I'm trying to understand this problem. Alice I attempting to send a 2 classical bit message to Bob using 1 qubit such that there are 4 states $\varphi_{00}$ $\varphi_{01}$ $\varphi_{10}$ $\varphi_{11}$ that represent $\varphi_{xy}$ and their values respectively are $|0\rangle$ $|+\rangle$ $|1\rangle$ $|-\rangle$.

This explains the basis that will let Eve measure $x$ from the given state $\varphi_{xy}$ with a probability of $\cos^{2}\left ( \frac{\pi }{8} \right )$, but I can't seem to conclude the same mathematically. enter image description here

Here's what I have derived so far using $P[|\phi_0\rangle$ when $|\varphi_{xy}\rangle$ is $|\varphi\rangle ] = |\langle\phi_0|\varphi\rangle|^{2}$:

  1. $P[|\phi_0\rangle \text{when} |\varphi_{xy}\rangle \text{is}|0\rangle ] = \cos^{2}\left ( \frac{\pi }{8} \right )$
  2. $P[|\phi_0\rangle \text{when} |\varphi_{xy}\rangle \text{is}|+\rangle ] = \frac{1}{2}(\cos\frac{\pi }{8}+\sin\frac{\pi }{8})^{2}$
  3. $P[|\phi_0\rangle \text{when} |\varphi_{xy}\rangle \text{is}|0\rangle ] = \sin^{2}\left ( \frac{\pi }{8} \right )$
  4. $P[|\phi_0\rangle \text{when} |\varphi_{xy}\rangle \text{is}|-\rangle ] = \frac{1}{2}(\cos\frac{\pi }{8}-\sin\frac{\pi }{8})^{2}$
  5. $P[|\phi_1\rangle \text{when} |\varphi_{xy}\rangle \text{is}|0\rangle ] = \sin^{2}\left ( \frac{\pi }{8} \right )$
  6. $P[|\phi_1\rangle \text{when} |\varphi_{xy}\rangle \text{is}|-\rangle ] = \frac{1}{2}(\cos\frac{\pi }{8}-\sin\frac{\pi }{8})^{2}$
  7. $P[|\phi_1\rangle \text{when} |\varphi_{xy}\rangle \text{is}|0\rangle ] = \cos^{2}\left ( \frac{\pi }{8} \right )$
  8. $P[|\phi_1\rangle \text{when} |\varphi_{xy}\rangle \text{is}|+\rangle ] = \frac{1}{2}(\cos\frac{\pi }{8}+\sin\frac{\pi }{8})^{2}$

How do we conclude that the probability of learning $x$ is $\cos^{2}\left ( \frac{\pi }{8} \right )$?

Thanks in advance.

  • 1
    $\begingroup$ what textbook/paper is this from? $\endgroup$
    – glS
    Commented Mar 1, 2022 at 9:48
  • $\begingroup$ Not clear how Alice sends 2 classical bits, or whatever it means. I guess she sends 1 bit per time. $\endgroup$
    – kludg
    Commented Jul 31, 2022 at 11:40

3 Answers 3


The way I have understood the problem is Alice is sending an encoded qubit to Bob and we have to determine with what probability Bob can successfully guess the first bit that Alice has encoded. Bob is performing the measurement on $|\phi_0\rangle$, $|\phi_1\rangle$ basis. I assign labels to the state $|\phi_0\rangle$ and $|\phi_1\rangle$ such that they corresponds to outcome $0$ and outcome $1$ of Bob's measurement.
If Bob receives the state $|\psi_{00}\rangle$ to successfuly guess the first bit(which is zero) he has to measure the probability that $|\psi_{00}\rangle$ state is in $|\phi_0\rangle$ state. So probability of success $= |\langle \phi_0|\psi_{00}\rangle|^2 = \cos^2(\pi/8)$.
If Bob receives the state $|\psi_{01}\rangle$to successfuly guess the first bit(which is zero) he has to measure the probability that $|\psi_{01}\rangle$ state is in $|\phi_0\rangle$ state. Probability of success $= |\langle \phi_0|\psi_{01}\rangle|^2 = \cos^2(\pi/8)$.
You can get the angle between states from the diagram.
For the state $|\psi_{10}\rangle$ success is to guess the first bit as $1$. Probability of success $= |\langle \phi_1|\psi_{10}\rangle|^2 = \cos^2(\pi/8)$.
For the state $|\psi_{11}\rangle$ success is to guess the first bit as $1$. Probability of success $= |\langle \phi_1|\psi_{11}\rangle|^2 =\cos^2(7\pi/8) = \cos^2(\pi/8) \approx 0.85$
So irrespective of which state Bob receives, he can successfully guess the first bit with probability 0.85


I believe this question is about BB84 quantum key distribution scheme. We have Alice sending qubits to Bob and Eve intercepting them, measuring and resending measured qubits to Bob.

Intuitively, Eve uses the basis $\{|\phi_0\rangle, |\phi_1\rangle\}$ (also called Breidbart basis) because it lies midway between the encoding bases $\{|0\rangle\ , |1\rangle\}$ and $\{|+\rangle\ , |-\rangle\}$. Essentially, Breidbart basis introduces the least amount of disturbance when Eve eavesdrops (intercepts and measures).

