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How do you find the basis for the Schmidt decomposition when given a state of multiple qubits? For example, if you have the systems $A,B$ and $C$, how do they correspond to eigenvectors of a density matrix?

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  • $\begingroup$ I follow up until the last sentence. What do you mean with Do they correspond to eigenvectors of a density matrix? $\endgroup$ – glS Oct 24 '18 at 12:50
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The Schmidt decomposition is nothing but the singular value decomposition (SVD) applied to the coefficients of a bipartite state.

Any matrix $A$ can be written, using the SVD, as $A=\sum_k s_k\lvert u_k\rangle\!\langle v_k\rvert$ for some $s_k\ge0$ and orthonormal bases $\{u_k\}_k$ and $\{v_k\}_k$. In terms of the matrix elements of the matrix, this reads $$A_{ij}=\sum_k s_k\langle i\rvert u_k\rangle\langle v_k\rvert j\rangle \equiv \sum_k s_k u_{ki} v^*_{kj}.$$ Note in particular that the SVD can be applied also to non-square matrices.

If you have a bipartite state, you can write it as $$\lvert\psi\rangle=\sum_{ij}c_{ij}\lvert i,j\rangle,$$ and thus applying the SVD to the matrix whose components are $c_{ij}$ you get the Schmidt decomposition of the state.

Now what if the state is multipartite? Not much changes really. You have a state generally written as $$\lvert\Psi\rangle=\sum_{i_1\cdots i_n} c_{i_1,...,i_n}\lvert i_1,...,i_n\rangle.$$ You can think of this state as "bipartite" by separating the set of indices in two parts anyway you like. For example, if I separate the first index from all the others, the SVD gives me $$c_{i_1...i_n}=\sum_{k\in\{0,1\}} s_k^{(1)}\langle i_1\rvert u^{(1)}_k\rangle\langle v^{(1)}_k\rvert i_2,...,i_n\rangle.$$ Note that here $\lvert u_k^{(1)}\rangle$ is a two-dimensional vector, while $\lvert v_k^{(1)}\rangle$ is a $2^{n-1}$-dimensional one.

Putting this back into the general expression of $\lvert\Psi\rangle$ gives you the Schmidt decomposition with respect to the first qubit.

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