BB84 protocol introduction
The BB84 protocol, conceived by Bennet and Brassard in 1984, is a key example of Quantum Key Distribution (QKD). QKD aims to establish secure communication channels by leveraging the principles of quantum mechanics. In practice, photon polarization is often used in QKD implementations.
In the BB84 protocol, Alice prepares photons with polarizations chosen from four states. These states are defined by two bases: the rectilinear basis (vertical |\mathbf V\rangle and horizontal |\mathbf H\rangle) and the diagonal basis (left |\mathbf L\rangle and right |\mathbf R\rangle).
We can represent these polarization states using vectors. In the horizontal/vertical basis, we have:
| \mathbf V \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad | \mathbf H \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
and the diagonal polarization states are superpositions of the rectilinear states:
| \mathbf L \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \quad | \mathbf R \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}
Measurements in these bases correspond to observables. For the vertical/horizontal basis, the observable is represented by \boldsymbol \sigma_z:
\boldsymbol \sigma_z = | \mathbf V \rangle \langle \mathbf V | - | \mathbf H \rangle \langle \mathbf H | = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
and for the right/left basis by \boldsymbol \sigma_x:
\boldsymbol \sigma_x = | \mathbf L \rangle \langle \mathbf L | - | \mathbf R \rangle \langle \mathbf R | = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
A key feature is that these observables do not commute, [\boldsymbol \sigma_z, \boldsymbol \sigma_x] \neq 0. This non-commutativity is central to the security of BB84.
When Bob receives a photon, he independently chooses to measure it in either the vertical/horizontal or the right/left basis. If Bob selects the same basis Alice used to prepare the photon, he will correctly determine Alice’s polarization state.
Consider a photon prepared in the right circular polarization state | \mathbf R \rangle. Let’s calculate the probabilities of Bob measuring it as vertically | \mathbf V \rangle or horizontally | \mathbf H \rangle polarized. We use the representation of right circular polarization as | \mathbf{R} \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix} in the horizontal/vertical basis.
The probability of measuring vertical polarization is given by |\langle \mathbf V | \mathbf R \rangle|^2. We compute the inner product:
\langle \mathbf V | \mathbf R \rangle = \begin{bmatrix} 0 & 1 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix} = \frac{-i}{\sqrt{2}}
The probability is:
P(\mathbf{V} | \mathbf{R}) = |\langle \mathbf V | \mathbf R \rangle|^2 = \left| \frac{-i}{\sqrt{2}} \right|^2 = \frac{1}{2}
Similarly, for horizontal polarization:
\langle \mathbf H | \mathbf R \rangle = \begin{bmatrix} 1 & 0 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix} = \frac{1}{\sqrt{2}}
and the probability is:
P(\mathbf{H} | \mathbf{R}) = |\langle \mathbf H | \mathbf R \rangle|^2 = \left| \frac{1}{\sqrt{2}} \right|^2 = \frac{1}{2}
This shows that measuring a right circularly polarized photon in the vertical/horizontal basis results in a 50% chance of obtaining either vertical or horizontal polarization.
For more insights into this topic, you can find the details here.