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An entangling gate is a unitary transformation $U$ on two quantum systems for which at least one pure product input has an entangled output:It need not create quantum entanglement on every product state. For example, the controlled-NOT gate leaves $|00⟩$ a product state but maps $|+⟩|0⟩$ to a Bell state, so it is an entangling gate.

A bipartite pure state is entangled exactly when it is not a product state: there are no normalized quantum states $|u{⟩}_A,|v{⟩}_B$ such thatEquivalently, its Schmidt rank exceeds one, or either reduced density matrix is a mixed state. This definition concerns quantum entanglement across the specified division into systems $A$ and $B$.

We prove the one-query constancy obstruction for phase oracles while allowing arbitrary quantum ancillas. By purification of a density operator, the quantum state immediately before the sole query can be taken asThe ancilla vectors need not be normalized or orthogonal. Classical randomness and earlier quantum measurements can be included in additional ancillas; allowing access to these systems only gives the proposed algorithm more power.

For the zero constant function, the phase oracle is $I$, leaving $|\Psi ⟩$ unchanged. For each $j$, consider the Boolean function that is one at $j$ and zero elsewhere. Since $M>2$, it is nonconstant, and its phase oracle producesAny later fixed unitary transformation preserves these inner products. An arbitrary final quantum measurement cannot distinguish the constant output from each nonconstant output with certainty unless they are orthogonal, by perfect discrimination of pure states requires orthogonality. Explicitly, an accepting POVM effect $E$ would satisfy $E|\Psi ⟩=|\Psi ⟩$ and $E|{\Psi }_j⟩=0$, forcing $⟨\Psi |{\Psi }_j⟩=0$.

Therefore exact classification requires $p_j=1/2$ for every $j$. Summing giveswhich contradicts $M>2$. Thus one query cannot decide constancy with certainty. The other constant function gives $-I$, differing only by a global phase, so it does not change the argument. For comparison, $M=2$ permits $p_0=p_1=1/2$; the impossibility arises from demanding the test work for every nonconstant Boolean function.

Use three-way qutrit sign-flip discrimination. Prepare the qutrit in the uniform superposition stateFor an unknown label $j\in \{0,1,2\}$, the supplied unitary operator is $U_j=I-2|j⟩⟨j|$. Apply it once and then apply the known inversion about the meanSince $D^{†}=D$ and $D^2=I$, this is an allowed unitary transformation. Using $⟨s|j⟩=1/\sqrt{3}$ givesPerform a computational-basis measurement and guess that its outcome is the unknown label. The Born rule givesEach incorrect outcome has probability $1/27$, verifying normalization. The success probability is $25/27$ separately for every label, so no assumption about their prior probabilities is needed.