What conditions must a matrix hold to be considered a valid density matrix?
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asked Jun 14, 2018 at 9:00
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2 Answers
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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 $1$.
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2$\begingroup$ The density matrix should be Hermitian as well $\endgroup$ Commented Jun 14, 2018 at 9:05
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$\begingroup$ Does not identity matrix satisfy all of these conditions? (say, $2x2$ identity matrix) $\endgroup$ Commented Jun 14, 2018 at 9:17
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$\begingroup$ 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. $\endgroup$ Commented Jun 14, 2018 at 9:20
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1$\begingroup$ @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. $\endgroup$ Commented Jun 14, 2018 at 9:23
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$\begingroup$ @DaftWullie You are right, but usually positivity is only defined in the context of Hermitian matricies. I will edit it for more clarity. $\endgroup$– biryaniCommented Jun 14, 2018 at 9:28
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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.
answered Jun 14, 2018 at 11:09
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$\begingroup$ Is it necessary that the initial possible states have to be orthogonal, for the density matrix to be diagonalizable? $\endgroup$ Commented Jun 15, 2018 at 4:34
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$\begingroup$ No, but that is required if those states are to be the eigenstates. $\endgroup$ Commented Jun 15, 2018 at 4:59
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$\begingroup$ 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. $\endgroup$ Commented Jun 15, 2018 at 5:03
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$\begingroup$ 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? $\endgroup$– genCommented May 3, 2019 at 18:59