Why are eigenvectors of the density matrix orthogonal pure states?

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?

• The density matrix is a hermitian operator, so it is diagonalizable in terms of an orthonormal basis of pure states.
– Zack
Commented Oct 27, 2023 at 19:55
• Instead of producing a new answer, I think the answer by Emilio Pisantry here is particularly complete and clear: physics.stackexchange.com/questions/404093/…. Commented Oct 28, 2023 at 0:05

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.

• Isn't the second expression that you've written for the entropy only true if you are working in the eigenbasis of the density matrix? It's not true for any decomposition in terms of projectors, just for the decomposition in terms of the eigenvectors of the density matrix (I think). Or, perhaps more generally, it's true for any projector-decomposition of the density matrix provided that that the projectors corresponds to a set of orthogonal states. Commented Oct 27, 2023 at 21:02
• @march, you are correct! But note that I made this comment (maybe not so explicit) by writing "This decomposition is typically not unique" and later "In the expression above, I've expressed $S(\rho)$ in terms of the eigenvalues $p_k$ of $\rho$".
– Alex
Commented Oct 27, 2023 at 21:15
• You're right; that's called "not readying carefully"! Commented Oct 27, 2023 at 21:33

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 $$Tr (\rho^2)<1$$

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).