Finding the optimal projective measurement to distinguish between two pure states

Question

I would like some help on what should be a simple computation that I'm failing to see through to the end. Suppose I have a qubit which can be in the state $|v\rangle$ with probability $p$, or $|w\rangle$ with probability $1-p$. I will choose some unitary basis $|a\rangle, |b\rangle$, and measure the qubit -- if I get $a$ I will guess the state was $v$, and if I get $b$ I will guess the state was $w$. My probability of success is $$S = p|\langle a|v \rangle|^2 + (1-p)|\langle b | w \rangle|^2$$ What I want is to maximize this probability. I can think of $a,b$ as functions of some parameter and apply calculus: $$S' = p 2 \Re(\langle a' | v \rangle \langle v | a \rangle) + (1-p)2\Re(\langle b' | w\rangle \langle w | b\rangle)$$ If we're setting this to $0$ we can ignore the factor of $2$.

Since $\langle a | a \rangle = 1$, differentiating we obtain $\Re\langle a' | a \rangle = 0$, and similarly $\Re\langle b'|b \rangle = 0$.

Writing for now $v = v_a a + v_b b$ we get $\langle v|a \rangle = \overline{v_a}$. Hence $$\Re(\langle a'|v\rangle \langle v|a \rangle) = \Re(v_a \overline{v_a} \langle a'|a \rangle + \overline{v_a}v_b \langle a'|b \rangle)$$ but since $v_a \overline{v_a} = |v_a|^2$ is real and $\langle a' | a \rangle$ is purely imaginary the first term dies and we obtain $\Re(\overline{v_a}v_b \langle a'|b \rangle)$. A similar computation goes for $w$, and in the end we can write

$$0 = S'/2 = \Re(p \overline{v_a}v_b \langle a'|b \rangle + (1-p) w_a \overline{w_b} (-\overline{\langle a'|b \rangle}))$$

where I used $\langle b'|a \rangle = - \overline{\langle a'|b \rangle}$ which is obtained by differentiating the relation $\langle a | b \rangle = 0$.

This is where I get stuck. It feels like I need to combine these terms somehow so some kind of cancellation between the $\langle a'|b \rangle$ and $- \overline{\langle a'|b \rangle}$ terms happen, but I don't see how to do it. Any help is greatly appreciated.

Edit: I realize now we can use $w_a \overline{w_b} = \overline{\overline{w_a}w_b}$ to obtain $$p \Re(\overline{v_a}v_b \langle a'|b \rangle) = (1-p) \Re(\overline{w_a}w_b \langle a'|b \rangle)$$ where I can ignore the conjugate over everything on the right hand side since we're taking the real part.

(What I wrote previously after this did not make sense)

Answer 1

$\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\PP}{\mathbb{P}}$Given an arbitrary state $\ket a$, let us write the corresponding density matrix/projector as $\PP_a$. Any such density matrix can be decomposed using a basis of Hermitian traceless operators (think the Bloch sphere representation for qubits) as $\PP_a\equiv 1/2(I + \sum_i a_i\sigma_i)$, with $\sigma_i$ the three Pauli matrices. The overlap $|\langle a|v\rangle|^2$ between two states reads, in this formalism, as $\operatorname{Tr}(\PP_a\PP_v)=\frac12(1 + \mathbf a\cdot \mathbf v)$, where $\mathbf a,\mathbf v$ are the vectors representing the states in the Bloch representation (Bloch sphere for qubits).

Defining $S=S(\mathbf a)$ as per the OP, we then have $$ 2S = p\left(1+ \mathbf a\cdot \mathbf v\right) + (1-p)\left(1- \mathbf a\cdot \mathbf w\right) \\ = 1 + \mathbf a\cdot(p\mathbf v - (1-p)\mathbf w). $$ We want to maximise this quantity with respect to $\mathbf a\in\mathbb R$ with $\|\mathbf a\|=1$. An easy way is to impose the gradient $\nabla S$ be proportional to $\nabla(\|\mathbf a\|^2-1)=2\mathbf a$ (that is, to use the Lagrange multiplier method). An even easier way is to notice that, given an arbitrary (real) vector $\mathbf x$, the quantity $\mathbf a\cdot\mathbf x$ is maximised for $\mathbf a\simeq\mathbf x$.

We conclude that the maximum of $S$ is achieved when $$\mathbf a = N[p\mathbf v-(1-p)\mathbf w],$$ where $N\in\mathbb R$ ensures the normalisation.

Notice that, if $\langle v|w\rangle=0$, you have $\mathbf w=-\mathbf v$, and thus $2S=1+\mathbf a\cdot\mathbf v$, which gives $S_{\max}=1$ for the choice $\mathbf a=\mathbf v$. This is of course what we should expect: orthogonal states can be reliably distinguished via suitable choice of measurement.