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Interaction of Mechanics and Mathematics
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Lecture notes for UCSD PHY 110A/B and PHY 200A
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Solid mechanics is a collection of physical laws, mathematical techniques, and computer algorithms that can be used to predict the behavior of a solid material that is subjected to mechanical or thermal loading. This book describes how solid mechanics can be used to solve practical problems." The remainder of the book contains a more detailed description of the physical laws that govern deformation and failure in solids, as well as the mathematical and computational methods that are used to solve problems involving deformable solids.
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1. AC = ACDB + ACBD, AD = ACBD + ABCD, CB = CDAB + CADB, and DB = CADB+ACDB. Therefore −AC+AD−CB+ DB = −ACDB + ABCD − CDAB + ACDB = ABCD − CDAB = [AB, CD] In preparing this solution manual, I have realized that problems 2 and 3 in are misplaced in this chapter. They belong in Chapter Three. The Pauli matrices are not even defined in Chapter One, nor is the math used in previous solution manual. – Jim Napolitano 2. (a) Tr(X) = a 0 Tr(1) + ï¿¿ ï¿¿ Tr(σ ï¿¿)a ï¿¿ = 2a 0 since Tr(σ ï¿¿) = 0. Also Tr(σ k X) = a 0 Tr(σ k) + ï¿¿ ï¿¿ Tr(σ k σ ï¿¿)a ï¿¿ = 1 2 ï¿¿ ï¿¿ Tr(σ k σ ï¿¿ + σ ï¿¿ σ k)a ï¿¿ = ï¿¿ ï¿¿ δ kï¿¿ Tr(1)a ï¿¿ = 2a k. So, a 0 = 1 2 Tr(X) and a k = 1 2 Tr(σ k X). (b) Just do the algebra to find a 0 = (X 11 + X 22)/2, a 1 = (X 12 + X 21)/2, a 2 = i(−X 21 + X 12)/2, and a 3 = (X 11 − X 22)/2. 3. Since det(σ · a) = −a 2 z − (a 2 x + a 2 y) = −|a| 2 , the cognoscenti realize that this problem really has to do with rotation operators. From this result, and (3.2.44), we write det ï¿¿ exp ï¿¿ ± iσ · ˆ nφ 2 ï¿¿ï¿¿ = cos ï¿¿ φ 2 ï¿¿ ± i sin ï¿¿ φ 2 ï¿¿ and multiplying out determinants makes it clear that det(σ · a ï¿¿) = det(σ · a). Similarly, use (3.2.44) to explicitly write out the matrix σ · a ï¿¿ and equate the elements to those of σ · a. Withˆn in the z-direction, it is clear that we have just performed a rotation (of the spin vector) through the angle φ. 4. (a) Tr(XY) ≡ ï¿¿ a ï¿¿a|XY |aï¿¿ = ï¿¿ a ï¿¿ b ï¿¿a|X|bï¿¿ï¿¿b|Y |aï¿¿ by inserting the identity operator. Then commute and reverse, so Tr(XY) = ï¿¿ b ï¿¿ a ï¿¿b|Y |aï¿¿ï¿¿a|X|bï¿¿ = ï¿¿ b ï¿¿b|Y X|bï¿¿ = Tr(Y X). (b) XY |αï¿¿ = X[Y |αï¿¿] is dual to ï¿¿α|(XY) † , but Y |αï¿¿ ≡ |βï¿¿ is dual to ï¿¿α|Y † ≡ ï¿¿β| and X|βï¿¿ is dual to ï¿¿β|X † so that X[Y |αï¿¿] is dual to ï¿¿α|Y † X †. Therefore (XY) † = Y † X †. (c) exp[if (A)] = ï¿¿ a exp[if (A)]|aï¿¿ï¿¿a| = ï¿¿ a exp[if (a)]|aï¿¿ï¿¿a| (d) ï¿¿ a ψ * a (x ï¿¿)ψ a (x ï¿¿ï¿¿) = ï¿¿ a ï¿¿x ï¿¿ |aï¿¿ * ï¿¿x ï¿¿ï¿¿ |aï¿¿ = ï¿¿ a ï¿¿x ï¿¿ï¿¿ |aï¿¿ï¿¿a|x ï¿¿ ï¿¿ = ï¿¿x ï¿¿ï¿¿ |x ï¿¿ ï¿¿ = δ(x ï¿¿ï¿¿ − x ï¿¿) 5. For basis kets |a i ï¿¿, matrix elements of X ≡ |αï¿¿ï¿¿β| are X ij = ï¿¿a i |αï¿¿ï¿¿β|a j ï¿¿ = ï¿¿a i |αï¿¿ï¿¿a j |βï¿¿ *. For spin-1/2 in the | ± zï¿¿ basis, ï¿¿+|S z = ¯ h/2ï¿¿ = 1, ï¿¿−|S z = ¯ h/2ï¿¿ = 0, and, using (1.4.17a), ï¿¿±|S x = ¯ h/2ï¿¿ = 1/ √ 2. Therefore |S z = ¯ h/2ï¿¿ï¿¿S x = ¯ h/2|. = 1 √ 2 ï¿¿ 1 1 0 0 ï¿¿ 6. A[|iï¿¿ + |jï¿¿] = a i |iï¿¿ + a j |jï¿¿ ï¿¿ = [|iï¿¿ + |jï¿¿] so in general it is not an eigenvector, unless a i = a j. That is, |iï¿¿ + |jï¿¿ is not an eigenvector of A unless the eigenvalues are degenerate.
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