Deutsch-Jozsa Paper
... However, it is possible to extract some joint of the m values, by measuring certain observables properties G[f(0),...,f(m-1)] which are not diagonal in the basis (1). This is called the method of computation by quantum parallelism and is possible only with computers whose computations are coherent q ...
... However, it is possible to extract some joint of the m values, by measuring certain observables properties G[f(0),...,f(m-1)] which are not diagonal in the basis (1). This is called the method of computation by quantum parallelism and is possible only with computers whose computations are coherent q ...
Quantum strategies
... Simon’s algorithm is a quantum strategy which is more successful than any mixed, i.e., probabilistic, one [1]. Similarly, in the problem of searching a database of size N , the locations in the database correspond to pure strategies; again we may imagine the oracle choosing a mixed strategy designed ...
... Simon’s algorithm is a quantum strategy which is more successful than any mixed, i.e., probabilistic, one [1]. Similarly, in the problem of searching a database of size N , the locations in the database correspond to pure strategies; again we may imagine the oracle choosing a mixed strategy designed ...
A solid disk with mass = 0
... 3) A figure skater with an initial moment of inertia of 40 kg.m2 spins at a rotational speed of 180 rpm. Assume there is no friction acting on the skater. a) What is the angular velocity of the skater (in SI units)? ...
... 3) A figure skater with an initial moment of inertia of 40 kg.m2 spins at a rotational speed of 180 rpm. Assume there is no friction acting on the skater. a) What is the angular velocity of the skater (in SI units)? ...
Physics Tutorial 19 Solutions
... 13. A scanning tunnelling microscope (STM) image that Lawrence Livermore National Laboratory of Silicon (100) took for a deposition of one monolayer of molybdenum has a resolution of 1 1010 m. (a) Briefly describe the application of quantum tunnelling to the probing tip of a STM and how this is us ...
... 13. A scanning tunnelling microscope (STM) image that Lawrence Livermore National Laboratory of Silicon (100) took for a deposition of one monolayer of molybdenum has a resolution of 1 1010 m. (a) Briefly describe the application of quantum tunnelling to the probing tip of a STM and how this is us ...
ROLLING, TORQUE, and ANGULAR MOMENTUM
... g of circular objects j and its relationshipp with friction -Redefinition of torque as a vector to describe rotational problems that are more complicated than the rotation of a rigid body about a fixed axis -Angular Momentum of single particles and systems or particles -Newton’s second law for rotat ...
... g of circular objects j and its relationshipp with friction -Redefinition of torque as a vector to describe rotational problems that are more complicated than the rotation of a rigid body about a fixed axis -Angular Momentum of single particles and systems or particles -Newton’s second law for rotat ...
Atomic Structure and Periodic Trends
... • The orbitals of a given sublevel (e.g. p, or d, or f) are degenerate (of the same energy). • The lowest energy state occurs with the maximum number of unpaired electrons. – Meaning…..electrons enter an empty orbital of a given sublevel before pairing up. ...
... • The orbitals of a given sublevel (e.g. p, or d, or f) are degenerate (of the same energy). • The lowest energy state occurs with the maximum number of unpaired electrons. – Meaning…..electrons enter an empty orbital of a given sublevel before pairing up. ...
Quantum mechanical modeling of the CNOT (XOR) gate
... the considerations are formally equivalent with a spin-1/2 system. It particularly means that the actual Hilbert space(s) reduces to a 2-dimensional space, and the corresponding algebra is the well known SU(2) algebra. And this notion points out the symmetry groups that should be considered. As with ...
... the considerations are formally equivalent with a spin-1/2 system. It particularly means that the actual Hilbert space(s) reduces to a 2-dimensional space, and the corresponding algebra is the well known SU(2) algebra. And this notion points out the symmetry groups that should be considered. As with ...
Quantum Physics 2005
... many wavelengths (momenta) be added together. • The act of measuring position by forcing a particle to pass through an aperture causes the particle wave to diffract. ...
... many wavelengths (momenta) be added together. • The act of measuring position by forcing a particle to pass through an aperture causes the particle wave to diffract. ...
Mutually unbiased bases, orthogonal Latin squares, and hidden
... unknown, i.e., the outcomes “spin-up” and “spin-down” occur with the same probability. The eigenbases of ˆ x, ˆ y, and ˆ z Pauli operators form so-called mutually unbiased bases 共MUBs兲: Every vector from one basis has equal overlap with all the vectors from other bases. MUBs encapsulate the conce ...
... unknown, i.e., the outcomes “spin-up” and “spin-down” occur with the same probability. The eigenbases of ˆ x, ˆ y, and ˆ z Pauli operators form so-called mutually unbiased bases 共MUBs兲: Every vector from one basis has equal overlap with all the vectors from other bases. MUBs encapsulate the conce ...
Quantum statistics: Is there an effective fermion repulsion or boson
... due only to statistics is positive for spinless fermions and negative for spinless bosons. Heer2 共similar to most other texts that treat the subject, including one authored by one of us3兲 says, ‘‘The quantum correction that is introduced by statistics appears as an attractive potential for Bose–Ein ...
... due only to statistics is positive for spinless fermions and negative for spinless bosons. Heer2 共similar to most other texts that treat the subject, including one authored by one of us3兲 says, ‘‘The quantum correction that is introduced by statistics appears as an attractive potential for Bose–Ein ...
11B Rotation
... the edge of a wheel free to rotate. The force acts for 0.002 s. What is the final angular velocity? I = mR2 = (2 kg)(0.4 m)2 I = 0.32 kg m2 ...
... the edge of a wheel free to rotate. The force acts for 0.002 s. What is the final angular velocity? I = mR2 = (2 kg)(0.4 m)2 I = 0.32 kg m2 ...