Density Matrix Description of NMR
BCMB/CHEM 8190
Operators in Matrix Notation
If we stay with one basis set, properties vary only because of
changes in the coefficients weighting each basis set function
x = (h/2)<| Ix |> - this is how we calculate observables
= c1  + c2  + c3  + c4  = j cj j
AB example:
1 = 1.0  + 0.0  + 0.0  + 0.0 
2 = 0.0  + 0.6  + 0.8  + 0.0 
3 = 0.0  + 0.8  - 0.6  + 0.0 
4 = 0.0  + 0.0  + 0.0  + 1.0 
<| Ix |> = j,k cj* ck <j| Ix |k>
We need to calculate <j| Ix |k> only once if we stay
with this basis set – these can be put in a n x n matrix.
Matrix equivalent: <| Ix |> =
(c1, c2, …)* 
 c1
 Ix   c2

  •
Density Matrices and Observables for an
Ensemble of Spins
Expectation values pertain to single spin properties; we observe net
behavior of an ensemble of spins. For a particular system (two spin
½ nuclei) operator representations are the same; all variations are in
basis set coefficients. We need to average over these coefficients.
<| Ix |> =
(c1, c2, …)* 
 c1
 Ix   c2

  •
= j,k cj* ck Ixjk
For a 2 spin system: (c1*,c2*) Ix11 Ix12  c1 = (c1*, c2*)  Ix11c1 + Ix12c2 
 Ix21 Ix22 c2
 Ix21c1 + Ix22c2 
= c1*c1 Ix11+ c1*c2 Ix12+ c2*c1 Ix21+ c2*c2 Ix22
Products of averaged coefficients can also be collected
and used in matrix form. This is called a density matrix, .
<| Ix |> = Tr |cjck*| | Ix | = Tr |  | | Ix|
For a 2 spin system: <| Ix |> = 11Ix11+ 12Ix21+ 21Ix12+ 22Ix22
Solving for the Time Dependence of 
Our observables are time dependent (magnetization precesses).
All time dependence can also be put in basis set coefficients, or
in coefficient products of the density matrix.
Schrodinger’s time dependent equation (H (t) = - i (h/2) d((t))/dt)
Allows us to solve for dcj/dt, d/dt (cjck*) = cj (dck*/dt) + (dcj/dt) ck*
Result in terms of a density matrix is Lioville – von Neuman eq.
d/dt |  | = i / (h/2) { |  | | H | - | H | |  | } = i / (h/2) [, H]
Implications: If we know  at any time (equilibrium at time 0) and
know the Hamiltonian (H), we can solve for  at any time, and
calculate any observable as the Tr |  | | O | (O is any operator).
Test: Does X Magnetization Precess in B0?
• What elements of  dictate x magnetization?
<Ix> = Tr [] [Ix] - examine spin ½ case
11 12   0 1 / 2
 1 / 212  1 / 2 21
= Tr 



21  22 1 / 2 0 
 0 12
• Therefore, consider (0) =
21 0 


0 
1 / 2
• What is H?, H = -(hB0/(2))Iz = -(hB0/(2)) 

0

1
/
2


• d / dt  0 12 
21 0 


  0 12 1 / 2
0  1 / 2
0   0 12 

i /  (   B 0 )  



  21 0  
21
0
0

1
/
2
0

1
/
2


 



Consider evolution of one element
• (d/dt)12(t) = iB0(-½ 12(t) – ½ 12(t)) = i0 12(t)
• Solution: 12(t) = 12(0) exp(-i0t) or:
= 12(0) (cos(0t) –i sin(0t)
• Same happens for 21(t)
• hence elements corresponding to x magnetization
precess
t=0
t=1
Density Matrix at Equilibrium
• Have a route to an observable:
<| Ix(t)|> = Tr | (t) | | Ix|
• Have a route to (t):
d/dt |(t)| = i /(h/2) [ |(t)|, |H| ]
• Need a place to start: (0)
• Diagonal elements: consider cncn* a probabliity
• cncn* = exp(-En/(kT))/Z  1/Z - En/(kT)/Z; (small En)
• nn = 1/Z - nn; Z  # of states
• Note: text uses  for density matrix
– really deviation matrix
Off-diagonal elements at equilibrium
•  must satisfy time dependent Schrodinger Eq.
H = (i2/h)/t = E; in limit of H = H0
• Solution is: n = nexp(i2/h)Ent
• But, the members of the ensemble do not have a common
origin in time.
• Hence add a random phase factor:
nm = nexp(i((2/h)Ent + m) = ncmn(t)
• cmn(t) = cos(n,m,t) + i sin(n,m,t)
• Density matrix element = cicj*(t) = cosicosj +
sinicosj +cosIsinj +sinIsinj
• if i = j,  P()(cosi2 + sini2)  P()
• If I  j, cosicosj etc = 0, and all off diagonals = 0
Examples for a spin ½ nucleus
• H = -h/(2)B0IZ ,
E = (+/-) h/(4)B0
 eq
0
exp( B 0 / 2kT)

