Solution Manual for Physical Mathematics, 2nd Edition by Kevin Cahill.

Solution Manual for Physical Mathematics, 2nd Edition by Kevin Cahill 1 2 3 ik ik ik k 4 4 k ik Solutions to the Exercises 1 Solutions to the Exercises on Linear Algebra 1. What is the most general function of three Grassmann numbers θ1, θ2, θ3? Solution: Grassmann numbers anticommute, that is {θi,θj }= θiθj +θj θi = 0 and θ 2 = θ 2 = θ 2 = 0. So the most general function of three Grassmann numbers is f (θ1, θ2, θ3) = a + b θ1 + c θ2 + d θ3 + e θ1θ2 + f θ1θ3 + g θ2θ3 + h θ1θ2θ3. 2. Derive the cyclicity (1.24) of the trace from Eq.(1.23). Solution: Since Tr(AB) = Tr(B A), it follows with B replaced by BC D that Tr(ABC D) = Tr(DABC) = Tr(CD AB) = Tr(BC DA). 3. Show that (AB) T = B T A T , which is Eq.(1.26). Solution: With a sum over the repeated index 4 understood, (AB) T ik = [(AB)]ki = Ak4 B4i = A T 4k B T i4 = B T i4 A T 4k = B T A T . 4. Show that a real hermitian matrix is symmetric. Solution: Aik = A † = A ∗ ki = Aki. 5. Show that (AB) † = B † A † , which is Eq.(1.29). Solution: With a sum over the repeated index 4 understood, (AB) † ik = (AB)∗T = (AB)∗ ki = Ak ∗ 4 B4 ∗ i = A ∗ 4 T Bi ∗T = Bi ∗T A4 ∗T = B †A † . 6. Show that the matrix (1.41) is positive on the space of all real 2-vectors but not on the space of all complex 2-vectors. 1 2 Solutions to the Exercises ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟θ = = = 0. ⎜0 0 0 −1 0⎟⎜ 0 0 0 0 0 0 −1⎟ ⎜ ⎟ ⎜ ⎝ ⎠ ⎝ ⎠ −1 1 b −1 1 b 0 0 0 0 0 0 0 0 i i 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 Solution: a b 1 1 a = a 2 + b 2 is positive if a and b are both real (and nonzero), but a¯ b¯ 1 1 a = a 2 + b 2 + a¯b − b¯a is complex when the imaginary part of a¯b ∗= 0. 7. Show that the two 4 × 4 matrices (1.46) satisfy Grassmann’s algebra (1.11) for n = 2. Solution: ⎛ 0 0 1 0 ⎞ ⎛ 0 0 1 0 ⎞ ⎛ 0 0 0 0 ⎞ θ 2 = ⎜0 0 0 −1⎟ ⎜0 0 0 0 0 = = 0. ⎝0 0 0 0 ⎠ ⎝0 ⎠ ⎝0 0 0 0⎠ ⎛ 0 1 0 0 ⎞ ⎛ 0 1 0 0 ⎞ ⎛ 0 0 0 0 ⎞ 2 0 0 0 0 0 0 0 0 0 0 0 0 2 ⎝0 0 0 1⎠ ⎝0 0 0 1⎠ ⎝0 0 0 0⎠ θ1θ2 + θ2θ1 = ⎛ 0 0 1 0 ⎞ ⎛ 0 ⎝0 0 0 0 ⎠ ⎝0 0 ⎞ 1 ⎟ ⎠ ⎛ 0 1 0 0 ⎞ ⎛ 0 0 ⎞ + ⎜ ⎝0 0 0 1 ⎟ ⎠ ⎜ ⎝0 0 ⎠ 0 0 0 0 0 0 ⎛ 0 0 0 1 ⎞ ⎛ 0 −1 ⎞ 0 0 0 0 0 = + ⎟ = 0. And since addition is cummutative, we also have θ2θ1 + θ1θ2 = θ1θ2 + θ2θ1 = 0. 8. Show that the operators ai = θi defined in terms of the Grassmann matrices (1.46) and their adjoints a † = θ † satisfy the anticommutation relations (1.47)