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 Iz and the single spin functions α and β - recall ˆ ˆ ˆ ˆ ˆ ˆ Ix α = 1 2 β Ix β = 1 2α Iy α = 1 2 iβ Iy β = −1 2 iα Iz α = +1 2α Iz β = −1 2 β α α = β β =1 α β = β α = 0 - recall the expectation value for an observable Q = ψ Qˆ ψ = ∫ ψ∗Qˆψ dτ Qˆ - some operator ψ - some wavefunction - the matrix representation is the possible expectation values for the basis functions α β α ⎡ α Iˆ α α Iˆ β ⎤ ⎢ z z ⎥ ⎢ ˆ ˆ ⎥ β ⎣ β Iz α β Iz β ⎦

⎡ α Iˆ α α Iˆ β ⎤ ⎡1 2 α α −1 2 α β ⎤ ⎡1 2 0 ⎤ ⎡1 0 ⎤ ˆ ⎢ z z ⎥ 1 Iz = = ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎢ ˆ ˆ ⎥ 1 2 β α −1 2 β β 0 −1 2 2 0 −1 ⎣ β Iz α β Iz β ⎦ ⎣⎢ ⎦⎥ ⎣ ⎦ ⎣ ⎦ • 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 Operators in Matrix Notation

• The matrices for Ix, Iy,and Iz are called the Pauli spin matrices ˆ ⎡ 0 1 2⎤ 1 ⎡0 1 ⎤ ˆ ⎡0 −1 2 i⎤ 1 ⎡0 −i ⎤ ˆ ⎡1 2 0 ⎤ 1 ⎡1 0 ⎤ Ix = ⎢ ⎥ = ⎢ ⎥ Iy = ⎢ ⎥ = ⎢ ⎥ Iz = ⎢ ⎥ = ⎢ ⎥ ⎣1 2 0 ⎦ 2 ⎣1 0⎦ ⎣1 2 i 0 ⎦ 2 ⎣i 0 ⎦ ⎣ 0 −1 2⎦ 2 ⎣0 −1 ⎦

• Express α , β , α and β as 1×2 column and 2×1 row vectors ⎡1⎤ ⎡0⎤ α = ⎢ ⎥ β = ⎢ ⎥ α = [1 0] β = [0 1] ⎣0⎦ ⎣1⎦

• Using matrices, the operations of Ix, Iy, and Iz on α and β , and the orthonormality relationships, are shown below

ˆ 1 ⎡0 1⎤⎡1⎤ 1 ⎡0⎤ 1 ˆ 1 ⎡0 1⎤⎡0⎤ 1 ⎡1⎤ 1 Ix α = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = β Ix β = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = α 2⎣1 0⎦⎣0⎦ 2⎣1⎦ 2 2⎣1 0⎦⎣1⎦ 2⎣0⎦ 2 ⎡1⎤ ⎡0⎤ α α = [1 0]⎢ ⎥ =1 β β = [0 1]⎢ ⎥ =1 0 1 ˆ 1 ⎡0 −i⎤⎡1⎤ 1 ⎡0⎤ 1 ˆ 1 ⎡0 −i⎤⎡0⎤ 1 ⎡−i⎤ 1 ⎣ ⎦ ⎣ ⎦ Iy α = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = i β Iy β = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = −i α 2⎣ i 0⎦⎣0⎦ 2⎣i ⎦ 2 2⎣ i 0⎦⎣1⎦ 2⎣ 0⎦ 2 ⎡0⎤ ⎡1⎤ α β = [1 0]⎢ ⎥ = 0 β α = [0 1]⎢ ⎥ = 0 ˆ 1 ⎡1 0 ⎤⎡1⎤ 1 ⎡1⎤ 1 ˆ 1 ⎡1 0 ⎤⎡0⎤ 1 ⎡ 0 ⎤ 1 ⎣1⎦ ⎣0⎦ Iz α = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = α Iz β = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = − β 2⎣0 −1⎦⎣0⎦ 2⎣0⎦ 2 2⎣0 −1⎦⎣1⎦ 2⎣−1⎦ 2 Operators in Matrix Notation • Likewise, we can express αα αβ βα ββ αα ⎡ αα Iˆ αα αα Iˆ αβ αα Iˆ βα αα Iˆ ββ ⎤ the two-spin operators in ⎢ Ax Ax Ax Ax ⎥ ⎢ ˆ ˆ ˆ ˆ ⎥ matrix form αβ αβ IAx αα αβ IAx αβ αβ IAx βα αβ IAx ββ ⎢ ⎥ - example: consider the ˆ ˆ ˆ ˆ βα ⎢ βα IAx αα βα IAx αβ βα IAx βα βα IAx ββ ⎥ expectation value for the Ix ⎢ ⎥ Iˆ Iˆ Iˆ Iˆ operator on the A spin in an ββ ⎣⎢ ββ Ax αα ββ Ax αβ ββ Ax βα ββ Ax ββ ⎦⎥ AX system in the two spin basis ˆ ˆ ˆ ˆ ˆ ˆ (recall I x α = 1 2 β I x β = 1 2 α I y α = 1 2 i β I y β = − 1 2 iα I z α = 1 2 α I z β = − 1 2 β )

