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.

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

2 Answers 2


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.


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).

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    $\begingroup$ To be precise, the freedom is really a partial isometry (or isometry, when starting from the eigenvalue decomposition) rather than a unitary. $\endgroup$ Dec 29, 2021 at 15:18
  • $\begingroup$ I agree, but "unitary freedom" is what Nielsen & Chuang call it in the book. $\endgroup$ Dec 29, 2021 at 15:43
  • $\begingroup$ 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). $\endgroup$ Dec 29, 2021 at 15:51
  • $\begingroup$ Yeah, padding the ensembles is exactly what they do. $\endgroup$ Dec 29, 2021 at 16:10
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    $\begingroup$ 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. $\endgroup$ 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).


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