This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Quantum Physics 2008/09 Solutions to Tutorial Sheet 7: Time Dependent Perturbations and Radiative Transitions For the first two problems, you may assume that the hydrogen eigenfunctions are: u 100 = ( a 3 ) 1 / 2 exp( r/a ) u 211 = a 3  1 / 2 r 8 a sin exp ( i ) exp ( r/ 2 a ) u 210 = 8 a 3  1 / 2 r 2 a cos exp ( r/ 2 a ) u 21 1 = a 3  1 / 2 r 8 a sin exp ( i ) exp ( r/ 2 a ) and Z exp( br ) r n d r = n ! /b n +1 , n > 1 1. A hydrogen atom is placed in a uniform but timedependent electric field of magnitude E = 0 for t < , E = E exp( t/ ) for t ( > 0) where E is a constant. At time t = 0 , the atom is in the ground ( 1 s ) state. Show that the probability, to lowest order in perturbation theory, that as t , the atom is in the 2 p state in which the component of the orbital angular momentum in the direction of the field is zero, is given by p 1 s 2 p =  c ( )  2 = 2 15 3 10 ( e E a ) 2 ( E 2 p E 1 s ) 2 + ( h/ ) 2 [ Hint: take the field direction to be the zdirection. Write down the potential energy of the electron in the given field and treat as a timedependent perturbation]. We will use timedependent perturbation theory. Since the field decays away exponen tially there is only a finite probability that the transition will occur. The approximation from TDPT is that there is no probablity of transition to 2p and back again. We take the direction of the electric field to be the zdirection, as suggested. Then the perturbation for t 0 is H ( t ) = e (E z exp( t/ )) and the transition probability amplitude is c 1 s 2 p ( t ) = ( i h ) 1 t Z 210  H ( t )  100 exp( it ) d t where = ( E 2 p E 1 s ) / h . Using the given eigenfunctions, the required matrix element is 210  H  100 = e E exp( t/ ) 210  r cos  100 = e E exp( t/ ) a 4 4 2 2 Z =0 Z =0 Z r =0 r 4 exp( 3 r/ 2 a ) d r cos 2 sin d d The integration just gives 2 , whilst the integration yields +1 Z 1 cos 2 d(cos ) = " cos 3 3 # +1 1 = 2 3 The radial integral is Z r 4 exp( 3 r/ 2 a ) d r = 2 a 3 5 4! from the given integral. Putting it all together 210  H ( t )  100 = A exp( t/ ) where A = 2 15 / 2 3 5 e E a Thus c 1 s 2 p ( t ) = A i h t Z exp( t / + it ) d t = A i h " exp( it t/ ) 1 i 1 / # Thus the desired probability is p 1 s 2 p =  c ( )  2 = 2 15 3 10 ( e E a ) 2 ( E 2 p E 1 s ) 2 + ( h/ ) 2 If we had numbers, for we should now check that this probability is indeed small  if so then we justify our use of TDPT, if not then we know we should solve the problem exactly....
View
Full
Document
This note was uploaded on 11/21/2010 for the course PHYSICS 137B taught by Professor Moore during the Spring '07 term at University of California, Berkeley.
 Spring '07
 MOORE
 mechanics, Quantum Physics

Click to edit the document details