Density Matrix Description of NMR BCMB/CHEM 8190

Density Matrix Description of NMR BCMB/CHEM 8190

Operators in Matrix Notation It will be important, and convenient, to express the commonly used operators in matrix form Consider the operator I z and the single spin functions α and - recall Î x α = 1 2 β Î x β = 1 2α Î y α = 1 2iβ Î y β = 1 2iα Î z α = +1 2α Î z β = 1 2 β - recall the expectation value for an observable Q = ψ ˆQ ψ = α α = β β =1 α β = β α = 0 ψ ˆQψ dτ ˆQ - some operator ψ - some wavefunction - the matrix representation is the possible expectation values for the basis functions α β α β α Îz α α Îz β β Îz α β Îz β Î z = α Îz α α Îz β β Îz α β Îz β = 1 2 α α 1 2 α β 1 2 β α 1 2 β β = 1 2 0 0 1 2 = 1 2 1 0 0 1 This is convenient, as the operator is just expressed as a matrix of numbers no need to derive it again, just store it in computer β

Î x = 0 1 2 = 1 1 2 0 2 Operators in Matrix Notation The matrices for I x, I y,and I z are called the Pauli spin matrices 0 1 1 0 0 1 2i Î y = = 1 1 2i 0 2 0 i Î z = 1 2 0 i 0 Express α, β, α and β as 1 2 column and 2 1 row vectors α = 1 0 β = 0 1 α = [1 0] β = [0 1] 0 1 2 = 1 2 1 0 0 1 Using matrices, the operations of I x, I y, and I z on α and β, and the orthonormality relationships, are shown below Î x α = 1 0 1 2 1 0 1 0 = 1 0 2 1 = 1 2 β Î x β = 1 0 1 2 1 0 0 1 = 1 1 2 0 = 1 2 α Î y α = 1 0 i 2 i 0 1 0 = 1 0 2 i = i 1 2 β Î y β = 1 0 i 2 i 0 0 1 = 1 i = i 1 2 0 2 α Îz α = 1 1 0 2 0 1 1 0 = 1 1 2 0 = 1 2 α Î z β = 1 1 0 2 0 1 0 1 = 1 0 = 1 2 1 2 β α α = [1 0] 1 0 =1 β β = [0 1] 0 1 =1 α β = [1 0] 0 1 = 0 β α = [0 1] 1 0 = 0

Operators in Matrix Notation Likewise, we can express the two-spin operators in matrix form - example: consider the expectation value for the I x operator on the A spin in an AX system in the two spin basis αα αβ βα ββ αα αα ÎAx αα αα ÎAx αβ αα ÎAx βα αα ÎAx ββ αβ αβ ÎAx αα αβ ÎAx αβ αβ ÎAx βα αβ ÎAx ββ βα βα ÎAx αα βα ÎAx αβ βα ÎAx βα βα ÎAx ββ ββ ββ ÎAx αα ββ ÎAx αβ ββ ÎAx βα ββ ÎAx ββ (recall Î x α = 1 2 β Î x β = 1 2α Î y α = 1 2iβ Î y β = 1 2iα Î z α = 1 2α Î z β = 1 2 β ) 1 2 αα βα 1 2 αα ββ 1 2 αα αα 1 2 αα αβ 1 2 αβ βα 1 2 αβ ββ 1 2 αβ αα 1 2 αβ αβ Î Ax = = 1 2 βα βα 1 2 βα ββ 1 2 βα αα 1 2 βα αβ 1 2 ββ βα 1 2 ββ ββ 1 2 ββ αα 1 2 ββ αβ 0 0 1 2 0 0 0 1 0 0 0 0 1 2 = 1 0 0 0 1 1 2 0 0 0 2 1 0 0 0 0 1 2 0 0 0 1 0 0 The matrices for I x, I y,and I z are 0 0 1 0 Î Ax = 1 0 0 0 1 2 1 0 0 0 0 1 0 0 0 1 0 0 Î Xx = 1 1 0 0 0 2 0 0 0 1 0 0 1 0 0 0 i 0 Î Ay = 1 0 0 0 i 2 i 0 0 0 0 i 0 0 0 i 0 0 Î Xy = 1 i 0 0 0 2 0 0 0 i 0 0 i 0 1 0 0 0 Î Az = 1 0 1 0 0 2 0 0 1 0 0 0 0 1 1 0 0 0 Î Xz = 1 0 1 0 0 2 0 0 1 0 0 0 0 1

