cross polarization (cp) - leibniz-fmp.de · barth-jan van rossum: solid-state nmr cross...
TRANSCRIPT
Barth-Jan van Rossum: Solid-State NMR
Cross polarization (CP)
this approach is valid as long:
1. system contains a large number of spins
2. strong 1H-1H dipolar couplings are present
‘classical’ description of cross-polarization
uses concept of spin temperature
thermodynamic approach:
- polarization exchange between two
reservoirs with different spin temperature
coupled to a large reservoir (‘lattice’)
- relaxation to equilibrium
13C
1H CP
CP
decoupling
Barth-Jan van Rossum: Solid-State NMR
Zeeman interaction
spins rotate with theLarmor frequency ω0 …
z
y
x
B0
…or, spin up and spin down have an energy difference of ω0
ω0
ω0
Barth-Jan van Rossum: Solid-State NMR
high Tlow T
low T à large population differences à high polarization
high T à small population differences à low polarization
“Boltzmann’s ingenious concept”
mhhh..
Barth-Jan van Rossum: Solid-State NMR
large splitting à large population differences à high polarizationsmall splitting à small population differences à low polarization
large splitting
smallsplitting
equaltemperature
“Boltzmann’s ingenious concept”
Barth-Jan van Rossum: Solid-State NMR
“Boltzmann’s ingenious concept”
à high magnetic fields give larger signal
à 1H (high γ) more sensitive than 13C or 15N (low γ) à CP
à e- (very high γ) used to enhance NMR signal à DNP
à low temperature larger signal
than high temperature
Barth-Jan van Rossum: Solid-State NMR
Cross polarization (CP)
like the Zeeman interaction, the
spin-lock pulse gives rise to a
splitting (spin up, spin down)
pulse :spins rotate
spin-lock pulse:spins are trapped
during CP, both spin-types (1H and 13C)
are ‘spin-locked’
Barth-Jan van Rossum: Solid-State NMR
Cross polarization (CP)
since B0, γH and γC are all
fixed, the Zeeman splitting
is different for 1H and 13C …
… however, the spin-lock
fields B1H and B1
C can be
chosen so that the splitting
for 1H and 13C becomes
equal
Hartmann-Hahn Matching
c
Barth-Jan van Rossum: Solid-State NMR
ω0
ω1
‘real’ Boltzmanndistribution
apparentBoltzmanndistribution
in equilibrium:
levels separated by ω0
à occupation of levels according
to real Boltzmann distribution
during spin lock:
levels separated by ω1
with ω1 << ω0
à levels are ‘compressed’
à occupation of levels looks like
a Boltzmann distribution, but
one for much lower T
high T
Cross polarization (CP)
low T
Barth-Jan van Rossum: Solid-State NMR
ω0
ω1
‘real’ Boltzmanndistribution
during spin lock:
relaxation back to the normal
temperature with T1ρ
high T
spin locking 1H
lowers the spin-temperature
low T
T1ρ
Cross polarization (CP)
apparentBoltzmanndistribution
Barth-Jan van Rossum: Solid-State NMR
(T1ρS ) T1ρH
Cross polarization (CP)
by matching energy splitting for 1H and 13C
(Hartmann-Hahn matching), polarization
can be exchanged with conservation of energyhigh T
S (13C)
low T
I (1H)
heteronuclear (1H - 13C) dipolar interaction
couples 1H and 13C polarization reservoirs
homonuclear interaction (mostly 1H- 1H)
provides coupling to lattice (T1ρ relaxation)
ambient T
equilibration lowers 13C spin temperature
à 13C polarization first increases (short CP)
coupling to lattice reduces spin temperature
à 13C polarization relaxes (long CP)
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
Overview
dipolar interaction
- Hamiltonian
- recoupling
- dipolar truncation
cross polarization (CP) - part II
DNP (dynamic nuclear polarization)
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
Dipolar interaction
α
zero for α = 90° max for α = 0°
the dipolar coupling is the interaction between
two magnetic moments µ1 and µ2
r r
same α !
