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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 ...
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. ...
Mathematical work on nonlinear Choquard equations like the above has been investigated in recent years, and the existence and multiplicity results for such type equations have been considered in many papers under some different assumptions on the potential and nonlinearity by using various variational arg...
If the combination of parameters Ω0/ω and Ω1/ω gives rise to unstable solutions of the Mathieu equation, the quasienergy spectrum of the parametrically driven oscillator becomes absolutely continuous37 so that the system can absorb an infinite amount of energy from the drive; this case is ...
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 ...
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 ...