Density Matrices and Observables for an Ensemble of Spins Expectation values give average results for an observable for a single spin or spin system - i.e. for a group of spins that all have the same wavefunction The density matrix/operator provides an avenue for average or net behavior of an ensemble of spins First, consider a single spin - write a wavefunction for a linear combination of the members of the basis set (α and β) ψ = α + β = c j φ j ψ = α + β = c jφ j j j We can substitute the vector equivalents for α, β, α and β to get vector expressions for ψ and ψ α = 1 0 β = 0 1 α = [1 0] β = [0 1] 1 ψ = α + β = 0 0 + 1 = c 1 ψ = α + β = c 1 [1 0]+ c 2 [0 1] = [c 1 c 2 ] So, the wavefunction (and its complex conjugate) can be written in terms of vectors of the coefficients only

The expectation value for some observable represented by the operator Q, operating on a wavefunction ψ can be written as Q = ψ ˆQ ψ Density Matrices and Observables for an Ensemble of Spins = α + β ˆQ α + β ˆQ ψ k = α ˆQ α + α ˆQ β + β ˆQ α + β ˆQ β = c j c k ψ j j Substitute the matrix/vector equivalents Q = ψ ˆQ ψ α ˆQ α α ˆQ β = [c 1 c 2 ] β ˆQ α β ˆQ β ψ = c 1 ψ = [c 1 c 2 ] = [c 1 c 2 ] α ˆQ α + α ˆQ β β ˆQ α + β ˆQ β α ˆQ ˆQ z α α ˆQ z β = ˆQ ψk β ˆQ z α β ˆQ z β = c 1 α ˆQ α + c 1 α ˆQ β + c 2 β ˆQ α + c 2 β ˆQ β = c c j k ψ j j ˆQ Note the expression for includes products of the coefficients, rather than the coefficients themselves, suggesting a way to represent the spin state (wavefunction) in terms of the products ψ ψ = c 1 [c 1 c 2 ] = c c 1 1 ψ = α + β = c j φ j ψ = α + β = c jφ j - note: NOT ψ ψ =1! j j

Density Matrices and Observables for an Ensemble of Spins This leads to a different way to formulate the expectation value - i.e. for a group of spins that all have the same wavefunction ˆQ = Tr{ ψ ψ ˆQ } c = Tr 1 α ˆQ α α ˆQ β β ˆQ α β ˆQ = Tr c 1 α ˆQ α + β ˆQ α α ˆQ β + β ˆQ β β α ˆQ α + β ˆQ α α ˆQ β + β ˆQ β = α ˆQ α + β ˆQ α + α ˆQ β + β ˆQ β = c 1 α ˆQ α + c 1 α ˆQ β + c 2 β ˆQ α + c 2 β ˆQ β = c c j k ψ j j Now.for an ensemble of spins, the magnitude of an observable is the sum of the expectation values for each spin ˆQ ψk Q = ψ 1 ˆQ ψ1 + ψ 2 ˆQ ψ2 + ψ 3 ˆQ ψ3 +... ψ N ˆQ ψn = Tr{ ( ψ 1 ψ 1 + ψ 2 ψ 2 + ψ 3 ψ 3 +... ψ N ψ N ) ˆQ } The average value of an observable for an ensemble is ( ) ψ i ˆQ ψi Q = 1 N N i=1 = Tr{ ( 1 N) ( ψ 1 ψ 1 + ψ 2 ψ 2 + ψ 3 ψ 3 +... ψ N ψ N ) ˆQ } - the overbar indicating the average over the ensemble

Density Matrices and Observables for an Ensemble of Spins To simplify this expression, we'll define the density operator ( ) ψ 1 ψ 1 + ψ 2 ψ 2 + ψ 3 ψ 3 ˆρ = 1 N ( ) +... ψ N ψ N = ψ ψ For an ensemble, the average value of the observable, and the average macroscopic observable, then are Q = Tr{ ˆρ ˆQ } Q macro = N Tr{ ˆρ ˆQ } What does this mean? - any observable can be ascertained from two spin operators - one of these represents the observable - the other represents the ensemble average of the wavefunctions (not necessary to know the wavefunctions for all N members of the ensemble) For two non-interacting spins, the matrix representation of the density operator (density matrix) is ˆρ = ψ ψ = c 1 c 2 = ρ 11 ρ 12 recall ψ ψ = c 1 ρ 21 ρ 22 - ρ 11 and ρ 22 are the populations of states 1 and 2 (i.e. α and β for spin ½) - ρ 12 and ρ 21 are the coherences

