# Can you distinguish between $|0\rangle, |1\rangle$, and $\frac{1}{\sqrt 2} (|0\rangle + |1\rangle)$?

(A beginner here; possibly a stupid question. Please be gentle. Sorry if I used a wrong tag.)

Suppose that I receive a (classically) random number, which is either $$1$$ or $$2$$ or $$3$$. Depending on this number, I will set a value of a qubit to either $$|0\rangle$$, or $$|1\rangle$$, or $$\frac {1}{\sqrt 2}( |0\rangle + |1\rangle)$$. Then I give the qubit to you.

Is it possible to figure out which of the three numbers I received, using the qubit?

(For example, if I measure the qubit and its value is $$0$$, I know it wasn't the second option, but I don't know if it was the first or the third option.)

If it is not possible, is it as least possible to distinguish whether it was the third state (the superposition) or it was some of the first two states (the pure states)?

• IIRC this exact case is explained as an example somewhere in Mike & Ike when talking about POVMs. Commented May 26, 2023 at 21:46
• p. 92 of Nielsen and Chuang gives a Positive Operator Valued Measurement (POVM) to distinguish $|0\rangle$ from $|+\rangle=\frac{1}{\sqrt 2}(|0\rangle+|1\rangle)$. Commented May 27, 2023 at 0:07
• So what is the correct procedure to maximize the probability of success in finding exactly which of the three numbers were received and not just the case of X or not X which is what Mike and Ike and the accepted answer handle? Commented May 28, 2023 at 19:22

If two quantum states $$|\psi\rangle$$ and $$|\phi\rangle$$ are such that $$\langle\phi|\psi\rangle \neq 0$$ (i.e. they are not orthogonal), then it will not be possible to determine which was given with complete certainty, regardless of what basis the measurements are performed on. The best probability of distinguishability for any strategy is given by $$1 - |\langle\phi|\psi\rangle|^2$$ (for mixed states, the more general fidelity formula is used). So there's no guaranteed way of distinguishing either $$|0\rangle$$ from $$|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$$ or $$|1\rangle$$ from $$|+\rangle$$, and thus no way of finding out which of the three it was with certainty.

However, it is possible to be able to obtain a piece of information that would assure you that you were given either $$|0\rangle$$ or $$|1\rangle$$ and not $$|+\rangle$$, though the procedure could fail. If you performed a Hadamard transformation on the unknown state then each state would become $$|+\rangle$$, $$|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle -|1\rangle)$$, and $$|0\rangle$$ respectively. If you then measured a $$1$$, you would be assured you were at first given either $$|0\rangle$$ or $$|1\rangle$$ and not $$|+\rangle$$, but, if you measured a 0, then it could have been any of the three and since the system has collapsed to $$|0\rangle$$ all possibility of distinguishing further will have been lost.

• Perfect, thank you, especially for the second part! Commented May 27, 2023 at 19:28

Suppose we are given 10 mystery qubits which had been prepared in the same unknown state $$\Psi = a|0\rangle + b|1\rangle$$. Then as mentioned above, performing Hadamard operation on each qubit will send it into the state

$$a(|0\rangle + |1\rangle) + b(|0\rangle - |1\rangle)$$ $$= a|0\rangle + a|1\rangle + b|0\rangle - b|1\rangle$$ $$= \color{red}{(a+b)}|0\rangle + \color{red}{(a-b)}|1\rangle$$

If both $$a,b \neq 0$$, then the probabilities have become unbalanced between $$|0\rangle$$ outcome and $$|1\rangle$$ outcome. Given 10 identically prepared state, we might measure 9 of them and check if there is statistically significant imbalance of outcomes. The tenth qubit can remain unmeasured with the probable state known to us.

On the other hand, if only 1 single qubit $$\Psi = a|0\rangle + b|1\rangle$$ is given to us, then there is no way we might clone $$a,b$$ using any physical means (by the no cloning theorem), and so the state will remain forever a mystery, or destroyed with some information gained about $$a,b$$ but no certainty.