*To learn more or modify/prevent the use of cookies, see our Cookie Policy and Privacy Policy.*

*To learn more or modify/prevent the use of cookies, see our Cookie Policy and Privacy Policy.*

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).

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.

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.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.PHYSICS 621 Fall Semester 2012 ODU Problem 5) d 2 y(x) m 2 y(x) 0 for real m. Problem 3) Consider the two vectors A 4 and B 6 in the space of the plane. Problem 4) Prove the triangle inequality V W V W for arbitrary vectors in any vector space with an inner product. Problem 5) Assume the two operators and are Hermitian. However, they are neither normalized nor orthogonal to each other.Make sure you find the most general 2 dx solution what are the Solve the differential equation Problem 6) Proof that for any complex numbers c, z with z exp ( c ) we have exp ( c Problem 7) Find the Fourier transform ( p) of the function 2 1 exp x : 2 1 ( p) f f (x) ( ) PHYSICS 621 Fall Semester 2012 ODU c. To turn them into an orthonormal set, first we have to normalize the first one: A 4 0.8 . 1 0 0 1 isin cos cos isin cos i cos sin sin 2 cos2 q.e.d. 0 The corresponding normalized eigenvectors are 0 , 1 , 0 . Calculate exp Answ.: 1 0 0 0 0 i 0 0 0 0 0 1 2 3 0 1 0 , 0 0 0 , 0 0 0 , 0 0 0 0 0 1 i 0 0 0 0 0 0 and from there on, it repeats.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: !V tells us how strongly the two states are as you can see looking at the equation, if you start out in either basis state, over time you will into the other one and 1) Write down the Hamiltonian in matrix form 2) Find both eigenvalues and (normalized) eigenstates of the Hamiltonian 3) The latter are the stable configurations of the system. How can they be distinguished, and what do you think will be the ground state of ammonia?4) Comment on the behavior of these two eigenstates under the Parity operation (which, in this case, simply interchanges with Are they eigenstates of the operation?Answ.: We show that we get the same result after integration of both sides: f (x x dx f (x f (x) (x x dx f (x x f (x x dx 0 dx f dx f f (x where the first line follows from the definition of the delta function. does it mean PHYSICS 621 Fall Semester 2013 ODU Graduate Quantum Mechanics Problem Set 4 Solution Problem 1) Assuming a particle is described with the usual cartesian coordinates (x,y,z) and momenta (px, py, pz).In the 2nd line, we make use of the fact that any function f(x) must converge to zero at and the definition PHYSICS 621 Fall Semester 2013 ODU Graduate Quantum Mechanics Problem Set 4 Problem 1) Assuming a particle is described with the usual cartesian coordinates (x,y,z) and momenta (px, py, pz). Problem 3) Use the vector potential representation of a constant magnetic field B along the from our first homework problem set. 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?

## Comments Quantum Mechanics Solved Problems