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Although the square of the density operator $\rho^2$ is positive, and thus permits a spectral decomposition, the eigenvalues associated with that spectral decomposition are not necessarily the squares of the original probabilities $p_i$ (used to construct the density operator), a assumption upon which the proof relies.
(This is because the spectral decomposition of the density operator has eigenvalues that do not necessarily coincide with the probabilities of the states used to construct the density operator, as shown in p.103 of the book.)
The text was updated successfully, but these errors were encountered:
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Potential Error in Exercise 2.71 Solution
Error in Exercise 2.71 Solution
May 7, 2020
Although the square of the density operator$\rho^2$ is positive, and thus permits a spectral decomposition, the eigenvalues associated with that spectral decomposition are not necessarily the squares of the original probabilities $p_i$ (used to construct the density operator), a assumption upon which the proof relies.
(This is because the spectral decomposition of the density operator has eigenvalues that do not necessarily coincide with the probabilities of the states used to construct the density operator, as shown in p.103 of the book.)
The text was updated successfully, but these errors were encountered: