# Importance of zero and non-zero eigenvalues of density matrix

What can we say about the quantum state from the number of zero and non-zero eigenvalues of the corresponding density matrix? Anything related to entanglement or any other properties? Does they vary with the nature of states such as it is pure or mixed?

• I understand you are talking of a mixed state $\rho$, and its density matrix? These eigenvalues will have nothing to do with entanglement. --- On the other hand, if you have a pure state with two parts and you consider the reduced density matrix of one part, its eigenvalues will tell you sth. about entanglement. – Norbert Schuch Jan 5 '15 at 15:47
• @NorbertSchuch : I don't know who gave me the -1, but the answer is correct. For identical entangled particles the total spin zero implies high symmetry. – Sofia Jan 5 '15 at 15:50
• @Dutta : your guess is correct. Please see my answer. I don't know why is there a -1. – Sofia Jan 5 '15 at 15:51
• @Sofia : As far as I can see, the question asks about eigenvalues of the density operator (which is the case of a pure state would be $\lvert\psi\rangle\langle\psi\rvert$, and thus would always have one eigenvalue one and the rest zeros. I don't see how the spin is related to that. But maybe Dutta could help to clarify the question? – Norbert Schuch Jan 5 '15 at 15:53
• @NorbertSchuch : I understand what you say. I believe indeed that she made a mistake. Well, I issue a question to Dutta. I am so sorry, if the question were I understood it (probably wrongly), it were an interesting question. I withdraw my answer for the moment - until clarification. – Sofia Jan 5 '15 at 16:05

What can we say about the quantum state from the number of zero and non-zero eigenvalues of the corresponding density matrix?

The number of zero eigenvalues has no significance, and is not really well defined anyway.

If the number of non-zero eigenvalues is not one, then there are many different ways to write the density matrix $\rho$ as a coherent decompositions of the form $\rho = \sum_k p_k|\psi_k\rangle\langle\psi_k|$ with $\langle\psi_k|\psi_k\rangle=1$ and $p_i\geq p_j \geq 0$ for $i \leq j$. Iff $\langle\psi_i|\psi_j\rangle=\delta_{ij}$, then this decomposition is an eigendecomposition. Because $\rho$ is Hermitian and positive, an eigendecomposition is also a singular value decomposition, and hence describes all optimal low rank approximations (with respect to the Euclidean norm) in a succinct form. Hence this decomposition is somtimes called optimal coherent decomposition by some communities.

For practical computations, one can just decompose the density matrix into a sum of pure states. The optimal way to do this (i.e. that you get the least error for the number of pure states that you use) is the optimal coherent decomposition, where you compute the eigenvalue decomposition of the density matrix. The dynamics of Schrödinger equations is such that any such decomposition stays valid (and optimal) during time propagation, i.e. you can just propagate each individual pure state.

The last sentence of this pragmatic explanation assumes that $\langle\psi_i(t)|\psi_j(t)\rangle=\langle\psi_i(t_0)|\psi_j(t_0)\rangle$ is preserved during time propagation, which is valid for "closed" systems.

Anything related to entanglement or any other properties? Does they vary with the nature of states such as it is pure or mixed?

As others pointed out, an entangled state is also a pure state. If you compute a partial trace over an entangled state, you get a mixed state, but this is not really related to the eigendecomposition. But this is an interesting observation nevertheless, because the optimal coherent decomposition for the corresponding subsystem won't be preserved in general during time propagation, and hence there can be some sort of quantum leap from the perspective of the subsystem in terms of the optimal coherent decomposition. But the optimal coherent decomposition is only unique if $p_i> p_j \geq 0$ for $i < j$ anyway.

• Thanks. I do not believe that the number of zero eigenvalues has no significance. But I do not know this subject much. – Dutta Jan 6 '15 at 17:25
• The number of zero-eigenvalues is certainly well-defined, because the eigenvalues of any matrix are well-defined. That does not mean that a decomposition into pure states needs to be well-defined (it isn't, if the states are not orthogonal or the eigenvalues have multiplicities). – Martin Jan 12 '15 at 13:47
• @Martin Well, you could say that there is exactly one zero-eigenvalue, namely "0", or you could say that there are infinitely many zero-eigenvalues (because the canonical commutator relations can only be satisfied in an infinite-dimensional Hilbert space), but both answers are pretty useless. That's what I mean by "The number of zero eigenvalues has no significance". You might object that the number of non-zero eigenvalues is infinite too. However, if we are satisfied with a certain accuracy (say 1%), then a "well-defined" number of non-zero eigenvalues is sufficient to get that accuracy. – Thomas Klimpel Jan 12 '15 at 22:23
• For example, a 193 nm ArF laser produces unpolarized quasi-monochromatic light with a small narrow bandwidth around 193.3 nm. At least two eigenvalues are needed to model the unpolarized light. Whether you need additional eigenvalues to model the quasi-monochromatic light (or its very narrow bandwidth) depends on what you do with it. If you spilt it up such that it travels on completely different paths, then you need additional eigenvalues for it. If the small difference in wavelength never has any appreciable consequences, then you also don't need to model it by additional eigenvalues. – Thomas Klimpel Jan 12 '15 at 22:35