~ cos αα
β
γ βγ
~ cos β
~ cos γ
r
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
Dipolar interaction
the dipolar coupling is the interaction between
two magnetic moments µ1 and µ2
(correspondence principle)
how does the dipolar coupling look in quantum mechanics?
r
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
Dipolar interaction
à ‘truncation’ or ‘quenching’
Only terms that commute with Zeeman interaction (Iz) remain
Zeeman interaction (~500 MHz) >> dipolar interaction (~50 kHz)
A, B – terms:
commute with Iz (“secular terms”) à no time evolution under Iz
C, D, E, F – terms:
do not commute with Iz (“non-secular terms”) à time evolution under Iz
à averaged to zero
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
Dipolar interaction
truncated
B-term only relevant when ωI ~ ωS
homonuclear case (e.g.1H-1H) à B-term
heteronuclear case (e.g. 1H-13C) à B-termr and θ : spatial parameters
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
θ
D∝1
r3 3cos2 θ −1( )
3cos2 θ −1= 0 in isotropic liquids
0 in solids or oriented media
Dipolar interaction
the dipolar coupling depends on :
- the distance between the spins (r)
- the angle θ between the magnetic field B
and the vector connecting the spins
r
B
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
Recoupling
The time-average of the dipolar coupling is zero
but not the instantaneous coupling, it is oscillating around zero
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
Overview
dipolar interaction
- Hamiltonian
- recoupling
- dipolar truncation
cross polarization (CP) - part II
DNP (dynamic nuclear polarization)
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
since B0, γH and γC are all
fixed, the Zeeman splitting
is different for 1H and 13C …
… however, the spin-lock
fields B1H and B1
C can be
chosen so that the splitting
for 1H and 13C becomes
equal
Hartmann-Hahn Matching
c
Cross polarization (CP) – part II
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
the heteronuclear dipolar interaction between 1H (I-spins) and 13C (S-spins)
drives the polarization transfer:
H ~ b · Iz · Sz
B1
z
y
xduring CP, the spin lock pulses are along x-axis…
Cross polarization (CP) – part II
…transform to a frame that rotates around the x-axis
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
in NMR, everything is rotating…
rotating frames
spins rotate with theLarmor frequency…
RF pulses are rotating withthe Larmor frequency…
z
y
x B1
z
y
x
B0
blerk…
observation from a frame that rotates itself, makes everything appear static
z
y
x B1
z
y
x
it is as if the Zeeman
interaction has ‘gone’…..
these kind of transformations are quite common in NMR (rotating frame, interaction frame…)
WTF…
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
the heteronuclear dipolar interaction between 1H (I-spins) and 13C (S-spins)
drives the polarization transfer:
H ~ b · Iz · Sz
B1
z
y
xduring CP, the spin lock pulses are along x-axis…
…transform to a frame that rotates around the x-axis
with ω1H (for the I spins) and ω1
C (for the S spins)
B1
z
y
xH’ ~ b’ ·( Iz · Sz + Iy · Sy ) cos (ω1H- ω1
C )t =
= b’ ·( Iz · Sz + Iy · Sy ) [ for ω1H = ω1
C ]
Hartmann-Hahn matching !
Cross polarization (CP) – part II
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
H’ ~ b’ ·( Iz · Sz + Iy · Sy )
the magnetization is along x-axis
à transform to a frame with the new z’ axis along the old x-axis
x à z’ or Ix à Iz
replace x,y,z as follows: y à y’ or Iy à Iy
z à -x’ or Iz à -Ix
H’’ ~ b’ ·( Ix · Sx + Iy · Sy )
z
y
x x'
y'
z'
Ix = ½ ( I+ + I– )
Iy = -½·i ( I+ – I– )rewrite, using raising and lowering operators:
Cross polarization (CP) – part II
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
H’’ = b’ ·( Ix · Sx + Iy · Sy )
H’’ = b’ [ ½ ( I+ + I– ) · ½ ( S + + S – ) + (-½·i) ( I+ – I– ) · (-½·i) ( S + – S – ) ]
H’’ = ½·b’ ( I +·S – + I –·S + )
I –
I +
I –
I +
0
0
I – S +
I + S –
I S I S
all other combinations give zero
Cross polarization (CP) – part II
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
1H
1H netpolarization random
13C
I – S +
I + S –
I S I S
net magnetization is ‘spin-up’ for 1H (I-spins)
Cross polarization (CP) – part II
Barth-Jan van Rossum: Solid-state NMR, advanced concepts I
1H
1H netpolarization random
13C
I – S +
I + S –
I S I S
net magnetization is ‘spin-up’ for 1H (I-spins)
Cross polarization (CP) – part II