⎡1 2 αα βα 1 2 αα ββ 1 2 αα αα 1 2 αα αβ ⎤ ⎡ 0 0 1 2 0 ⎤ ⎡0 0 1 0⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1 2 αβ βα 1 2 αβ ββ 1 2 αβ αα 1 2 αβ αβ ˆ ⎢ ⎥ ⎢ 0 0 0 1 2⎥ 1 ⎢0 0 0 1⎥ IAx = ⎢ ⎥ = = 1 2 βα βα 1 2 βα ββ 1 2 βα αα 1 2 βα αβ ⎢ 1 2 0 0 0 ⎥ 2⎢1 0 0 0⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 0 1 2 0 0 0 1 0 0 ⎣1 2 ββ βα 1 2 ββ ββ 1 2 ββ αα 1 2 ββ αβ ⎦ ⎣ ⎦ ⎣ ⎦

• The matrices for Ix, Iy,and Iz are

⎡0 0 1 0⎤ ⎡0 1 0 0⎤ ⎡0 0 −i 0⎤ ⎡0 −i 0 0⎤ ⎡1 0 0 0 ⎤ ⎡1 0 0 0 ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ 1 0 0 0 1 1 1 0 0 0 1 0 0 0 −i 1 i 0 0 0 1 0 1 0 0 1 0 −1 0 0 Iˆ = ⎢ ⎥ Iˆ = ⎢ ⎥ Iˆ = ⎢ ⎥ Iˆ = ⎢ ⎥ Iˆ = ⎢ ⎥ Iˆ = ⎢ ⎥ Ax 2⎢1 0 0 0⎥ Xx 2⎢0 0 0 1⎥ Ay 2⎢i 0 0 0⎥ Xy 2⎢0 0 0 −i⎥ Az 2⎢0 0 −1 0⎥ Xz 2⎢0 0 1 0 ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎣0 1 0 0⎦ ⎣0 0 1 0⎦ ⎣0 i 0 0⎦ ⎣0 0 i 0⎦ ⎣0 0 0 −1⎦ ⎣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 ( and ) α β ∗ ∗ ∗ ψ = c1 α + c2 β = ∑cjφ j ψ = c1 α + c2 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 ⎡1⎤ ⎡0⎤ α = ⎢ ⎥ β = ⎢ ⎥ α = [1 0] β = [0 1] ⎣0⎦ ⎣1⎦

⎡1⎤ ⎡0⎤ ⎡c1 ⎤ ∗ ∗ ∗ ∗ ∗ ∗ ψ = c1 α + c2 β = c1 ⎢ ⎥+ c2 ⎢ ⎥ = ⎢ ⎥ ψ = c1 α + c2 β = c1 [1 0]+ c2[0 1] = [c1 c2 ] ⎣0⎦ ⎣1⎦ ⎣c2 ⎦ • So, the wavefunction (and its complex conjugate) can be written in terms of vectors of the coefficients only Density Matrices and Observables for an Ensemble of Spins • The expectation value for some observable represented by the operator Q, operating on a wavefunction ψ can be written as

Q = ψ Qˆ ψ ∗ ∗ ∗ ψ = c1 α + c2 β = ∑cjφ j ψ = c1 α + c2 β = ∑c jφ j j j ⎡ ∗ ∗ ⎤ ˆ ⎡ ⎤ = ⎣c1 α + c2 β ⎦ Q ⎣c1 α + c2 β ⎦ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ = c1 c1 α Q α + c1 c2 α Q β + c2c1 β Q α + c2c2 β Q β = ∑c jck ψ j Q ψk j • Substitute the matrix/vector equivalents ⎡c ⎤ ˆ 1 ∗ ∗ Q = ψ Q ψ ψ = ⎢ ⎥ ψ = [c1 c2 ] ⎣c2 ⎦ ⎡ α Qˆ α α Qˆ β ⎤⎡c ⎤ ⎡c α Qˆ α + c α Qˆ β ⎤ ∗ ∗ ⎢ ⎥ 1 ∗ ∗ ⎢ 1 2 ⎥ = [c1 c2 ] ⎢ ⎥ = [c1 c2 ] ⎡ ˆ ˆ ⎤ α Qz α α Qz β ⎢ ˆ ˆ ⎥⎣c2 ⎦ ⎢ ˆ ˆ ⎥ ⎣ β Q α β Q β ⎦ ⎣c1 β Q α + c2 β Q β ⎦ Qˆ = ⎢ ⎥ ⎢ ˆ ˆ ⎥ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ ⎣ β Qz α β Qz β ⎦ = c1 c1 α Q α + c1 c2 α Q β + c2c1 β Q α + c2c2 β Q β = ∑c jck ψ j Q ψk j • Note the expression for Q ˆ includes products of the coefficients, rather than the coefficients themselves, suggesting a way to represent the spin state (wavefunction) in terms of the products

⎡c ⎤ ⎡c c∗ c c∗ ⎤ ψ ψ = 1 [c∗ c∗ ] = ⎢ 1 1 1 2 ⎥ - note: NOT ψ ψ = 1 ! ⎢ ⎥ 1 2 ∗ ∗ ⎣c2 ⎦ ⎣c2c1 c2c2 ⎦ 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 Qˆ = Tr{ ψ ψ Qˆ}

