I am given one copy of one of two quantum states -

  1. $\frac{| 0 \rangle + e^{i\theta} | 1 \rangle}{\sqrt{2}} $, for some unknown fixed $\theta$.
  2. One of $| 0 \rangle/|1 \rangle$ - don't know which one, but one of the two.

I need to guess which one of the above two cases it is. Note that one can always guess this with probability $1/2$. So, to make the problem non-trivial we want to guess with any probability better than $1/2$ - i.e. with $1/2 + \epsilon$, for any $\epsilon > 0$.

Here are some more details on the constraints/things I have tried, but don't work.

  • The state need not be reusable after this guess - so we can perform any kind of measurement we want. The only constraint is that we are only given one copy of the quantum state.
  • This problem is different from quantum state distinguishability problem. Why? because here we are only given one copy of the quantum state. In addition, we need to tell apart the two cases with any probability better than $1/2$.
  • This can also be thought of geometrically in the bloch sphere - in the first case, the quantum state lies in the X-Y plane and in the second case, it lies along the Z-axis.
  • Note also that if $\theta$ were known in the first case - like if the first case were $|+\rangle$, then we could just measure in the $\{|+\rangle,|-\rangle\}$ basis - then if we were given $|+\rangle$, we always observe $|+\rangle$ outcome, while if we were given $|0\rangle / |1\rangle$, we would observe $|-\rangle$ as well and we could tell the two cases apart with probability $3/4$.
  • I have also tried to think of if we can use POVMs to do this - this paper - https://arxiv.org/pdf/quant-ph/0604189.pdf - gives a way to visualize POVMs geometrically - but I can't think of how to use it either.
  • Note that I can peform a CNOT on a $|0\rangle$ state with the given state as control - and in that case, if it were case 1, then I get an entangled state, while in case 2, I get a product state. Hence, this problem can also be seen as telling apart an entangled state from product state but using a single copy. However, it seems much easier than general single copy entanglement detection, because here I only need to tell apart the cases with any probability $1/2 + \epsilon$. Looking up some research literature on single copy entanglement detection, I found this - https://www.nature.com/articles/s41534-017-0055-x.pdf - but it seems the paper's results are on special kinds of states and not general quantum states.
  • A final way which I haven't exhausted is applying a unitary and doing multiple measurements somehow - even though the state collapses after each measurement, maybe there is a way I can count how many measurements I need till I get some particular state, which is detectable. But can't think of anything in this line.

Any help/leads would be very appreciated / any thoughts on if this is not possible as well.

  • $\begingroup$ What is the probability distribution for $\theta$? $\endgroup$ Commented Jun 22, 2021 at 16:38
  • 1
    $\begingroup$ Do you know the Holevo-Helstrom theorem? (see theorem 3.4) $\endgroup$
    – Rammus
    Commented Jun 22, 2021 at 16:43
  • $\begingroup$ $\theta$ comes from an arbitrary distribution - unknown, but fixed. $\endgroup$
    – nishkr
    Commented Jun 22, 2021 at 17:44
  • $\begingroup$ So is this like determining whether we have $|0\rangle$ - $\frac{|0\rangle + e^{i \theta}|1\rangle}{\sqrt{2}}$ pair or $|1\rangle$ - $\frac{|0\rangle + e^{i \theta}|1\rangle}{\sqrt{2}}$ pair? Or we just have a single qubit which can be in $|0\rangle / |1\rangle$ or $\frac{|0\rangle + e^{i \theta}|1\rangle}{\sqrt{2}}$ ? $\endgroup$ Commented Jun 23, 2021 at 5:35
  • 2
    $\begingroup$ @Rammus If you don't know anything about the distribution, and you only get one sample, then you're stuck using the max entropy prior for $\theta$ and the problem is trivially impossible. Your knowledge of the state in both cases corresponds to the maximally mixed density matrix, and states with identical density matrices are indistinguishable. I suspect this sounds confusing to you because it sounds like you're thinking in frequentist terms ("one sample from an unknown but fixed distribution") instead of Bayesian terms ("I don't know anything about the distribution; what's my prior"). $\endgroup$ Commented Jun 24, 2021 at 7:48

1 Answer 1


No, there is no way to do this with probability better than 1/2. Basically, you are saying that, ideally, you'd like a protocol where $$ |0\rangle\mapsto|0\rangle,\quad |1\rangle\mapsto|0\rangle,\quad \frac{|0\rangle+e^{i\theta}|1\rangle}{\sqrt2}\mapsto|1\rangle\ \forall\theta. $$ We can get an upper bound on the achievable fidelity using the operator $$ R=\sum_ip_i|\psi^{in}_i\rangle\langle\psi^{in}_i|^T\otimes|\psi^{out}_i\rangle\langle\psi^{out}_i|, $$ where $p_i$ is the probability that $|\psi^{in}_i\rangle$ is given. If $\lambda$ is the maximum eigenvalue of $R$ then the maximum fidelity is $F\leq d\lambda$, where $d$ is the Hilbert space dimension of the input states.

In this case, we have $d=2$ and $$ R=\frac14|00\rangle\langle 00|+\frac14|10\rangle\langle 10|+\frac{1}{2}\frac{1}{2\pi}\int_0^{2\pi}d\theta \frac12(|0\rangle+ e^{-i\theta}|1\rangle)(\langle 0|+e^{i\theta}\langle 1|)=\frac14 I. $$ Thus, $\lambda=\frac14$, and $F\leq \frac12$. Since you know that you can achieve $F=1/2$, this is the optimal.

  • $\begingroup$ how do you know that this operator's maximum eigenvalue leads to the max fidelity? $\endgroup$ Commented Aug 6, 2023 at 21:58
  • $\begingroup$ well, there's a derivation to do, but it's all related with the Choi map: if you know a Choi map $\rho$, then how to you calculate the average fidelity for a set of inputs and corresponding target outputs? You'll find it's $d\text{Tr}(R\rho)$. So, if I don't know $\rho$, construct $R$ and find the $\rho$ that maximises it. $\endgroup$
    – DaftWullie
    Commented Aug 15, 2023 at 1:52

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