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My question is specifically from the book Quantum Computing for Everyone by Chris Bernhardt.

In the last section (The Ekert Protocol for Quantum Key Distribution) from Chapter 5 (Bell's Inequality). I am trying to understand the derivation for why if Eve, an eavesdropper, is measuring one of the qubits the proportion of times Alice and Bob agree changes. The proportion of times they agree will increase to 3/8.

Background: The Ekert protocol proposed by Artur Ekert in 1991 is a method for quantum key distribution. The protocol is based on the principle of quantum entanglement and a clever use of the identities prescribed by Bell's test to entangled qubits. I am trying to understand the derivations to detect an eavesdropper, Eve, when Alice and Bob are trying to exchange keys for secure communication.

There are three bases for the three directions of measurement, $0^{0},120^{0},240^{0}$.

\begin{equation} 0^{0}=\left(\left|\uparrow\right\rangle ,\left|\downarrow\right\rangle \right)=\left(\left[\begin{array}{c} 1\\ 0 \end{array}\right],\left[\begin{array}{c} 0\\ 1 \end{array}\right]\right) \end{equation} \begin{equation} 120^{0}=\left(\left|\searrow\right\rangle ,\left|\nwarrow\right\rangle \right)=\left(\left[\begin{array}{c} \frac{1}{2}\\ -\frac{\sqrt{3}}{2} \end{array}\right],\left[\begin{array}{c} \frac{\sqrt{3}}{2}\\ \;\;\;\frac{1}{2} \end{array}\right]\right) \end{equation} \begin{equation} 240^{0}=\left(\left|\swarrow\right\rangle ,\left|\nearrow\right\rangle \right)=\left(\left[\begin{array}{c} \frac{1}{2}\\ \;\;\frac{\sqrt{3}}{2} \end{array}\right],\left[\begin{array}{c} \frac{\sqrt{3}}{2}\\ -\frac{1}{2} \end{array}\right]\right) \end{equation}

Alice receives one qubit and Bob the other from a steam of entangled qubits. The entangled qubits are in the state,

\begin{equation} \frac{1}{\sqrt{2}}\left|\uparrow\right\rangle \left|\uparrow\right\rangle +\frac{1}{\sqrt{2}}\left|\downarrow\right\rangle \left|\downarrow\right\rangle \end{equation}

We know that 3$n$ qubits have been sent. Alice and Bob make measurements on these qubits by randomly using one of the three bases.

We know that $3n$ qubits have been sent. Alice and Bob make measurements on these qubits by randomly using one of the three bases. After they make the measurements they reveal the sequence of bases that they chose. They are only revealing the basis they chose, not the result of the measurements. They will agree on approximately $n$ of them. We assume that $n$ is a fairly large number. In each place they have chosen the same basis they will have made the same measurement. They will both either have 0, or both have 1. They will get either a 0 or a 1 with equal probability but they will both get exactly the same answer. This string of $n$ 0s and 1s will be their key if Eve is not listening in.

They test for whether Eve is eavesdropping or not. If Eve is listening in, she will have to make measurements and whenever a measurement is done, the entangled states become unentangled. They check the other 2$n$ strings of 0 and 1 that come from the times when they chose different bases. Depending on whether Eve is eavesdropping or not gives different values for the proportion of qubits that they agree on. The two properties, depending on whether there is an eavesdropper or not, are shown below.

  • If the states are entangled when Alice and Bob make their measurements, then from the Bell Inequality they know that in each place they should agree only 1/4 of the time. Is the below correct?

We know that there is a probability of 1/2 of getting either 0 or 1 if the pair of qubits are entangled. This is because there is an equal probability of getting 0 or 1 as seen from this equation:

\begin{equation} \frac{1}{\sqrt{2}}\left|\uparrow\right\rangle \left|\uparrow\right\rangle +\frac{1}{\sqrt{2}}\left|\downarrow\right\rangle \left|\downarrow\right\rangle \end{equation}

We also know that Alice and Bob are measuring using different bases since we are only comparing those qubits that were measured using different bases. When they measure with different bases, the probability of getting 0 or 1 is 1/4. We can see this from the set of equations shown below that express each basis in terms of the other basis and by looking at the coefficients of the vector that represents 0 or 1.