⎧ ∗ ∗ ⎡ ˆ ˆ ⎤⎫ ⎡ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ ⎤ ⎪⎡c c c c ⎤ α Q α α Q β ⎪ c1c1 α Q α + c1c2 β Q α c1c1 α Q β + c1c2 β Q β = Tr⎨⎢ 1 1 1 2 ⎥⎢ ⎥⎬ = Tr⎢ ⎥ c c∗ c c∗ ⎢ ˆ ˆ ⎥ ⎢ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ ⎥ ⎩⎪⎣ 2 1 2 2 ⎦⎣ β Q α β Q β ⎦⎭⎪ ⎣c2c1 α Q α + c2c2 β Q α c2c1 α Q β + c2c2 β Q β ⎦ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ = c1c1 α Q α + c1c2 β Q α + c2c1 α Q β + c2c2 β Q β ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ ∗ ˆ = c1 c1 α Q α + c1 c2 α Q β + c2c1 β Q α + c2c2 β Q β = ∑c jck ψ j Q ψk j • Now….for an ensemble of spins, the magnitude of an observable is the sum of the expectation values for each spin ˆ ˆ ˆ ˆ Q = ψ1 Q ψ1 + ψ2 Q ψ2 + ψ3 Q ψ3 +...... ψN Q ψN

Tr ...... Qˆ = {( ψ1 ψ1 + ψ2 ψ2 + ψ3 ψ3 + ψN ψN ) } • The average value of an observable for an ensemble is N ˆ Q = (1 N)∑ ψi Q ψi i=1 Tr ⎡ 1 N ...... ⎤Qˆ = {⎣( )( ψ1 ψ1 + ψ2 ψ2 + ψ3 ψ3 + ψN ψN )⎦ } - 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 ˆ ρ = (1 N)( ψ1 ψ1 + ψ2 ψ2 + ψ3 ψ3 ) +...... ψN ψN = ψ ψ • For an ensemble, the average value of the observable, and the average macroscopic observable, then are ˆ ˆ ˆ ˆ Q = Tr{ρQ} Qmacro = N Tr{ρQ} • 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

⎡ ∗ ∗ ⎤ ⎡ ∗ ∗ ⎤ c1c1 c1c2 ⎡ρ11 ρ12 ⎤ c1c1 c1c2 ρˆ = ψ ψ = ⎢ ⎥ = ⎢ ⎥ recall ψ ψ = ⎢ ⎥ ∗ ∗ ∗ ∗ ⎢ ⎥ ⎣ρ21 ρ22 ⎦ c c c c ⎣c2c1 c2c2 ⎦ ⎣ 2 1 2 2 ⎦

- ρ11 and ρ22 are the populations of states 1 and 2 (i.e. α and β for spin ½) - ρ12 and ρ21 are the coherences Solving for the Time Dependence of ρˆ • Our observables are time dependent (for example, bulk magnetization precesses with time) hence requiring Schrodinger's time-dependent equation c d(ψ(t)) dt = −i ! Hˆψ(t) ψ = ∑ jϕ j j • We'll consider a single spin with two possible states

ψ(t) = c1(t) ψ1 + c2 (t) ψ2 ψ1 ψ1 = ψ2 ψ2 =1 ψ1 ψ2 = ψ2 ψ1 = 0

ˆ ⎡ ⎤ d(c1(t) ψ1 + c2 (t) ψ2 ) dt = −i ! H⎣c1(t) ψ1 + c2 (t) ψ2 ⎦ ˆ ˆ d(c1(t)) dt ψ1 + d(c2 (t)) dt ψ2 = −i !c1(t) H ψ1 −i !c2 (t) H ψ2 ⎡ ˆ ˆ ⎤ ψ1 [d(c1(t)) dt] ψ1 + ψ1 [d(c2 (t)) dt] ψ2 = −i !⎣c1(t) ψ1 H ψ1 + c2 (t) ψ1 H ψ2 ⎦ ⎡ ˆ ˆ ⎤ [d(c1(t)) dt] ψ1 ψ1 + [d(c2 (t)) dt] ψ1 ψ2 = −i !⎣c1(t) ψ1 H ψ1 + c2 (t) ψ1 H ψ2 ⎦ ⎡ ˆ ˆ ⎤ d(c1(t)) dt = −i !⎣c1(t) ψ1 H ψ1 + c2 (t) ψ1 H ψ2 ⎦ multipled on left by ψ1 ⎡ ˆ ˆ ⎤ d(c2 (t)) dt = −i !⎣c1(t) ψ2 H ψ1 + c2 (t) ψ2 H ψ2 ⎦ multipled on left by ψ2 ˆ d(ck (t)) dt = −i !∑cn (t) ψk H ψn n ∗ ∗ ˆ d(ck (t)) dt = i !∑cn (t) ψn H ψk without proof n Solving for the Time Dependence of ρˆ • These equations tell us how the wavefunctions (coefficients) change with time ˆ ∗ ∗ ˆ d(ck (t)) dt = −i !∑cn (t) ψk H ψn d(ck (t)) dt = i !∑cn (t) ψn H ψk n n • Thus, we can write a general expression for the change with time of the density matrix d ψ ρˆ ψ d(c c∗ ) d(c∗ ) d(c ) k m = k m = c m + k c∗ dt dt k dt dt m ∗ ˆ ˆ ∗ = i ! ∑ckcn ψn H ψm − i ! ∑ ψk H ψn cncm n n ˆ ˆ ˆ ˆ = i ! ∑ ψk ρ ψn ψn H ψm − i ! ∑ ψk H ψn ψn ρ ψm n n ⎡ ˆ ˆ ˆ ˆ ⎤ = i !⎣ ψk ρH ψm − ψk Hρ ψm ⎦ dρˆ(t) ˆ ˆ ˆ = i ! ⎣⎡ρˆ⎦⎤⎡H⎤−⎡H⎤⎣⎡ρˆ⎦⎤ ≡ i !⎡ρˆ(t), H⎤ (commutator) dt { ⎣ ⎦ ⎣ ⎦ } ⎣ ⎦ • The last expression is called the Lioville - von Neuman equation - knowing ρ ˆ at some given time (i.e. at equilibrium, t = 0), can solve for the time dependence of ρ ˆ (i.e. ρ ˆ at any other time), so can calculate observables Q Tr ⎡ ˆ⎤⎡Qˆ⎤ Q N Tr ⎡ ˆ⎤⎡Qˆ⎤ = {⎣ρ⎦⎣ ⎦} macro = {⎣ρ⎦⎣ ⎦} Example: Precession of x Magnetization in B0 • The Bloch equation analysis demonstrated how x magnetization (Mx) precesses with time about the B0 axis

