BB84 protocol
The BB84 protocol, developed by Bennet and Brassard in 1984, is an example of QKD.
QKD can theoretically use any quantum object with a two-valued observable, but in practice, photon polarization is favored.
Alice uses a source of polarized photons and chooses each photon’s polarization from four options. These polarizations are defined by two linear polarization bases: vertical/horizontal (| \mathbf V\rangle and |\mathbf H\rangle) and right/left (|\mathbf R \rangle and |\mathbf L\rangle) at 45^\circ.
Each polarization represents two orthogonal polarizations, corresponding to the two output channels of a polarizing beam splitter.
We can use the representation of polarization states in quantum mechanics using vectors in a 2-dimensional complex vector space. The standard basis for polarization is typically chosen to be horizontal and vertical polarization.
| \mathbf V \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
represents vertical polarization and:
| \mathbf H \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
represents horizontal polarization.
For the polarization left and right, we need to consider them as superpositions of horizontal and vertical polarizations. Left polarization (| \mathbf L \rangle) is at a 45^\circ degree angle with respect to the horizontal and vertical axes, and right polarization (|\mathbf R \rangle) is at a -45^\circ angle.
The left polarization is:
| \mathbf L \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ 1 \end{bmatrix}
A left polarized photon is an equal superposition of horizontal and vertical polarizations. The factor \frac{1}{\sqrt{2}} is for normalization. We can verify that:
| \mathbf L \rangle = \frac{1}{\sqrt{2}} (| \mathbf H \rangle + | \mathbf V \rangle)
The right polarization is:
|\mathbf R \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -1 \end{bmatrix}
A right polarized photon is also an equal superposition of horizontal and vertical polarizations, but with a phase difference (specifically, a minus sign). We can verify that:
|\mathbf R \rangle = \frac{1}{\sqrt{2}} (|\mathbf H \rangle - | \mathbf V \rangle)
these vector representations are based on expressing different polarization states as linear combinations of the basis states, which are typically chosen to be horizontal and vertical polarizations. With these basis:
| \mathbf V \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \quad | \mathbf H \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}
and:
| \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}
We can represents the observable associated with measurements in the vertical/horizontal polarization basis as:
\boldsymbol \sigma_z = | \mathbf V \rangle \langle \mathbf V - | \mathbf H \rangle \langle \mathbf H | = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}
The one in the right/left polarization basis:
\boldsymbol \sigma_x = | \mathbf L \rangle \mathbf \langle \mathbf L | - | \mathbf R \rangle \langle \mathbf R | = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
We can then compute the commutator:
\begin{aligned} & [\boldsymbol \sigma_z, \boldsymbol \sigma_x] = \boldsymbol \sigma_z \boldsymbol \sigma_x - \boldsymbol \sigma_x \boldsymbol \sigma_z \\ & \boldsymbol \sigma_z \boldsymbol \sigma_x = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \\ & \boldsymbol \sigma_x \boldsymbol \sigma_z = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \\ & [\boldsymbol \sigma_z, \boldsymbol \sigma_x] = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} - \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} \neq 0 \end{aligned}
The observables do not commute.
When Bob receives a photon, randomly selects either the vertical/horizontal or the right/left basis to measure its polarization. If Bob’s basis matches Alice’s sending basis, his measurement accurately reflects Alice’s polarization choice.
Let’s consider a photon prepared in a right circular polarization state, which we represent as | \mathbf R \rangle. We want to calculate the probabilities of measuring this photon to be in either a vertical polarization state | \mathbf V \rangle or a horizontal polarization state | \mathbf H \rangle.
We use the vector representations for right circular, vertical, and horizontal polarization in the horizontal/vertical basis:
| \mathbf R \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix}, \quad | \mathbf V \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad | \mathbf H \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
To find the probability of measuring a photon in state | \mathbf R \rangle to be in state | \mathbf V \rangle, we calculate the probability of transition from state | \mathbf R \rangle to state | \mathbf V \rangle. This probability is given by the squared modulus of the projection of state | \mathbf R \rangle onto state | \mathbf V \rangle, |\langle \mathbf V | \mathbf R \rangle|^2.
We first calculate the inner product \langle \mathbf V | \mathbf R \rangle:
\langle \mathbf V | \mathbf R \rangle = \begin{bmatrix} 0 & 1 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix} = \frac{1}{\sqrt{2}} (0 \times 1 + 1 \times (-i)) = \frac{-i}{\sqrt{2}}
The probability of measuring the photon to be vertically polarized when it is initially right circularly polarized is the squared modulus of this inner product:
P(\mathbf{V} | \mathbf{R}) = |\langle \mathbf V | \mathbf R \rangle|^2 = \left| \frac{-i}{\sqrt{2}} \right|^2 = \left( \frac{|-i|}{\sqrt{2}} \right)^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}
Similarly, to find the probability of measuring a photon in state | \mathbf R \rangle to be in state | \mathbf H \rangle, we calculate the probability of transition from state | \mathbf R \rangle to state | \mathbf H \rangle. This probability is given by the squared modulus of the projection of state | \mathbf R \rangle onto state | \mathbf H \rangle, |\langle \mathbf H | \mathbf R \rangle|^2.
We first calculate the inner product \langle \mathbf H | \mathbf R \rangle:
\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}} (1 \times 1 + 0 \times (-i)) = \frac{1}{\sqrt{2}}
The probability of measuring the photon to be horizontally polarized when it is initially right circularly polarized is the squared modulus of this inner product:
P(\mathbf{H} | \mathbf{R}) = |\langle \mathbf H | \mathbf R \rangle|^2 = \left| \frac{1}{\sqrt{2}} \right|^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}
So, if we have a photon in the right circular polarization state | \mathbf R \rangle and we measure its polarization in the vertical/horizontal basis, the probability of finding it vertically polarized is 1/2 and the probability of finding it horizontally polarized is also 1/2. This means that there is an equal chance of measuring a right circularly polarized photon to be either vertically or horizontally polarized.
Let’s now consider a photon prepared in a right circular polarization state | \mathbf{R} \rangle. We want to calculate the probabilities of measuring this photon to be in either a vertical polarization state | \mathbf{V} \rangle or a horizontal polarization state | \mathbf{H} \rangle.
We need the vector representations for right circular, vertical, and horizontal polarization in the horizontal/vertical basis:
| \mathbf{R} \rangle = \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix}, \quad | \mathbf{V} \rangle = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad | \mathbf{H} \rangle = \begin{bmatrix} 1 \\ 0 \end{bmatrix}
To find the probability of measuring a photon in state | \mathbf{R} \rangle to be in state | \mathbf{V} \rangle, we need to calculate the probability P(\mathbf{V} | \mathbf{R}) = |\langle \mathbf{V} | \mathbf{R} \rangle|^2.
We first calculate the inner product \langle \mathbf{V} | \mathbf{R} \rangle:
\langle \mathbf{V} | \mathbf{R} \rangle = \begin{bmatrix} 0 & 1 \end{bmatrix} \frac{1}{\sqrt{2}} \begin{bmatrix} 1 \\ -i \end{bmatrix} = \frac{1}{\sqrt{2}} (0 \times 1 + 1 \times (-i)) = \frac{-i}{\sqrt{2}}
The probability of measuring the photon to be vertically polarized when it is initially right circularly polarized is the squared modulus of this inner product:
P(\mathbf{V} | \mathbf{R}) = |\langle \mathbf{V} | \mathbf{R} \rangle|^2 = \left| \frac{-i}{\sqrt{2}} \right|^2 = \frac{1}{2}
Similarly, to find the probability of measuring a photon in state | \mathbf{R} \rangle to be in state | \mathbf{H} \rangle, we calculate P(\mathbf{H} | \mathbf{R}) = |\langle \mathbf{H} | \mathbf{R} \rangle|^2. The inner product \langle \mathbf{H} | \mathbf{R} \rangle is calculated as:
\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}} (1 \times 1 + 0 \times (-i)) = \frac{1}{\sqrt{2}}
The probability of measuring the photon to be horizontally polarized when it is initially right circularly polarized is the squared modulus of this inner product:
P(\mathbf{H} | \mathbf{R}) = |\langle \mathbf{H} | \mathbf{R} \rangle|^2 = \left| \frac{1}{\sqrt{2}} \right|^2 = \frac{1}{2}
If we have a photon in the right circular polarization state | \mathbf{R} \rangle and we measure its polarization in the vertical/horizontal basis, the probability of finding it vertically polarized is 1/2 and the probability of finding it horizontally polarized is also 1/2.
To identify these correct cases, the perform the process called reconciliation: Bob publicly announces his chosen bases, and Alice reveals which of Bob’s choices were correct.
Alice and Bob then discard the mismatched cases and retain only the data from correctly aligned bases, resulting in identical keys. The communication about basis choices occurs over a public channel.
Now consider Eve eavesdropping. To learn the key, Eve intercepts photons, measures their polarization, and send a new photon to Bob with the measured polarization. However, Eve doesn’t know Alice’s basis choice and must guess it.
The no-cloning theorem (demonstrated here) prevents Eve from copying the unknown polarization state, Eve must perform a measurement to gain information, and send a new photon in a polarization chosen by Eve, not the original polarization chosen by Alice.
If Eve guesses the wrong basis (50\% chance), quantum mechanics dictates that measuring polarization in a non-orthogonal basis will have a different expectation value.
When Eve send a photon based on a measurement in a wrong basis, the photon sent has a polarization uncorrelated to Alice’s original polarization in those mismatched basis cases. Consequently, measuring in the correct basis (matching Alice’s original choice for these bits during reconciliation), Bob statistically will get the wrong result 50\% of the time in these “wrong basis” interception cases by Eve (the percentage is due to the particular orientation of the basis chosen, I will describe later for a generic angle \theta).
To detect Eve, Alice and Bob sacrifice a random subset of their reconciled key. Bob publicly announces the measured values of these sacrificed bits. Alice reveals the original sent bits for these positions and they compare these bits.
If Eve is absent, the error rate is low, due to channel noise and imperfections inherent in practical implementations; in an idealized scenario with perfect components and no channel noise, the error rate would be exactly zero.
However, if Eve is present and performing the measure-and-resend attack, the errors introduced by the wrong-basis measurements will increase the error rate significantly. This increased error rate alerts Alice and Bob to Eve’s presence, and they discard the potentially compromised key.
The security of BB84 relies on:
- quantum measurement disturbance: Measuring polarization in a non-orthogonal basis changes the state,
- no-cloning theorem: Unknown quantum states cannot be perfectly copied, forcing Eve to measure and potentially disturb the signal,
- statistical detection: Errors introduced by Eve’s measurements in wrong bases are detectable by Alice and Bob during error checking.
Optimal angle
The 50\% probability of error in BB84 when Eve intercepts and measures in a basis mismatched to Alice’s transmission is not accidental; it’s a designed feature to maximize the security of the protocol.
This maximal error rate is achieved by Alice and Bob using two mutually unbiased bases oriented at 45^\circ to each other. This specific angle is chosen to ensure that any eavesdropping attempt by Eve, particularly when guessing the wrong basis, introduces the maximum possible disturbance, thereby maximizing the likelihood of detecting the presence.
Let’s analyze why a 45^\circ angle between the chosen bases is optimal for maximizing eavesdropping detection by examining how the error probability changes with a generic angle \boldsymbol\theta between potential measurement bases used by an eavesdropper.
