Chapter 6: Product operators - The James Keeler Group

( ) −{ } ( )≡ { − }+ { }exp −iθI I exp iθI cosθ I sin θ – Iy z y z x≡−cosθIz−sinθIxFinally, note that a rotation of an operator about its own axis has no effect e.g. arotation of I xabout x leaves I xunaltered.6.1.5 Shorthand notationTo save writing, the arrow notation is often used. In this, the term Ht is writtenover an arrow which connects the old and new density operators. So, forexample, the followingσ( tp)= exp( −iω1tpIx) σ( 0) exp( iω1tpIx)is writtenω1tI pσ( 0)⎯ x→σtFor the case where σ(0) = I z⎯⎯ ( p)1 p xI ⎯⎯⎯ →cos ω t I – sinωt Izω tI6.1.6 Example calculation: spin echo1 p z1 pa b e f90 x delayτ180 x delayτacquire°( ) °( )At a the density operator is –I y. The transformation from a to b is freeprecession, for which the Hamiltonian is ΩI z; the delay τ therefore correspondsto a rotation about the z-axis at frequency Ω. In the short-hand notation this isΩτI−I⎯z→σby⎯ ( )To solve this diagram I above is needed with the angle = Ωτ; the "new operator"yis I xΩτIz−I ⎯⎯→− cosΩτI + sin ΩτIyIn words this says that the magnetization precesses from –y towards +x.The pulse about x has the Hamiltonian ω 1I x; the pulse therefore correspondsto a rotation about x for a time t psuch that the angle, ω 1t p, is π radians. In theshorthand notationω− cosΩτ I + sin Ωτ I ⎯⎯⎯ 1p tI xyx→σ()e[6.2]Each term on the left is dealt with separately. The first term is a rotation of yabout x; the relevant diagram is thus IIω tI1 p x−cosΩτ I ⎯⎯⎯→−cosΩτcosω t I −cosΩτsinωt Iyyx1 p y1 pHowever, the flip angle of the pulse, ω 1t p, is π so the second term on the right iszero and the first term just changes sign (cos π = –1); overall the result isπIx−cosΩτI ⎯⎯→cosΩτIyThe second term on the left of Eqn. [6.2] is easy to handle as it is unaffected byyz6–5