Quantum Mechanics Solved Problems

Tags: View On America EssayResearch Paper On Green MarketingArgumentative Essay On Police BrutalityDrunk Driving EssayThesis On Icecream AdvertisingDescriptive Essay Dominant ImpressionFree Business Plan Software Uk

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.

By continuing to use this site, you consent to the use of cookies.

We use cookies to offer you a better experience, personalize content, tailor advertising, provide social media features, and better understand the use of our services.CORRECT Problem 5) The most general solution is y(x) A exp(mx) B which can be shown plugging it in (as a 2nd order differential equation, there must be two integration constants, A and B). Problem 5) Assume the two operators and are Hermitian. 1 b 1 f (ax b)dx f (x) a (x a )dx a b f( ) a where we integrated the r.h.s.Since and we can solve for A and B in terms of the initial conditions at Problem 6) z exp(c) exp(Re(c) i Im(c)) exp(Re(c)) ( cos(Im(c)) exp(Re(c)) ( cos(Im(c)) Im(c)) Problem 7) See next recitation PHYSICS 621 Fall Semester 2013 ODU Graduate Quantum Mechanics Problem Set 2 Problem 1) Do continuous functions defined on the interval and that vanish at the end points x 0 and x L form a vector space? How about all functions with If the functions do not qualify, list the things that go wrong. What can you say about i) Answ.: The product is not necessarily Hermitian since the 2 operators necessarily commute: PHYSICS 621 Fall Semester 2013 ODU ii) iv) Answ.: This is Hermitian: , Answ.: This is not Hermitian (unless the commutator is zero). following the standard rules for the For the left hand side, we make a variable substitution: u ax, du adx.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) The ammonia molecule NH3 has two different possible configurations: One (which we will call 1 ) where the nitrogen atom is located above the plane spanned the three H atoms, and the other one (which we will call 2 ) where it is below.(These two states span the Hilbert space in our simple example).In both states, the expectation value of the energy n H n is the same, E (n 1,2).On the other hand, the two states are not eigenstates of the in fact, we have 2 H 1 1 H 2 (where V is some positive value).the rate of change F of the momentum: Problem 2) r Show that a vector potential given (in cylindrical coordinates) z) b corresponds to a 2 constant magnetic field B A b along the (You may use the formula sheet) Problem 3) What is the force of a magnetic field of 0.1 Tesla in on an electron instantaneously moving along the with of the speed of light? What kind of motion does it describe (give all the numeric parameters, e.g. Problem 4) For each of the following statements about Quantum Mechanics, indicate whether you believe them to be correct or wrong.Give a sentence explanation for each of your responses: a.


Comments Quantum Mechanics Solved Problems

The Latest from uralmso.ru ©