M x (t) = M x,0 cos(ω0t) − M y,0 sin(ω0t) • Here, we'll see if we can predict this behavior (roughly) using the QM methods we've outlined - recall ⎡ 0 1 2⎤ ˆ ⎡ ˆ⎤ ˆ ⎡ ˆ⎤ ˆ Q = Tr ⎣⎡ρ⎦⎤ Q Qmacro = N Tr ⎣⎡ρ⎦⎤ Q Ix = ⎢ ⎥ { ⎣ ⎦} { ⎣ ⎦} ⎣1 2 0 ⎦ - ask, "What elements of the density matrix are important for Ix?"

- determine I x (spin ½) ⎧⎡ρ ρ ⎤⎡ 0 1 2⎤⎫ ⎡ρ × 0 + ρ ×1 2 ρ ×1 2 + ρ × 0 ⎤ I Tr ⎡ ˆ⎤⎡Iˆ ⎤ Tr 11 12 Tr 11 12 11 12 1 2( ) x = {⎣ρ⎦⎣ x ⎦} = ⎨⎢ ⎥⎢ ⎥⎬ = ⎢ ⎥ = ρ12 + ρ21 ⎩⎣ρ21 ρ22 ⎦⎣1 2 0 ⎦⎭ ⎣ρ21 × 0 + ρ22 ×1 2 ρ21 ×1 2 + ρ22 × 0⎦

• This says x-magnetization is proportional to ρ12 and ρ21 (only those density matrix elements are relevant for x-magnetization • If we like, we can write a density matrix/operator with only elements ρ12 and ρ21 to analyze x-magnetization ⎡ 0 ρ ⎤ ˆ 12 ρx = ⎢ ⎥ ⎣ρ21 0 ⎦ Example: Precession of x Magnetization in B0

• How do the elements of ρ ˆ and ρˆ x vary with time? ˆ d⎣⎡ρx ⎦⎤ ⎡1 2 0 ⎤ = i !⎡ρˆ , Hˆ ⎤ = i ! ⎡ρˆ ⎤⎡Hˆ ⎤−⎡Hˆ ⎤⎡ρˆ ⎤ ˆ ˆ ⎣ x ⎦ {⎣ x ⎦⎣ ⎦ ⎣ ⎦⎣ x ⎦} H = −γ!B0 Iz = −γ!B0 ⎢ ⎥ dt ⎣ 0 −1 2⎦

⎧⎡ 0 ρ12 ⎤⎡1 2 0 ⎤ ⎡1 2 0 ⎤⎡ 0 ρ12 ⎤⎫ = i !(−γ!B0 )⎨⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎬ ⎩⎣ρ21 0 ⎦⎣ 0 −1 2⎦ ⎣ 0 −1 2⎦⎣ρ21 0 ⎦⎭

⎧⎡ 0 −1 2 ρ12 ⎤ ⎡ 0 1 2 ρ12 ⎤⎫ ⎡ 0 −ρ12 ⎤ = −iγB0 ⎨⎢ ⎥−⎢ ⎥⎬ = −iγB0 ⎢ ⎥ ⎩⎣1 2 ρ21 0 ⎦ ⎣−1 2 ρ21 0 ⎦⎭ ⎣ρ21 0 ⎦ ⎡ ˆ⎤ d⎣ρ⎦ ˆ ˆ ˆ = i !⎡ρˆ, H⎤ = i ! ⎣⎡ρˆ⎦⎤⎡H⎤−⎡H⎤⎣⎡ρˆ⎦⎤ dt ⎣ ⎦ { ⎣ ⎦ ⎣ ⎦ }

⎧⎡ρ11 ρ12 ⎤⎡1 2 0 ⎤ ⎡1 2 0 ⎤⎡ρ11 ρ12 ⎤⎫ = i !(−γ!B0 )⎨⎢ ⎥⎢ ⎥−⎢ ⎥⎢ ⎥⎬ ⎩⎣ρ21 ρ22 ⎦⎣ 0 −1 2⎦ ⎣ 0 −1 2⎦⎣ρ21 ρ22 ⎦⎭

