Ann has two ortho-normal bases at her disposal for encoding cbits into qbits:
\begin{align*}
\lvert A_{0} \rangle &= \cos(\theta_{a}) \lvert 0 \rangle - \sin(\theta_{a}) \lvert 1 \rangle & \lvert A_{1} \rangle &= \sin(\theta_{a}) \lvert 0 \rangle + \cos(\theta_{a}) \lvert 1 \rangle \\
\lvert B_{0} \rangle &= \cos(\theta_{b}) \lvert 0 \rangle - \sin(\theta_{b}) \lvert 1 \rangle & \lvert B_{1} \rangle &= \sin(\theta_{b}) \lvert 0 \rangle + \cos(\theta_{b}) \lvert 1 \rangle
\end{align*}
Note that for \(\theta_{a} = 0\), the \(A\) basis becomes
\begin{align*}
\lvert A_{0} \rangle &= \lvert 0 \rangle & \lvert A_{1} \rangle &= \lvert 1 \rangle
\end{align*}
and for \(\theta_{b} = \pi/4\), the \(B\) basis becomes
\begin{align*}
\lvert B_{0} \rangle &= \lvert {-} \rangle & \lvert B_{1} \rangle &= \lvert {+} \rangle
\end{align*}
Bob uses these two bases for decoding. Eve can use a third ortho-normal basis:
\begin{align*}
\lvert C_{0} \rangle &= \cos(\theta_{c}) \lvert 0 \rangle - \sin(\theta_{c}) \lvert 1 \rangle & \lvert C_{1} \rangle &= \sin(\theta_{c}) \lvert 0 \rangle + \cos(\theta_{c}) \lvert 1 \rangle
\end{align*}
You can check that
\begin{align*}
\lvert A_{0} \rangle \langle A_{0} \rvert &= \begin{bmatrix}
\cos^{2}(\theta_{a}) & -\cos(\theta_{a}) \sin(\theta_{a}) \\
-\cos(\theta_{a}) \sin(\theta_{a}) & \sin^{2}(\theta_{a})
\end{bmatrix} \\
\lvert A_{1} \rangle \langle A_{1} \rvert &= \begin{bmatrix}
\sin^{2}(\theta_{a}) & \cos(\theta_{a}) \sin(\theta_{a}) \\
\cos(\theta_{a}) \sin(\theta_{a}) & \cos^{2}(\theta_{a})
\end{bmatrix}
\end{align*}
Let
\begin{align*}
\theta_{ab} &\equiv \theta_{a} - \theta_{b} & \theta_{ac} &\equiv \theta_{a} - \theta_{c} &\theta_{bc} &\equiv \theta_{b} - \theta_{c}
\end{align*}
Then, for example:
\begin{align*}
\langle C_{0} \vert A_{0} \rangle &= \cos(\theta_{c})\cos(\theta_{a}) + \sin(\theta_{c}) \sin(\theta_{a}) = \cos(\theta_{ac}) \\
\langle C_{0} \vert A_{1} \rangle &= \cos(\theta_{c})\sin(\theta_{a}) - \sin(\theta_{c}) \cos(\theta_{a}) = \sin(\theta_{ac}) \\
\langle C_{1} \vert A_{0} \rangle &= \sin(\theta_{c})\cos(\theta_{a}) - \cos(\theta_{c}) \sin(\theta_{a}) = -\sin(\theta_{ac}) \\
\langle C_{1} \vert A_{1} \rangle &= \sin(\theta_{c})\sin(\theta_{a}) + \cos(\theta_{c}) \cos(\theta_{a}) = \cos(\theta_{ac})
\end{align*}
We are going to consider three scenarios where Ann and Bob share a message and Eve eavesdrops:
- Eve does not eavesdrops
- Eve eavesdrops with a random choice of basis \(A\) or \(B\)
- Eve eavesdrops with basis \(C\)
In this post we will go over scenario 3, where Eve eavesdrops with basis \(C\).
Ann flips a fair coin and chooses an encoding basis from \(A\) or \(B\). Bob flips another coin and chooses a decoding basis from \(A\) or \(B\). If they do not make the same basis choice, they discard that event. Eve always uses the same eavesdropping basis (\(C\)). Now there are two possible events:
- Ann encodes with \(A\), Eve eavesdrops with \(C\), Bob decodes with \(A\)
- Ann encodes with \(B\), Eve eavesdrops with \(C\), Bob decodes with \(B\)
Consider event 1. Ann prepares the state \(\lvert A_{0} \rangle\). The density matrix is
\begin{equation*}
\rho = \lvert A_{0} \rangle \langle A_{0} \rvert
\end{equation*}
Then Eve eavesdrops with the \(C\) basis. The probability that Eve gets \(\lvert C_{0} \rangle\) is
\begin{equation*}
p(C_{0} \vert \rho) = \langle C_{0} \vert \rho \vert C_{0} \rangle = \langle C_{0} \vert A_{0} \rangle \langle A_{0} \vert C_{0} \rangle = \cos^{2}(\theta_{ac})
\end{equation*}
After the measurement, the density matrix is given by
\begin{equation*}
\rho_{0} = \lvert C_{0} \rangle \langle C_{0} \rvert
\end{equation*}
The probability that Eve gets \(\lvert C_{1} \rangle\) is
\begin{equation*}
p(C_{1} \vert \rho) = \langle C_{1} \vert \rho \vert C_{1} \rangle = \langle C_{1} \vert A_{0} \rangle \langle A_{0} \vert C_{1} \rangle = \sin^{2}(\theta_{ac})
\end{equation*}
Right after the measurement, the density matrix is given by
\begin{equation*}
\rho_{1} = \lvert C_{1} \rangle \langle C_{1} \rvert
\end{equation*}
Thus, the density matrix that Bob receives is
\begin{equation*}
\rho = \cos^{2}(\theta_{ac}) \lvert C_{0} \rangle \langle C_{0} \rvert + \sin^{2}(\theta_{ac}) \lvert C_{1} \rangle \langle C_{1} \rvert
\end{equation*}
Bob then decodes with the \(A\) basis. The probability that Bob gets \(\lvert A_{0} \rangle\) is
\begin{equation*}
p(A_{0} \vert \rho) = \langle A_{0} \vert \rho \vert A_{0} \rangle = \cos^{4}(\theta_{ac}) + \sin^{4}(\theta_{ac})
\end{equation*}
The probability that Bob gets \(\lvert A_{1} \rangle\) is
\begin{equation*}
p(A_{1} \vert \rho) = \langle A_{1} \vert \rho \vert A_{1} \rangle = 2\cos^{2}(\theta_{ac}) \sin^{2}(\theta_{ac})
\end{equation*}
Thus, the probability of success for event 1 is
\begin{equation*}
p_{1} = \cos^{4}(\theta_{ac}) + \sin^{4}(\theta_{ac})
\end{equation*}
In a very similar manner, you can show that the probability of success for event 2 is
\begin{equation*}
p_{2} = \cos^{4}(\theta_{bc}) + \sin^{4}(\theta_{bc})
\end{equation*}
Thus, the overall probability of success is
\begin{equation*}
P = \frac{p_{1} + p_{2}}{2} = \frac{\cos^{4}(\theta_{ac}) + \sin^{4}(\theta_{ac}) + \cos^{4}(\theta_{bc}) + \sin^{4}(\theta_{bc})}{2}
\end{equation*}
(Yet again, we have assumed that there is no noise.)
