# What is the probability of error in quantum fingerprinting using cswap test?

In quantum fingerprinting, the states are sent to a 3rd party in order to verify if both qubits match or not. To do so, a swap test is needed by the 3rd party.

I read some information on swap test including this link here and I noticed that the probability of cswap test is:

If $$|\phi _{x}\rangle =\ |\phi _{y}\rangle$$ then we observe a 0 with $$prob(0) = 1$$ according to wikipedia.

So I was wondering what are the probabilities of a false positive and false negative in quantum fingerprinting?

According to what I saw I'm guessing:

• Probability to get a false positive (a 0 even if both states don't match) is 1 - P(getting 0) so it's 0?
• Probability of getting a false negative (a 1 even if both states match) is 1/2?

But I'm not sure if my conclusion is correct, could someone please confirm this and give some explanation?

I'll just sketch the conclusion. The probability that the measurement of the control qubit is in the $$|0\rangle$$ state is $$p_0=\frac{1}{2}+\frac{1}{2}F(|\phi\rangle,rho)$$, where $$|\phi\rangle$$ is the state vector of one state, and $$\rho$$ is the density matrix. Thus the fidelity of the two states is $$F(|\phi\rangle, rho)=2*p_0-1$$.
Since by its definition, fidelity can be a number varting continuously from $$0$$ to $$1$$, just a few measurements(say three or four times) is never sufficient. If sufficiently many measurements are performed, then we'll have confidence in estimating its value, e.g., if we performed 100 measurements then we know that the decile of the reduced value of fidelity should be reasonable.