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PHYSICS 116A Homework 7 Solutions 1. Boas, Ch. 3, §6, Qu. 6

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PHYSICS 116A Homework 7 Solutions 1. Boas, Ch. 3, 6, Qu. 6. § The Pauli spin matrices in quantum mechanics are 0 1 0 i 1 0 A = , B = − , C = . 1 0 i 0 0 1 − Show that A2 = B2 = C2 = I (the unit matrix) Also show that any of these two matrices anticommute, that is AB = BA, etc. Show that the commutator of A and B is 2iC and − similarly for the other pairs in cyclic order. 2 0 1 0 1 0 0 + 1 1 0 1 + 1 0 1 0 A = = · · · · = , 1 0 1 0 1 0 + 0 1 1 1 + 0 0 0 1 · · · · 2 0 i 0 i 0 0+( i) i 0 ( i)+( i) 0 1 0 B = − − = · − · · − − · = , i 0 i 0 i 0 + 0 i i ( i) + 0 0 0 1 · · · − · 1 0 1 0 1 0 C2 = = , 0 1 0 1 0 1 − − 0 1 0 i 0 0 + 1 i 0 ( i) + 1 0 i 0 AB = − = · · · − · = , 1 0 i 0 1 0 + 0 i 1 ( i) + 0 0 0 i · · · − · − 0 i 0 1 i 0 BA = − = − = AB, i 0 1 0 0 i − 2i 0 AB BA = = 2iC, − 0 2i − 0 i 1 0 0 i BC = − = , i 0 0 1 i 0 − 1 0 0 i 0 i CB = − = − = BC, 0 1 i 0 i 0 − − − 0 2i BC CB = = 2iA, − 2i 0 1 0 0 1 0 1 CA = = , 0 1 1 0 1 0 − − 0 1 1 0 0 1 AC = = − = CA, 1 0 0 1 1 0 − − 0 2 CA AC = = 2iB. − 2 0 − 2. Boas, Ch. 3, 6, Qu. 7. § Find the matrix product of 1 4 1 2 3 − − . 2 1 2 − By evaluating this in two ways, verify the associative law for multiplication, A(BC)=(AB)C, which justifies our writing it as ABC (i.e. without brackets). 1 Writing 1 4 1 A = 2 3 , B = − , C = − 2 1 2 − we have 1 4 AB = 2 3 − = 4 5 , 2 1 − so 1 (AB)C = 4 5 − = 6. 2 Similarly 1 4 1 9 BC = − − = , 2 1 2 4 − − so 9 A(BC)= 2 3 = 6, 4 − =(AB)C. 3. Boas, problem 3.6–16. Find the inverse of 2 0 1 − A = 1 1 2 − 3 1 0 by employing eq. (6.13) on p. 119 of Boas and also by using Gauss-Jordan elimination. Eq. (6.13) pm p. 119 of Boas is T 1 C A− = , (1) det A where CT is the transpose of the matrix whose elements are the cofactors of A. First we compute the determinant of A using the expansion of cofactors based on the elements of the third column 2 0 1 − 1 1 2 0 det A = 1 1 2 = (1) − (2) − = (1)(4) (2)( 2) = 8 . − 3 1 − 3 1 − − 3 1 0 Next, we evaluate the matrix of cofactors and take the transpose: 1 2 0 1 0 1 − 1 0 − 1 0 1 2 − 2 1 1 − T 1 2 2 1 2 1 C = − − = 6 3 5 . − 3 0 3 0 − 1 2 − 4 2 2 1 1 2 0 2 0 − − − 3 1 − 3 1 1 1 − 2 The inverse of A is given by Eq. (1), so it follows that 2 1 1 − 1 1 A− = 6 3 5 . (2) 8 − 4 2 2 You should check this calculation by verifying that AA 1 = I, where I is the 3 3 identity matrix. − × Next we solve the problem by Gauss-Jordan elimination in which we perform the same row oper- ations on the matrix A and the inverse matrix I. This proceeds as follows: 2 0 1 1 0 0 − 1 1 2 0 1 0 − 3 1 0 0 0 1 R(3) R(3) 3 R(1) 2 0 1 1 0 0 + 2 − → 1 1 2 0 1 0 −−−−−−−−−−−→ − 3 3 0 1 2 2 0 1 1 R(2) R(2)+ R(1) 2 0 1 1 0 0 → 2 − 5 1 0 1 2 2 1 0 −−−−−−−−−−−→ − 3 3 0 1 2 2 0 1 2 0 1 1 0 0 R(3) R(3)+R(2) − 5 1 → 0 1 1 0 −−−−−−−−−−→ − 2 2 0 0 4 2 1 1 R(3) 1 R(3) 2 0 1 8 0 0 4 − 5 1 → 0 1 4 8 0 −−−−−−−−→ − 2 8 0 0 1 4 2 2 R(2) R(2) 5 R(3) 2 0 1 8 0 0 2 − 1 → − 0 1 0 6 3 5 −−−−−−−−−−−→ − 8 − − 0 0 1 4 2 2 2 0 1 8 0 0 R(2) R(2) − 1 →− 0 1 0 6 3 5 −−−−−−−−→ 8 − 0 0 1 4 2 2 2 0 0 4 2 2 R(1) R(1) R(2) − 1 − − → − 0 1 0 6 3 5 −−−−−−−−−−→ 8 − 0 0 1 4 2 2 R(1) 1 R(1) 1 0 0 2 1 1 2 1 − →− 0 1 0 6 3 5 . (3) −−−−−−−−−→ 8 − 0 0 1 4 2 2 The left hand matrix, which started as A, has been transformed into the identity matrix, so the 1 right hand matrix, which started as the identity matrix, has been transformed into A− . Thus 2 1 1 1 1 − A− = 6 3 5 , (4) 8 − 4 2 2 3 in agreement with Eq. (2). The Gauss-Jordan method can be efficiently implemented on a com- puter and requires many fewer operations than using Eq. (1) except for small matrices. By hand, the Gauss-Jordan method involves a lot of copying from line to line which is not needed in the computer program. 