Why are eigenvectors of the density matrix orthogonal pure states?

Question

I am a first year PhD student studying condensed matter theory. I'm working with entanglement entropy and am having some trouble understanding the diagonal form of the density matrix. I get why we have

$$ S(\rho) = -tr[\rho \ln(\rho)] = -\sum _{i} \lambda_i ln(\lambda_i) $$

where $\lambda_i$ are the eigenvalues of $\rho$.

What I'm having issues understanding is why the eigenvectors of the density matrix are necessarily pure states that can't be represented in terms of the other pure states possible for a given ensemble. Is there some way of characterizing an orthonormal basis of pure states?

Answer 1

I believe there's a misunderstanding and a mix-up of concepts in what you've written. Let me try to clarify.

Every density operator, $\rho$, can be represented as the convex combination of one-dimensional projectors:

$$\rho = \sum_k p_k |\psi_k \rangle \langle \psi_k | \quad \text{where} \sum_k p_k =1 \quad \text{and} \quad \forall_k:p_k \geq 0.$$

This decomposition is typically not unique. With that in mind, the von Neumann entropy is defined as:

$$S(\rho) = -\text{tr}(\rho \log \rho) = -\sum_k p_k \log p_k.$$

In the expression above, I've expressed $S(\rho)$ in terms of the eigenvalues $p_k$ of $\rho$. If the system is in a pure state, given by $\rho = |\psi \rangle \langle \psi |$, then its eigenvalues are such that one $p_k = 1$ and all others are zero. As a result, the entropy of a pure state is zero.

Answer 2

If you consider the explicit form of the trace of $\rho^2$ after the insertion of the identity in the complete eigenstate set of the hamiltonian you'll get $$Tr(\rho^2)=\sum_{n}\sum_{ij}p_ip_j\langle\psi_i|\psi_j\rangle\langle\psi_j|n\rangle\langle n|\psi_i\rangle$$ The states shoud be different from each other (even if they're not orthogonal) and the criterion is that for pure states $Tr (\rho^2)=1$ while for mixed states $Tr (\rho^2)<1$

Answer 3

I am a first year PhD student studying condensed matter theory. I'm working with entanglement entropy... $$ S(\rho) = -tr[\rho \ln(\rho)] $$

... What I'm having issues understanding is why the eigenvectors of the density matrix are necessarily pure states

If you write a density matrix in "any old basis" it will of course have off-diagonal terms.

However, if you write the density matrix in a basis $\{|\chi_i\rangle\}$ in which it is diagonal, then it will have the form: $$ \hat \rho = \sum_i p_i |\chi_i\rangle\langle\chi_i|\;,\tag{1} $$ where $0\le p_i\le1$ and $\sum_i p_i = 1$.

You can read off the eigenvectors from Eq. (1) directly. The eigenvectors are the $\{|\chi_i\rangle\}$ and their corresponding eigenvalues are the $\{p_i\}$.

To put it another way: $$ \hat \rho |\chi_j\rangle = p_j|\chi_j\rangle $$

Any given $|\chi_i\rangle$ is just a ket in Hilbert space, which, by definition it describes a pure state, since the corresponding density matrix for just this pure state is $\hat \rho_i = |\chi_i\rangle\langle\chi_i|$, which is also a diagonal matrix in this basis, but has only a single non-zero eigenvalue (1).