Solving for the Time Dependence of Our observables are time dependent (for example, bulk magnetization precesses with time) hence requiring Schrodinger's time-dependent equation ψ = c j ϕ j d(ψ(t)) dt = i! Ĥψ(t) We'll consider a single spin with two possible states ψ(t) = (t) ψ 1 + (t) ψ 2 ψ 1 ψ 1 = ψ 2 ψ 2 =1 ψ 1 ψ 2 = ψ 2 ψ 1 = 0 j ˆρ d( (t) ψ 1 + (t) ψ 2 ) dt = i! Ĥ c (t) ψ + c (t) ψ 1 1 2 2 d( (t)) dt ψ 1 + d( (t)) dt ψ 2 = i! (t) Ĥ ψ 1 i! (t) Ĥ ψ 2 ψ 1 [ d( (t)) dt] ψ 1 + ψ 1 [ d( (t)) dt] ψ 2 = i! (t) ψ 1 Ĥ ψ 1 + (t) ψ 1 Ĥ ψ 2 [ d( (t)) dt] ψ 1 ψ 1 + [ d( (t)) dt] ψ 1 ψ 2 = i! (t) ψ 1 Ĥ ψ 1 + (t) ψ 1 Ĥ ψ 2 d( (t)) dt = i! (t) ψ 1 Ĥ ψ 1 + (t) ψ 1 Ĥ ψ 2 d( (t)) dt = i! (t) ψ 2 Ĥ ψ 1 + (t) ψ 2 Ĥ ψ 2 d(c k (t)) dt = i! c n (t) ψ k Ĥ ψ n n d(c k (t)) dt = i! c n (t) ψ n Ĥ ψ k without proof n multipled on left by ψ 1 multipled on left by ψ 2

Solving for the Time Dependence of ˆρ These equations tell us how the wavefunctions (coefficients) change with time d ψ k ˆρ ψ m dt d(c k (t)) dt = i! c n (t) ψ k Ĥ ψ n d(c k (t)) dt = i! c n (t) ψ n Ĥ ψ k n Thus, we can write a general expression for the change with time of the density matrix d ˆρ(t) dt = d(c kc m ) d(c = c m ) k + d(c k ) c m dt dt dt = i! c k c n ψ n Ĥ ψ m i! ψ k Ĥ ψ n c n c m n = i! ψ k ˆρ ψ n ψ n Ĥ ψ m i! ψ k Ĥ ψ n ψ n ˆρ ψ m n = i! ψ k ˆρĤ ψ m ψ k Ĥ ˆρ ψ m { } i! ˆρ(t), Ĥ = i! ˆρ Ĥ Ĥ ˆρ n n (commutator) The last expression is called the Lioville - von Neuman equation - knowing ˆρ at some given time (i.e. at equilibrium, t = 0), can solve for the time dependence of ˆρ (i.e. ˆρ at any other time), so can calculate observables Q = Tr{ ˆρ ˆQ } Q macro = N Tr ˆρ ˆQ { } n

Example: Precession of x Magnetization in B 0 The Bloch equation analysis demonstrated how x magnetization (M x ) precesses with time about the B 0 axis M x (t) = M x,0 cos(ω 0 t) M y,0 sin(ω 0 t) Here, we'll see if we can predict this behavior (roughly) using the QM methods we've outlined - recall Q = Tr{ ˆρ ˆQ } Q macro = N Tr ˆρ ˆQ { } Î x = 0 1 2 1 2 0 - ask, "What elements of the density matrix are important for Ix?" - determine I x (spin ½) I x = Tr ˆρ ρ { Îx} = Tr 11 ρ 12 0 1 2 ρ 21 ρ 22 1 2 0 = Tr ρ 0 + ρ 11 12 1 2 ρ 1 11 2 + ρ 0 12 = 1 2(ρ 12 + ρ 21 ) ρ 21 0 + ρ 22 1 2 ρ 21 1 2 + ρ 22 0 This says x-magnetization is proportional to ρ 12 and ρ 21 (only those density matrix elements are relevant for x-magnetization If we like, we can write a density matrix/operator with only elements ρ 12 and ρ 21 to analyze x-magnetization ˆρ x = 0 ρ 12 ρ 21 0

