BB84 error rate: angle dependence in eavesdropping
In previous posts (here and here) about the BB84 protocol, I mentioned the 50\% error probability when an eavesdropper, Eve, measures in a basis different from Alice’s. This isn’t a coincidence; it’s a deliberate design to enhance the protocol’s security. This maximal error rate is achieved because BB84 uses two mutually unbiased bases oriented at 45^\circ. This specific angle maximizes the disturbance Eve introduces when guessing the wrong basis, making the presence more easily detectable.
Let’s examine why this 45^\circ angle is optimal for detecting eavesdropping by analyzing how the error probability changes with a generic angle \theta between potential measurement bases used by Eve.
Assume Alice and Bob use two bases. Alice’s primary basis B_1 is composed of horizontal |\mathbf{H}\rangle and vertical |\mathbf{V}\rangle polarization states. Eve chooses to measure in a basis Y = \{|\boldsymbol\theta\rangle, |\boldsymbol\theta + \pi/2\rangle\}, where \theta is the angle of |\boldsymbol\theta\rangle relative to the horizontal axis.
If Alice sends a horizontally polarized photon |\mathbf{H}\rangle, and Eve measures in basis Y, the probability of Eve measuring |\boldsymbol\theta\rangle is:
|\langle \boldsymbol\theta | \mathbf{H} \rangle|^2 = \cos^2 \left(\theta\right)
And the probability of Eve measuring |\boldsymbol\theta + \pi/2\rangle is:
|\langle \boldsymbol\theta + \pi/2 | \mathbf{H} \rangle|^2 = \sin^2 \left(\theta\right)
If Eve measures |\boldsymbol\theta\rangle and resends |\boldsymbol\theta\rangle, the probability of Bob (measuring in basis B_1) getting vertical polarization |\mathbf{V}\rangle is:
|\langle \mathbf{V} | \boldsymbol\theta \rangle|^2 = \sin^2 \left(\theta\right)
If Eve measures |\boldsymbol\theta + \pi/2\rangle and resends |\boldsymbol\theta + \pi/2\rangle, the probability of Bob getting vertical polarization |\mathbf{V}\rangle is:
|\langle \mathbf{V} | \boldsymbol\theta + \pi/2 \rangle|^2 = \cos^2 \left(\theta\right)
The overall probability of Bob measuring vertical polarization |\mathbf{V}\rangle (an error if Alice sent horizontal |\mathbf{H}\rangle) is:
\begin{aligned} P(\text{error}) = & P(\text{Eve measures } |\boldsymbol\theta\rangle) \times P(\text{Bob gets } |\mathbf{V}\rangle | \text{Eve sent } |\boldsymbol\theta\rangle) \\ & + P(\text{Eve measures } |\boldsymbol\theta + \pi/2\rangle) \times P(\text{Bob gets } |\mathbf{V}\rangle | \text{Eve sent } |\boldsymbol\theta + \pi/2\rangle) \\ & = \cos^2 \left(\theta\right) \times \sin^2 \left(\theta\right) + \sin^2 \left(\theta\right) \times \cos^2 \left(\theta\right) \\ & = 2 \sin^2 \left(\theta\right) \cos^2 \left(\theta\right) = \frac{1}{2} \sin^2 (2\theta) \end{aligned}
Considering different angles \theta:
- \theta = 0^\circ, P(\text{error}) = 0\%. Eve measuring in the same basis introduces no errors, making detection impossible,
- \theta = 30^\circ, P(\text{error}) = \frac{1}{2} \sin^2 (60^\circ) = 37.5\%. The error rate is lower than the maximum,
- \theta = 45^\circ, P(\text{error}) = \frac{1}{2} \sin^2 (90^\circ) = 50\%. This is the maximum error probability,
- \theta = 60^\circ, P(\text{error}) = \frac{1}{2} \sin^2 (120^\circ) = 37.5\%. Again, a lower error rate,
- \theta = 90^\circ, P(\text{error}) = 0\%. Similar to 0^\circ, no errors are introduced in this specific scenario.
The error probability P(\text{error}) = \frac{1}{2} \sin^2 (2\theta) is maximized when \sin^2 (2\theta) is maximized, which occurs when 2\theta = 90^\circ or \theta = 45^\circ. The choice of 45^\circ angle between the bases in BB84 is therefore optimal.
It maximizes the error rate introduced by an eavesdropper who guesses the wrong basis, thus maximizing the chance of detecting eavesdropping attempts and ensuring the highest possible security for quantum key distribution.
For more insights into this topic, you can find the details here.