851HW4_Solutions09 - PHYS851 Quantum Mechanics I Fall 2009...

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PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 4: Solutions 1. The 2-Level Rabi Model: The standard Rabi Model consists of a bare Hamiltonian H 0 = 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 0 + V . Write down the 2x2 Hamiltonian matrix in the {| 1 ) , | 2 )} basis and then compute the ‘dressed-state’ 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 parenrightbigg parenleftbigg 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 dressed-state 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
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For positive detuning, Δ > 0, according to our answer for (a), we want lim Ω 0 ( 2 | g ) = 0 and lim Ω 0 ( 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 (Δ + radicalbig Δ 2 + | Ω | 2 ) 2 + | Ω | 2 (12) For negative detuning, Δ < 0, the limits are reversed, so we need to use (9) for | g ) and (10) for | e ) , and multiply each by 1 (a nonphysical global phase factor), giving | g ) = Ω * | 1 ) + ( | Δ | + radicalbig Δ 2 + | Ω | 2 ) | 2 ) radicalBig ( | Δ | + radicalbig Δ 2 + | Ω | 2 ) 2 + | Ω | 2 (13) | e ) = ( | Δ | + radicalbig Δ 2 + | Ω | 2 ) | 1 ) + Ω || 2 ) radicalBig ( | Δ | + radicalbig Δ 2 + | Ω | 2 ) 2 + | Ω | 2 (14) In the {| g ) , | e )} basis, the Hamiltonian is H = ω g | g )( g | + ω e | e )( e | (15) while in the {| 1 ) , | 2 )} basis, this becomes H = Δ 2 ( | 2 )( 2 | − | 1 )( 1 | ) + Ω 2 | 2 )( 1 | + Ω * 2 | 1 )( 2 | (16) (d) Sketch the energy spectrum versus Ω for the case of fixed Δ > 0. What are ω g and ω e at Ω = 0? What
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