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proofsmallestThe Kochen–Specker theorem states that noncontextual hidden variable theories are incompatible with quantum mechanics.We provide a state-independent proof of the Kochen–Specker theorem using the smallest number of projectors,i.e.,thirty rank-2 projectors,associated with the Mermin pentagram...
Measures whose values describe the state, energy, or distribution of a property of an observed system. From: Encyclopedia of Physical Science and Technology (Third Edition), 2003 About this pageSet alert Also in subject area: Computer ScienceDiscover other topics ...
In this case, ρi> 0 for all values of i, and the absolute appearance probability is irrespective of the initial condition. We now show the proof for this statement in the following. We consider a Markov chain with the total M microscopic states, and use the notation py(m) for the ...
This theorem improves substantially upon the result in36, which requires N=nO(1/ϵ). The proof idea is given in Section “Methods”, and the detailed proof is given in Supplementary Sections 1, 2, 3. Using the proof techniques presented in this work, one can show that the sample ...
Specifically, the paper is organized as follows: in the next section, we give preliminaries, and the second section is divided into two subsections. In Sect. 2.1, we study the case q< p<2^{*} when N<4+\alpha . In Sect. 3, we give the proof of Theorem 1.1....
butter to quantum information theorists; moreover the language of functional analysis was not foreign to me either and so I believed I ought to have more than it should take to make the explanations clear; in my mind I had a clear theorem, together with an airtight proof, to communicate. ...
Even in the absence of a rigorous quantum advantage, special-purpose electronic and photonic machine learning circuits are being pursued in order to increase the speed and energy-efficiency of well-established classical machine learning models [1–4]. Therefore, here we will focus on implementing ...
Proof. If there is plane stress state at every point in the thickness, then [6.35] immediately follows from the equivalences [6.34] and from definition [3.50] of the shear forces Qα, Q(1)α. Finally, inserting [6.35] in [3.48] gives S = 0. We can also directly prove [6.36] by ...
Theorem: Let │ψAB∈HA ⊗ HB. Then ∃ orthonormal |kA〉∈ HA and |kB〉∈ HB such that │ψAB=∑kλk|kA〉|kB〉, where ∑kλk2=1 and ∀k, λk≥0.λk are known as the Schmidt coefficients for |ψA B〉. Proof: In terms of basis ...