$\renewcommand{\ket}[1]{|#1\rangle}$ A "false" (equally superimposed qubit) is created by mechanically firing with 50/50 probability a resonance photon at a Hydrogen atom qubit in the ground state. This qubit is sent to Alice and it now has 50/50 probability of being in state $\ket{0}$ or state $\ket{1}$, but it is not in a quantum mechanical superimposed state! Alice is also sent a "real" 50/50 superimposed qubit created by a Rabi process.

Is there any way that Alice can perform gates and measurements on the two qubits to determine which is which?

  • 2
    $\begingroup$ Nope. With a single instance of the system Alice cannot possibly tell the difference. $\endgroup$
    – DanielSank
    Commented Aug 16, 2015 at 0:02
  • $\begingroup$ One cannot conclude with 2 qbits. You must use 2 flows of qbits of each. In this case the theory allows you to determine which flow is which. Experiments results show that it is a little more difficult. After 34 years, there is not only one experiment that can begin to show the capacity to detect third listeners $\endgroup$
    – user46925
    Commented Sep 21, 2015 at 2:28

2 Answers 2


Not with certainty. However, Alice can probabilistically tell them apart by performing a measurement that asks "is the state of the system $(|0\rangle + |1\rangle)/\sqrt{2}$?" For example, if the qubit were a spin $1/2$ particle and $|0\rangle$ and $|1\rangle$ were spin up and spin down, this would correspond to measuring spin along the $\hat{x}$ direction and asking if the direction is $+\hat{x}$. The real qubit will satisfy this 100% of the time, while the fake one will get the right answer half of the time.

This is the central idea behind the BB84 quantum key distribution algorithm; you can tell (probabilistically) if somebody has looked at the qubits in your message because you can do this test; if the qubits were measured, they would become like the "fake" qubits in your question and fail the test half of the time.

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    $\begingroup$ Thanks Kevin. I didn't know about the BB84 quantum key distribution. But, I gather (as I suspected) that there is no way to determine which qubit is which, without making multiple measurements, which would require an ensemble of the two qubits. $\endgroup$ Commented Aug 17, 2015 at 16:44

The minimum probability of error for correctly identifying one of two states $\rho$ and $\sigma$ prepared with equal probability (where minimization is over POVMs and the pair of states is fixed) is related to the trace distance $\|\rho-\sigma\|_1$ between them via the Helstrom bound $p_\text{min}=\frac{1}{2}-\frac{1}{4}\|\rho-\sigma\|_1$. For some reason, it's very hard to find this actually stated anywhere, but it's definitely in Helstrom's book on quantum estimation theory. In particular, the minimum error probability is zero if and only if the two density operators have orthogonal support, when $\|\rho-\sigma\|_1=2$. The two density matrices in your example are $\rho=(1,1;1,1)/2$ and $\sigma=(1,0;0,1)/2$, so $\|\rho-\sigma\|_1=1$. The optimal measurement therefore has $p_\text{min}=1/4$. A measurement satisfying this bound is easy to find: just measure in the $\left\vert\pm\right\rangle$ basis, and if you measure $\left\vert+\right\rangle$ report $\rho$ and if you measure $\left\vert-\right\rangle$ report $\sigma$. With $p=1/2$, $\rho$ will be prepared. You'll always be right in this case. With $p=1/2$, $\sigma$ will be prepared, and then you have $p=1/2$ to measure $\left\vert+\right\rangle$ and incorrectly report that you received $\rho$, for an overall error probability of $\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$.


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