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The vector model of nmr spectroscopy is a simple, very visual way of comprehending the results of pulses and delay times in many modern nmr experiments. We start by considering the state of the spins in solution when the sample is inserted into the magnet.

Just after the sample is put into the magnet the spins begin precessing about the external magnetic field, which is usually taken as the z-axis. In a sample of normal concentration there are something on the order of 10¹⁶ spins in the sample, all precessing out of phase with their magnetic moments pointing in random directions:

At this point, addition of these magnetic moment vectors will result in a net magnetic moment (or magnetisation, as it is frequently called) of zero since there are so many of them pointed in random directions ... we say that the spin distribution is isotropic. Over time, however, the axis about which the magnetic moments are precessing will change so that there are slightly more spins with a component of their magnetic moment aligned with the external field than against the field. Even though the spins are still precessing with random phases, these component vectors along the field axis add up to give a net magnetisation. At this point we say that the spin distribution is anisotropic.

In order to simplify the road to understanding the vector model we need an understanding of the rotating coordinate frame. Once having done this we can proceed:

In the rotating frame, if we choose the correct rotational frequency, the magnetic moment vector will appear to be motionless. Application of an rf pulse in the transverse (xy) plane will result in an oscillating magnetic field, B₁, which can viewed (in the laboratory frame) as a rotating vector. If we view this in the rotating frame this magnetic field will appear to be motionless, however, if the rotating frame frequency is chosen to be identical to that of B₁ ... also the effective external magnetic field will be zero. It is as if the external field vanishes and is replaced by B₁ which is at 90 degrees to the external field. What will happen? Well, what always happens with spin magnetic moments ... they will precess about the field that they find themselves in. If we are starting at equilibrium , there will be a net magnetisation aligned along the z-axis parallel to the external field, B. In the rotating frame, the external field vanishes and is replaced by B₁ ... the equilibrium magnetisation will begin to rotate around the B₁ field vector towards the -y axis:

To determine the direction of rotation we use the right-hand rule: line up the thumb along the B₁ axis and the direction of rotation is indicated by the fingers.

If we leave the B₁ field on long enough the magnetisation vector will reach the transverse or xy plane. In this case we say that we have issued a 90 degree pulse. How do we know how long to leave the transmitter on to do this? Well, the signal that we see from the magnetisation is only the portion that is projected onto the transverse plane because of the orientation of the receiver coils. Thus, if the equilibrium magnetisation is tipped by a small amount, only a small portion of this will be detectable in the form of a small signal. The further we tip the equilibrium magnetisation towards the transverse plane the larger the projection on the transverse plane and the larger the signal we will observe. Obviously it will be at a maximum when the tip angle is 90 degrees. Any more than that, and the signal strength will begin to decrease:

What now? With the magnetisation tipped towards the transverse plane what happens? Well, let's take the simplest case in which the magnetisation is tipped 90 degrees into the plane. If the larmor frequency of the nucleus is identical to that of the B₁ field the transverse magnetisation vector will be motionless and will simply decay back to the equilibrium condition. That is, the transverse magnetisation will decrease in size and the z-axis magnetisation will increase. If, on the other hand, the Larmor frequency of the nucleus is not the same as that of the B₁ field, but is say, 100 Hz more, then the effective magnetic field will be nonzero but small. This means that the transverse magnetisation will precess about B_eff with frequency gB_eff. This is known as chemical shift evolution:

Again, the right-hand rule applies to the direction of rotation. If the Larmor frequency is 100 Hz less than the rotating frame frequency then the magnetisation vector will rotate in the opposite direction to that in the figure, corresponding to a negative B_eff.

What about scalar coupling? Let's look at a doublet with a coupling constant of 20Hz. If we issue a pulse at the frequency of the chemical shift of the doublet then one of the peaks will rotate clockwise at 10Hz relative to the rotating frame frequency and the other will rotate counterclockwise at -10Hz, again relative to the rotating frame frequency:

We now have the tools needed to analyse the effects of pulses and evolution times on pulse sequences. Let's give it a try.

