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 2, where Eve eavesdrops with a random choice of basis \(A\) or \(B\).
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 also flips a coin and chooses an eavesdropping basis from \(A\) or \(B\). Now there are four possible events:
- Ann encodes with \(A\), Eve eavesdrops with \(A\), Bob decodes with \(A\)
- Ann encodes with \(A\), Eve eavesdrops with \(B\), Bob decodes with \(A\)
- Ann encodes with \(B\), Eve eavesdrops with \(A\), Bob decodes with \(B\)
- Ann encodes with \(B\), Eve eavesdrops with \(B\), Bob decodes with \(B\)
For events 1 and 4, the probability that Bob successfully decodes the qubit that Ann encoded is 1:
\begin{equation*}
p_{4} = p_{1} = 1
\end{equation*}
However, when Eve uses a basis for eavesdropping that is different from the encoding/decoding basis, the probability of success is less than 1.
Consider event 2. 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 \(B\) basis. The probability that Eve gets \(\lvert B_{0} \rangle\) is
\begin{equation*}
p(B_{0} \vert \rho) = \langle B_{0} \vert \rho \vert B_{0} \rangle = \langle B_{0} \vert A_{0} \rangle \langle A_{0} \vert B_{0} \rangle = \cos^{2}(\theta_{ab})
\end{equation*}
After the measurement, the density matrix is given by
\begin{equation*}
\rho_{0} = \lvert B_{0} \rangle \langle B_{0} \rvert
\end{equation*}
The probability that Eve gets \(\lvert B_{1} \rangle\) is
\begin{equation*}
p(B_{1} \vert \rho) = \langle B_{1} \vert \rho \vert B_{1} \rangle = \langle B_{1} \vert A_{0} \rangle \langle A_{0} \vert B_{1} \rangle = \sin^{2}(\theta_{ab})
\end{equation*}
Right after the measurement, the density matrix is given by
\begin{equation*}
\rho_{1} = \lvert B_{1} \rangle \langle B_{1} \rvert
\end{equation*}
Thus, the density matrix that Bob receives is
\begin{equation*}
\rho = \cos^{2}(\theta_{ab}) \lvert B_{0} \rangle \langle B_{0} \rvert + \sin^{2}(\theta_{ab}) \lvert B_{1} \rangle \langle B_{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_{ab}) + \sin^{4}(\theta_{ab})
\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_{ab}) \sin^{2}(\theta_{ab})
\end{equation*}
A similar argument holds true for the case of Ann sending \(\lvert A_{1} \rangle\). Thus, the probability of success in event 2 is
\begin{equation*}
p_{2} = \cos^{4}(\theta_{ab}) + \sin^{4}(\theta_{ab})
\end{equation*}
A similar argument shows that the probability of success in event 3 is the same:
\begin{equation*}
p_{3} = p_{2} = \cos^{4}(\theta_{ab}) + \sin^{4}(\theta_{ab})
\end{equation*}
Thus, the overall probability of success is
\begin{equation*}
P = \frac{p_{1} + p_{2} + p_{3} + p_{4}}{4} = \frac{1 + \cos^{4}(\theta_{ab}) + \sin^{4}(\theta_{ab})}{2}
\end{equation*}
(Again, we have assumed that there is no noise.)
