By Sandibek B. Nurushev, Mikhail F. Runtso, Mikhail N. Strikhanov

This e-book is dedicated to the polarization (spin) physics of excessive strength debris and includes 3 components. the 1st half provides the theoretical prefaces of polarization within the particle physics for interpretations, predictions and bases for knowing the next components. the second one a part of the ebook offers the outline of the basic polarization experiments together with the hot ones. This half is dedicated to the cutting edge instrumentations, provides the parameters of the polarized beams, ambitions, polarized gasoline jets and polarimeters. The 3rd a part of the publication concentrates at the very important achievements in polarization physics. The publication can be utilized in lectures on nuclear and particle physics and and nuclear tools and strategies. As supplementary examining this publication comes in handy for researchers operating in particle and nuclear physics.

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**Example text**

82) where the vectors A and B are independent of the spin variables. As can be verified by direct transformation of the Pauli matrices, the operator σˆ is Hermitian; therefore, its eigenvalue is a real number. The mean value of the operator σˆ between the spin states of the particle is called the polarization vector P : P = σˆ . 84) Since the eigenvalue of the square of the spin operator is s 2 = 14 · σ 2 = 34 . At the same time, it can be shown that the length of the polarization vector P is always smaller than unity, |P | ≤ 1 (Bilen’kii et al.

233)); if the left-hand side is vector, then the righthand side should also be vector with the same property under space inversion: if one side is covariant, the other side should also be covariant. If the left-hand side of the equation has a free index in a certain position (upper or lower), then the righthand side of the equation should have the same index in the same position. Some examples illustrating these statements are given below. 46 1 Spin and Its Properties The examples of correctly written equations are aμ bμ = c, aμ b μ cλ = dλ , aμ bμ = D λρ cλ kρ .

129) Two remaining equations contain mixed amplitudes a2 and a3 . The sum and difference of the second and third of Eq. 130) E(a2 − a3 ) = −3A(a2 − a3 ). 131) and These two equations have two solutions a2 = a 3 , E = A and a2 = −a3 , E = −3A. 133) EIV = −3A. , are degenerate. The state |IV has the energy −3A, and, as a 30 1 Spin and Its Properties result, the sum of the energies of four states is zero. This corresponds to our choice E0 = 0 in the definition of the Hamiltonian Hss . The energy A remains indefinite.