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Given a composite system with $N$ qubits represented by some $2^N$-dimensional vector, how would I get the quantum state of an individual qubit?

Note that I understand some states are not separable given quantum entanglement. My question is whether there is a set of steps, an algorithm, for generally determining whether a qubit's state is separable, and then getting the state itself if it is separable.

The wikipedia entry provides a somewhat brief answer to this question but doesn't expand on the general case for separating quantum states.

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    $\begingroup$ Does the partial trace answer what you're looking for? $\endgroup$ – keisuke.akira Aug 1 at 1:47
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    $\begingroup$ are you asking "how do I get the state of an individual qubit?" or whether there is an algorithm to determine whether "a qubit state is entangled"? These are very different questions $\endgroup$ – glS Aug 1 at 18:35
  • $\begingroup$ @glS Definitely the former. I just thought it might involve the latter. $\endgroup$ – dhjtricks Aug 2 at 19:27
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Given a separable bipartite state as $|\psi\rangle\otimes|\phi\rangle$, you "get" the states of the single systems but taking only the corresponding state, e.g. here $|\psi\rangle$ or $|\phi\rangle$.

More generally, you might not know the structure of the state, and you might have entanglement between the different subsystems, in which case the reduced states are not pure, and you need to use density matrices to properly take this into account.

Given an $N$-partite state $\rho$, get the corresponding reduced states via the partial trace operation. For example, if you want the state of the first $N-1$ qubits, you do $$\operatorname{Tr}_N(\rho) \equiv (I\otimes\operatorname{Tr})\rho \equiv \sum_k(I\otimes\langle k\rvert)\rho(I\otimes \lvert k\rangle).$$ More explicitly, if the matrix elements of $\rho$ are written as $\rho_{i_1,...,i_N;j_1,...,j_N}$, we have $$[\operatorname{Tr}_N(\rho)]_{i_1,...,i_{N-1};j_1,...,j_{N-1}} \equiv \sum_{k}\rho_{i_1,...,i_{N-1},k;j_1,...,j_{N-1},k}.$$

To get the state of a single qubit, you do the above for all degrees of freedom except the first one.

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