CHEM 991J Lecture 321 11/18/2011: T1 relaxation

1. The equation of motion We now do an ensemble average on this over all values of the random Hamiltonian. We can eliminate the first term, because if it gave a non-zero average value, this would appear as a frequency and be eliminated in the interaction representation. We also assume that the time scale of the change of the density matrix is very slow compared with the time scale of the motions, and so replace the limits of integration with 0 to ∞ Now we turn to the Hamiltonian. In the static frame we can write it as: CHEM 991J Lecture 321 11/18/2011: T1 relaxation Two adjustments. First, only the R part is time dependent (on the time scale of molecular motion). Second, only the J part is changed on going to the interaction representation. So This follows from Wigner’s definition of an irreducible tensor operator. Note that this allows for unlike I and S spins: m = mI + mS .

2. The equation of motion If I and S are the same nucleus, we get Inserting this in the equation of motion of the density matrix We get: Again, ρ and J are time independent on the time scale of molecular motion, so we can pull them out of the integral and out of the ensemble average. We can also remove the R functions from the commutators since they commute with the spin operators. CHEM 991J Lecture 321 11/18/2011: T1 relaxation , where θ and ϕ are polar angles, unless m = –m′, because R are irreducible and therefore orthogonal functions, so we can remove one sum.

3. The equation of motion Let’s look at the integral First, we can exclude the exponential term from the ensemble average, since it doesn’t vary over the sample. We define the part inside the angle brackets as an autocorrelation function. If we set τ = 0, and integrate over all orientations to get an average, we obtain, for all five rank-two spatial functions: We assume the functional form of f2m is a decaying exponential with a time constant τc. For random isotropic motion, it turns out this time constant in independent of m. Putting this back in above, we find the integral is now a Fourier transform CHEM 991J Lecture 321 11/18/2011: T1 relaxation G(mω) is usually given the designation J(mω) and is called a spectral density function.

4. T1 relaxation So now we go back to ...and deal with the double commutator Let ρ′(0) = Iz + Sz. Then [J20,[J20, ρ′(0)]] = 0. [J2-1,[J21,(Iz + Sz)]] = [J2-1,[J21, J10]] = –[J2-1,[J10, J21]] = –[J2-1,J21] –[J2-1,J21] = –2[(I–1Sz + IzS–1), (I+1Sz + IzS+1)] = –2([I–1Sz, I+1Sz] +[I–1Sz, IzS+1] + [IzS–1, I+1Sz] + [IzS–1, IzS+1]) [I–1, I+1] = -½([(Ix – iIy),(Ix + iIy)] = Iz [I–1, I+1]Sz2 = ¼Iz Iz2[S–1, S+1] = ¼Sz [AB, CD] = A[B,C]D + AC[B,D] + [A,C]DB + C[A,D]B [I–1Sz, I+1Sz] = I–1[Sz, I+1]Sz +I–1I+1[Sz, Sz] + [I–1, I+1]SzSz + I+1[I–1, Sz]Sz = 0 +0 + IzSz2 + 0 = ¼Iz CHEM 991J Lecture 321 11/18/2011: T1 relaxation [I–1Sz, IzS+1] = I–1[Sz, Iz]S+1 +I–1Iz[Sz,S+1] + [I–1, Iz]S+1Sz + Iz[I–1, S+1]Sz = 0 +I–1IzS+1 + I–1S+1Sz + 0 = ½(I–1S+1 – I–1S+1) = 0 [IzS–1, I+1Sz] = Iz[S–1, I+1]Sz + IzI+1[S–1,Sz] + [Iz, I+1]SzS–1 + I+1[Iz,Sz]S–1 = 0 + IzI+1S–1 + I+1SzS–1 + 0 = ½(I+1S–1 – I+1S–1) = 0 [IzS–1, IzS+1] = Iz[S–1,Iz]S+1 + IzIz[S–1, S+1] + [Iz, Iz]S+1S–1 + Iz[Iz, S+1]S–1 = 0 + Iz2Sz + 0 + 0 = ¼Sz Total: [J2-1,[J21,(Iz + Sz)]] = –2(¼Iz + ¼Sz) = –½(Iz + Sz)

5. T1 relaxation So now we go back to ...and deal with the double commutator [J2-2,[J22,(Iz + Sz)]] = [J2-2,[J22, J10]] = –[J2-2,[J10, J22]] = –2[J2-2,J22] = –8[I–1S–1, I+1S+1] [AB, CD] = A[B,C]D + AC[B,D] + [A,C]DB + C[A,D]B [I–1S–1, I+1S+1] = I–1[S–1, I+1]S+1 + I–1I+1[S–1,S+1] + [I–1,I+1]S+1S–1+I+1 [I–1,S+1]S–1 = 0 + I–1I+1Sz + Iz S+1S–1 + 0 = I–1I+1Sz + Iz S+1S–1 I–1I+1 = –½(Ix – iIy)(Ix + iIy) = –½(Ix2 + Iy2 + i[Ix,Iy]) = –½(Ix2 + Iy2 – Iz) = –½(½– Iz) S+1S–1 = –½(Sx + iSy)(Sx – iSy) = –½(Sx2 + Sy2 – i[Sx,Sy]) = –½(Sx2 + Sy2 + Sz) = –½(½+Sz) CHEM 991J Lecture 321 11/18/2011: T1 relaxation –8(I–1I+1Sz + Iz S+1S–1) = –8(–½(½– Iz)Sz –½Iz(½+Sz) = –8(–¼Sz –¼Iz) = ½(Iz + Sz) Likewise [J22,[J2–2,(Iz + Sz)]] = ½(Iz + Sz)