# Is there a circuit which certainly distinguishes between $|0\rangle$ and $|+\rangle$ states?

Suppose I have a qubit $| \theta \rangle$ which equals to $| 0 \rangle$ or to $| + \rangle = \frac{(|0 \rangle + |1 \rangle )}{\sqrt{2}}$.

Is it possible to build a quantum circuit C:

$C(| 0 \rangle \otimes | \textbf{x} \rangle) = |0 \rangle \otimes | \textbf{y} \rangle$

$C(|+ \rangle \otimes | \textbf{x} \rangle) = |1 \rangle \otimes | \textbf{z} \rangle$

so that I could measure the first qubit of an outcome and know for sure which state $| \theta \rangle$ was in?

($|\textbf{x} \rangle$ are a few extra qubits supplied of my choice, $| \textbf{y} \rangle$ and $| \textbf{z} \rangle$ are junk qubits that the circuit produces)

If no, what would be a proof that this is not possible?

• Is $|\mathbf{x}\rangle$ from the same Hilbert space, i.e. only a linear combination of $|0\rangle$ and $|1\rangle$?
– noah
May 10, 2017 at 14:20
• @Michael, yes. For example, $| \textbf{x} \rangle = |0010101 \rangle$ May 10, 2017 at 14:56
• This is essentially a specific case of the no-cloning theorem: en.wikipedia.org/wiki/No-cloning_theorem (since states that may be perfectly distinguished may be copied) May 10, 2017 at 18:39

No, this is not possible. You are proposing a protocol that will take a system state $|s\rangle$ (where $s=0,+$), couple it with some ancilla state $|a\rangle$, and then evolve it to $$|s\rangle\otimes |a\rangle \mapsto U |s\rangle\otimes |a\rangle$$ through some global unitary $U$ such that \begin{align} U |0\rangle\otimes |a\rangle & = |0\rangle\otimes |b\rangle \\ U |+\rangle\otimes |a\rangle & = |1\rangle\otimes |c\rangle. \end{align} However, because $U$ needs to be unitary, it needs to preserve the scalar product $$(\langle +|\otimes \langle a|)( |0\rangle\otimes |a\rangle)=\frac{1}{\sqrt{2}},$$ whereas your target states have $$(\langle 1|\otimes \langle c|)( |0\rangle\otimes |b\rangle)=0.$$ Moreover, since any arbitrary quantum channel is an incoherent mixture of unitaries-plus-projective-measurements, and each individual component of that mixture is limited as above, any quantum channel is similarly limited.