⎧⎡1 2 ρ11 −1 2 ρ12 ⎤ ⎡ 1 2 ρ11 1 2 ρ12 ⎤⎫ ⎡ 0 −ρ12 ⎤ = −iγB0 ⎨⎢ ⎥−⎢ ⎥⎬ = −iγB0 ⎢ ⎥ ⎩⎣1 2 ρ21 −1 2 ρ22 ⎦ ⎣−1 2 ρ21 −1 2 ρ22 ⎦⎭ ⎣ρ21 0 ⎦

• So, both ρ12 and ρ21 change with time, and we can solve for their time dependence d (t) [ρ12 ] (iω0t) = −iγB (−ρ (t)) = iω ρ (t) ρ (t) = ρ (0)e = ρ (0)(cos(ω t) + i sin(ω t)) dt 0 12 0 12 12 12 12 0 0

d[ρ21(t)] (−iω t) = −iγB (ρ (t)) = −iω ρ (t) ρ (t) = ρ (0)e 0 = ρ (0)(cos(ω t)− i sin(ω t)) dt 0 21 0 21 21 21 21 0 0 Example: Precession of x Magnetization in B0 • So, both ρ12 and ρ21 change with time, and we can solve for their time dependence d (t) [ρ12 ] (iω0t) = −iγB (−ρ (t)) = iω ρ (t) ρ (t) = ρ (0)e = ρ (0)(cos(ω t) + i sin(ω t)) dt 0 12 0 12 12 12 12 0 0

d[ρ21(t)] (−iω t) = −iγB (ρ (t)) = −iω ρ (t) ρ (t) = ρ (0)e 0 = ρ (0)(cos(ω t)− i sin(ω t)) dt 0 21 0 21 21 21 21 0 0

• 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 ⎡ρˆ⎤⎡Qˆ⎤ for instance I = Tr ⎡ρˆ⎤⎡Iˆ ⎤ {⎣ ⎦⎣ ⎦} x {⎣ ⎦⎣ x ⎦} - we have a route to the time dependence of the density matrix dρˆ(t) ˆ ˆ ˆ = i ! ⎣⎡ρˆ⎦⎤⎡H⎤−⎡H⎤⎣⎡ρˆ⎦⎤ ≡ i !⎡ρˆ(t), H⎤ dt { ⎣ ⎦ ⎣ ⎦ } ⎣ ⎦ • So…..what don't we have - a place to start, for instance, at equilibrium (ρ at equilibrium, ρeq) - values for density matrix elements • Assume ρ at equilibrium, ρeq - presumably, we can decide what all elements of ρ are at some time, so choose t = 0 (equilibrium)

• What are the diagonal elements (ρnn)? - ρnn = cncn* = a probability of being in a particular state (i.e. α or β) - these we can get from Boltzman factors/statistics ∗ (−En (kBT )) (−En /(kBT )) for basis and small ρnn = cncn = e Z ≅1 Z − En (ZkBT) Z = ∑e =1+1= 2 α , β En - so, Z ≅ number of states n • For convenience, we define a deviation matrix, σ, with elements

ρnn ≅1 Z − En (ZkBT) =1 Z −σ nn Density Matrix at Equilibrium

• What are off-diagonal elements (ρnm) 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

∗ (−En (kBT )) ˆ ˆ ρnn = cncn = e Z ≅1 Z − En (ZkBT) =1 Z −σ nn H = −γ!B0 Iz E = −mγ!B0 ⎡1 2 0 ⎤ Z e(−En /(kBT )) 2 for spin 1 2 ˆ ˆ = ∑ = H = −γ!B0 Iz = −γ!B0 ⎢ ⎥ n ⎣ 0 −1 2⎦ • The density matrix, and deviation matrix, at equilibrium are: ⎡e(−En (kBT )) Z 0 ⎤ 1 ⎡e(−En (kBT )) 0 ⎤ 1 ⎡e(γ!B0 (2kBT )) 0 ⎤ ρˆ = = = eq ⎢ ( E (k T )) ⎥ ⎢ ( E (k T )) ⎥ ⎢ ( B (2k T )) ⎥ ⎣⎢ 0 e − n B Z⎦⎥ 2⎣⎢ 0 e − n B ⎦⎥ 2⎣⎢ 0 e −γ! 0 B ⎦⎥ ⎡−E Zk T 0 ⎤ ⎡−E k T 0 ⎤ ⎡γ!B 2k T 0 ⎤ ˆ n B 1 n B 1 0 B σ eq = ⎢ ⎥ = ⎢ ⎥ = ⎢ ⎥ ⎣ 0 −En ZkBT⎦ 2⎣ 0 −En kBT⎦ 2⎣ 0 −γ!B0 2kBT⎦ • There should be no time dependence of these at equilibrium: d⎡σˆ ⎤ ⎣ eq ⎦ ⎡ ˆ ˆ ⎤ ⎡ ˆ ⎤⎡ ˆ ⎤ ⎡ ˆ ⎤⎡ ˆ ⎤ = i ! σ eq, H = i ! ⎣σ eq ⎦ H − H ⎣σ eq ⎦ dt ⎣ ⎦ { ⎣ ⎦ ⎣ ⎦ }

