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.


1 Answer 1


The reason why they go from the pure state to the individual density matrices of the subsystems is that this is going to help you find the Schmidt coefficients and the orthonormal bases for both sites. If I give you some arbitrary pure, two qubit state, it is not obvious what the Schmidt coefficients and bases are. However, find the reduced density matrices, and the spectral decomposition immediately gives you everything.

  • $\begingroup$ 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 $\rho_{A}$ $\endgroup$
    – Physkid
    Oct 9, 2023 at 10:15
  • 1
    $\begingroup$ 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 $|\psi\rangle$. $\endgroup$
    – DaftWullie
    Oct 9, 2023 at 11:33

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