\begin{equation} \left|\uparrow\right\rangle =\frac{1}{2}\left|\searrow\right\rangle +\frac{\sqrt{3}}{2}\left|\nwarrow\right\rangle \end{equation} \begin{equation} \left|\downarrow\right\rangle =-\frac{\sqrt{3}}{2}\left|\searrow\right\rangle +\frac{1}{2}\left|\nwarrow\right\rangle \end{equation} \begin{equation} \left|\searrow\right\rangle =\frac{1}{2}\left|\uparrow\right\rangle -\frac{\sqrt{3}}{2}\left|\downarrow\right\rangle \end{equation} \begin{equation} \left|\nwarrow\right\rangle =\frac{\sqrt{3}}{2}\left|\uparrow\right\rangle +\frac{1}{2}\left|\downarrow\right\rangle \end{equation} \begin{equation} \left|\uparrow\right\rangle =\frac{1}{2}\left|\swarrow\right\rangle +\frac{\sqrt{3}}{2}\left|\nearrow\right\rangle \end{equation} \begin{equation} \left|\downarrow\right\rangle =\frac{\sqrt{3}}{2}\left|\swarrow\right\rangle -\frac{1}{2}\left|\nearrow\right\rangle \end{equation} \begin{equation} \left|\swarrow\right\rangle =\frac{1}{2}\left|\uparrow\right\rangle +\frac{\sqrt{3}}{2}\left|\downarrow\right\rangle \end{equation} \begin{equation} \left|\nearrow\right\rangle =\frac{\sqrt{3}}{2}\left|\uparrow\right\rangle -\frac{1}{2}\left|\downarrow\right\rangle \end{equation} \begin{equation} \left|\searrow\right\rangle =-\frac{1}{2}\left|\swarrow\right\rangle +\frac{\sqrt{3}}{2}\left|\nearrow\right\rangle \end{equation} \begin{equation} \left|\nwarrow\right\rangle =\frac{\sqrt{3}}{2}\left|\swarrow\right\rangle +\frac{1}{2}\left|\nearrow\right\rangle \end{equation} \begin{equation} \left|\swarrow\right\rangle =-\frac{1}{2}\left|\searrow\right\rangle +\frac{\sqrt{3}}{2}\left|\nwarrow\right\rangle \end{equation} \begin{equation} \left|\nearrow\right\rangle =\frac{\sqrt{3}}{2}\left|\searrow\right\rangle +\frac{1}{2}\left|\nwarrow\right\rangle \end{equation}

When Eve is not eaves dropping when Alice or Bob make a measurement the qubits are entangled. Hence the probability that Alice and Bob will get the same qubits, or that they agree, when the qubits are entangled is given by,

\begin{equation} \left(\frac{1}{2}\right)\left(\frac{1}{4}\right)+\left(\frac{1}{2}\right)\left(\frac{1}{4}\right)=\frac{1}{4}\label{eq:Agree-Entangled} \end{equation}

  • However if Eve is measuring one of the qubits the proportion of times they agree changes. The proportion of times they agree will increase to 3/8. Why is this true?

PS: If any additional clarifications are needed, please let me know. If this question is already answered somewhere or if it is too simple, please let me know and I can delete it. I am new to this forum. Just joined it today and could not find the answer I was looking for.

Related Questions: In the E91 protocol, how can Alice and Bob detect Eve if she waits to measure until A&B publish their bases?

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We assume Eve, along with Alice and Bob, picks her basis at random. We are given that Alice and Bob chose different bases. So there is a $\frac23$ chance that Eve picks the same basis as either Alice or Bob, and a $\frac13$ chance that she picks the remaining basis.

If Eve picks the same one as either Alice or Bob, the odds remain at $\frac14$.

But what happens if she picks the remaining basis? She does her measurement, which disentangles the qubits, but gives them the same value. Alice now does her measurement. She has a 3/4 chance of getting a different answer from Eve and a 1/4 chance of getting the same answer as Eve. Likewise for Bob. So the probability that Alice and Bob get the same result is:

$\frac34 \cdot\frac34 + \frac14 \cdot \frac14 = \frac58$.

So all three have equal probability, the result is: $\frac23(\frac14) + \frac13(\frac58) = \frac38$.

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