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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 = Δ 2 (  2 )( 2  −  1 )( 1  ) and a coupling term V = Ω * 2  1 )( 2  + Ω 2  2 )( 1  . (a) What is the energy, degeneracy, and state vector of the bare ground state for Δ > 0, Δ = 0, and Δ < 0? For Δ > 0, the energy of the ground state is − Δ / 2, the degeneracy is 1, and the state vector is  1 ) . For Δ = 0, the energy of the ground state is 0, the degeneracy is 2 and the degenerate subspace is { 1 ) ,  2 )} . For Δ < 0, the energy of the ground state si Δ / 2 = − Δ  / 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 Ω negationslash = 0. Use ω g for the lowest eigenvalue, and ω e for the highest (in energy). The matrix representation of H in the { 1 ) ,  2 )} basis is: H = parenleftbigg − Δ 2 Ω * 2 Ω 2 Δ 2 parenrightbigg (1) The characteristic equation is then: det  H − ωI  = − parenleftbigg Δ 2 + ω parenrightbiggparenleftbigg Δ 2 − ω parenrightbigg −  Ω  2 4 = ω 2 − 1 4 ( Δ 2 +  Ω  2 ) = 0 (2) the solutions are then ω g = − 1 2 radicalbig Δ 2 +  Ω  2 (3) ω e = 1 2 radicalbig Δ 2 +  Ω  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  Δ  → ∞ ), determine the normalized dressedstate eigenvectors. Label the state corresponding to ω 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 − ωI )  ω ) = 0. Hitting this with ( 1  and inserting the projector I =  1 )( 1  +  2 )( 2  , then doing the same for ( 2  , gives ( ( 1  H  1 )− ω ) ( 1  ω ) + ( 1  H  2 )( 2  ω ) = 0 (5) ( 2  H  1 )( 1  ω ) + ( ( 2  H  2 ) − ω ) ( 2  ω ) = 0 (6) putting in the matrix elements and multiplying by 2 gives − (Δ + 2 ω ) ( 1  ω ) + Ω * ( 2  ω ) = 0 (7) Ω ( 1  ω ) + (Δ − 2 ω ) ( 2  ω ) = 0 (8) The first equation gives, before normalization,  ω ) = Ω *  1 ) + (Δ + 2 ω )  2 ) . (9) The second gives,  ω ) = (Δ − 2 ω )  1 )− Ω  2 ) . (10) 1 For positive detuning, Δ > 0, according to our answer for (a), we want lim Ω → ( 2  g ) = 0 and lim Ω → ( 1  e ) → 0, thus we should use (10) for  ω g ) and (9) for  ω e ) . This gives  g ) = (Δ + radicalbig Δ 2 +  Ω  2 )  1 ) − Ω  2 ) radicalBig (Δ + radicalbig Δ 2 +  Ω  2 ) 2 +  Ω  2 (11)  e ) = Ω *  1 ) + (Δ + radicalbig Δ 2 +  Ω  2 )  2 ) radicalBig...
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This note was uploaded on 12/04/2011 for the course PHY 7070 taught by Professor Smith during the Spring '11 term at University of Wisconsin.
 Spring '11
 Smith
 Physics, Energy, Work

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