Compute the squared overlap between different given qubit states

Question

I was checking this problem from a book. And here is an example, but I think it's wrong. If it is not wrong can you please explain how did they derive it?

As per my workout, it should be $1$. But It seems they are doing something fishy here.

Problem 7. The qubit trine is defined by the following states \begin{align*} \lvert \psi_0 \rangle = \lvert 0 \rangle, \quad\lvert \psi_1 \rangle = -\frac{1}{2} \lvert 0 \rangle - \frac{\sqrt{3}}{2} \lvert 1 \rangle, \quad\lvert \psi_2 \rangle = -\frac{1}{2} \lvert 0 \rangle + \frac{\sqrt{3}}{2} \lvert 1 \rangle \end{align*} where $\{\lvert 0 \rangle, \lvert 1 \rangle\}$ is an orthonormal basis set. Find \begin{align*} \lvert \langle \psi_0 \lvert \psi_1 \rangle \rvert^2, \quad \lvert \langle \psi_1 \lvert \psi_2 \rangle \rvert^2, \quad \lvert \langle \psi_2 \lvert \psi_0 \rangle \rvert^2. \end{align*}

Solution 7. Using $\langle 0 \lvert 0 \rangle = 1$, $\langle 1 \lvert 1 \rangle = 1$, and $\langle 0 \lvert 1 \rangle = 0$, we find \begin{align*} \lvert \langle \psi_0 \lvert \psi_1 \rangle \rvert^2 = \frac{1}{4}, \quad \lvert \langle \psi_1 \lvert \psi_2 \rangle \rvert^2 = \frac{1}{4}, \quad \lvert \langle \psi_2 \lvert \psi_0 \rangle \rvert^2 = \frac{1}{4}. \end{align*}

To be honest I would be surprised if this book is giving wrong solutions.

Answer 1

Let's decompose each of the calculation.

By having in mind that $\langle 0 | 0\rangle = \langle 1 | 1\rangle = 1$ and $\langle 0 | 1\rangle =\langle 1 | 0\rangle = 0$, we have: \begin{align*} &\langle \psi_0 | \psi_1\rangle = \frac{-1}{2} \langle 0 | 0\rangle - \underbrace{\frac{\sqrt{3}}{2}\langle 0 | 1\rangle}_{=0} = -\frac{1}{2} \\ & \Longrightarrow |\langle \psi_0 | \psi_1\rangle|^2 = \left( \frac{-1}{2} \right)^2 = \frac{1}{4} \end{align*} You do the exact same thing for $|\langle \psi_0 | \psi_2\rangle|^2$, the only thing changing being the sign in front of $\frac{\sqrt{3}}{2}$, but since it "becomes" 0, nothing changes.

As for the last one :

\begin{align*} \langle \psi_1 | \psi_2\rangle &= \left( \frac{-1}{2} \langle 0 | - \frac{\sqrt{3}}{2}\langle 1 |\right) \left( \frac{-1}{2} | 0 \rangle + \frac{\sqrt{3}}{2} |1 \rangle \right) \\ &= \frac{1}{4}\underbrace{\langle 0 | 0\rangle}_{=1} - \frac{3}{4}\underbrace{\langle 1 | 1\rangle}_{=1} -\underbrace{\frac{\sqrt{3}}{4}\langle 0 | 1\rangle + \frac{\sqrt{3}}{4}\langle 1 | 0\rangle}_{=0} \\ &= -\frac{1}{2} \\ & \Longrightarrow |\langle \psi_1 | \psi_2\rangle|^2 = \left( \frac{-1}{2} \right)^2 = \frac{1}{4} \end{align*}

Please tell me if there is something you don't understand in this, I hope this helps ! :)