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

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?

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 $$\psi$$).
• Thank you. How does the eigenvalue $\lambda_{i}$ in $$\sum_{i} \lambda_{i} |\phi_{i} \rangle \langle \phi_{i} |$$ leads to the probability $p_{i}$ in definition of the density matrix $\rho$? Aug 30 at 14:43
• @DaftWullie Understood. I wonder if my confusion stems from the use of notation here. The mod square of an eigenvalue yields a probability so, to be explicit, should $\lambda_{i}$ be $\lambda_{i}^{2}$ instead? Aug 31 at 1:33
• No. The mod-square of a probability amplitude yields a probability. Here, by virtue of being in a density matrix, you have already implicitly taken the mod-square, so it really is the $\lambda_i$ that are the probabilities. Aug 31 at 6:37