\theta_2 \otr{11}{11} \\
& \; &
+ \sin\theta_1\cos\theta_1 \sin\theta_1\cos\theta_2 \otr{11}{00} \\
& \; &
+ \mbox{ 12 other terms }
}$$
Notice that we get a collection of outer-products in all kinds
of combinations.
For reasons that will make sense shortly, we're only focusing on the
terms that involve both \(\kt{00}\) and \(\kt{11}\) in various
outer-product combinations.
Next, we want to compute $$ \swich{\psi}{P_{v_1} \otimes P_{v_2}}{\psi} $$ where $$ \ksi \eql \isqt{1} \parenl{ \kt{00} + \kt{11} } $$ That is, $$ \frac{1}{2} \swichh{ \br{00} + \br{11} }{P_{v_1} \otimes P_{v_2}}{\kt{00} + \kt{11} } $$ Now we see that the when the projector is applied rightwards, the two \(\kt{00},\kt{11}\) terms in the vector will eliminate all outer-products that do NOT have either of these on the right, such as \(\kt{00},\kt{11}\)
Similarly, when that result multiplies leftwards into \(\br{00} + \br{11}\), all outerproducts do not have either \(\br{00},\br{11}\) such as \(\otr{{\bf 10}}{01}\), will get eliminated.
The result is $$\eqb{ \swich{\psi}{P_{v_1} \otimes P_{v_2}}{\psi} & \eql & \frac{1}{2} \sin^2\theta_1 \sin^2\theta_2 + \cos^2\theta_1 \cos^2\theta_2 + 2 \sin\theta_1\cos\theta_1 \sin\theta_2\cos\theta_2 \\ & \vdots & \\ & \eql & \frac{1}{2} \cos^2 \parenl{\theta_1 - \theta_2} }$$