The spin-echo pulse sequence is simple but most useful in a multitude of nmr experiments. The sequence is:

(p/2)_-x - delay - p_x - delay - acquire

where p/2 and p refer to 90 degree and 180 degree pulses respectively. The x and y subscripts refer to the phases of the pulses. This can be viewed as the axis along which the B₁ field is applied. Now, consider what will happen to our doublet. During the delay after the initial (p/2)_-x pulse the counter-rotating vectors travel through an angle at their precession frequencies. The angle is determined by the length of the delay but, as we shall see, this time is unimportant to the result. The p_x pulse rotates each vector through 180 degrees about the x-axis. Then, during the next delay which is equal to the first delay, the vectors continue their precession through the same angle as they travelled during the first delay. They end up colinear or on top of on another on the -y axis. What is interesting here is that this result is not dependent on the value of J or the delay times. This will always be the case for any doublet, as long as the delay times are equal. (The first vector diagram here represents the state of the system after the initial (p/2)_-x pulse)

What will happen if the B₁ field is not at the same frequency as the chemical shift of the doublet? The two vectors will rotate, each with their own frequency as before. This time however, their frequencies relative to the rotating frame will be different ... they may not even be counter-rotating from the rotating frame point of view. This being the case one, labelled a in the diagram, will rotate faster than the other, labelled b. As before, they will each rotate through some angle during the first delay and then the p_x pulse will rotate them 180 degrees. Finally, the during the second delay, the vectors will rotate towards the -y axis and end up colinear, just as in the first case:

This is very interesting. It means that no matter where in the spectrum the double appears the two vectors that represent the doublet peaks will always realign at the end of the pulse sequence. In other words, the result of the experiment is chemical shift independent. This is of such importance that this particular pulse sequence is built in to many larger pulse sequences as a 'building block' that will get rid of chemical shift effects. Why would we want to do this? Well, in a long sequence of pulses and delays lasting perhaps milliseconds chemical shift evolution will contribute to frequency dependent phase errors in the resulting spectrum. The simple expedient of including a spin-echo building block will get rid of this effect.

Let's look at another pulse sequence, the APT (Attached Proton Test) or J-modulation experiment. The pulse sequence looks like:

¹³C: (p/2)_x - d - p_x - d - acquire

¹H: - - - - - - - - - decoupler on

Of particular interest is the situation when the delay, d, is equal to 1/J. Without using the ¹H decoupler, the pulse sequence would be identical to those above ... a simple spin-echo. With it, interesting things happen. Let's again consider the fate of a doublet and for simplicity, we assume that the rotating frame frequency is set at the chemical shift of the doublet. For the first part of the sequence, prior to the application of the decoupler things will be as they were above. Each doublet vector will rotate in a direction opposite to the other a J/2 Hz. The equation for calculating the rotation angle of such a vector is:

where J is rotational frequency of the vector and t_d is the time of rotation.

If the delay time is 1/J then the angle that the doublet vectors will rotate through is 2p(1/J)(J/2) or p radians. The application of ¹H decoupling will cause the ¹H-¹³C coupling to collapse into a singlet (which doesn't evolve in the rotating frame since we are using a pulse which is on the chemical shift of the doublet):

where {¹H} represents the ¹H decoupler.

So, the coupling evolution takes place during the first delay but not the second. Let's now look at the situation for a triplet. Again, for simplicity, we assume that the transmitter frequency (and therefore the rotating frame frequency) is set exactly on the chemical shift. Of the three vectors representing the triplet, on will rotate clockwise at J Hz, one will remain motionless and one will rotate counterclockwise at J Hz. If we use a t_d value of 1/J again, the angle of rotation will be 2p(1/J)J or 2p radians. Also, during the second delay time, with the ₁H decoupler on the vectors will collapse to one vector that is motionless in the rotating frame:

This is interesting. For the doublet, the vector ends up lying along the +y axis and for the triplet it lies along the -y axis. This corresponds to signals that are 180 degrees out of phase. When we do our phase correction of the spectrum the CH peaks will be positive and the CH₂ peaks will be negative. Further analysis will show the CH₃ and C peaks to be positive and negative respectively. What a great way to separate out the differing types of carbons in a complicated spectrum!