Lecture Density Matrix Formalism

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Lecture Density Matrix Formalism l used to describe the state of a spin ensemble, as well as its evolution in alue X-component of the magnetic moment of nucleus A: ve function and is a linear combination of the eigenstates of the form: solution of the time-independent Schroedinger equation. The “bra, <n| and the angular momentum operator can be written in the matrix form as:

description

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: - PowerPoint PPT Presentation

Transcript of Lecture Density Matrix Formalism

Page 1: Lecture  Density Matrix Formalism

Lecture Density Matrix FormalismA 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:

IXA =

Thus:

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The N2 terms can be put in the matrix form as follow:

Where = d*mn, i.e. D is a Hermitian matrix

Thus, = NoA

The angular momentum operators for spin ½ systems are:

For spin 1:

For a coupled A(½ )X(½) system

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Using the expression: = -(4/p)MoA(d11/2 – d22/2 + d33/2 – d44/2)

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

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For an isolated spin ½ system:

For A(½)X(½) system:

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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: 1QA: 12 (|++> |-+>)= A + J/2; 24 (|-+> |--

>)= X - J/2; 1Qx: 13 (|++> |+->)= X + J/2; 34 (|

+-> |-->)= A - J/2;

Other unobservable transitions are: Double quantum transition 2QAX (Flip-flip): |++> |--

> (flop-flop): |--> |++>Zero quantum transitions (flip-flop): ZQAX: |+-> |-+> 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)

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1. Equilibrium populations:

At 4.7 T:

For a CH system, A = 13C and X = 1H and x 4C q 4p

Thus, Hence: where 4

Therefore, Unitary matrix

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2. The first pulse: where

The pulse created 1QX (proton) (non-vanishing d13 and d24)

3. Evolution from t(1) to t(2):

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To calculate D(2) we need to calculate the evolution of only the non-vanishing elements, i.e. d13 and d24

in the rotating frame.

are the rotating frame resonance frequencies of spinA 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. 13C):

D(3) = R180XCD(2)R-1180XC =

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

+

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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

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D(7) = R180XCD(2)R-1180XC = D(5)

Proton magnetization, d13 and d24 has been transferred to the carbon magnetization, d12 and d34 with and

8. The role of 2: Signal is proportional to d12+d34 we can’t detect signal at this time otherwise s will cancel out.The effect of 2 is as follow (Only non-vanishing elements, d12 and d13 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 13C evolve. Thus,

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As described in Appendix B, in a quasrature detectin mode the total magnetization MTC is:

if we reintroduce the p/4 factor.

Thus,

This is the final signal to be detected. The 13C signal evolve during detection time, td at a freqquency H and is amplitude modulated by proton evolution The. Fourier transform with respect to td and te 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 180o pulse during te we will see spectrum I.3.b If there is also no 1H decoupling is during detection we will get spectrum I.3.c.

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