This problem has arisen in my research and reduces to the following.

Alice tells Bob she has secretly picked A or B. Depending on which she has picked she will send him states from a set. Bob can ask for as many states as he likes, but Alice will send a random member from the set, not the same state many times.

Set A consists of simple product states such as |0010> and |1100>

Set B consists of simple equal superpositions of two of the above states such as:

$$ \frac{1}{\sqrt{2}} (|0001>+|0101>) $$

Is it possible for Bob to determine which selection Alice has made?

Things I note/have tried:

For larger numbers of qubits, states from set B are almost certainly entangled. This doesn't necessarily make it possible to distinguish them from unentangled states in A.

One way to confirm that they cannot be distinguished would be to show that their density operators are equal. In a similar problem where Alice sends 2-qubit product states OR Bell basis states, it is easy to prove the density operators are the same and the states cannot be distinguished. I have tried this approach but found it sensitive to the probability distribution over the set.

  • $\begingroup$ Perhaps you can clarify the aim here. Are you proposing a protocol for secure communication? If so, is it proposed to solve a problem that exists in the known protocols? $\endgroup$ May 17 at 8:03
  • $\begingroup$ This problem came up when attempting to design a quantum algorithm. So far I can get an routine that produces a state from set A if the answer is FALSE and set B is the answer is TRUE. Therefore, if it is possible to distinguish them (which I suspect it isn't) then the algorithm would be complete. $\endgroup$ May 17 at 11:20
  • $\begingroup$ Product states make this a straightforward generalization of physics.stackexchange.com/q/645150 $\endgroup$ May 17 at 21:38
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    $\begingroup$ Of course it will depend on the probability distribution -- simply consider the case of a "totally biased" distribution, where you always only send one fixed product state or one fixed superposition; these are clearly distinguishable. $\endgroup$ May 22 at 12:09

2 Answers 2


You have to specify whether there is freedom in choosing the measurement. In general, the optimal probability to discriminate between two states $\rho_1,\rho_2$ with priors $p_1,p_2$ is $\frac12(1+\|p_1 \rho_1-p_2\rho_2\|_1)$, and there is always a projective measurement saturating this bound. Generalisations to more general cases are possible. Look up quantum state discrimination. See also https://quantumcomputing.stackexchange.com/q/16290/55 and links therein.

  • $\begingroup$ This is the single-shot probability; the OP seems to be interested in the case of multiple copies. $\endgroup$ May 22 at 12:10

The answer is no. In the case of one qubit instead of 4, which is your examples, your scenario is what's going on with the BB84 protocol. In BB84, can Bob determine what Alice selects? Certainly no. Otherwise, Eve would be able to and the BB84 protocol is broken. Bob has to guess whether Alice selects A or B to set up measurement before comparing notes with Alice on the correctness of his guess.


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