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.
DEPENDS if a particle is in an eigenstate of an observable, I can predict the outcome of a measurement of that observable precisely. The Heisenberg Uncertainty principle means that nothing can be measured precisely. 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. Problem 4) Prove the triangle inequality V W V W for arbitrary vectors in any vector space with an inner product. Answ.: Since both sides are clearly positive, it is sufficient to show V 2 ( V W ) 2 V V V 2 W V V W W V W W V V V W W V 2 Re ( V W V 2 V W W W The last line follows from the Schwarz Inequality since the real part of any complex number is less or equal to its absolute value. Prove that the matrix isin cos cos (interchanging columns and rows and taking the Answ.: The adjoint matrix is cos complex conjugate). Find its eigenvalues and eigenvectors Answ.: The eigenvalues are 0, (solutions to the characteristic equation 0). cos(1) 0 isin(1) ( ( 0 1 0 ( (( isin(1) 0 cos(1) ( Problem 2) Show that (ax b) 1 b (x ) evaluating a a f (ax b)dx for an arbitrary function f(x). Answ.: We show that we get the same result after integration for both forms: !
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.