QMGreensite_merged

In order for
2 to be an eigenfunction, the two terms in parentheses
following the last equal sign must be identical. Thus,

!I
!S
JISþ 2 JIStan¼
!Iþ!S
JISþ 2 JIS=tan,

2 tan!ðÞ
I
!S 2 JIS 1 
tan^2 



¼0,

2 tan

1 
tan^2 

¼

2 JIS

ðÞ!I
!S

,

tanð 2 Þ¼

2 JIS

ðÞ!I
!S

,

½ 2 : 162 Š

which completes the demonstration, becauseis defined according to
[2.157]. By inspection,

E 2 ¼^12 ðÞ!I
!S
JISþ 2 JIStan, ½ 2 : 163 Š

which is easily shown to be equal to [2.159] by solving [2.157] for tan.
By comparing [2.138] and [2.156], the transformation matrixUthat
converts the product basis into the strong coupling basis (and
diagonalizes the Hamiltonian) is given by

U¼

10 00

0 cos sin 0

0 
sin cos 0

00 01

2

(^66)
(^64)
3
(^77)
75 : ½^2 :^164 Š
In the limit of weak scalar coupling,¼0 and the wavefunctions of
the two energy levels
(^) and
(^) are independent. The weak coupling
Hamiltonian simplifies to
H¼!IIzþ!SSzþ 2 JISIzSz: ½ 2 : 165 Š
To calculate evolution of the density operator under the weak coupling
Hamiltonian, the effect of the operation
ðtÞ¼exp½Š
i 2 IzSzð 0 Þexp½Š ½i 2 IzSz 2 : 166 Š
for ¼JISt must be calculated. The derivation is similar to the
derivation of the rotation operators; thus,
exp½Š¼i 2 IzSz Eþi 2 IzSz
^122 ðÞ 2 IzSz^2 þ...: ½ 2 : 167 Š
64 CHAPTER 2 THEORETICALDESCRIPTION OFNMR SPECTROSCOPY