851HW4_Solutions09

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

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Unformatted text preview: 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 = 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 dressed-state 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 dressed-state 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...
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851HW4_Solutions09 - PHYS851 Quantum Mechanics I, Fall 2009...

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