Mathematically, we can verify that $$\tag{1} \cos^2(\pi/8) = \frac{2 + \sqrt{2}}{4} = \frac{\left(\cos(\pi/8) + \sin(\pi/8)\right)^2}{2}.$$ Therefore, the squared overlap of $|\phi_0\rangle$ with $|0\rangle$ and $|+\rangle$ is $\cos^2(\pi/8)$: $$|\langle\phi_0 |0\rangle|^2 = |\langle\phi_0 |+\rangle|^2 = \cos^2(\pi/8).$$

By Equation (1), the probability that Eve measures $|\phi_0\rangle$ when she receives $|0\rangle$ or $|+\rangle$ is: \begin{align*} &\Pr(|\phi_0\rangle \ \textrm{given} \ |0\rangle) = \cos^2(\pi/8),\\ &\Pr(|\phi_0\rangle \ \textrm{given} \ |+\rangle) = \frac{\left(\cos(\pi/8) + \sin(\pi/8)\right)^2}{2}=\cos^2(\pi/8). \end{align*}

Given that the probability of receiving $|0\rangle$ or $|+\rangle$ is 1/2 we conclude that Eve learns $x=0$ correctly with the following probability: \begin{align*} \Pr(x=0 \ \textrm{is correct}) = \Pr(|\phi_0\rangle \ \textrm{given} \ |0\rangle)1/2 + \Pr(|\phi_0\rangle \ \textrm{given} \ |+\rangle)1/2 = \cos^2(\pi/8). \end{align*}

Similarly, \begin{align*} &\Pr(|\phi_1\rangle \ \textrm{given} \ |1\rangle) = \cos^2(\pi/8),\\ &\Pr(|\phi_1\rangle \ \textrm{given} \ |-\rangle) = \frac{\left(-\cos(\pi/8) - \sin(\pi/8)\right)^2}{2}=\cos^2(\pi/8). \end{align*}

It is straightforward to deduce that the probability of learning the correct $x$ is $\cos^2(\pi/8)$.


So, just to clarify the problem....

Alice has two bits that she wants to send to Bob, so she encodes a qubit into one of 4 states. Of course, these 4 states cannot be orthogonal to each other, so there's some overlap and possibility for misinterpretation.

However, Bob is not actually interested in learning both bits. He only wants to know one bit (but Alice does not know which one). So the plan is that Bob will measure in two different bases depending on which bit he wants to learn. We'll assume these are the Z and X bases (I think the original question used slightly different bases: I'm not finding the setup entirely clear, possibly because some of it is missing), and the two outcomes of each measurement correspond to the two different bit values.

To answer...

The first question is what states should Alice use to encode her bits? Alice needs to calculate the probability of success, assuming that Bob picks his measurement basis 50:50 at random. Hence if she wants to communicate the two bit values 00, the success probability is $$ p^S_{00}=\frac{1}{2}(|\langle 0|\psi_{00}\rangle|^2+|\langle +|\psi_{00}\rangle|^2), $$ i.e. half the time Bob picks the Z basis, and we want the outcome to be $|0\rangle$, and half the time Bob picks the X basis, and we want the outcome to be $|+\rangle$. There are three similar quantities, \begin{align*} p^S_{01}&=\frac{1}{2}(|\langle 0|\psi_{01}\rangle|^2+|\langle -|\psi_{01}\rangle|^2) \\ p^S_{10}&=\frac{1}{2}(|\langle 1|\psi_{10}\rangle|^2+|\langle +|\psi_{10}\rangle|^2) \\ p^S_{11}&=\frac{1}{2}(|\langle 1|\psi_{11}\rangle|^2+|\langle -|\psi_{11}\rangle|^2) \end{align*} These are all independent, because it's just an independent optimisation of the different states $|\psi_{ij}\rangle$. Of course, we don't want to repeat lots of work. So, assume we'll be able to find the best choice of $|\psi_{00}\rangle$. Observe that is we set \begin{align*} |\psi_{01}\rangle&=Z|\psi_{00}\rangle \\ |\psi_{10}\rangle&=X|\psi_{00}\rangle \\ |\psi_{11}\rangle&=Y|\psi_{00}\rangle, \end{align*} then those added Paulis change the different measurement basis outcomes into those of $|\psi_{00}\rangle$, and so each achieves the same probability, which must be equal.

So, we want to find a state $|\psi_{00}\rangle$ that maximises $$\frac12(|\langle 0|\psi_{00}\rangle|^2+|\langle +|\psi_{00}\rangle|^2)$$. Let's write this as $$ \frac12\langle\psi_{00}|(|0\rangle\langle 0|+|+\rangle\langle +|)|\psi_{00}\rangle. $$ The best choice of $|\psi_{00}\rangle$ is clearly the maximum eigenvector of $\frac12(|0\rangle\langle 0|+|+\rangle\langle +|)$, while the outcome is just the maximum eigenvalue.

So if you find the eigenvalues and eigenvectors of the matrix $$ \frac14\left(\begin{array}{cc} 3 & 1 \\ 1 & 1 \end{array}\right), $$ you'll find the eigenvector is $\cos\frac{\pi}{8}|0\rangle+\sin\frac{\pi}{8}|1\rangle$ and the eigenvalue is $\cos^2\frac{\pi}{8}$. By construction, all the other success probabilities must be the same.


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