An exercise question (9.7) from Quantum computation and Quantum Information by Michael E. Nielson and Isaac L. Chuang says that I can write the difference of any 2 arbitrary density operators $\rho,\sigma$ as a spectral decomposition: $$ \rho - \sigma = U D U^\dagger $$ But for this to be true, isn't it necessary that the density matrices must commute? From what I understand, I will need an orthonormal basis in which both will be diagonalizable. Based on the Simultaneous diagonalizable theorem this is only the case when $\rho$ and $\sigma$ commute.

What am I missing?


1 Answer 1


If the density matrices commute then there exists a joint spectral decomposition of both $\rho$ and $\sigma$. I.e. there exists a unitary $U$ and diagonal positive semidefinite matrices $D_{\rho}$ and $D_{\sigma}$ such that $$ \rho = U D_{\rho} U^\dagger \qquad \text{and} \qquad \sigma = U D_{\sigma} U^\dagger. $$ Note that here you are diagonalizing two different matrices using the same unitary.

The statement you're actually concerned with is different. You're asking whether a single matrix $X := \rho - \sigma$ can be diagonalized. But this is true as $X$ is Hermitian and so the spectral theorem applies.


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