Download Homework 3 Solutions - Quantum Mechanics I | PHYS 622 and more Assignments Quantum Mechanics in PDF only on Docsity! Physics 622 Spring 2000 - Dr. D. Fivel Homework 3 - Solutions Part I - Chapter 1 - Problems 11-18 √ Exercise 11:: Being a sum of projectors with real coefficients ρ is Hermitian. The trace is trivially equal to unity. If |x〉 is any vector then 〈x|ρ|x〉 = ∑ j νj〈x|π(xj)|x〉 = ∑ j νj |〈xj |x〉|2. Evidently this is a positive number. Hence ρ is a positive operator (by definition an operator whose expectation value is positive in every state). Since the coefficients of the νj are between 0 and 1, the sum is bounded by the sum of the νj ’s which is unity. Hence 0 < 〈x|ρ|x〉 ≤ 1. If |x〉 is a unit eigenvector of ρ with eigenvalue λ, then 〈x|ρ|x〉 = λ〈x|x〉 = λ. Hence the eigenvalues of ρ lie between 0 and 1. √ Exercise 12: The endpoints of a diameter of the Poincaré sphere S will be a pair of orthogonal states. Hence a diagonal mixture will correspond to a point P on a diameter D of S. Let C be the circle on the sphere cut by the plane through P perpendicular to D. Clearly the endpoints of any diameter d of C form the constituents of the required balanced mixture. √ Exercise 13: Here is the easy way to do it: Half the mixture is an equal mixture of left and right circularly polarized light and hence is unpolarized. Hence it is represented by the unit operator. The remainder of the mixture is described by the projector π for some linearly polarized state |x〉. Let π′ be the projector for the orthogonal state. Since an unpolarized mixture can be expressed as (1/2)π + (1/2)π′ we see that the mixture as a whole is (1/2)((1/2)π + (1/2)π′) + (1/2)π = (3/4)π + (1/4)π′. This gives the diagonal form. The balanced form can now be gotten by the method of exercise 12. √ Exercise 14:: Let |xj〉 be an eigenbasis of W and λj the corresponding eigenvalues. Then for any unit-vector |x〉 we have 〈x|W |x〉 = ∑ j λjαj (∗ ∗ ∗) HW-3 - 1 where αj = |〈xj |x〉|2. The absolute value of the right side is largest when αj is unity for the j for which |λj | is largest. But then αj must be zero for all the others since the αj ’s have sum unity. But in this case |x〉 coincides with |xj〉, proving the assertion. √ Exercise 15: (π(x) − π(y))2 = π(x) + π(y) − π(x)π(y) − π(y)π(x) (π(x) − π(y))3 = (π(x) − π(y))(π(x) + π(y) − π(x)π(y) − π(y)π(x)) = π(x)+π(x)π(y)−π(x)π(y)−π(x)π(y)π(x)−π(y)π(x)−π(y)+π(y)π(x)π(y)+π(y)π(x) = π(x) − π(y) − π(x)π(y)π(x) + π(y)π(x)π(y). But π(x)π(y)π(x) = |〈x|y〉|2π(x) π(y)π(x)π(y) = |〈x|y〉|2π(y) Hence with p(x, y) = |〈x|y〉|2 we obtain W 3 = (1 − p(x, y))W. Hence if λ is an eigenvalue of W we have: λ3 = λ(1 − p) so the largest absolute value is |1 − p|1/2 as asserted. √ Exercise 16: Let the {|xj〉} be a basis, let U be a unitary operator, and let |x〉 =∑ j αj |xj〉. Then defining |x̃〉 = ∑ j α∗jU |xj〉 and |ỹ〉 = ∑ k β∗kU |xk〉 we have 〈ỹ|x̃〉 = ∑ jk βkα ∗ jδjk = 〈x|y〉. HW-3 - 2