⎧1 ⎡γ!B0 2kBT 0 ⎤⎡1 2 0 ⎤ 1 ⎡1 2 0 ⎤⎡γ!B0 2kBT 0 ⎤⎫ = i !(−γ!B0 )⎨ ⎢ ⎥⎢ ⎥− ⎢ ⎥⎢ ⎥⎬ ⎩2⎣ 0 −γ!B0 2kBT⎦⎣ 0 −1 2⎦ 2⎣ 0 −1 2⎦⎣ 0 −γ!B0 2kBT⎦⎭

⎧1 ⎡1 2γ!B0 2kBT 0 ⎤ 1 ⎡ 1 2γ!B0 2kBT 0 ⎤⎫ ⎡0 0⎤ = −iγB0 ⎨ ⎢ ⎥− ⎢ ⎥⎬ = ⎢ ⎥ = 0 ⎩2⎣ 0 1 2γ!B0 2kBT⎦ 2⎣ 0 1 2γ!B0 2kBT⎦⎭ ⎣0 0⎦ - so, no variation with time, as expected (diagonal matrices commute: AB=BA, so AB-BA=0 for diagonal matrices A and B) What about Mz? γ!⎡1 0 ⎤ ˆ ˆ ˆ ˆ ˆ ˆ Q = Tr{ρQ} Qmacro = N Tr{ρQ} µz = γ!Iz = ⎢ ⎥ 2 ⎣0 −1 ⎦ • At thermal equilibrium, the equilibrium bulk magnetization along the z-axis, Mz, is equal to what we have defined as M0 2 2 I 2 2 2 2 Nγ ! B 2 Nγ ! B I(I +1) Nγ ! B 0 0 for spin 0 M 0 ≈ ∑ m ≈ 1 2, M 0 ≈ kBT(2I +1) m=−I 3kBT 4kBT

• We can calculate this observable ⎧1 ⎡γ!B 2k T 0 ⎤ ⎡1 0 ⎤⎫ ⎡ ˆ ⎤ ˆ 0 B γ! M z,eq = M 0 = N Tr{⎣σ eq ⎦[µz ]} = N Tr⎨ ⎢ ⎥ ⎢ ⎥⎬ ⎩2⎣ 0 −γ!B0 2kBT⎦ 2 ⎣0 −1 ⎦⎭ 2 2 ⎧γ!⎡γ!B0 2kBT 0 ⎤⎫ Nγ ! B = N Tr⎨ ⎢ ⎥⎬ = 0 ⎩ 4 ⎣ 0 γ!B0 2kBT⎦⎭ (4kBT)

• Curie's law (Curie-Weiss law) for magnetic susceptibility in paramagnetic compounds N Aµ0 2 χ m = µ 3kBT - the dimensionless quantity χm is the molar magnetic susceptibility, µ0 is the permeability of free space (proportional to B0), µ is the Bohr magneton (J/T) What about Mx? γ!⎡0 1⎤ ˆ ˆ ˆ ˆ ˆ ˆ Q = Tr{ρQ} Qmacro = N Tr{ρQ} µx = γ!Ix = ⎢ ⎥ 2 ⎣1 0⎦ • At thermal equilibrium, no net, bulk magnetization is expected in the transverse plane (Mx or My) - this is confirmed by the following calculation

⎧1 ⎡γ!B 2k T 0 ⎤ ⎡0 1⎤⎫ ⎡ ˆ ⎤ ˆ 0 B γ! M x,eq = N Tr{⎣σ eq ⎦[µx ]} = N Tr⎨ ⎢ ⎥ ⎢ ⎥⎬ ⎩2⎣ 0 −γ!B0 2kBT⎦ 2 ⎣1 0⎦⎭

⎧γ!⎡ 0 γ!B0 2kBT⎤⎫ = N Tr⎨ ⎢ ⎥⎬ = 0 ⎩ 4 ⎣−γ!B0 2kBT 0 ⎦⎭

• Same result for My at equilibrium (zero)

• So, these methods reproduce our expectations based on what we know already about bulk equilibrium magnetization What elements give Mx? • 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?"

⎧⎡ρ11 ρ12 ⎤⎡ 0 1 2⎤⎫ ⎡ρ11 × 0 + ρ12 ×1 2 ρ11 ×1 2 + ρ12 × 0 ⎤ I Tr ⎡ ˆ⎤⎡Iˆ ⎤ Tr Tr 1 2( ) x = {⎣ρ⎦⎣ x ⎦} = ⎨⎢ ⎥⎢ ⎥⎬ = ⎢ ⎥ = ρ12 + ρ21 ⎩⎣ρ21 ρ22 ⎦⎣1 2 0 ⎦⎭ ⎣ρ21 × 0 + ρ22 ×1 2 ρ21 ×1 2 + ρ22 × 0⎦

- ask again, for Mx

⎧⎡σ σ ⎤ ⎡0 1⎤⎫ ⎧ ⎡σ σ ⎤⎫ ⎡ ˆ⎤ ˆ 11 12 γ! γ! 12 11 M x = Tr{⎣σ ⎦[µx ]} = Tr⎨⎢ ⎥ ⎢ ⎥⎬ = Tr⎨ ⎢ ⎥⎬ = γ! 2(σ 12 +σ 21) ⎩⎣σ 21 σ 22 ⎦ 2 ⎣1 0⎦⎭ ⎩ 2 ⎣σ 22 σ 21⎦⎭

