# How to get the state of an individual qubit in a composite system?

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.

• Does the partial trace answer what you're looking for? – keisuke.akira Aug 1 at 1:47
• 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 – glS Aug 1 at 18:35
• @glS Definitely the former. I just thought it might involve the latter. – dhjtricks Aug 2 at 19:27

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 $$|\phi\rangle$$.
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 $$[\operatorname{Tr}_N(\rho)]_{i_1,...,i_{N-1};j_1,...,j_{N-1}} \equiv \sum_{k}\rho_{i_1,...,i_{N-1},k;j_1,...,j_{N-1},k}.$$