I think I understand 1D Proton NMR with the Fourier Transform, qualitatively. However, I am struggling to understand 2D COSY NMR.

I am stuck on a few things;

  1. In 1D Proton FT-NMR, the "Pulse" I understand to be the application of a transverse magnetic field, applied at a mixture of frequencies simultaneously, like "noise". This is so that the chemically different protons in the molecule, who each have a different precession frequency, can all resonate from a single pulse. Then, we turn off the transverse field and measure the FID at a certain sampling rate.

So my first question is - in 2D COSY NMR, are both "pulses" the same? i.e., do we subject the sample to a "noise" pulse, then a delay, and then a second "noise" pulse, and then measure the FID?

I think we do, because as far as I have read, the only thing we are varying in the second dimension in the experiment is the length of the delay between pulses.

  1. So to my mind, in 2D COSY, during the delay time, the protons are undergoing FID the same as they do in 1D NMR, even if we are not measuring it at this time.

How, then, does varying the delay time lead to the discovery of "correlation" between protons? What is it about applying a second pulse to partially relaxed protons, and then measuring the FID, that leads us to discover how the protons are correlated, and to the presence or absence of cross-peaks in our 2D spectrum?

If anyone has a link that explains this, that would be fine. All my searches on the Internet so far just talk about how to interpret the COSY spectrum, with nothing about how/why it actually works.


3 Answers 3


In the 1D-proton experiment a hard 90° pulse is applied to the sample before the FID is acquired over some time, t. This set of events is known as the ‘pulse sequence’, and the resulting data is in the time domain, and needs to subjected to a Fourier transform in order to get a 1D spectrum in the frequency domain.

2D experiments are measured much in the same way, except instead of running the pulse sequence once, we run it multiple times, changing some variable and collecting all of the data into one set which will later become the 2D spectrum. [*]

The standard COSY experiment is one of the simplest 2D experiments to understand. It’s pulse sequence differs from the 1D-proton experiment only by an additional hard 90° pulse. We can explain most (though not all) properties of the COSY experiment using a vector model with no mathematics.

Consider a single magnetisation vector, M, which isn't coupled to anything. $\ce{CHCl3}$ would be a good example of this, and leads to a single cross-peak along the diagonal of the resulting COSY spectrum (a trivial result, but easier to explain).

enter image description here

The experiment consists of 4 steps:

  1. An initial 90° pulse is applied, which transfers M from the z-axis down to the y-axis. This is no different to the pulse used when acquiring a 1D-proton spectrum.

  2. M is allowed to precess for some time, t1. During this time, M ends up somewhere in the xy-plane (having both an x and a y component). The exact position will depend on the length of t1, which is the variable we will differ during the acquisition of our COSY.

  3. A second 90° pulse is applied. This 90° pulse transfers the y-component back onto the z-axis (-z to be precise). The x-component is unaffected (this requires more than a qualitative explanation).

  4. Immediately following the second 90° pulse, we detect. The FID is acquired during t2, which remains constant (just as multiple scans of a 1D proton are acquired for the same length of time). What we are actually acquiring is the residual magnetisation in the x-axis (which wasn't affected by the previous 90° pulse), similar to the standard 1D-proton experiment. If we applied a Fourier transform at this stage, we would indeed just end up with a 1D spectrum. FID in memory to be combined with subsequenc

  5. The above procedure (steps 1-4) is repeated, changing t1 and keeping all other variables constant. For each value of t1, the FID is recorded during t2 - this is the basis of the COSY experiment. The data is combined into a single set, and a Fourier transform applied twice which gives us the 2D-COSY spectrum in the frequency domain. In this simple case, we'd end up with a single cross-peak along the diagonal corresponding to the chemical shift of the proton (in $\ce{CHCl3}$) as expected:

enter image description here

In more complex systems, where multiple protons are present, we observe additional cross peaks in the COSY spectrum off the diagonal corresponding to the coupling between pairs of spins.

The appearance of these additional cross peaks cannot be explained qualitatively using a vector diagram (as per above), but is a consequence of the second 90° pulse that we apply prior to acquisition of the FID.

Where coupling exists, the second 90° pulse doesn't just change the position of the magnetisation vector M, but also causes a population change for the transitions of the spin system through a phenomenon known as coherence transfer.

[*]: There is some potential confusion here, as even in a 1D experiment, we typically run the pulse sequence multiple times in order to improve signal:noise ratio. In this case, all of the variables are kept constant. In a 2D experiment, we run the pulse sequence multiple times, changing some variable in order to build up a 2D data set. We may also run the set of pulse sequences multiple times, to improve signal:noise ratio as per the 1D experiment.


