Chapter 6: Product operators - The James Keeler Group

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quantum mechanics and the solution to them are already all know. The identityrequired here to solve this equation is( ) ( )≡ −exp −iβI I exp iβI cosβ I sin β I[6.1]x z x z yThis is interpreted as a rotation of I zby an angle β about the x-axis. By puttingβ= ω 1t pthis identity can be used to solve Eqn. [6.1]σ( tp)= cosω1tp Iz−sinω1tpIyThe result is exactly as expected from the vector model: a pulse about the x-axisrotates z-magnetization towards the –y-axis, with a sinusoidal dependence on theflip angle, β.6.1.4 Standard rotationsGiven that there are only three operators, there are a limited number of identitiesof the type of Eqn. [6.1]. They all have the same form( ) ( )exp −iθIa{old operator} exp iθIa≡ cosθ{old operator} + sinθ{new operator}where {old operator}, {new operator} and I aare determined from the threepossible angular momentum operators according to the following diagrams; thelabel in the centre indicates which axis the rotation is aboutI II IIIxzz–y zyyx–y–x yx–x–z–zAngle of rotation = Ωt for offsets and ω 1 t p for pulsesFirst example: find the result of rotating the operator I yby θ about the x-axis,that is( ) ( )exp −iθI I exp iθIx y xFor rotations about x the middle diagram II is required. The diagram shows thatI y(the "old operator") is rotated to I z(the "new operator"). The required identityis therefore( ) ( )≡ +exp −iθI I exp iθI cosθ I sinθISecond example: find the result ofx y x y z( y){ z} ( y)exp −iθI – I exp iθIThis is a rotation about y, so diagram III is required. The diagram shows that –I z(the "old operator") is rotated to –I x(the "new operator"). The required identityis therefore6–4
( ) −{ } ( )≡ { − }+ { }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
Page 1 and 2: 6 Product Operators †The vector m
Page 3: and z-components of the magnetizati
Page 7 and 8: means that the two lines of the spi
Page 9 and 10: IVxVy-yzzzyzxzzz-xz-xangle = πJtFo
Page 11 and 12: 6.1.3 Spin echoes in heteronuclear
Page 13 and 14: ( I I + I I )= ( I + iI )( I + iI )
Page 15 and 16: 6.9 Multiple -quantum coherence6.9.

Chapter 6: Product operators - The James Keeler Group

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