Transcript 投影片 1 - Academia Sinica

Chapter 6 Product Operator
Product operator is a complete and rigorous quantum mechanical description of
NMR experiments and is most suited in describing multiple pulse experiments.
Wave functions: Describe the state of a system. One can calculate all
properties of the system from its wavefunction.
Operators: Represents an observable which operate on a function to give a
new function.
Angular Momentum: A measure of the ability of an object to continue
rotational motion. Spin angular momentum Ix, Iy, Iz etc.
Hamiltonian: The operator of energy if a system. One Hamiltonian to describe
a particular interaction. The evolution of a Hamiltonian
determines the state of the spins and the signal we detect.
The operator of a single spin (Ix, Iy, and Iz): The density operator of a
spin-1/2 system:
(t) = a(t)Tx + b(t)Iy + c(t)Iz
At equilibrium only Iz is non-vanishing. Thus, (t)eq = Iz with c(t) = 1 .
Hamiltonian for free precession (During delay time):
H = ·Iz, where  is the rotational frequency.
Hamiltonian of an X-pulse:
H = 1·Ix. Similarly, H = 1·Iy is the Hamiltonian of a Y-pulse.
Equation of motion of a density operator:
(t) = exp(-iHt) (0)exp(iHt)
Example: The effect of X-pulse to the spin in equilibrium. (Considered to be
very short so that evolution caan be ignored):
H = 1Ix; (0) = Iz. ,
Thus, (t) = exp(-iHt) (0)exp(iHt) = exp(-i 1tpIx)Izexp(i 1tpIx)
= Iz Cos1tp – Iy Sin1tp
Standard rotations:
exp(-iIa)(old operator)exp(iIa) = cos (Old operator) + sin (new operator)
Example: exp(-iIx)Iyexp(iIx)
Old operator = Iy and new operator = Iz
 Find new operator 
xp(-iIx)Iyexp(iIx) = cos Iy + sin Iz
Example 2: exp(-iIy){-Iz}exp(iIy) = -cos Iz - sin Ix
Shorthand notation: (tp) = exp(-i1tpIx) (0) exp(i1tpIx) 
For the case where (0) = Iz,
Example: Calculate the results of spin echo
90x
180x

Time: 0 a
At time 0: (0) = Iz;

b
c
d
At time 0  a: Rotation about x by 90o : (0) = - Iy
At time a  b: Free precession (Operator = IZ)
(Cosidered as rotation wrt Z-axis)
At b  c rotation wrt x by 180o : The second term is not affected
Or
Thus:
C  d: Free precession. Again, consider the two terms separately we got:
First term:
Second s=term:
Collecting together the terms in Ix and Iy we got
(coscos + sinsin)Ty + (cossin  - sin cos ) Ix = Iy
Or
Independent of  and   The magnetization is refocused.
 - - 1800 -  - refocus the magnetization and is equivalent
to -1800 pulse.
Product operator for two spins: Cannot be treated by vector model
Two pure spin states: I1x, I1y, I1z and I2x, I2y, I2z
I1x and I2x are two absorption mode signals and
I1y and I2y are two dispersion mode signals.
These are all observables (Classical vectors)
Coupled two spins: Each spin splits into two spins
Anti-phase magnetization: 2I1xI2z, 2I1yI2z, 2I1zI2x, 2I1zI2y
(Single quantum coherence)
(Not observable)
Double quantum coherence: 2I1xI2x, 2I1xI2y, 2I1yI2x, 2I1yI2y
(Not directly observable)
Zero quantum coherence: I1zI2z (Not directly observable)
Including an unit vector, E there are a total of 16 product operators in a
weakly-coupled two-spin system. Understand the operation of these 16
operators is essential to understand multiple NMR expts.
Example 1: Free precession of spin I1x of a coupled two-spin system:
Hamiltonian: Hfree = 1I1z + 2I2z
No effect
= cos1tI1x + sin1tI1y
Example 2: The evolution of 2I1xI2z under a 90o pulse about the y-axis
applied to both spins:
Hamiltonian: Hfree = 1I1y + 1I2y
Evolution under coupling:
Hamiltonian: HJ = 2J12I1zI2z
Causes inter-conversion of in-phase and
anti-phase magnetization according to the
Diagram, i.e. in  anti and anti according
to the rules:
Must have one
component in the
X-Y plane !!!
Useful identify:
Spin echo in homonuclear-coupled two spins:
Non-selective pulse:
Assuming only Ix present at the beginning: Since chemical shift is refocused in
spin-echo expt we consider only effect of coupling and 180o pulse:
Coupling:
180o pulse:
 No effect on the magnetization if both spins are flipped by 180o !!!
 The final results
 When  = 1/4J Ix completely converts to antiphase 2IyIz.
 Used in HSQC experiment.
Inter-conversion of in-phase and anti-phase
magnetizations:
In  Anti:
Anti  in:
Heteronuclear coupling: In this case one can apply the
pulse to either spins such as in the sequence a  c.
Sequence a is similar to that of homonuclear coupling.
In sequence b the 180o pulse apply only to spin 1:
During second delay the coupling effect gives:
Collecting terms results in only Ix left
 J-coupling has been refocused (So is sequence c)
(No transfer of magnetization or decoupling)
180X
Coherence order:
Raising and lowering operators: I+ = ½(Ix + iIy); I- = 1/2 (Ix –i-Iy)
Coherence order of I+ is p = +1 and that of I- is p = -1
 Ix = ½(I+ + I-); Iy = 1/2i (I+ - I-) are both mixed states contain order
p = +1 and p = -1
For the operator: 2I1xI2x we have:
2I1xI2x = 2x ½(I1+ + I1-) x ½(I2+ + I2-) = ½(I1+I2+ + I1-I2-) + ½(I1+I2- + I1-I2+)
P = +2
P = -2
P=0
P=0
The double quantum part, ½(I1+I2+ + I1-I2-) can be rewritten as:
(Pure double quantum state)
Similar the zero quantum part can be rewritten as:
½(I1+I2- + I1-I2+) = ½ (2I1xI2x – 2I1yI2y)
(Pure zero quantum state)
Multiple Quantum Coherence:
Active spins: Spins that contains transverse components, Ix or Iy.
Passive spins: Spins that contain only the longitudinal component, Iz.
Evolution of Multiple Quantum Coherence:
Chemical shift evolution: Analogous to that of Ix and Iy except that
it evolves with frequency of 1 + 2 for p = ±2 and 1 - 2 for p = 0
Coupling: