Ex5 3

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 5.3}$

Suppose Eve attacks the BB84 quantum key distribution of Section 2.4 as follows. For each qubit she intercepts, she prepares a second qubit in state $\ket 0$, applies a $C_{not}$ from the transmitted qubit to her prepared qubit, sends the first qubit on to Bob, and measures her qubit. How much information can she gain, on average, in this way? What is the probability that she is detected by Alice and Bob when they compare $s$ bits? How do these quantities compare to those of the direct measure-and-transmit strategy discussed in Section 2.4?

page revision: 0, last edited: 18 Nov 2012 18:36

Let Alice's prepared qubit be in the state $\ket{\psi} = a\ket{0} + b\ket{1}$. Eve's qubit begins in state $\ket{0}$, so the combined two-qubit state is $\ket{\psi} \otimes \ket{0} = a \ket{00} + b \ket{10}$, where the first (left) qubit is from Alice and the second (right) qubit is from Eve.

Now, Eve applies a CNOT with the first qubit as the control qubit and the second qubit as the target qubit. Since CNOT takes $\ket{00} \rightarrow \ket{00}$ and $\ket{10} \rightarrow \ket{11}$, this results in a combined two-qubit state of $a \ket{00} + b \ket{11}$.

Next, the first qubit is sent to Bob, who measures it, and Eve measures the second qubit. (It's not clear from the problem statement whether Bob or Eve actually measures first, but the result works out the same either way.) Recall that in BB84, we know that the measurements will occur in either the standard or Hadamard basis. So Bob and Eve will each (randomly) choose one of these bases to measure in, just as Alice (randomly) chose one to prepare in.

We consider two cases:

1. Alice prepared in the standard basis.This means that after Eve's CNOT, the combined two-qubit state is either $\ket{00}$ or $\ket{11}$. If Eve measures in the standard basis, she obtains Alice's original bit with certainty, and if Eve measures in the Hadamard basis, she obtains Alice's original bit only 50% of the time. But because the two qubits are unentangled, Bob's measurement is completely unaffected by Eve's presence. (That is, Eve has not changed the state of the transmitted qubit — it is still in the state $\ket{0}$ or $\ket{1}$ as it was prepared by Alice.)So in this case, Eve learns 50% of the original bits with certainty, and 0% of Bob's measurements are corrupted.

(1)2. Alice prepared in the Hadamard basis.This means that after Eve's CNOT, the combined two-qubit state is either $\ket{\Phi^+}=\tfrac{1}{\sqrt2}(\ket{00}+\ket{11})$ or $\ket{\Phi^-}=\tfrac{1}{\sqrt2}(\ket{00}-\ket{11})$. Now, the two qubits are entangled. Whether Bob measures in the standard basis or the Hadamard basis, he will have equal probability of measuring a 0 or 1, regardless of what Alice's original bit was. For example, let's calculate the probability of Bob's Hadamard-basis measurement in the case that Alice prepared a $\ket{+}$:The same holds for the case where Alice prepares $\ket{-}$. That is, even though Alice and Bob are both using the Hadamard basis, Bob now only gets the same result as Alice half the time, because of the CNOT gate applied by Eve. (Note that exactly the same calculation applies to Eve's measurement as well — she will have a 50% probability of getting the correct result, regardless of which basis she measures in. This follows because both states $\ket{\Phi^+}$ and $\ket{\Phi^-}$ are symmetric in the two qubits.)

So in this case, Eve learns 0% of the original bits with certainty, and 50% of Bob's measurements are corrupted.

(2)In summary:Assuming Alice prepares in each basis 50% of the time on average, Eve will learn 25% of the original bits with certainty, and 25% of Bob's measurements will be corrupted (that is, for each bit, there is a 75% chance that Eve goes undetected). So when Alice and Bob compare $s$ bits afterward, the probability that Eve goes undetected is:This is exactly the same result as the direct measure-and-transmit strategy discussed in section 2.4.

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