NMR Techniques at Liquid Helium Temperatures - Low ...

2.3 NMR in 3 He 21
An exactly equal situation is the one where the transverse frequency fβ is swept
while the field H has a constant value HL. Then we have fL = γHL and
f 2 β = f 2 B + γ2H2
L f 2
B + γ
+
2
2H2 L
2
2
− γ 2 f 2 B H2 L cos2 β. (2.45)
Finally, we want to express f β via H β from the equations above assuming that
cosβ is the same. The resulting equation is
f 2 ⎛
1
β = ⎝ f
2
2 B + f 2 rf +
( f 2 B − f 2 rf )(( f 2 B + 3 f 2 rf )H2 β − 4 f 2 rf H2 L )
H 2 β
⎞
⎠. (2.46)
Let us mark the field corresponding to cos2 = 1 5 (which corresponds to β ≈ 63.4◦ )
as He and Eq. (2.43) transforms to
f 2 B = 5 f 2 rfH2 e − f 2 rfH2
L
. (2.47)
H 2 e − 5H 2 L
Combining the two above equations we get
f 2 β =
f 2 rf
3H 2 e − 5H 2 L
− 2He
5H 4 L
H 2 β
+ H 2 e
H 2 e − 5H 2 L
2 − H2 L
H2
β
− 5H2
L
. (2.48)
Simplifying this we get the approximate expression (omitting all 3rd order terms
and higher) between the frequency and field scales:
fβ − frf
= δ f
HL − Hβ Hβ
=
. (2.49)
frf
frf
The difference between the exact transformation (Eq. (2.48)) and the approximate
expression (Eq. (2.49)) is at maximum less than 1 0/00. Figure 2.6 shows the same
data as Fig. 2.4 as a function of the frequency difference δ f from the Larmor
frequency instead of the current used to create field H.
2.3.5 Temperature measurement from the NMR spectrum
The longitudinal resonance frequency of 3 He-B, fB(p,T ), depends on pressure
and temperature. We can write Eq. (2.42) in the form
x =
( fB
fL )2 − 1
H 2 L
cos 2 β · ( fB
fL )2 x − 1
, (2.50)