- so, σ12 and σ21 (ρ12, ρ21) 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) dρˆ(t) ˆ ˆ ˆ = i ! ⎣⎡ρˆ⎦⎤⎡H⎤−⎡H⎤⎣⎡ρˆ⎦⎤ ≡ i !⎡ρˆ(t), H⎤ (commutator) dt { ⎣ ⎦ ⎣ ⎦ } ⎣ ⎦ dσˆ(t) ˆ ˆ ˆ = i ! ⎣⎡σˆ⎦⎤⎡H⎤−⎡H⎤⎣⎡σˆ⎦⎤ ≡ i !⎡σˆ(t), H⎤ (commutator) dt { ⎣ ⎦ ⎣ ⎦ } ⎣ ⎦ • The general solution to this latter equation (without proof) is: ˆ ˆ σˆ(t) = e(−(i !)Ht)σˆ(0) e((i !)Ht) = Rˆ (t)σˆ(0)Rˆ −1(t) - can write in terms of rotation operators ( Rˆ , R ˆ − 1) that include the Hamiltonian ˆ ˆ • The R z operator describes precession (B0 interacting with I z ) ˆ - the matrix elements of R z are ˆ ˆ ⎡1 2 0 ⎤ ˆ ⎡1 2 0 ⎤ H = −γ!B0 Iz = −γ!B0 ⎢ ⎥ Iz = ⎢ ⎥ ⎣ 0 −1 2⎦ ⎣ 0 −1 2⎦ ˆ ˆ ˆ ˆ ˆ (−(i !)H(t)) (−(i !)(−γ!B0Iz )t) (iγB0tIz ) (iωtIz ) Rz (t) = e = e = e = e ˆ - the matrix representation of R z is then ⎡ (iΔωt 2) ⎤ ˆ e 0 Rz = ⎢ ⎥ ⎣ 0 e(−iΔωt 2) ⎦ - if basis elements are eigenfunctions of then is diagonal and easy to evaluate

- for convenience, we'll write the deviation matrix for Mx as chemical shift offsets (δ) in σ12 and σ21 (elements involved in precession)

MX and My Precess at Δω

• For convenience, we'll write the deviation matrix for Mx as chemical shift offsets (δ) in σ12 and σ21 (elements involved in precession) ⎡0 δ⎤ σˆ(0) = ⎢ ⎥ ⎣δ 0⎦ ˆ • We can then calculate σ ˆ (t ) from σ ˆ ( 0 ) under the R z operator ⎡e(iΔωt 2) 0 ⎤⎡0 δ⎤⎡e(−iΔωt 2) 0 ⎤ ˆ ˆ ˆ ˆ −1 σ (t) = Rz (t)σ (0)Rz (t) = ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ 0 e(−iΔωt 2) ⎦⎣δ 0⎦⎣ 0 e(iΔωt 2) ⎦ ⎡e(iΔωt 2) 0 ⎤⎡ 0 δe(iΔωt 2) ⎤ ⎡ 0 δe(iΔωt) ⎤ = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ ⎣ 0 e(−iΔωt 2) ⎦⎣δe(−iΔωt 2) 0 ⎦ ⎣δe(−iΔωt) 0 ⎦ • We can write the complex exponentials as cosines and sines (Euler's formula) eix = cos x + isin x eiωt = cosωt + isinωt ⎡ 0 δe(iΔωt) ⎤ ⎡ 0 cos(Δωt)⎤ ⎡ 0 isin(Δωt)⎤ σˆ(t) = ⎢ ⎥ = δ⎢ ⎥+δ⎢ ⎥ ⎣δe(−iΔωt) 0 ⎦ ⎣cos(Δωt) 0 ⎦ ⎣−isin(Δωt) 0 ⎦ • Results are expected (same as predicted by Bloch equations) ˆ σ 12 (t) = cos(Δωt)+ isin(Δωt) M x (t) = M x,0 cos(ω0t) − M y,0 sin(ω0t)

σˆ 21(t) = cos(Δωt)− isin(Δωt) M y (t) = M y,0 cos(ω0t) + M x,0 sin(ω0t) 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 (B1) in the x-direction • In the rotating frame, the Hamiltonian is ˆ ˆ Hʹ = −(γB1 !)Iʹx ˆ • The R xoperator describes rotation around the x-axis (RF pulse along the x-axis) (−(i !)Hˆ ʹt) ((i !)Hˆ ʹt) ˆ ˆ −1 σˆ(t) = e σˆ(0) e = Rxσˆ(0)Rx

• Problem: the Ix operator doesn't commute with the Hamiltonian that gave rise to our basis set functions -can't simply insert Ix in the exponential operator to evaluate matrix elements (Ix mixes spin states, creates transverse magnetization, coherences) ˆ • Operations of R x operators don't simply cause elements of σ(t) to oscillate, but convert diagonal elements (z-magnetization, σ11, σ22) to off-diagonal elements (x- and y-magnetization, coherences) in a sin(B1γt) dependent manner Rotation operators for an RF pulse along x-axis in the rotating frame

• The Rx operator in the rotating frame is (for spin ½) ⎡cos( t 2) isin( t 2)⎤ ⎡cos( t 2) isin( t 2)⎤ ˆ ω1 − ω1 ˆ −1 ω1 ω1 Rx = ⎢ ⎥ Rx = ⎢ ⎥ ⎣−isin(ω1t 2) cos(ω1t 2)⎦ ⎣isin(ω1t 2) cos(ω1t 2)⎦

