# Does no-cloning theorem implies a no-comparison theorem?

I was reading about no cloning theorem and it arose a thought experiment, if there were a way of compare quantum states (for being equal) then you could build a pseudocloning machine that searches for quantum states till it finds an equal state, so I think there could be something like a No-Comparison theorem.

Does exist something like that?

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## 3 Answers

You can indeed test two quantum states for being equal, but the results are not 100% guaranteed accurate: you measure the eigenvalue of the SWAP operator (which swaps the two quantum states). If they are equal, then you have a 100% chance of getting the +1 eigenvalue. If they are orthogonal, then you have a 50% chance of getting either the +1 and -1 eigenvalue.

This test (a) destroys the quantum state if you test it against a state that it is not equal to and (b) only yields the correct answer half the time if the answer is "no". These two drawbacks mean that you cannot use it to clone. However, this is still a very useful test as a subroutine in designing some quantum algorithms.

I don't know whether anybody has proved a theorem saying that you cannot test equality better than the swap test, but it is definitely true, as the OP speculates, that there is no perfect test for equality of quantum states.

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Your recipe applies to testing of two either orthogonal or coinciding states, doesn't it? Otherwise the  sharp jump from 100% to 50% as the one state  turns from being equal  to slightly unequal to the other becomes disturbing. – Slaviks Mar 15 '12 at 13:54
That's right ... fixed. – Peter Shor  Mar 15 '12 at 15:19

The question boils down to whether it is possible to compare a given unknown state with a known state, without disturbing it. The answer is clearly a no. Because, doing so is equivalent to determining the complete state.

The measure used to say how close is a state to a given state is fidelity. If $\psi$ is the unknown arbitrary state, and $\phi$ is a known state, then the fidelity of $\psi$ with respect to $\phi$ is $|\langle\psi|\phi\rangle|^2$, which is $Tr[|\psi\rangle\langle\psi|\ |\phi\rangle\langle\phi|]$(that is the expectation value of $|\phi\rangle\langle\phi|$, if you like). If it were possible to measure this quantity using a single copy of $\psi$, it means that it is possible to measure the expectation value of any observable; since every hermitian operator is a sum of such projections, weighted with the eigen values.

And since measurement is just a comparison, finding the closest state to a given state equivalent to measuring it's fidelity with every state.

So, just like cloning, comparison is also impossible.

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You can make use of entanglement to make a measurement. Check out this article on QND (Quantum Non-demolition measurements) for a start. physorg article

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But that has little to do with what the question is asking, has it? – leftaroundabout Mar 9 '12 at 14:43
How exactly can you obtain a result of your operation without a measurement? – Antillar Maximus Mar 9 '12 at 14:49
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