 1 / 2



B
kT

0
exp(
/
2
)
0


 eq
0
 B 0 / 2kT)

 1 / 2




B
kT
0
/
2
)
0


• No time variation as expected:
• (d/dt) = (i2/h)[ , H ] = [ H] – [H ] = 0
(product of diagonal matrices)
What about MZ and MX?
• Curies law for susceptibility
M Z  Tr([][ Z ])
  B 0

2 2
0
1
0



 B0
4kT
 (  / 2)
)
 Tr( 

0  1
4kT
  B 0




0
4kT

• Expect no X magnetization at equilibrium
0
1


1
1
I X 

M X  Tr ([][ X ]); [ X ] 
2
2 1 0
1 0  0 1
)0
M X  Tr ( 



0  1 1 0
What elements do give MX?
  11
M X  Tr  
  21
12  1
0 1   1
 
  2 ( 12   21 )


2
 22 
1 0 
• We have argued that these are related to
transition probabilities
• We have shown that these precess in a
magnetic field
Rotation operators – a more general
description of rotation
• (d/dt)(t) = i /(h/2) [(t), H]
• General solution:
(t) = exp(-(i2/h)Ht)(0)exp((i2/h)Ht) = R(t)(0)R-1(t)
• If basis set elements are eigen functions of H, R is
diagonal and easy to evaluate

• Chemical shift evolution: RZ = exp(-iB0cst IZ)
0
exp( it / 2)

0 
;  ( 0)  
RZ  


0
exp(

i


t
/
2
)

0




MX and MX Precess at 
0
0
 0  exp( it / 2)

exp( it / 2)
( t )  
  0 




0
exp(
i
t
/
2
)
0
exp(
i
t
/
2
)






0
0
 exp( it / 2)
exp( it / 2)

( t )  

0
exp( it / 2)  exp( it / 2)
0


0
 exp( it )

( t )  



exp(
i
t
)
0




0
cos( t )
0
i sin( t )


 
 ( M X  M Y )
( t )   


0
0
cos( t )

  i sin( t )

Rotation operators for an RF pulse
along X axis in the rotating frame
• Hamiltonian in the rotating frame: H’ = -(2B1/h)I’X
• Formal solution at end of pulse (time t) is given as:
(t)= exp(-(i2/h)H’t)(0)exp((i2/h)H’t)= RX(0)RX-1
• Cannot simply insert I’X in exponential operator to
evaluate matrix elements; I’X mixes spin states
• Elements of (t) do not just oscillate, but convert one
element into another – for example (11- 22) to (12+
12) something we associate with z magnetization to
what we associate with y magnetization in a sin(B1 t)
dependent way.
X pulse with 1t = , /2
For one spin 1/2
R X ,
 cos( 1t / 2) i sin( 1t / 2)
RX  

i sin( 1t / 2) cos(1t / 2) 
0 i  1
 0  i
; R X ,  
;



 i 0
 i 0 
R X , / 2
 2
1 i  1
 1  i
2
; R X , / 2 
;




2  i 1
2  i 1 
Similar arguments lead to Y pulse operations
The One Pulse FID
(t) = RZ(t) RX,/2 (0) R-1X,/2 RZ-1(t)
90X
 0 
 M Z (eq )
 ( 0)  

0  
1 i   0 
 1  i
(2 / 2 )
( t p )  ( 2 / 2 ) 




1
0
1



i
i





1 i     i
 0  i
 (1 / 2) 
( t p )  (1 / 2) 




1
0





i
i
i





t
Evolution of the FID
• First point in the FID:
• (t1) = RZ(dw)(tp)RZ-1(dw); (dw = dwell time); (t1) =
 id W
id W




exp(
)
0
exp(
)
0



 0  i 
2
2
(1 / 2) 



 id W 

i

d
0
i

W


0
exp(
)
0
exp(
)

2
2




•
•
•
•
•
(t2) = RZ(dw)(t1)RZ-1(dw); (second point in FID)


Calculating the FID: FID = Tr{[(ti)] [MX + MY]}
Programs like GAMMA, SPINEVOLUTION work this way
Density Matrix Simulation of 2nd Order Spectra
• For two spin basis set is: , , , 
11 12 13
21  22 .

.
.
31

.
.
 .
.
.

.

.
12+ 21 34+ 43
13+ 31 24+ 42
• 12,21, … are associated with lines in an AX
spectrum. 1:1 for first order spectrum
• Don’t have to have a 1:1 association if  is not an
eigen function of H.
• ciI + ciI may evolve coherently as one line – ie
12,13, are mixed
• Calculation of Mx still works for a second order
system