Example: Precession of x Magnetization in B 0 How do the elements of ˆρ and ˆρ x vary with time? d ˆρ x = i! ˆρ x, dt Ĥ = i! ˆρ x Ĥ Ĥ { ˆρ x } 0 ρ = i!( γ!b 0 ) 12 1 2 0 1 2 0 0 ρ 12 ρ 21 0 0 1 2 0 1 2ρ 21 0 0 1 2 ρ = iγb 12 0 0 1 2 ρ 12 1 2 ρ 21 0 1 2 ρ 21 0 = iγb 0 ρ 12 0 ρ 21 0 d ˆρ dt [ ] d ρ 12 (t) dt d ρ 21 (t) = i! ˆρ, Ĥ = i! ˆρ Ĥ Ĥ { ˆρ } ρ = i!( γ!b 0 ) 11 ρ 12 1 2 0 ρ 21 ρ 22 0 1 2 1 2 0 0 1 2 ρ 11 ρ 12 ρ 21 ρ 22 1 2 ρ = iγb 11 1 2 ρ 12 0 1 2 ρ 11 1 2 ρ 12 1 2 ρ 21 1 2 ρ 22 1 2 ρ 21 1 2 ρ 22 = iγb 0 ρ 12 0 ρ 21 0 1 2 0 Ĥ = γ!b 0 Î z = γ!b 0 0 1 2 So, both ρ 12 and ρ 21 change with time, and we can solve for their time dependence [ ] dt = iγb 0 ( ρ 12 (t)) = iω 0 ρ 12 (t) ρ 12 (t) = ρ 12 (0)e (iω 0t) = ρ 12 (0)(cos(ω 0 t) + i sin(ω 0 t)) = iγb 0 (ρ 21 (t)) = iω 0 ρ 21 (t) ρ 21 (t) = ρ 21 (0)e ( iω 0t) = ρ 21 (0)(cos(ω 0 t) i sin(ω 0 t))

Example: Precession of x Magnetization in B 0 So, both ρ 12 and ρ 21 change with time, and we can solve for their time dependence [ ] d ρ 12 (t) dt d ρ 21 (t) [ ] dt = iγb 0 ( ρ 12 (t)) = iω 0 ρ 12 (t) ρ 12 (t) = ρ 12 (0)e (iω 0t) = ρ 12 (0)(cos(ω 0 t) + i sin(ω 0 t)) = iγb 0 (ρ 21 (t)) = iω 0 ρ 21 (t) ρ 21 (t) = ρ 21 (0)e ( iω 0t) = ρ 21 (0)(cos(ω 0 t) i sin(ω 0 t)) Results echo Bloch equation results Elements that represent x-magnetization precess

Density Matrix at Equilibrium So..what do we have? - we have a route to an observable: Q = Tr{ ˆρ ˆQ } for instance I x = Tr ˆρ Îx { } - we have a route to the time dependence of the density matrix d ˆρ(t) dt { } i! ˆρ(t), Ĥ = i! ˆρ Ĥ Ĥ ˆρ So..what don't we have - a place to start, for instance, at equilibrium (ρ at equilibrium, ρ eq ) - values for density matrix elements Assume ρ at equilibrium, ρ eq - presumably, we can decide what all elements of ρ are at some time, so choose t = 0 (equilibrium) What are the diagonal elements (ρ nn )? - ρ nn = c n c n * = a probability of being in a particular state (i.e. α or β) - these we can get from Boltzman factors/statistics ρ nn = c n c n = e ( E n (k B T )) - so, Z number of states Z 1 Z E n (Zk B T ) Z = e ( E n /(k B T )) n =1+1= 2 for α, β basis and small E n For convenience, we define a deviation matrix, σ, with elements ρ nn 1 Z E n (Zk B T ) =1 Z σ nn

Density Matrix at Equilibrium What are off-diagonal elements (ρ nm ) at equilibrium? - off-diagonal elements represent coherences - these are only finite when there are spins that have transverse polarization vectors (wavefunction is linear combination of basis set elements) AND there is a net alignment of these in the x-y plane - at equilibrium, there is no net bulk magnetic moment in the x-y plane, so these elements are zero

Examples for a spin ½ nucleus ˆρ eq = e( E n (k B T )) Z 0 0 e ( E n (k B T )) Z = 1 2 0 0 e ( E n (k B T )) = 1 2 e ( E n (k B T )) Ĥ = γ!b 0 Î z E = mγ!b 0 ρ nn = c n c n = e ( E n (k B T )) Z 1 Z E n (Zk B T ) =1 Z σ nn Z = e ( E n /(k B T )) 1 2 0 = 2 for spin 1 2 Ĥ = γ!b 0 Î z = γ!b 0 n 0 0 e ( γ!b 0 ( T )) e (γ!b 0 ( T )) σˆ eq = E n Zk B T 0 = 1 E n k B T 0 = 1 γ!b 0 T 0 0 E n Zk B T 2 0 E n k B T 2 0 γ!b 0 T d ˆ dt σ eq = i! σ ˆ eq, Ĥ = i! σˆ eq Ĥ Ĥ { σ ˆ eq } = i!( γ!b 0 ) 1 γ!b 0 T 0 1 2 0 2 0 γ!b 0 T 0 1 2 1 2 1 2 0 0 1 2 0 1 2 The density matrix, and deviation matrix, at equilibrium are: There should be no time dependence of these at equilibrium: 1 1 2γ!B = iγb 0 T 0 0 1 1 2γ!B 0 T 0 2 0 1 2γ!B 0 T 2 0 1 2γ!B 0 T = 0 0 = 0 0 0 γ!b 0 T 0 0 γ!b 0 T - so, no variation with time, as expected (diagonal matrices commute: AB=BA, so AB-BA=0 for diagonal matrices A and B)

M 0 Nγ 2! 2 B 0 k B T(2I +1) What about M z? Q = Tr{ ˆρ ˆQ } Q macro = N Tr ˆρ ˆQ I m 2 Nγ 2! 2 B 0 I(I +1) m= I 3k B T M z,eq = M 0 = N Tr σˆ eq 1 γ!b { [ ˆµ z ]} = N Tr 0 T 0 γ! 2 0 γ!b 0 T 2 = N Tr γ! 4 { } ˆµ z = γ!îz = γ! 2 1 0 0 1 At thermal equilibrium, the equilibrium bulk magnetization along the z-axis, M z, is equal to what we have defined as M 0 We can calculate this observable for spin 1 2, M 0 Nγ 2! 2 B 0 4k B T 1 0 0 1 γ!b 0 T 0 0 γ!b 0 T = Nγ 2! (4k B T ) Curie's law (Curie-Weiss law) for magnetic susceptibility in paramagnetic compounds χ m = N Aµ 0 3k B T µ 2 - the dimensionless quantity χ m is the molar magnetic susceptibility, µ 0 is the permeability of free space (proportional to B 0 ), µ is the Bohr magneton (J/T) 2 B 0

What about M x? Q = Tr{ ˆρ ˆQ } Q macro = N Tr ˆρ ˆQ M x,eq = N Tr σˆ eq 1 γ!b { [ ˆµ x ]} = N Tr 0 T 0 γ! 2 0 γ!b 0 T 2 = N Tr γ! 4 { } ˆµ x = γ!îx = γ! 2 0 1 1 0 0 γ!b 0 T γ!b 0 T 0 = 0 0 1 1 0 At thermal equilibrium, no net, bulk magnetization is expected in the transverse plane (M x or M y ) - this is confirmed by the following calculation Same result for M y at equilibrium (zero) So, these methods reproduce our expectations based on what we know already about bulk equilibrium magnetization

What elements give M x? The equilibrium density matrix (or deviation density matrix) indicates no x-magnetization at equilibrium How about not at equilibrium? The elements of the density matrix can be associated with particular properties (especially for first order systems) - previously, we asked, "What elements of the density matrix are important for Ix?" I x = Tr ˆρ ρ { Îx} = Tr 11 ρ 12 0 1 2 ρ 21 ρ 22 1 2 0 = Tr ρ 0 + ρ 11 12 1 2 ρ 1 11 2 + ρ 0 12 = 1 2(ρ 12 + ρ 21 ) ρ 21 0 + ρ 22 1 2 ρ 21 1 2 + ρ 22 0 - ask again, for M x M x = Tr{ σˆ [ ˆµ σ ]} = Tr 11 σ 12 γ! x σ 21 σ 22 2 0 1 1 0 = Tr γ! 2 σ 12 σ 11 σ 22 σ 21 = γ! 2(σ +σ ) 12 21 - so, σ 12 and σ 21 (ρ 12, ρ 21 ) are associated with x-magnetization - and, as we've demonstrated previously, these precess in a magnetic field - these are also associated with transition probabilities: connect α and β for single spin cases

Rotation operators a more general description of rotation (precession) We previously wrote an expression for the time dependence of the density matrix (same for deviation matrix) d ˆρ(t) dt dσ ˆ(t) dt { } i! ˆρ(t), Ĥ = i! ˆρ Ĥ Ĥ ˆρ { } i! ˆ = i! σˆ Ĥ Ĥ σˆ σ (t), Ĥ (commutator) (commutator) The general solution to this latter equation (without proof) is: σ ˆ(t) = e ( (i!)ĥt) σ ˆ(0) e ((i!)ĥt) = ˆR(t) σ ˆ(0) ˆR 1 (t) - can write in terms of rotation operators ( ˆR, ˆR 1 ) that include the Hamiltonian ˆR z Î z - the matrix elements of ˆR z are 1 2 0 Ĥ = γ!b 0 Î z = γ!b 0 Î z = 1 2 0 0 1 2 0 1 2 The operator describes precession (B 0 interacting with ) ˆR z (t) = e ( (i!)ĥ(t)) = e ( (i!)( γ!b 0Îz )t) = e (iγb 0tÎz ) = e (iωtîz ) - the matrix representation of ˆR z is then ˆR z = e(iδωt 2) 0 0 e ( iδωt 2)

M X and M y Precess at Δω For convenience, we'll write the deviation matrix for M x as chemical shift offsets (δ) in σ 12 and σ 21 (elements involved in precession) σ ˆ(t) = ˆR z (t) σ ˆ(0) ˆR 1 z (t) = e(iδωt 2) 0 e = e(iδωt 2) 0 e σ ˆ(0) = 0 δ δ 0 σ ˆ(t) σ ˆ(0) We can then calculate from under the ˆRz operator 0 ( iδωt 2) 0 ( iδωt 2) 0 δ δ 0 e( iδωt 2) 0 e 0 δe δe ( iδωt 2) 0 (iδωt 2) 0 (iδωt 2) 0 δe (iδωt) = δe ( iδωt) 0 We can write the complex exponentials as cosines and sines (Euler's formula) e ix = cos x + isin x ˆ σ (t) = e iωt = cosωt + isinωt 0 δe (iδωt) 0 cos(δωt) 0 isin(δωt) = δ +δ δe ( iδωt) 0 cos(δωt) 0 isin(δωt) 0 Results are expected (same as predicted by Bloch equations) σˆ 12 (t) = cos(δωt)+ isin(δωt) σˆ 21 (t) = cos(δωt) isin(δωt) M x (t) = M x,0 cos(ω 0 t) M y,0 sin(ω 0 t) M y (t) = M y,0 cos(ω 0 t) + M x,0 sin(ω 0 t)

Rotation operators for an RF pulse along x-axis in the rotating frame In the rotating frame, an RF pulse, say along the x-axis, produces a magnetic field (B 1 ) in the x-direction In the rotating frame, the Hamiltonian is ˆR x H ˆ ʹ = (γb1 The operator describes rotation around the x-axis (RF pulse along the x-axis)!) I ˆʹ x σ ˆ(t) = e ( (i!) H ˆ ʹt) σ ˆ(0) e ((i!) H ˆ ʹ t) = ˆR x σ ˆ(0) ˆR 1 x Problem: the I x operator doesn't commute with the Hamiltonian that gave rise to our basis set functions -can't simply insert I x in the exponential operator to evaluate matrix elements (I x mixes spin states, creates transverse magnetization, coherences) ˆR x Operations of operators don't simply cause elements of σ(t) to oscillate, but convert diagonal elements (z-magnetization, σ 11, σ 22 ) to off-diagonal elements (x- and y-magnetization, coherences) in a sin(b 1 γt) dependent manner

Rotation operators for an RF pulse along x-axis in the rotating frame The R x operator in the rotating frame is (for spin ½) ˆR 1 x = cos(ω t 2) isin(ω t 2) 1 1 ˆR x = cos(ω t 2) isin(ω t 2) 1 1 isin(ω 1 t 2) cos(ω 1 t 2) isin(ω 1 t 2) cos(ω 1 t 2) - here ω 1 is the frequency (radians/sec) of rotation due to the B 1 field - ω 1 t is the pulse angle (radians): ω 1 t = π/2 (90 pulse), ω 1 t = π (180 pulse) The most common pulses in NMR are 90 and 180 pulses, so and we can easily write R x operators for these - substituting cos(ω 1 t 2) = cos((π 2) 2) = ± (1+ cos(π 2)) 2 = ± 1 2 = ± 2 2 sin(ω 1 t 2) = sin((π 2) 2) = ± (1 cos(π 2)) 2 = ± 1 2 = ± 2 2 cos(ω 1 t 2) = cos(π 2) = 0 sin(ω 1 t 2) = sin(π 2) =1 ˆR x,π /2 = 2 2 1 i i 1 1 ˆR x,π /2 = 2 2 1 i i 1 ˆR x,π = 0 i i 0 The R y operators are derived in the same manner ˆR 1 x,π = 0 i i 0

One Pulse and FID Example: using the R x and R z operators, analyze the effect of a 90 pulse along the x-axis on equilibrium magnetization, and then monitoring precession of the transverse magnetization σ ˆ(t) = ˆR z (t) ˆR x σ ˆ(0) ˆR 1 x ˆRz 1 (t) First, using the R x operators, apply a 90 pulse along the x-axis to equilibrium magnetization Î y = 1 0 i Recall and Îy = 1 2 i 0 2 σ ˆ(t) = e ( (i!) H ˆ ʹt) σ ˆ(0) e ((i!) H ˆ ʹ t) = ˆR x σ ˆ(0) ˆR 1 x Here, at t = 0 (equilibrium), the equilibrium deviation density matrix (σ(0)=σ eq ) can be written as σ ˆ(t) = ˆR 1 x,π /2 σˆ eq ˆR x,π /2 σˆ eq = 1 γ!b 0 T 0 1 δ 0 M z (eq) 2 0 γ!b 0 T 2 0 δ So, σ(t) (time just after the pulse) is = 2 2 1 i 1 δ 0 i 1 2 0 δ 2 2 1 i = 1 1 i δ iδ = 1 0 2iδ = δ 1 0 i i 1 4 i 1 iδ δ 4 2iδ 0 2 i 0 0 i i 0 - so, the result gives -y magnetization (-I y )

Evolution of the FID Starting just after the pulse (call this t 0 ) we can calculate the density matrix at each point in the FID (t 1, t 2, t 3, etc., where the time between points, Δt, is the dwell time) using R z operators σ ˆ(t 1 ) = ˆR z (t 1 ) σ ˆ(t 0 ) ˆR 1 z (t 1 ) σ ˆ(t 2 ) = ˆR z (t 2 ) σ ˆ(t 1 ) ˆR 1 z (t 2 ) σ ˆ(t 3 ) = ˆR z (t 3 ) σ ˆ(t 2 ) ˆR 1 z (t 3 ) σ ˆ(t 1 ) = ˆR z (t 1 ) σ ˆ(t 0 ) ˆR 1 z (t 1 ) = e(iδωt 1 2) 0 0 iδ 0 e ( iδωt 1 2) iδ 0 0 0 e (iδωt 1 2) e( iδωt1 2) Once the density matrix for each point in the FID is known, the expectation value for the amplitude of the magnetization can be calculated at each point (thus, we have calculated the FID) { } M x (t) = N Tr{ σ ˆ(t) [ ˆµ x ]} M y (t) = N Tr σ ˆ(t) ˆµ y Programs like GAMMA, SPINEVOLUTION work this way - Smith, S. A., Levante, T. O., Meier, B. H., and Ernst, R. R. (1994) J. Magn. Reson. A 106, 75-105 - Veshtort, M., and Griffin, R. G. (2006) J. Magn. Reson. 178, 248-282

Density Matrix Simulation of 2nd Order Spectra For two spin basis set is: αα, αβ, βα, ββ ρ = ρ11 ρ21 ρ31. ρ12 ρ22.. ρ13....... ρ 12 +ρ 21 ρ 34 +ρ 43 ρ 13 +ρ 31 ρ 24 +ρ 42 ρ 12,ρ 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 ψ is not an eigen function of H. c i φ I + c i φ I may evolve coherently as one line ie ρ 12,ρ 13, are mixed Calculation of Mx still works for a second order system