1. Is there a clear and intuitive meaning to the eigenvectors and eigenvalues of a density matrix?

  2. Does a density matrix always have a a basis of eigenvectors?  


In general, the density matrix of a given system can always be written in the form $$ \rho = \sum_i p_i |\phi_i\rangle\langle\phi_i|, \tag 1 $$ representing among other things a probabilistic mixture in which the pure state $|\phi_i\rangle$ is prepared with probability $p_i$, but this decomposition is generally not unique. The clearest example of this is the maximally mixed state on, say, a two-level system with orthonormal basis $\{|0⟩,|1⟩\}$, $$ \rho = \frac12\bigg[|0⟩⟨0|+|1⟩⟨1|\bigg], $$ which has exactly the same form on any orthonormal basis for the space.

Generally speaking, though, the eigenvalues and eigenvectors of a given density matrix $\rho$ provide a set of states and weights such that $\rho$ can be written as in $(1)$ - but with the added guarantee that the $|\phi_i⟩$ are orthogonal.

This does not uniquely specify the states in question, because if any eigenvalue $p_i$ is degenerate then there will be a two-dimensional (or bigger) subspace within which any orthonormal basis is equally valid, but that kind of undefinedness is just an intrinsic part of the structure.

  • $\begingroup$ Will the eigenvalues in the base of eigenvectors will be as the probabilities specified in $\rho$'s definition? $\endgroup$ – proton Dec 30 '17 at 14:08
  • $\begingroup$ @proton The expansion in $(1)$ isn't really a definition - it is just one possible mathematical expansion that, depending on context, may or may not have physical significance. The eigenvalues & eigenvectors do admit the interpretation of a probabilistic mixture of the given states with the given probabilities provided by a purely classical RNG. However, this is not the only way to produce density matrices, and you can get mixed states by e.g. tracing out one half of an entangled system that's otherwise in a pure state. $\endgroup$ – Emilio Pisanty Dec 30 '17 at 21:01

The density operator is defined as

$$\rho = \sum_{i=1}^N p_i |X_i\rangle\langle X_i|$$

for a set of $N$ states $|X_i\rangle$ that occur with probability $p_i$. These states form an orthonormal basis to ensure the normalization condition $Tr(\rho)=1$.

We can see from this definition that the eigenvalues are the probabilities $p_i$ that are real numbers; the operator assigns simply the probability to some pure state (that is not in a superposition with other states) without changing this state. Therefore, we know from linear algebra that if eigenvalues are real, then the density matrix must be hermitean.

Another way to see hermiticity is to check $\langle b|\rho = (\rho|b\rangle)^*$ for any state $|b\rangle$.

From hermiticity follows that eigenvalues and eigenvectors exists (spectral theorem).

  • $\begingroup$ What if $\rho$ is defined for a set $ \left\{ \lvert X_i \rangle \right\}$ that is not an orthonormal base? $\endgroup$ – proton Dec 30 '17 at 11:52
  • $\begingroup$ Then, you can expand $|X_i>$ as a superposition of some orthonormal states $|\alpha_{ij}>$ such that $|X_i> = \sum_j b_{ij}|\alpha_{ij}>$ and you will get also a hermitean matrix (you can prove hermiticity e.g. by applying the matrix on a conjugate state and check whether it coincides with the application of the matrix on the original state) that can be diagonalized. $\endgroup$ – kryomaxim Dec 30 '17 at 12:03
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    $\begingroup$ Are you trying to imply that any matrix with real eigenvalues is Hermitian? Because that's false. $\endgroup$ – Federico Poloni Dec 30 '17 at 13:13
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    $\begingroup$ These states form an orthonormal basis to ensure the normalization condition Tr(ρ)=1. --- This is, for the very least, misleading. Normalization of the X_i is all that is required. $\endgroup$ – Norbert Schuch Dec 30 '17 at 14:10
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    $\begingroup$ As written the answer is slightly misleading: if $\vert\pm\rangle_k$ are eigenstates of $\sigma_k$ for instance, it is perfectly possible to have $\rho=\frac{1}{2}\vert + \rangle_z +\frac{1}{2}\vert - \rangle_x$. Thus the density operator is surely not defined in terms of an orthonormal basis. $\endgroup$ – ZeroTheHero Dec 30 '17 at 16:19

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