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Solving Problem 2.4 in Ballentine: Nonnegativeness Derivation
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[QUOTE="strangerep, post: 6862643, member: 70760"] For this type of problem, it's often more efficient to work with the eigenvalues directly, rather than a generic matrix. For this problem, we are [B]given[/B] that: $$Tr(\rho^2) ~\le 1 ~,~~~~ Tr(\rho) ~=~ 1 ~,~~~~ \rho = \rho^\dagger ~.$$For the 2D case, there are 2 eigenvalues, ##\rho_1## and ##\rho_2##, say, hence the eigenvalues of ##\rho^2## are the squares of these. Self-adjointness of ##\rho## implies both the ##\rho_i## are real, hence ##\,\rho_i^2 \ge 0##. The trace of a matrix is the sum of its eigenvalues, so we have 2 conditions: $$\rho_1 + \rho_2 ~=~ 1 ~,~~~~ \rho^2_1 + \rho^2_2 ~\le~ 1 ~.$$Squaring the 1st equation gives $$\rho_1^2 + \rho_2^2 + 2 \rho_1 \rho_2 ~=~ 1 ~,$$and using this in conjunction with the 2nd equation implies... what? I leave it to you to figure out the rest of the proof, including the follow-on of why it doesn't work for higher dimensional matrices. :oldbiggrin: [/QUOTE]
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Solving Problem 2.4 in Ballentine: Nonnegativeness Derivation
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