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I am trying to solve a problem in the measurement and identification of quantum states with a promise as to what states it could be. Here is the problem. Imagine a system that produces qubits in one of four states $S = \{a,b,c,d \}$, evenly distributed. In one shot, I can receive $K$ copies of a state in $S$. However, a random unitary, evenly distributed in $SU(2)$ has been applied, so, at the detectors I receive $S^{\prime} = \{Ha,Hb,Hc,Hd \}$. I have a detector system that includes $2M$ detectors and they represent the projection operators onto any pair of basis vectors in any basis (thus we can choose which $M$ bases we want to use the detectors in). All I want to do is decide if the state is $a,b,c,$ or $d$. How many copies of the state do I need (ie what is a minimum for $K$)? What is the smallest $M$?

Another, more general question is this: what should I use to represent the state before I do a measurement? Should it not be a density matrix which is an integral over all states? Since there is a random unitary applied, that means that I can receive, in a fixed basis, any state with equal probability. What would be the update rule for this density matrix after the $j^{th}$ measurement result?

I realize there are pieces missing in my question. I will revise it this post this evening, but hopefully it will give the general idea.

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I don't understand the problem. Once the state is randomly rotated there is no way to tell if it was $a$,$b$,$c$ or $d$. –  Piotr Migdal Feb 28 '12 at 16:51
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I'm not quite sure I understand: If $H$ is chosen randomly then $Ha$ is uncorrelated with $a$. Averaging over $H$ you get $\rho = \sum_H p(H) (H a)^{\otimes K} = \sum_H p(H) (H b)^{\otimes K} = \sum_H p(H) (H c)^{\otimes K} = \sum_H p(H) (H d)^{\otimes K}$, so no matter how many copies you have you can't distinguish the initial state, since these states are related by unitary operators (and you've just applied a random unitary). –  Joe Fitzsimons Feb 28 '12 at 16:53
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@PiotrMigdal: Looks like we were typing at the same time about the same thing! –  Joe Fitzsimons Feb 28 '12 at 16:54
    
I guess I should have paid more attention in QInformation class.... :(....let me see if I can make this an interesting question. –  Ben Sprott Feb 28 '12 at 17:18
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What if instead of receiving $K$ copies of a state, you can receive $k$ copies of $N$ states chosen randomly from $S$. That is to say, you receive a string of qubits, and you get $K$ copies of each qubit. Given some setup, you will accumulate 4 different probability distributions because you are also allowed to know when a qubit has been passed. Can you not then use this added information to decide on which distribution corresponds to which original state? There is certainly information embedded in the promise of 4 different distributions and I guess I am trying to squeeze that out. –  Ben Sprott Feb 28 '12 at 18:23
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I think that it may be quite difficult to find the most efficient solution, which would involve collective measurements involving several of the inputs at once (although in your question you seem to rule out this possibility by mentioning measurements of single qubits at a time). Also, one should consider POVMs rather than projective measurements.

Assuming that you measure only one qubit at a time, I can answer your question of how to represent them. The first qubit is as you suspect the fully mixed state. After that you should use Bayes' rule to update your probability distributions for S (if S is not assumed to always be uniformly random) and S'. This is not my expertise, but this part of it is a classical problem not a quantum problem. Use these updated probability distributions to form the density operator of the next state you receive. You then need to decide which type of measurement is best given your current state of knowledge about the probabilities. I can't help you there.

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Collective measurements tend to do better. But as mentioned in the comments the original question asks about distinguishing states which are indistinguishable: they have the same reduced density matrix. –  Joe Fitzsimons Mar 2 '12 at 10:30
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