How do you test a pair of unknown qubits for orthogonality with certainty?

Question

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.

glS · Answer 1 · 2021-07-22T17:52:17.040

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.

score 1 · Answer 2 · answered Mar 06 '24 at 15:49

Though gIS' answer concerns deterministic measurements, its conclusion doesn't change when considering probabilistic measurements.

Though it's not a formal proof, we can use picos to look for such a measurement. That is, we want it to look for $M_0$ and $M_1$ such that:

Both are positive semi-definite;
$M_0+M_1=I$
For all $|\psi\rangle$ and $|\varphi\rangle$ such that $\langle\psi|\varphi\rangle=0$, $\mathrm{tr}\left(M_0|\psi,\varphi\rangle\!\langle\psi,\varphi|\right)=1$. This corresponds to the fact that the measurement identifies orthogonal states perfectly.
The probability of getting the $0$ outcome on non-orthogonal states is as low as possible.

The last condition is not very formal, and the penultimate one is quite hard to test, so we'll solve a simpler problem: find $M_0$ and $M_1$ such that:

Both are positive semi-definite;
$M_0+M_1=I$
We obtain the outcome $0$ with certainty on input $|0,1\rangle\!\langle0,1|$, $|1,0\rangle\!\langle1,0|$, $|+,-\rangle\!\langle+,-|$, $|-,+\rangle\!\langle-,+|$, $|i,-i\rangle\!\langle i,-i|$ and $|-i,i\rangle\!\langle -i,i|$.
We want to minimize the average probability of getting the outcome $0$ on input $|0,0\rangle\!\langle0,0|$, $|1,1\rangle\!\langle1,1|$, $|+,+\rangle\!\langle+,+|$, $|-,-\rangle\!\langle-,-|$, $|i,i\rangle\!\langle i,i|$ and $|-i,-i\rangle\!\langle -i,-i|$.

Clearly, if a non-trivial measurement exists for the problem we care about, it will also be a non-trivial solution for the second one. Conversely, if there is no non-trivial measurement that solves the second problem, then no non-trivial measurement solves the first.

We thus input this to picos:

import numpy as np
import picos as pc
sdp = pc.Problem(verbosity = 0)
M0 = pc.HermitianVariable("M0", (4, 4))
M1 = pc.HermitianVariable("M1", (4, 4))
sdp.add_constraint(M0 >> 0)
sdp.add_constraint(M1 >> 0)
sdp.add_constraint(M0 + M1  == np.eye(4))
zero_state = np.array([[1], [0]])
one_state = np.array([[0], [1]])
plus_state = np.array([[1], [1]]) / np.sqrt(2)
minus_state = np.array([[1], [-1]]) / np.sqrt(2)
i_state = np.array([[1], [1j]]) / np.sqrt(2)
minus_i_state = np.array([[1], [-1j]]) / np.sqrt(2)
certainty_states = [
    np.kron(zero_state, one_state),
    np.kron(one_state, zero_state),
    np.kron(plus_state, minus_state),
    np.kron(minus_state, plus_state),
    np.kron(i_state, minus_i_state),
    np.kron(minus_i_state, i_state),
]
extremal_states = [
    np.kron(zero_state, zero_state),
    np.kron(one_state, one_state),
    np.kron(plus_state, plus_state),
    np.kron(minus_state, minus_state),
    np.kron(i_state, i_state),
    np.kron(minus_i_state, minus_i_state),
]
certainty_states = [s @ s.conj().T for s in certainty_states]
extremal_states = [s @ s.conj().T for s in extremal_states]
for state in certainty_states:
    sdp.add_constraint(pc.trace(M0 * state) == 1)
obj = pc.trace(M0 * sum(extremal_states))
sdp.set_objective("min", obj)
sdp.solve()
print(sdp.value / 6)
print(np.array(M0))

This code yields:

0.9999999999605058
[[ 1.00000000e+00+0.00000000e+00j  1.97355128e-16-3.66797943e-16j
  -3.26236171e-16+2.02131752e-16j  7.63261303e-16+4.46637958e-21j]
 [ 1.97355128e-16+3.66797943e-16j  1.00000000e+00+0.00000000e+00j
   1.29011297e-11-3.66985535e-21j -3.23831307e-16-3.85180042e-16j]
 [-3.26236171e-16-2.02131752e-16j  1.29011297e-11+3.66985535e-21j
   1.00000000e+00+0.00000000e+00j  1.98606930e-16+1.52525270e-16j]
 [ 7.63261303e-16-4.46637958e-21j -3.23831307e-16+3.85180042e-16j
   1.98606930e-16-1.52525270e-16j  1.00000000e+00+0.00000000e+00j]]

Thus, up to inaccuracies, the optimal measurement is the trivial one that always return $0$.

As a sanity check, we can invert the role of extremal_states and certainty_states in the code above in order to check that we get the POVM associated to the SWAP test (or something at least as good). Doing so yields:

0.499999987905889
[[1.00000000e+00+0.00000000e+00j 1.00930041e-16-3.04715104e-16j
  2.41912746e-16-1.61029373e-16j 3.27055113e-16+6.68833130e-21j]
 [1.00930041e-16+3.04715104e-16j 4.99999980e-01+0.00000000e+00j
  4.99999997e-01+3.36471447e-18j 3.22435776e-16-2.47496372e-17j]
 [2.41912746e-16+1.61029373e-16j 4.99999997e-01-3.36471447e-18j
  4.99999980e-01+0.00000000e+00j 1.41739720e-16-7.93076212e-17j]
 [3.27055113e-16-6.68833130e-21j 3.22435776e-16+2.47496372e-17j
  1.41739720e-16+7.93076212e-17j 1.00000000e+00+0.00000000e+00j]]

which is what we expected.

All in all, there is no measurement other than the trivial one that behaves deterministically on orthogonal states.

How do you test a pair of unknown qubits for orthogonality with certainty?

2 Answers2