Quantum Mechanics Solved Problems

Quantum Mechanics cannot predict with certainty the result of any particular measurement on a single particle. A 32 4 2 Then, we determine the orthogonal part of the 2nd vector: 6 j) (1.2 0.8 j) 3.12 B ( B A) As the final step, we have to normalize this vector: B 3.12 0.6 5.2 and form an orthonormal basis. PHYSICS 621 Fall Semester 2013 ODU Graduate Quantum Mechanics Problem Set 3 Solution Problem 1) 0 0 1 Consider the matrix 0 0 0 . Answ.: Yes it is identical to its adjoint (swapping rows and columns). 1 0 1 2 2 Verify that U is diagonal, U being the matrix formed using each normalized eigenvector as one of its columns. ) Answ.: 1 1 1 1 0 0 2 2 2 2 U 0 1 0 0 1 0 U, 1 0 1 1 0 1 2 2 2 2 q.e.d 1 1 1 1 0 0 2 2 2 1 0 0 2 U 0 1 0 0 0 0 0 0 0 1 0 0 1 0 1 1 0 2 2 2 2 iv) 1 n (where, for any matrix, M0 1). Collecting all terms, we find PHYSICS 621 Fall Semester 2013 ODU 1 1 1 0 16 120 2 24 0 1 0 1 1 0 1 1 6 120 2 24 which clearly is a unitary matrix.

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Give a sentence explanation for each of your responses: a.Immediately afterwards, the physical observable corresponding to B is measured, and again immediately after that, the one corresponding to A is remeasured (independently from the result of the 2nd measurement).What is the probability of obtaining a1 a second time?Write down the x, y and z components of the angular momentum operator in terms of these canonical variables. Given our interpretation of Lz as of rotations around the z axis, can you interpret your result in terms of the transformation of the vector L under the coordinate transformation generated Lz? Then follow the explicit procedure (Legendre transformation) in the lecture to find the corresponding Hamiltonian. The Hamiltonian for this case in cylindrical coordinates z) with canonical momenta , , Pz ) is given 2 2 2 r ( Pz 2m writing down equations of motion, give an interpretation (in terms of the momenta or velocities) of , , Pz ) . Answ.: Lx ypz zpy , Ly zpx xpz , Lz xpy ypx L x , Lz 0 0 pz x zpx 0 0 Similarly, , Lz .Problem 2) Write down the Lagrangian for two equal masses m at positions x1 and x2 (each measured relative to the equilibrium position), coupled to each other and (on their other sides) to two fixed walls with springs with constant k but otherwise free to move along the If the system is in equilibrium, all three springs are relaxed is exactly the set up in Example 1.8.6 in book, p. Since Lz is the generator of rotations around the z axis, any change of a variable under such a rotation an infinitesimally small angle is given Lz . PHYSICS 621 Fall Semester 2013 ODU q r 2 b ) 2 ( which is true if either of the two expressions in parantheses is zero. This means that if the charge is momentarily moving only in radial and (no tangential motion), than instantaneously its radial momentum is conserved.With what eigenvalues (called PHYSICS 621 Fall Semester 2013 ODU Graduate Quantum Mechanics Problem Set 5 Solution Problem 1) An operator A, corresponding to a physical observable, has two normalized eigenstates and with eigenvalues a1 and a2, respectively.Immediately afterwards, the physical observable corresponding to B is measured, and again immediately after that, the one corresponding to A is remeasured.Problem 2) Consider the vector space V spanned real 2x2 matrices. Therefore, if a 1 b f (x) (ax b)dx 1 a f ( ua (u b)du a f ( a ) where we use and a for positive a, q.e.d.For negative a, the only change is that the factor in front of the integral becomes negative and therefore equal to However, the limits of integration are multiplied with a negative number at the same time, so that the integral over u would now go from to Changing the order back interchanging upper and lower integration limit yields another minus sign which cancels the first one, yielding the desired result. Show that (x x dx 0 else multiplying both the l.h.s. with an arbitrary function f(x) and integrating over all x.The Heisenberg Uncertainty principle means that nothing can be measured precisely e. cos PHYSICS 621 Fall Semester 2013 ODU Answ.: The 2 vectors are linearly independent (you write one as a multiple of the other).The and components of any angular momentum cannot simultaneously be measured with arbitrary precision. The time evolution of a quantum mechanical wave function is described a unitary operator. If not, follow the construction in the book (p.15) and the lecture to turn them into an othomormal set. Since the space has only 2 dimensions, they therefore form a basis.