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 1, where Eve does not eavesdrops.
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. Otherwise, there are two possible events:
- Ann encodes with basis \(A\), Bob decodes with basis \(A\)
- Ann encodes with basis \(B\), Bob decodes with basis \(B\)
Lets consider the first event. Ann can send Bob the state \(\lvert A_{0} \rangle\). The density matrix is
\begin{equation*}
\rho = \lvert A_{0} \rangle \langle A_{0} \rvert
\end{equation*}
Bob receives this density matrix and uses the \(A\) basis for decoding. The probability that Bob measures the state \(\lvert A_{0} \rangle\) is
\begin{equation*}
p(A_{0} \vert \rho ) = \langle A_{0} \vert \rho \vert A_{0} \rangle = 1
\end{equation*}
and the probability that Bob measures the state \(\lvert A_{1} \rangle\) is
\begin{equation*}
p(A_{1} \vert \rho ) = \langle A_{1} \vert \rho \vert A_{1} \rangle = 0
\end{equation*}
Thus, the probability that Bob successfully decodes the state that Ann encoded is 1. A similar argument holds when Ann sends Bob the \(\lvert A_{1} \rangle\) state: Bob will successfully decode the state with probability 1.
The same story follows for event 2. Thus, as long as Ann and Bob use the same basis for encoding and decoding, they will always agree on the message. (We have assumed that there is no noise.)
