Lecture Density Matrix Formalism A tool used to describe the state of a spin ensemble, as well as its evolution in time. The expectation value X-component.

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Lecture Density Matrix Formalism A tool used to describe the state of a spin ensemble, as well as its evolution in time. The expectation value X-component of the magnetic moment of nucleus A: Where  is the wave function and is a linear combination of the eigenstates of the form: Where |n> are the solution of the time-independent Schroedinger equation. The “bra, <n| and “ket, |n>”, and the angular momentum operator can be written in the matrix form as: I XA = Thus:

The N 2 terms can be put in the matrix form as follow: Where = d* mn, i.e. D is a Hermitian matrix Thus, = N o  A The angular momentum operators for spin ½ systems are: For spin 1: For a coupled A(½ )X(½) system

Using the expression: = -(4/p)M oA (d 11 /2 – d 22 /2 + d 33 /2 – d 44 /2) Where And Remember and Similarly: and In modern NMR spectrometers we normally do quadrature detection, i.e. For nucleus A we have: Similarly, for nucleus X: The density matrix at thermal equilibrium: Thus, if n ≠ m and Evolution of the density matrix can be obtained by solving the Schoedinger equation to give: Effect of radiofrequency pulse: Where R is the rotation matrix

For an isolated spin ½ system: For A(½)X(½) system:

Density matrix description of the 2D heteronuclear correlated spectroscopy For a coupled two spin ½ system, AX there are four energy states (Fig. I.1); (1) |++>; (2) |-+>; (3) |+->, and (4) |-->. The resonance frequencies for observable single quantum transitions (flip or flop) among these states are: 1Q A: 12 (|++>  |-+>)= A + J/2; 24 (|-+>  |-- >)= X - J/2; 1Q x : 13 (|++>  |+->)= X + J/2; 34 (|+->  |-->)= A - J/2; Other unobservable transitions are: Double quantum transition 2Q AX (Flip-flip): |++>  |--> (flop-flop): |-->  |++> Zero quantum transitions (flip-flop): ZQ AX : |+->  |-+> or |-+>  |+-> Density matrix of the coupled spin system is shown on Table I.1. The diagonal elements are the populations of the states. The off-diagonal elements represent the probabilities of the corresponding transitions. (uncoupled)(coupled) (1) (2) (3) (4)

1. Equilibrium populations: At 4.7 T: For a CH system, A = 13 C and X = 1 H and x  4 C  q  4p Thus, Hence: where  4 Therefore, Unitary matrix

2. The first pulse: where  The pulse created 1Q X (proton) (non-vanishing d 13 and d 24 ) 3. Evolution from t(1) to t(2):

To calculate D(2) we need to calculate the evolution of only the non-vanishing elements, i.e. d 13 and d 24 in the rotating frame. are the rotating frame resonance frequencies of spin A and X, respectively, and  TrH is the transmitter (or reference) frequency. Hence: where B* and C* are the complex conjugates of B and C, respectively. 4. The second pulse (rotation w.r.t. 13 C): D(3) = R 180XC D(2)R XC = 5. Evolution from t(3) to t(4): (  is lab frame and  is rotating frame resonance frequency) Substituting B and C into the equations we get:  J is absent  Decoupled due to spin echo sequence +

6. The role of  1 (Evolution with coupling): and Let  = 1/2J and We have: Let Thus, 7. The third and fourth pulses: Combine the two rotations into one and

D(7) = R 180XC D(2)R -1180XC = D(5)  Proton magnetization, d 13 and d 24 has been transferred to the carbon magnetization, d 12 and d 34 with and 8. The role of  2 : Signal is proportional to d 12 +d 34 we can’t detect signal at this time otherwise s will cancel out. The effect of  2 is as follow (Only non-vanishing elements, d 12 and d 13 need to be considered: Hence:  For  2 = 0 the terms containing s cancel.  For  2 = 1/2J we have: 9. Detection: During this time proton is decoupled and only 13 C evolve. Thus,

As described in Appendix B, in a quasrature detectin mode the total magnetization M TC is: if we reintroduce the p/4 factor. Thus,  This is the final signal to be detected. The 13 C signal evolve during detection time, t d at a freqquency  H and is amplitude modulated by proton evolution  The. Fourier transform with respect to t d and t e results in a 2D HETCOR spectrum as shown on Fig. I.3a. The peak at -  H is due to transformation of sine function due to The negative peak can be removed by careful placing the reference frequency and the spectral width or by phase cycling.  If there is no 180 o pulse during t e we will see spectrum I.3.b  If there is also no 1 H decoupling is during detection we will get spectrum I.3.c.