What conditions must a matrix hold to be considered a valid density matrix?


2 Answers 2


If a matrix has unit trace and if it is positive semi-definite (and Hermitian) then it is a valid density matrix. More specifically check if the matrix is Hermitian; find the eigenvalues of the matrix , check if they are non-negative and add up to $1$.

  • 2
    $\begingroup$ The density matrix should be Hermitian as well $\endgroup$
    – DaftWullie
    Jun 14, 2018 at 9:05
  • $\begingroup$ Does not identity matrix satisfy all of these conditions? (say, $2x2$ identity matrix) $\endgroup$ Jun 14, 2018 at 9:17
  • $\begingroup$ its eigenvalue is $1$ (hence non-negative and add up to $1$, or do I have to consider it as $2$ ones?), also it is Hermitian. $\endgroup$ Jun 14, 2018 at 9:20
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    $\begingroup$ @ArchilZhvania The identity matrix does not have eigenvalues that sum to 1, because all its eigenvalues are 1, and there's more than one of them. However, for a $d$ level system, $\mathbb{I}/d$ is a valid density matrix, known as the maximally mixed state. $\endgroup$
    – DaftWullie
    Jun 14, 2018 at 9:23
  • $\begingroup$ @DaftWullie You are right, but usually positivity is only defined in the context of Hermitian matricies. I will edit it for more clarity. $\endgroup$
    – biryani
    Jun 14, 2018 at 9:28

Suppose someone has prepared your quantum system in one of an orthogonal set of states $\{|\psi_j\rangle\}$. You don't know which of these states they've prepared it in, but you do know that they prepared state $|\psi_j\rangle$ with probability $p_j$. Your system is then described by the density matrix,

$\rho = \sum_j \, p_j \, |\psi_j\rangle \langle\psi_j|$.

There are some properties that will apply to any density matrix of this form.

  • Clearly it is diagonalizable, since it is explicitly written in terms of its eigenvalues $p_j$ and eigenstates $|\psi_j\rangle$.

  • Since the $|\psi_j\rangle \langle\psi_j|$ are Hermitian, and since probabilities are real numbers, the density matrix is Hermitian.

  • Since probabilities are all either zero or positive, the density matrix is positive semidefinite.

  • Since all probabilities must sum to 1, and the trace is a sum of eigenvalues, the density matrix must have a trace of 1.

These are exactly the properties required of all density matrices. Hopefully this derivation of them gives a bit of understanding of why they are required.

  • $\begingroup$ Is it necessary that the initial possible states have to be orthogonal, for the density matrix to be diagonalizable? $\endgroup$ Jun 15, 2018 at 4:34
  • $\begingroup$ No, but that is required if those states are to be the eigenstates. $\endgroup$ Jun 15, 2018 at 4:59
  • $\begingroup$ Thanks. It would be nice to have a single thread containing all these (and other) mathematical theorems regarding density matrices along with their proofs. These things seem to confuse so many beginners (including me). Maybe I'll make such a thread, soon. $\endgroup$ Jun 15, 2018 at 5:03
  • $\begingroup$ This explains why these conditions hold for mixed states. Density operators can also describe pure states. Why should any of these conditions hold for an operator describing a pure state? $\endgroup$
    – gen
    May 3, 2019 at 18:59

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