NRQM10midsol

NRQM10midsol - Physics 216 Solutions to the Midterm Exam...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Physics 216 Solutions to the Midterm Exam Spring 2010 1. Suppose we define G ( t ) integraldisplay dxK ( x, t, ; x, 0) (1) where K ( x, t ; x , t ) is the propagator. Assume that the system has a time-independent Hamiltonian and a discrete energy level spectrum. (a) Prove that the Fourier transform of G , tildewide G ( E ) lim i planckover2pi1 integraldisplay G ( t ) e iEt/ planckover2pi1 e t dt (2) has poles at all the discrete energy levels of the system. Take to be a positive infinitesimal quantity. Assuming that the Hamiltonian is time-independent, the time evolution operator, defined by | ( t ) ) = U ( t, t ) | ( t ) ) , is given by U ( t, t ) = e iH ( t t ) / planckover2pi1 . The propagator is defined by K ( x, t ; x , t ) = ( x | U ( t, t ) | x ) . We also assume that the spectrum of H consists of a discrete set of energy eigenvalues. Denoting the corresponding energy eigenstates by | n ) , K ( x, t ; x , t ) = summationdisplay n ( x | e iH ( t t ) / planckover2pi1 | n )( n | x ) = summationdisplay n e iE n ( t t ) / planckover2pi1 ( x | n )( n | x ) , where H | n ) = E n | n ) . Using G ( t ) = integraldisplay dxK ( x, t ; x, 0) = summationdisplay n e iE n t/ planckover2pi1 integraldisplay dx |( n | x )| 2 = summationdisplay n e iE n t/ planckover2pi1 , (3) since the last integral is equal to 1 by the normalization condition of the energy eigenstates. Hence, using eq. (3) it follows that for any positive infinitesimal , tildewide G ( E ) lim i planckover2pi1 integraldisplay G ( t ) e iEt/ planckover2pi1 e t dt = lim i planckover2pi1 summationdisplay n integraldisplay e i ( E E n ) t/ planckover2pi1 e t dt = summationdisplay n 1 E n E , which has poles precisely at the discrete energy levels of the system. 1 (b) Consider the harmonic oscillator with a charge e in one dimension. You are asked to calculate the discrete energy levels of the system, when it is placed in a uniform electric field of strength E . Compute the levels in three independent ways: (i) using the WKB approximation; (ii) using the path integral technique; (iii) solving the Schrodinger equation exactly. Do you expect the same result from all three methods? The Lagrangian for the system is L = 1 2 m x 2 1 2 m 2 x 2 + e E x. (i) To compute the energy levels in the WKB approximation, we must compute the turning points of the potential, V ( x ) = 1 2 m 2 x 2 e E x. (4) The turning points are determined by setting V ( x ) = E , where E is the energy, and solving for x . The resulting equation is: x 2 2 e E x m 2 eE m 2 = 0 ....
View Full Document

Page1 / 11

NRQM10midsol - Physics 216 Solutions to the Midterm Exam...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online