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?


1 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 $\psi$).

  • $\begingroup$ 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 $\rho$? $\endgroup$
    – Physkid
    Aug 30, 2023 at 14:43
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    $\begingroup$ it is a probability: for a density matrix, the trace is 1, so the eigenvalues sum to 1 (like a probability) and each is bounded between 0 and 1. $\endgroup$
    – DaftWullie
    Aug 30, 2023 at 15:03
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    $\begingroup$ I want to add one comment to this, which is that there are infinitely many ensembles which correspond to the same density matrix. The mapping of ensembles to density matrix is a surjective map. $\endgroup$
    – FDGod
    Aug 30, 2023 at 19:56
  • $\begingroup$ @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? $\endgroup$
    – Physkid
    Aug 31, 2023 at 1:33
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    $\begingroup$ 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. $\endgroup$
    – DaftWullie
    Aug 31, 2023 at 6:37

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