Can a density operator be written equivalently as $\rho=\sum_i p_i|\psi_i〉\!\langle\psi_i|$ and $\rho=\sum_i\lambda_i|\psi_i\rangle\!\langle\psi_i|$?

Question

My doubt arises from page 99, 101 of the book Quantum Computation and Quantum Information by Michael A.Nielson and Issac L.Chung.

Let {${p_{i}, | \psi_{i} \rangle }$} be an ensemble of pure states.

The density operator for this system can be defined by the equation:

$$\rho = \sum_{i} p_{i} |\psi_{i} \rangle \langle \psi_{i} |$$

On page 101, the author expressed the definition for a density operator like

$$\rho = \sum_{i} \lambda_{i} |\psi_{i} \rangle \langle \psi_{i} |$$

where $\lambda_{i}$ are the eigenvalues of the eigenstate $|\psi_{i} \rangle$.

The only relevant knowledge I do have on hand is the fact that the square of eigenvalue $\lambda_{i}$ yields the probability $p_{i} = |\lambda_{i}|^{2}$ that the measurement outcome of an eigenstate $|\psi_{i} \rangle$ is $\lambda_{i}$. Can someone shed some light on this please?

Answer 1

The statement $\rho=\sum_ip_i|\psi_i\rangle\langle \psi_i|$ is a very general way of writing down the density matrix. It must be noted that if you are simply presented with the matrix $\rho$, there are many different ensembles $\{p_i,|\psi_i\rangle\}$ that correspond to that matrix. Critically, $\rho$ is Hermitian, meaning that it has a spectral decomposition. This guarantees that one way of writing it (indeed, the one using the minimal number of terms) is in terms of the eigenvalues and eigenvectors $\{\lambda_i,|\phi_i\rangle\}$ (I've used $\phi$ here for the eigenvectors to distinguish from the general case of arbitrary vectors $\psi$).