# Are the states in the convex decomposition of a density matrix necessarily orthogonal?

In Nielsen and Chuang's QC&QI, I do not see a statement one way or another. In Steeb and Hardy's Problems and Solutions, orthogonality is asserted. If the $$p_i$$ in $$\sum_i p_i |\psi_i\rangle\langle\psi_i|$$ were guaranteed to be distinct, then the eigenvalues would be distinct and orthogonality would be assured. In the general case, I don't know, and I wonder if this question is important given that the $$\psi_i$$ are not uniquely determined.

• You can use |\psi_k\rangle to write a ket. Dec 26, 2021 at 7:09

No. In the ensemble interpretation of a density matrix

$$\rho=\sum_k p_k|\psi_k\rangle\langle \psi_k|$$

the states $$|\psi_k\rangle$$ are not necessarily orthogonal. Also, the probabilities $$p_k$$ are not necessarily eigenvalues.

## Ensembles vs eigendecomposition

A density matrix $$\rho$$ encodes multiple ensembles $$\{p_k, |\psi\rangle_k\}$$, see theorem $$2.3$$ on page $$103$$ in Nielsen & Chuang about the unitary freedom. Eigendecomposition of $$\rho$$ is one of those ensembles - one with orthogonal states $$|\psi_k\rangle$$ - but generally not the only one.

## Example

In fact, Nielsen & Chuang give an explicit example on page $$103$$

$$\rho=\frac12|a\rangle\langle a|+\frac12|b\rangle\langle b| = \frac34|0\rangle\langle 0|+\frac14|1\rangle\langle 1|\tag{2.165}$$

where $$|a\rangle\equiv\sqrt{\frac34}|0\rangle+\sqrt{\frac14}|1\rangle$$ and $$|b\rangle\equiv\sqrt{\frac34}|0\rangle-\sqrt{\frac14}|1\rangle$$ are not orthogonal. Note that even though $$\rho$$'s eigenvalues $$\frac34$$ and $$\frac14$$ are distinct the ensemble it represents is not unique (unlike eigendecomposition which is unique when eigenvalues are distinct).

• To be precise, the freedom is really a partial isometry (or isometry, when starting from the eigenvalue decomposition) rather than a unitary. Dec 29, 2021 at 15:18
• I agree, but "unitary freedom" is what Nielsen & Chuang call it in the book. Dec 29, 2021 at 15:43
• Wasn't meant as a criticism, I just felt it might be good to point this out. "Conceptually", memorizing this as a unitary DoF is indeed fine (and also correct when allowing to pad ensemble decompositions with zeros). Dec 29, 2021 at 15:51
• Yeah, padding the ensembles is exactly what they do. Dec 29, 2021 at 16:10
• The issue with the "unitary + padding" characterization rather than the "(partial) isometry" characterization is that it does not specify the relation of two given ensemble decomposition, but only of ensemble decompositions up to padding. Dec 29, 2021 at 16:12

To build on the other answer, we can in fact characterise the set of coefficients that can fit into a convex decomposition of a density matrix. Given $$\rho$$, we can write $$\rho = \sum_k a_k |u_k\rangle\!\langle u_k|$$ for some set of (not necessarily orthogonal) states $$\{|u_k\rangle\}$$ if and only if $$\mathbf a\preceq \boldsymbol{\lambda}(\rho)$$, meaning that the vector of coefficients $$\mathbf a\equiv(a_1,a_2,...)$$ is majorised by the vector of eigenvalues of $$\rho$$. See e.g. Watrous' book for a proof (section 4.3, page 245).