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EDIT: This is a computer programming / coding exercise

The states $\left|\psi\right>$ and $\left|\phi\right>$ are defined as

$∣ϕ⟩=\cos(θ_ϕ)\left|0\right>+\sin(θ_ϕ) \left|1\right>$ with prior $p(\phi)$

$∣\psi⟩=\cos(θ_\psi)\left|0\right>+\sin(θ_\psi) \left|1\right>$ with prior $p(\psi)$

With one measurement, what is the best average case probability ​that one can distinguish these two states?

Here is how I am solving it, and getting stuck

  • If the measurement is $0$, then we guess $\phi, \psi$ with probability $e$ and $1-e$, respectively. If the measurement is $1$, then we guess $\phi, \psi$ with probability $e_1$ and $1-e_1$, respectively
  • Probability of success when guessing $\phi$ is $p_s(\phi):=p(\phi)e\cos^2\theta_\phi+e_1\sin^2\theta_\phi)$. Probability of success when guessing $\psi$ is $p_s(\psi):=p(\psi)[(1-e)\cos^2\theta_\psi+(1-e_1)\sin^2\theta_\psi]$

It is tempting to say the answer is $\max(p_s(\phi),p_s(\psi))$, but I think if we are talking about average case probability, then we need to weighted sum $p_s(\cdot)$.

The task seems to be finding maximum $e, e_1$ and plug that to $P$.

\begin{align*} P&=p(\phi)(e\cos^2\theta_\phi+e_1\sin^2\theta_\phi)+p(\psi)((1-e)\cos^2\theta_\psi+(1-e_1)\sin^2\theta_\psi)\\ &=p(\phi)e\cos^2\theta_\phi+p(\phi)e_1\sin^2\theta_\phi+\\&p(\psi)\cos^2\theta_\psi -p(\psi)e\cos^2\theta_\psi+p(\psi)\sin^2\theta_\psi-p(\psi)e_1\sin^2\theta_\psi\\ &=p(\psi)+e(p(\phi)\cos^2\theta_\phi-p(\psi)\sin^2\theta_\psi)+e_1(p(\phi)\sin^2\theta_\phi+p(\psi)\sin^2\theta_\psi) \end{align*}

But now coeff of $e, e_1$ are constants, and the maximum is achieved when $e=e_1=1$ which doesn't make any sense.

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    $\begingroup$ If I understand correctly, we have $p(\psi)=1-p(\phi)$? In what context are you interested in such an exercise? Is it voluntary that you're only considering measurements in the computational basis? $\endgroup$
    – Tristan Nemoz
    Commented Dec 21, 2023 at 15:47
  • $\begingroup$ @TristanNemoz Yes $p(ψ)=1−p(ϕ)$. This is a coding exercise. I could consider the measurements in any basis. $\endgroup$
    – Minh Triet
    Commented Dec 21, 2023 at 19:02
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    $\begingroup$ Oh it's a coding exercise! In which case you probably want to optimize over the different POVM and select the one leading to the largest probability. The reason why I was asking about the context was precisely in case it's an exercise: how much help do you want? Just get some clues, an actual solution, etc...? Also, you should definitely mention that the exercise is meant to be a coding one, it's a totally different way to approach the question. $\endgroup$
    – Tristan Nemoz
    Commented Dec 21, 2023 at 19:27
  • $\begingroup$ I edited the question. Think that you already gave me a hint so I would like to try my hand at this once more before asking for more. Thank you and Merry Christmas $\endgroup$
    – Minh Triet
    Commented Dec 24, 2023 at 21:48

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