3
$\begingroup$

The early-90's, pre-teleportation work of Peres and Wootters studied the now-called double-trine or Mercedes-Benz states of two qubits in one of three product states:

$$|A\rangle\otimes|B\rangle=|s_i\rangle\otimes|s_i\rangle,$$

with $i\in\{0,1,2\},$ $|s_i\rangle=U^i|0\rangle$, and $U=\exp(-\frac{\pi\sqrt{-1}}{3}\sigma_y)$. On the complex plane these points make the Mercedes-Benz logo.

With two such qubits it's an interesting state-discrimination problem to deduce $i$, insofar as a referee can give the first qubit to Alice on the Earth and the second qubit to Bob on Mars; Alice and Bob can engage in local operations and classical communication (LOCC) to try their best to figure out $i$, but if they meet in the middle and can perform an entangling measurement between their qubits they can get an even better chance of deducing $i$.

This is kind of dual to the CHSH game, where Alice and Bob are given a-priori entangled qubits and aren't allowed classical communication thereafter; with the double-trine game, Alice and Bob are given unentangled states and have better-than-classical chances of winning only if they can meet up to entangle their qubits.

But, what is the best circuit that Alice and Bob can run to deduce $i$ given their two separate qubits, when they can meet up to perform entangling CNOT etc. gates between them? Is it as simple as the circuit to prepare the singlet state $|\Psi^-\rangle=\frac{1}{\sqrt 2}(|01\rangle-|10\rangle)$, starting from $|00\rangle$?

$\endgroup$
0

2 Answers 2

1
$\begingroup$

There's no simple answer to what is the "best". You need to define what you mean a bit more carefully. There are two common scenarios under which one optimises:

  • minimum error: we might get the wrong answer sometimes, but the average probability of success is maximised
  • unambiguous discrimination: we never get the answer wrong.

I'm going to do the calculation for unambiguous discrimination (because I'm lazy, and am expecting the maths for this case to be simpler. That said, the circuit may be more complicated at the end.)

Let $|\psi_i\rangle$ for $i\in\{0,1,2\}$ be the three states we are trying to discriminate. It is certainly true that these live in the symmetric subspace of two qubits, and are hence orthogonal to the singlet state $|\Psi^-\rangle$. So that is certainly a hint towards what we'll need to do, but it's not everything. Also, let $|\phi_i\rangle$ be orthogonal to $|\psi_{i+1\text{ mod }3}\rangle$ and $|\psi_{i+2\text{ mod }3}\rangle$. This lets us define POVM elements $$ M_i=x|\phi_i\rangle\langle\phi_i|. $$ (Technically, $x$ can depend on $i$, but it would become clear very quickly by symmetry that we should set them equal.)

If we make this measurement and get the answer $i$, we know that we definitely had the state $|\psi_i\rangle$.

What's the probability of success? $\langle\psi_i|M_i|\psi_i\rangle$ (which is equal for all $i$). We just have to choose $x$ to maximise this, under one constraints: we need $\sum_iM_i=I$. Thus, we introduce an $M_3=I-M_0-M_1-M_2$, which we require to be positive semi-definite. Working this through, we have $$ M_3=\left( \begin{array}{cccc} 1-\frac{9 x}{10} & 0 & 0 & \frac{3 x}{10} \\ 0 & 1-\frac{3 x}{5} & -\frac{3 x}{5} & 0 \\ 0 & -\frac{3 x}{5} & 1-\frac{3 x}{5} & 0 \\ \frac{3 x}{10} & 0 & 0 & 1-\frac{9 x}{10} \\ \end{array} \right) $$ This has eigenvalues $$ \left\{1,\frac{1}{5} (5-3 x),\frac{1}{5} (5-6 x),\frac{1}{5} (5-6 x)\right\}. $$ The largest value of $x$ that we can select is $x=\frac56$. In this case, we have $\langle\psi_i|M_i|\psi_i\rangle=\frac34$.

What does the circuit look like? We need to work out how to implement the POVM. One thing we know for sure: we'll need an ancilla qubit.

