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: PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 4: Solutions 1. The 2Level Rabi Model: The standard Rabi Model consists of a bare Hamiltonian H = 2 (  2 )( 2   1 )( 1  ) and a coupling term V = * 2  1 )( 2  + 2  2 )( 1  . (a) What is the energy, degeneracy, and state vector of the bare ground state for > 0, = 0, and < 0? For > 0, the energy of the ground state is / 2, the degeneracy is 1, and the state vector is  1 ) . For = 0, the energy of the ground state is 0, the degeneracy is 2 and the degenerate subspace is { 1 ) ,  2 )} . For < 0, the energy of the ground state si / 2 =   / 2, the degeneracy is 1, and the state vector is  2 ) . (b) Let the full Hamiltonian be H = H + V . Write down the 2x2 Hamiltonian matrix in the { 1 ) ,  2 )} basis and then compute the dressedstate energy levels for the case negationslash = 0. Use g for the lowest eigenvalue, and e for the highest (in energy). The matrix representation of H in the { 1 ) ,  2 )} basis is: H = parenleftbigg 2 * 2 2 2 parenrightbigg (1) The characteristic equation is then: det  H I  = parenleftbigg 2 + parenrightbiggparenleftbigg 2 parenrightbigg   2 4 = 2 1 4 ( 2 +   2 ) = 0 (2) the solutions are then g = 1 2 radicalbig 2 +   2 (3) e = 1 2 radicalbig 2 +   2 (4) (c) Following the method shown in lecture (i.e. treating positive and negative detunings separately, and matching the limiting values of the dressed and bare eigenstates in the limits   ), determine the normalized dressedstate eigenvectors. Label the state corresponding to g as  g ) and the other state as  e ) . Using Dirac notation, express the Full Hamiltonian as an operator in terms of the kets  g ) and  e ) and the corresponding bras, and then again using the kets  1 ) and  2 ) and the corresponding bras. The eigenvalue equation is ( H I )  ) = 0. Hitting this with ( 1  and inserting the projector I =  1 )( 1  +  2 )( 2  , then doing the same for ( 2  , gives ( ( 1  H  1 ) ) ( 1  ) + ( 1  H  2 )( 2  ) = 0 (5) ( 2  H  1 )( 1  ) + ( ( 2  H  2 ) ) ( 2  ) = 0 (6) putting in the matrix elements and multiplying by 2 gives ( + 2 ) ( 1  ) + * ( 2  ) = 0 (7) ( 1  ) + ( 2 ) ( 2  ) = 0 (8) The first equation gives, before normalization,  ) = *  1 ) + ( + 2 )  2 ) . (9) The second gives,  ) = ( 2 )  1 )  2 ) . (10) 1 For positive detuning, > 0, according to our answer for (a), we want lim ( 2  g ) = 0 and lim ( 1  e ) 0, thus we should use (10) for  g ) and (9) for  e ) . This gives  g ) = ( + radicalbig 2 +   2 )  1 )  2 ) radicalBig ( + radicalbig 2 +   2 ) 2 +   2 (11)  e ) = *  1 ) + ( + radicalbig 2 +   2 )  2 ) radicalBig...
View Full
Document
 Spring '11
 Smith
 Physics, Energy, Work

Click to edit the document details