Assume Alice and Bob use two bases. Alice’s primary basis B_1 consists of polarization states |\mathbf{H}\rangle (horizontal) and |\mathbf{V}\rangle (vertical).
Consider an eavesdropper Eve who chooses to measure in a basis Y defined by polarization directions |\boldsymbol\theta\rangle and |\boldsymbol\theta + \pi/2\rangle, where \theta is the angle of |\boldsymbol\theta\rangle relative to the horizontal axis. Note that |\boldsymbol\theta + \pi/2\rangle is orthogonal to |\boldsymbol\theta\rangle.
Consider the scenario where Alice sends a horizontally polarized photon |\mathbf{H}\rangle. If 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 a photon in state |\boldsymbol\theta\rangle, when Bob measures in Alice’s original basis B_1, the probability of getting vertical polarization |\mathbf{V}\rangle (an error if Alice sent horizontal |\mathbf{H}\rangle) is:
|\langle \mathbf{V} | \boldsymbol\theta \rangle|^2 = \sin^2 \left(\theta\right)
If Eve measures |\boldsymbol\theta + \pi/2\rangle and resends a photon in state |\boldsymbol\theta + \pi/2\rangle, when Bob measures in Alice’s original basis B_1, the probability of 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 getting a vertical polarization |\mathbf{V}\rangle (error) when Alice sent horizontal |\mathbf{H}\rangle and Eve measured in a basis with angle \theta (and then Bob measures in the original basis B_1) 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) \end{aligned}
Substituting the probabilities as function of \theta:
\begin{aligned} P(\text{error}) & = \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} \left[2 \sin \left(\theta\right) \cos \left(\theta\right)\right]^2 = \frac{1}{2} \sin^2 (2\theta) \end{aligned}
For \theta = 45^\circ = \pi/4, \frac{1}{2} \sin^2 (2 \times 45^\circ) = \frac{1}{2} \sin^2 (90^\circ) = \frac{1}{2} \times 1^2 = \frac{1}{2} = 50\%.
For a generic angle \theta, the error probability is given by \frac{1}{2} \sin^2 (2\theta), and considering various angles demonstrates 45^\circ is optimal for security:
- If \theta = 0^\circ, error probability is 0\%. If Eve measures in the same basis as Alice (and Bob’s measurement basis), there is no detectable errors introduced, making eavesdropping undetectable.
- if \theta = 30^\circ, error probability is \frac{1}{2} \sin^2 (60^\circ) = \frac{1}{2} (\frac{\sqrt{3}}{2})^2 = \frac{3}{8} = 37.5\%. This lower error rate compared to 45^\circ reduces the chances of detecting Eve.
- If \theta = 45^\circ, error probability is 50\%. This is the maximal disturbance achievable, providing the highest probability of detecting Eve’s intervention.
- if \theta = 60^\circ, error probability is \frac{1}{2} \sin^2 (120^\circ) = \frac{1}{2} \sin^2 (60^\circ) = 37.5\%. Again, a reduced error rate, less effective for detecting eavesdropping.
- if \theta = 90^\circ, error probability is 0\%. If Eve measures in a basis orthogonal to a potential basis Alice might have used, the error becomes zero in this specific scenario, and again, eavesdropping becomes harder to detect on average.
The 50\% error rate at \theta = 45^\circ is the maximum possible error introduced by a measurement in a mismatched basis. Alice and Bob’s choice of mutually unbiased bases at 45^\circ in BB84 is therefore not arbitrary but specifically designed to maximize \frac{1}{2} \sin^2 (2\theta).
This maximization ensures that any eavesdropping attempt by Eve, particularly when guessing incorrectly about the basis, has the highest possible chance of being detected through an increased error rate in the quantum key distribution process.
By maximizing the potential disturbance Eve introduces, the 45^\circ angle maximizes the security and eavesdropping detection capability of the BB84 protocol, making it the optimal choice for basis selection.