A quick glance at quantum cryptography by Lomonaco S.J.

March 9, 2017 | Quantum Physics | By admin | 0 Comments

By Lomonaco S.J.

The hot program of the foundations of quantum mechanics to cryptography has ended in a striking new measurement in mystery communique. because of those new advancements, it truly is now attainable to build cryptographic conversation structures which realize unau- unauthorized eavesdropping should still it happen, and which provide a warrantly of no eavesdropping may still it no longer ensue.

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Nz = 0, ±1, . . 1) obey the following orthonormality relation: d3 xϕ∗k (x)ϕk (x) = δk,k . 3) In order to represent the Hamiltonian in second-quantized form, we need the matrix elements of the operators that it contains. 4a) and the matrix element of the single-particle potential is given by the Fourier transform of the latter: ϕ∗k (x)U (x)ϕk (x)d3 x = 1 Uk −k . 5a) and also its inverse V (x) = 1 V Vq eiq·x . 5b) 26 1. Second Quantization For the matrix element of the two-particle potential, one then finds p , k | V (x − x ) |p, k 1 = 2 d3 xd3 x e−ip ·x e−ik ·x V (x − x )eik·x eip·x V 1 = 3 Vq d3 x d3 x e−ip ·x−ik ·x +iq·(x−x )+ik·x +ip·x V q = 1 V3 Vq V δ−p +q+p,0 V δ−k −q+k,0 .

B) For spin- 21 particles, determine, in the momentum representation, the spindensity operator, S(x) = N X δ(x − xα )Sα , α=1 in second quantization. 7 Consider electrons on a lattice with the single-particle wave function localized at the lattice point Ri given by ϕiσ (x) = χσ ϕi (x) with ϕi (x) = φ(x − Ri ). A Hamiltonian, operaPN H = T + V , consisting of a spin-independent P single-particle (2) 1 (xα , xβ ) can be tor T = α=1 tα and a two-particle operator V = 2 α=β V represented in the basis {ϕiσ } by XX 1 X X H= tij a†iσ ajσ + Vijkl a†iσ a†jσ alσ akσ , 2 i,j,k,l i,j σ σ,σ where the matrix elements are given by tij = i | t | j and Vijkl = ij | V (2) | kl .

IN = S− |i α j|α |i1 , i2 , . . , iN α α ˛ = nj (1 − ni )S− |i1 , i2 , . . , iN ˛j→i . The symbol |j→i implies that the state |j is replaced by P |i . In order to bring the P i into the right position, one has to carry out k j. 20 1. 15): a†i aj |. . , ni , . . , nj , . . = nj (−1) = ni (1 − ni )(−1) P kj nk a†i |. . , ni , . . , nj − 1, . . |. . , ni + 1, . . , nj − 1, .

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