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If you want to check if a pair of unknown qubits are the same, a standard test is the controlled SWAP test. This gives a result of 0 with certainty if the states are the same and 1 with a 50% chance if the states are orthogonal. The resulting probability distribution can also be used to approximate fidelity, etc.

Is there a/what is the equivalent test to determine orthogonality with certainty, i.e. giving a result of 0 with certainty if the states are orthogonal and 1 with some probability if the states are identical?

Alternatively, is there a more balanced test, which gives the same (lower than 50%) error in detecting either orthogonality or similarity?

Obviously, one opton is to carry out state tomography on each qubit to get their mathematical descriptions and calculate the orthogonality/fidelity accordingly, but this may not be the best solution in terms of copies required for a given accuracy.

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It is not possible for a measurement to deterministically give one outcome or the other depending on whether two states are equal or orthogonal. Such a measurement would be some two-outcome POVM $\mu$ such that $$\langle\mu(\text{yes}),\rho\otimes\rho\rangle=1, \qquad \langle\mu(\text{no}),\rho\otimes\sigma\rangle=1,$$ for all states $\rho$ and $\sigma$ with $\langle\sigma,\rho\rangle=0$.

This would imply, in particular, that $$\langle \mu(\text{yes}),\mathbb P_0\otimes \mathbb P_0\rangle =\langle \mu(\text{yes}),\mathbb P_1\otimes \mathbb P_1\rangle =1,\\ \langle \mu(\text{yes}),\mathbb P_0\otimes \mathbb P_1\rangle =\langle \mu(\text{yes}),\mathbb P_1\otimes \mathbb P_0\rangle =0, \\ 1 = \langle \mu(\text{yes}),\mathbb P_+\otimes \mathbb P_+\rangle = \frac12 + \frac14 \langle\mu(\text{yes}),I\otimes X+X\otimes I\rangle, \\ 1 = \langle \mu(\text{yes}),\mathbb P_-\otimes \mathbb P_-\rangle = \frac12 - \frac14 \langle\mu(\text{yes}),I\otimes X+X\otimes I\rangle,$$ where I'm using the shorthand notation $\mathbb P_\psi\equiv|\psi\rangle\!\langle\psi|$, and $X\equiv|0\rangle\!\langle1|+|1\rangle\!\langle0|$.

The last two conditions are clearly inconsistent, as summing them we'd get $2=1$, thus no such measurement can exist.

A measurement that always gives the outcome "yes" when the input has the form $\rho\otimes \rho$ is possible, for example, the trivial "measurement" that always returns "yes". But this clearly will not correctly classify orthogonal input states.

There's probably a better way to show this though.

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  • $\begingroup$ Thanks, I get that the measurement can't be deterministic in both cases. In fact, I think the existence of such a measurement would violate the no-cloning principle. The SWAP test is deterministic on identical states and probabilistic on orthogonal ones. I want the opposite: a test that's deterministic on orthogonal states and probabilistic on identical states. You're right, I can make a trivial measurement! So perhaps a better way of phrasing the question is: what measurement is deterministic for orthogonal states and has 50:50 probability for identical states (opposite of the SWAP). $\endgroup$ Jul 22 at 23:09
  • $\begingroup$ @Jason How could you use it to build a cloner? $\endgroup$ Jul 23 at 22:48

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