Let's say we have a quantum system with 2 qubits, which are in some linear combination of Bell basis. $\vert \psi_{j} \rangle = \alpha_{j} \vert {T_1} \rangle + \beta_{j} \vert {T_2} \rangle + \gamma_{j} \vert {T_3} \rangle + \epsilon_{j} \vert {T_4} \rangle$ for j = 1, 2


Given these 2 qubits, how do I know which qubit I have? To me, this sounds like a good example of a case when we cannot distinguish these states unambigously. My understanding is that in such a scenario, we can try to construct a POVM set that will allow us to distinguish the states without any false identification (although, we will not be able to ascertain which qubit we have at times).

But how does one go about constructing such a POVM set?

Bell basis:

  • $\vert{T_1}\rangle = \frac{1}{\sqrt{2}} (\vert{10}\rangle - \vert{01}\rangle)$
  • $\vert{T_2}\rangle = \frac{1}{\sqrt{2}} (\vert{10}\rangle + \vert{01}\rangle)$
  • $\vert{T_3}\rangle = \frac{1}{\sqrt{2}} (\vert{00}\rangle + \vert{11}\rangle)$
  • $\vert{T_4}\rangle = \frac{1}{\sqrt{2}} (\vert{00}\rangle - \vert{11}\rangle)$

1 Answer 1


We can do this for general quantum states instead. Suppose we have two states $|\phi\rangle$ and $|\psi\rangle$, which are not orthogonal. We want to construct the POVM that best distinguishes these two states, and makes no false positives.

As a simpler example, we can choose $|\phi\rangle = |0\rangle$, and $|\psi\rangle = 1/\sqrt{2}(|0\rangle+|1\rangle)$.

We begin by finding the states in the space that are orthogonal to each state, and call them $|\phi^\perp\rangle$, and $|\psi^\perp\rangle$. In our case, they would be $|1\rangle$ and $1/\sqrt{2}(|0\rangle-|1\rangle)$. Then, we simply use the POVM containing 3 elements,

$$E_1 = \frac{1}{1-|\langle \phi | \psi \rangle |} |\psi^\perp\rangle\langle\psi^\perp|$$ $$E_2 = \frac{1}{1-|\langle \phi | \psi \rangle |} |\phi^\perp\rangle\langle\phi^\perp|$$ $$E_3 = I - E_1 - E_2$$

Using this construction, you will never measure $E_1$ if the state is $|\psi\rangle$, and similarly for $E_2$. But, there is a small chance that $E_3$ is measured for both of these states. This will work for any pair of states.

  • $\begingroup$ Great and succinct answer. Thanks! $\endgroup$
    – skittish
    Commented May 4, 2020 at 6:25

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