Solutions to PC4130 AY0809: Quantum Mechanics Problems, Exams of Quantum Mechanics

Download Solutions to PC4130 AY0809: Quantum Mechanics Problems and more Exams Quantum Mechanics in PDF only on Docsity! Solutions to PC4130 AY0809 Paper 1(i) Dynamical phase,  t nn dttEt 0 ')]'([ 1 )(   1(ii) Let  n ti nn nettct )()()()(  , put into Schrodinger equation                 n n ti nn i nnn n nnn n ti nn n ti nn nn nn eHtceic dt d cci etctHetc t i ttHt t i )( )()( )()( )()())(()()( )())(()(           Since  t nn dttEt 0 ')]'([ 1 )(   ,   )( nn E            n in n n i nn n i nnn n i nnn n in nnn nn nnn e dt d cec eEtceEce dt d cci          )( Taking inner product with m ,     n in mnm n in mn i m n in mn n i nmn mnnm nn e dt d cce dt d cec e dt d cec )(       Reexpress dt d n m  : From )()()()(  nnn EH  , dt d EE dt d E dt dH dt d E dt dE dt d H dt dH n mnmnn n mmnm nnn n nn   )()()( )( )()()( )( )()()( )(    For nm  , mn nm n m EE dt dH dt d     ( )( )(       mn mn nm i n m mmm EE dt dH ec dt d cc mn   ( )( )( )( For adiabatic approximation, assume 0i 0 )()(  i i id dH dt dH                     t m mmm m mm m m mm mn mn nm i n m mmm dt dt d ctc dt d c dt dc dt d c EE dt dH ec dt d cc mn 0 )( ' ' exp)0()( ( )( )(      Express )()0()( timm mectc  ,  t m mm dtdt d it 0 ' ' )(  2 2)( cxxV  (even) To get first excited state, let 2 )( bxAxex    bbA bb AdxexAdxx bx 2 4 12 !1 !2 8 1 8 )( 222222 2       )64()( )2()( )( 2 )( 2 32 2 2 2 2 222 222 bxxbAex dx d ebxAAex dx d xV dx d m xV m p H bxbxbx      m b bb b bb bA m dxebxxbA mm p bx 2 3 4 8 1 8 62 !2 !4 8 1 8 4 2 )64( 22 22 22 2 22422 22 2                        b c bb AcdxexAcV bx 4 3 2 !2 !4 8 1 8 2 2242 2            m c mc cmc m H b c db Hd mc b b c mdb Hd b c m b H 2 3 2 4 3 22 3 (minimum) 0 2 3 2 0 4 3 2 3 4 3 2 3 2 2 2 3 2 22 2 2           3(i)    02 )()sin( 2 ),( rVrrdr m f     , ) 2 sin(2  k )()( 0 arVrV  