4. Boas, problem 3.6–21. Solve the following set of equations by the method of finding the inverse of the coefficient matrix. x + 2z = 8 , 2x y = 5 , − − x + y + z = 4 . The coefficient matrix is: 1 0 2 M = 2 1 0 . 1− 1 1 ! 1 One can compute the inverse M − by using 1 1 T M − = C . (5) detM First, we compute det M via the expansion of cofactors. Using the first row, 1 0 2 2 1 0 = (1)( 1) + 2[(2)(1) ( 1)(1)] = 5 . − − − − 1 1 1 Next, we compute the transpose of the matrix of cofactors, 1 0 0 2 0 2 − 1 1 − 1 1 1 0 − 1 2 2 − T 2 0 1 2 1 2 C = = 2 1 4 . − 1 1 1 1 − 2 0 − − 3 1 1 2 1 1 0 1 0 − − − 1 1 − 1 1 2 1 − Thus, Eq. (5) yields: 1 2 2 − 1 1 M − = 2 1 4 . 5 − − 3 1 1 − − Finally, the solution to the set of equations above is: x 1 2 2 8 2 − − 1 y = 2 1 4 5 = 1 . 5 − − − z 3 1 1 4 5 − − That is, the solution is x = 2, y = 1 and z = 5. − 4 0 i iθB 5. Boas, problem 3.6–32. For the Pauli spin matrix, B = i −0 , find e and show that your result is a rotation matrix. Repeat the calculation for e iθB. − First, we observe that: 2 0 i 0 i 1 0 B = − − = = I, i 0 i 0 0 1 where I is the 2 2 identity matrix. Thus, × n n B2 = I, and B2 +1 = B, for any non-negative integer n. By definition, the matrix exponential is defined via its power series, n 2 2 3 3 4 4 5 5 iθB ∞ (iθB) θ B θ B θ B θ B e = = I + iθB i + + i + n=0 n! − 2! − 3! 4! 5! ··· X θ2 θ4 θ3 θ5 = I 1 + + + iB θ + + − 2! 4! ··· − 3! 5! ··· n 2n n 2n+1 ∞ ( 1) θ ∞ ( 1) θ = I − + iB − = I cos θ + iB sin θ n=0 (2n)! n=0 (2n + 1)! X X 1 0 0 1 cos θ sin θ = cos θ + sin θ = , (6) 0 1 1 0 sin θ cos θ − − iθB which is a rotation matrix [cf. Eq.(6.14) on p. 120 of Boas]. The computation of e− follows precisely the same steps, with θ replaced everywhere by θ. Hence, − iθB cos θ sin θ e− = I cos θ iB sin θ = − . − sin θ cos θ 6. Boas, problem 3.7–27. The following matrix, 1 1 1 R − − , ≡ √2 1 1 − represents an active transformation of vectors in the x–y plane (axes fixed, vectors rotated or reflected). Show that the above matrix is orthogonal, find its determinant, and find the rotation angle, or find the line of reflection. The matrix R is orthogonal, since T 1 1 1 1 1 1 0 RR = − − − = = I . 2 1 1 1 1 0 1 − − − In addition, the determinant of R is 1 1 √2 √2 1 1 det R − 1 − 1 = 2 + 2 = 1 , ≡ √2 √2 − 5 which means that R is a proper rotation matrix. Comparing this with the general form for the matrix representation of an active rotation in two-dimensions, cos θ sin θ − . sin θ cos θ we see that sin θ = cos θ = 1/√2, which implies that θ = 3π/4 = 135◦. − 7. Boas, problem 3.7–34. For the matrix, 1 0 0 L = 0 0 1 , − 0 1 0 − find its determinant to see whether it produces a rotation or a reflection. If a rotation, find the axis and angle of rotation. If a reflection, find the reflecting plane and the rotation (if any) about the normal to this plane. First we compute the determinant of L by using the expansion of cofactors based on the elements of the third row.
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