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

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Unformatted text preview: Physics 216 Solutions to the Midterm Exam Spring 2010 1. Suppose we define G ( t ) ≑ integraldisplay ∞ βˆ’βˆž dxK ( x, t, ; x, 0) (1) where K ( x, t ; x β€² , t β€² ) 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 Η« β†’ i planckover2pi1 integraldisplay ∞ G ( t ) e iEt/ planckover2pi1 e βˆ’ Η«t dt (2) has poles at all the discrete energy levels of the system. Take Η« to be a positive infinitesimal quantity. Assuming that the Hamiltonian is time-independent, the time evolution operator, defined by | ψ ( t ) ) = U ( t, t ) | ψ ( t ) ) , is given by U ( t, t ) = e βˆ’ iH ( t βˆ’ t ) / planckover2pi1 . The propagator is defined by K ( x, t ; x β€² , t β€² ) = ( x β€² | 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 β€² , t β€² ) = summationdisplay n ( x β€² | e βˆ’ iH ( t βˆ’ t β€² ) / planckover2pi1 | n )( n | x ) = summationdisplay n e βˆ’ iE n ( t βˆ’ t β€² ) / planckover2pi1 ( x β€² | n )( n | x ) , where H | n ) = E n | n ) . Using G ( t ) = integraldisplay ∞ βˆ’βˆž dxK ( x, t ; x, 0) = summationdisplay n e βˆ’ iE n t/ planckover2pi1 integraldisplay ∞ βˆ’βˆž dx |( n | x )| 2 = summationdisplay n e βˆ’ 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 Η« , tildewide G ( E ) ≑ lim Η« β†’ i planckover2pi1 integraldisplay ∞ G ( t ) e iEt/ planckover2pi1 e βˆ’ Η«t dt = lim Η« β†’ i planckover2pi1 summationdisplay n integraldisplay ∞ e i ( E βˆ’ E n ) t/ planckover2pi1 e βˆ’ Η«t dt = summationdisplay n 1 E n βˆ’ 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 Λ™ x 2 βˆ’ 1 2 mΟ‰ 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 mΟ‰ 2 x 2 βˆ’ 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 βˆ’ 2 e E x mΟ‰ 2 βˆ’ eE mΟ‰ 2 = 0 ....
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