# Why does Schmidt decomposition (2 qubits) requires density matrix of each system?

This is in reference to Nielson & Chang (page 109).

Schmidt decomposition: suppose $$|\psi\rangle$$ is a pure state of a composite syste, AB. Then there exists orthonormal state $$|i_{A}\rangle$$ for system $$A$$, and orthonormal state $$|i_{B} \rangle$$ of system $$B$$ such that $$|\psi\rangle = \sum_{i}^{d}\lambda_{i} |i_{A} \rangle |i_{B} \rangle$$ where $$\lambda_{i}$$ are non - negative real numbers with the condition that $$\sum_{i}^{d} \lambda_{i}^{2} = 1$$

N&C proceed to mention

Let $$|\psi\rangle$$ be a pure state of a composite system, $$AB$$. Then by the Schmidt decomposition $$\rho^{A} = \sum_{i}^{d} \lambda_{i}^{2}|i_{A}\rangle \langle i_{A}|$$ and $$\rho^{B} = \sum_{i}^{d} \lambda_{i}^{2}|i_{B}\rangle \langle i_{B}|$$.

I'm struggling to follow why the author made the leap from $$|\psi\rangle$$ to $$\rho^{A}$$ and $$\rho^{B}$$. Any help to help in my understanding is greatly appreciated. It seems like the author recast the $$|\psi\rangle$$ from state vector to density operator.

• Am I correct to understand that given $|\psi\rangle$: 1. convert $|\psi\rangle$ to density matrix $\rho$ 2. do $Tr_{A}[\rho] = \rho_{B}$, do $Tr_{B}[\rho] = \rho_{A}$ 3. perform Schimdt decomposition on $\rho_{B}$ and also on $\rho_{A}$ Oct 9, 2023 at 10:15
• 3. Diagonalise $\rho_A$ and $\rho_B$. 4 take eigenvalues as squares of Schmidt coefficients and diagonal bases as Schmidt basis to give Schmidt decomposition of $|\psi\rangle$. Oct 9, 2023 at 11:33