Finding the optimal projective measurement to distinguish between two pure states

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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)

Pedro

Posted 2019-12-24T21:53:36.250

Reputation: 351

To clarify, you don't assume anything about $\langle v \mid w \rangle$? They could both be the same state or very close and you'd be SOL and that would be allowed? – AHusain – 2019-12-24T22:19:11.780

@AHusain, I think for the question to make sense we should assume they're distinct (i.e. non-collinear) states. Other than that I don't think it matters how close together they are. If they're very close my probability of success will be low, but the problem of maximizing that probability is still mathematically the same. – Pedro – 2019-12-24T23:50:38.117

This doesn’t take into account the probabilities, but is essentially the formalism you want: https://quantumcomputing.stackexchange.com/a/4172/1837

– DaftWullie – 2019-12-25T06:41:54.507

@Pedro it might help to introduce a parametrisation to write the derivatives $|a'\rangle$ etc explicitly. E.g. you can write $|a\rangle=\cos\theta|0\rangle+e^{i\varphi}\sin\theta|1\rangle$ and $|b\rangle=\sin\theta e^{-i\varphi}|0\rangle - \cos\theta|1\rangle$ (assuming you are dealing with qubits). This also makes it explicit that these states have more than one degree of freedom, so you need to impose vanishing derivatives wrt each parameter – glS – 2019-12-29T12:27:03.070

@glS This sounds like it will be a huge mess! – Norbert Schuch – 2019-12-29T14:15:03.300

@NorbertSchuch you are right, there seems to be a much simpler way, see my answer – glS – 2019-12-30T15:24:42.493

Answers

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$\newcommand{\ket}[1]{\lvert #1\rangle}\newcommand{\PP}{\mathbb{P}}$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.

glS

Posted 2019-12-24T21:53:36.250

Reputation: 12 247