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0004008274
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We notice that a problem of determination of the quantum state (whether it is pure or mixed) is usually nontrivial, Springer, Berlin 1932
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We notice that a problem of determination of the quantum state (whether it is pure or mixed) is usually nontrivial; J. von Neumann: Mathematische Grundlagen der Quanten- mechanik (Springer, Berlin, 1932);
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Mathematische Grundlagen der Quanten-mechanik
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Von Neumann, J.1
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0004150286
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see for example (Kluwer Academic London 1998). For example there is a famous deceptive question known as Pauli problem "Does the measurement of the distributions of position and momentum determine its quantum state?" [W. Pauli: in Handbuch der Physik, ed. H. Geiger and K. Scheel (Springer, Berlin, 1933) Vol 24, Pt 1, p 98], which is known to be incorrect
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A. Peres: Quantum Theory: Concepts and Methods (Kluwer Academic, London, 1998). For example, there is a famous deceptive question, known as Pauli prob-lem: "Does the measurement of the distributions of position and momentum determine its quantum state? [W. Pauli: in Handbuch der Physik, ed. H. Geiger and K. Scheel (Springer, Berlin, 1933) Vol. 24, Pt. 1, p. 98], which is known to be in- correct;
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Quantum Theory: Concepts and Methods
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Peres, A.1
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0000902540
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see for example, S. Weigert: Phys. Rev. A 45 (1992) 7688.
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(1992)
Phys. Rev. A
, vol.45
, pp. 7688
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Weigert, S.1
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0042162967
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the details of the present article are given in this reference
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G. Kimura: Phys. Lett. A 314 (2003) 339; the details of the present article are given in this reference.
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(2003)
Phys. Lett. A
, vol.314
, pp. 339
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Kimura, G.1
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note
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Suppose that (i) p̂ ≥ 0, i.e., ∀χ)ε HN, (χρ̂|χ)≥0. Then, ρ̂† = ρ̂ which can be shown by using the polarization identity for ρ̂†(i-2) ρi ≥ 0 because of ρi = (ρi|ρ\ρi)≥ 0in which |ρi) is the corresponding eigenstate. Conversely, suppose that and (i-2) hold, then |χ)εHN, (χ|ρ|χ)≥ 0 which can be easily checked by expanding |χ) = ∑iχi|ρi).
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16
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70349366256
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note
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For a simple proof of the Newton's formulas, it is convenient to make use of a function f (t) = π =i(1 - txi) = ∑Nj-0(-1)jajtj (t ε ℝ). By differentiating this with respect to t,we obtain df(t)/dt = ∑i=1N j(-1)jajtj - On the other hand, t holds that [df (t)/dt]/f(t)= d log f (t)/dt = σsum;ni=1n/(txi-1 Compare the coefficients of t between them, one can obtain the Newton's formula.
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