- here ω1 is the frequency (radians/sec) of rotation due to the B1 field - ω1t is the pulse angle (radians): ω1t = π/2 (90° pulse), ω1t = π (180° pulse) • The most common pulses in NMR are 90° and 180° pulses, so and we can easily write Rx operators for these

cos(ω1t 2) = cos((π 2) 2) = ± (1+ cos(π 2)) 2 = ± 1 2 = ± 2 2

sin(ω1t 2) = sin((π 2) 2) = ± (1− cos(π 2)) 2 = ± 1 2 = ± 2 2

cos(ω1t 2) = cos(π 2) = 0 sin(ω1t 2) = sin(π 2) =1

- substituting

⎡ 1 i⎤ ⎡1 i⎤ ⎡ 0 i⎤ ⎡0 i⎤ ˆ 2 − ˆ −1 2 ˆ − ˆ −1 Rx,π /2 = ⎢ ⎥ Rx,π /2 = ⎢ ⎥ Rx,π = ⎢ ⎥ Rx,π = ⎢ ⎥ 2 ⎣−i 1⎦ 2 ⎣i 1⎦ ⎣−i 0⎦ ⎣i 0⎦

• The Ry operators are derived in the same manner

Similar arguments lead to Y pulse operations One Pulse and FID

• Example: using the Rx and Rz operators, analyze the effect of a 90° pulse along the x-axis on equilibrium magnetization, and then monitoring precession of the transverse magnetization ˆ ˆ ˆ −1 ˆ −1 σˆ(t) = Rz (t)Rxσˆ(0)Rx Rz (t)

• First, using the Rx operators, apply a 90° pulse along the x-axis to equilibrium magnetization (−(i !)Hˆ ʹt) ((i !)Hˆ ʹt) ˆ ˆ −1 σˆ(t) = e σˆ(0) e = Rxσˆ(0)Rx • Here, at t = 0 (equilibrium), the equilibrium deviation density matrix (σ(0)=σeq) can be written as ⎡γ!B 2k T 0 ⎤ ⎡δ 0⎤ ˆ 1 0 B 1 σ eq = ⎢ ⎥ ≡ ⎢ ⎥∝ M z (eq) 2⎣ 0 −γ!B0 2kBT⎦ 2⎣0 −δ⎦ • So, σ(t) (time just after the pulse) is ⎡1 −i⎤ ⎡δ 0⎤ ⎡1 i⎤ ⎡ 1 −i⎤⎡ δ iδ ⎤ ⎡ 0 2iδ⎤ ⎡ 0 i ⎤ ˆ ˆ ˆ ˆ −1 2 1 2 1 1 1 σ (t) = Rx,π /2σ eqRx,π /2 = ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ = ⎢ ⎥⎢ ⎥ = ⎢ ⎥ = δ ⎢ ⎥ 2 ⎣−i 1⎦2⎣0 −δ⎦ 2 ⎣i 1⎦ 4⎣−i 1⎦⎣−iδ −δ⎦ 4⎣−2iδ 0 ⎦ 2⎣−i 0⎦

ˆ 1 ⎡0 −i⎤ ˆ 1 ⎡ 0 i ⎤ • Recall I y = ⎢ ⎥ and −Iy = ⎢ ⎥ 2 ⎣i 0⎦ 2 ⎣−i 0⎦ - so, the result gives -y magnetization (-Iy)

Evolution of the FID

• Starting just after the pulse (call this t0) we can calculate the density matrix at each point in the FID (t1, t2, t3, etc., where the time between points, Δt, is the dwell time) using Rz operators

ˆ ˆ −1 ˆ ˆ ˆ ˆ −1 ˆ ˆ ˆ ˆ −1 σˆ(t1) = Rz (t1)σˆ(t0 )Rz (t1) σ (t2 ) = Rz (t2 )σ (t1)Rz (t2 ) σ (t3 ) = Rz (t3 )σ (t2 )Rz (t3 )

⎡e(iΔωt1 2) 0 ⎤⎡ 0 iδ⎤⎡e(−iΔωt1 2) 0 ⎤ ˆ ˆ ˆ ˆ −1 σ (t1) = Rz (t1)σ (t0 )Rz (t1) = ⎢ ⎥⎢ ⎥⎢ ⎥ ⎣ 0 e(−iΔωt1 2) ⎦⎣−iδ 0⎦⎣ 0 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)

⎡ ˆ ⎤ ˆ ⎡ ˆ ⎤⎡ ˆ ⎤ M x (t) = N Tr{⎣σ (t)⎦[µx ]} M y (t) = N Tr{⎣σ (t)⎦⎣µy ⎦} • 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 ⎡ρ11 ρ12 ρ13 .⎤ ⎢ρ21 ρ22 . .⎥ ρ = ⎢ ⎥ ⎢ ρ 31 . . . ⎥ + + + + ⎢ ⎥ ρ12 ρ 21 ρ34 ρ 43 ρ13 ρ 31 ρ24 ρ 42 ⎣ . . . .⎦ • ρ12,ρ21, … 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.

• ciφI + ciφI may evolve coherently as one line – ie ρ12,ρ13, are mixed • Calculation of Mx still works for a second order system