I haven't gone as far as working out a detailed circuit. It's messy, and there's lots of freedom (so, possibly, there are some options that are less messy!). However, let me describe some of the basic steps. First, we define the following orthogonal states: \begin{align*} |\Psi_0\rangle&=\frac{\sqrt{3}}{2}|\phi_0\rangle|0\rangle+\frac12|\phi_0\rangle|1\rangle \\ |\Psi_1\rangle&=\frac{\sqrt{3}}{2}|\phi_1\rangle|0\rangle+\frac12\left(\frac35|\phi_0\rangle+\frac45|\psi_1\rangle\right)|1\rangle \\ |\Psi_2\rangle&=\frac{\sqrt{3}}{2}|\phi_2\rangle|0\rangle+\frac12\left(\frac35|\phi_0\rangle+\frac{2}{85}((12-\sqrt{59})|\psi_1\rangle+(3+4\sqrt{59})|\psi_2\rangle)\right)|1\rangle \\ |\Psi_3\rangle&=\frac{1}{\sqrt{2}}(|01\rangle-|10\rangle)|0\rangle \\ \end{align*} Now we need to find a unitary $V$ such that $$ V|\Psi_i\rangle=|0\rangle|i\rangle. $$ (It is probably easier to work with $V^\dagger$! This shows that the first 4 rows of the $8\times 8$ matrix $V$ are just $|\Psi_i\rangle^T$.) The circuit for $V$ is the one we need to implement. We take our initial state $|\psi_i\rangle$ and introduce a single-qubit ancilla $|0\rangle$. Then across the three-qubit state, apply $$ V|\psi_i\rangle|0\rangle, $$ and measure all 3 qubits in the standard basis. If the first qubit is 0, our measurement has succeeded and the state of the last two qubits tells us which result we got. Otherwise, our measurement failed, corresponding to the measurement operator $M_3$.

$\endgroup$
1
  • $\begingroup$ Hey I think I understood a lot of that! I hadn't thought of adding in an ancilla to get a good POVM. I think Peres and Wootters just wanted to show that a global measurement beats the best LOCC, but in your first sense in terms of minimizing error. I'm also (slowly) learning about the ZX calculus and I wonder if it makes sense to ask what a ZX-calculus spider diagram would look like for the referee to prepare the double trine state and for Alice and Bob to perform your measurement.. $\endgroup$ Commented Jul 10, 2023 at 21:13
1
$\begingroup$

Following on from my previous answer about unambiguous discrimination, there is also an optimal solution for the minimum error probability, which I found here.

The trick is to introduce a state $$|\Phi\rangle=\frac{1}{\sqrt{6}}((\sqrt{2}+1)|00\rangle+(1-\sqrt{2})|11\rangle.$$ This has the property that $\{|\Phi\rangle,U^1\otimes U^1|\Phi\rangle,U^2\otimes U^2|\Phi\rangle\}$ form an orthonormal basis. You can then take these as measurement projectors, associating the outcome from projector onto $U^i\otimes U^i|\Phi\rangle$ with holding state $|s_i\rangle|s_i\rangle$ originally. The success probability is then just $$ |\langle\Phi|s_0s_0\rangle|^2=\frac12+\frac{\sqrt{2}}{3}. $$ To implement this measurement as a unitary $V$ followed by projective measurements in the standard basis, you could find the unitary $$ V=\left( \begin{array}{cccc} \frac{1}{\sqrt{3}}+\frac{1}{\sqrt{6}} & 0 & 0 & \frac{1}{\sqrt{6}}-\frac{1}{\sqrt{3}} \\ \frac{\sqrt{2}-1}{2 \sqrt{3}} & \frac{1}{2} & \frac{1}{2} & \frac{1}{6} \left(\sqrt{3}+\sqrt{6}\right) \\ \frac{\sqrt{2}-1}{2 \sqrt{3}} & -\frac{1}{2} & -\frac{1}{2} & \frac{1}{6} \left(\sqrt{3}+\sqrt{6}\right) \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\ \end{array} \right). $$ This might look pretty awful, but it helps to transform it into the $Y$-basis, at which point it looks more like $$ V\equiv\left(\begin{array}{cccc} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{e^{2i\pi/3}}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{e^{-2i\pi/3}}{\sqrt{3}} \\ \frac{e^{-2i\pi/3}}{\sqrt{3}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & \frac{e^{2i\pi/3}}{\sqrt{3}} \\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \end{array}\right). $$

$\endgroup$
3
  • $\begingroup$ Thanks! V is 4x4 - does this mean no ancilla is used here? $\endgroup$ Commented Jul 11, 2023 at 11:05
  • 1
    $\begingroup$ Yes. It's a projective measurement on just the 2 qubits. $\endgroup$
    – DaftWullie
    Commented Jul 11, 2023 at 11:37
  • 1
    $\begingroup$ Oh - I think I see. Alice and Bob feed their photons through a circuit realizing $V$. WLOG call Alice's qubit the top qubit (and Bob the bottom). If the referee gave them $|s_0\rangle|s_0\rangle$ then they are most likely to measure $|00\rangle$; if instead they were given $|s_1\rangle|s_1\rangle$ then the top qubit would most likely measure $|0\rangle$ while the bottom qubit measures $|1\rangle$, etc. Compiling $V$ into a good circuit (using Clifford+T, e.g.,) might or might not introduce ancilla, but if so we'd need to uncompute those before measurement. $\endgroup$ Commented Jul 11, 2023 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.