# Finding basis of Schmidt decomposition

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?

• I follow up until the last sentence. What do you mean with Do they correspond to eigenvectors of a density matrix? – glS Oct 24 '18 at 12:50

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.