A bit of pedanticism first. Some of your terminology usage is inaccurate, I'd just like to take the opportunity to clear some of it up.

  • Nuclear spins do not "undergo a FID", they undergo free precession. The free induction decay (FID) is the signal that is measured when spins undergo free precession.

  • "Noise" in spectroscopy refers to unwanted, random signals arising from processes apart from the magnetisation of the spins. It's exactly analogous as the crackling noise you get if you try listening to a radio underground. The music you want to hear is the signal, but the crackling sound is the noise.

    From what I understand, you are referring to a pulse that is unselective in terms of the frequencies it can excite. Such a pulse is generally called "strong" or "hard", because it obeys the condition $B_1 \gg \Delta B_0$ (the strength of the pulse is much larger than the reduced external magnetic field).

Many 2D experiments are designed to measure correlations between two nuclear spins, commonly denoted I and S. In the case of COSY both I and S are the same nucleus (typically $\ce{^1H}$, but there are versions with other nuclei). The correlation that is being detected is through-bond coupling.

The pulse sequence of COSY looks easy, but the maths is actually more complicated than it is for something like HSQC. So, if you don't want a mathematical answer, quite a bit of detail has to be omitted. The best non-mathematical way I can explain it is as follows:

  • The first 90° pulse leads to excitation of spin I.
  • During the $t_1$ period, spin I undergoes free precession. As $t_1$ is increased, spin I will precess through a larger and larger angle. The rate at which spin I precesses is related to its resonance frequency. So, since we repeat the experiment for many different values of $t_1$, information about the resonance frequency of spin I is encoded in the data set we obtain.
  • Up till now, it is mostly the same as in a 1D experiment. The difference of course lies in the second 90° pulse. It turns out that the combined effect of the $t_1$ period, as well as the second 90° pulse, also leads to some transfer of the excitation from spin I to spin S. This transfer only occurs if spins I and S have a through-bond coupling.
  • In the detection period, both spins I and S have been excited and will therefore precess at their respective resonance frequencies. Both can be detected simultaneously (since they are the same nucleus).

Overall, what information do we have? From the $t_1$ incrementation, we know about the resonance frequency of spin I. This means that, after Fourier transformation (which converts from a time domain $t_1$ to a frequency domain $\omega_1$), we will have a peak, centred at $\omega_I$ (where $\omega$ is the resonance frequency).

From the FID obtained in $t_2$, we know about the resonance frequencies of both spins I and S. So, in the $\omega_2$ dimension we will have two peaks, centred at $\omega_I$ and $\omega_S$.

In the 2D spectrum, then, we will see two peaks. One is centred at $(\omega_I, \omega_I)$, and is known as the diagonal peak. The other one is centred at $(\omega_I, \omega_S)$, and is known as the cross peak. Now, the cross peak only appears if there has been transfer of magnetisation from spin I to spin S, which in turn can only occur if spins I and S have a through-bond coupling.

"But wait! Aren't there peaks at $(\omega_S,\omega_I)$ and $(\omega_S,\omega_S)$?" Yes, absolutely. That's because the first 90° pulse also leads to excitation of spin S; after all, spins I and S are the same nucleus, and an unselective pulse necessarily excites both. In an exactly analogous manner to that described above, this gives rise to the other two peaks (you can just switch the labels I and S in my description to see how this happens).

Finally, to answer some of your questions directly:

in 2D COSY NMR, are both "pulses" the same?

Yes, they are exactly the same. (This depends on how advanced an answer you want, though. For the purposes of 2D data processing, e.g. the States method, the phases of the two pulses may be different - i.e. one may be aligned along the x-axis, and the other along the y-axis. But I suppose you can ignore this for now.)

in 2D COSY, during the delay time, the protons are undergoing FID the same as they do in 1D NMR, even if we are not measuring it at this time

As I mentioned above it is free precession and not FID, but yes, this is exactly correct!

How, then, does varying the delay time lead to the discovery of "correlation" between protons?

The thing about varying the delay time, is that we are trying to measure frequencies. These can be either resonance frequencies (i.e. chemical shift), or coupling frequencies (recall that coupling constants are expressed in Hz).

I offer you the analogy of a clock. Let's say you have a clock, and you want to measure the frequency at which the minute hand rotates. If you simply look at the clock for one instant and record the position of the minute hand, you can't tell how fast it is rotating - or whether it is even rotating at all! You need to look at it continuously over a period of time, or you need to look at it at constant intervals and jot down how its position changes with time.

Using only one value of $t_1$, then, is analogous to just looking at the "nuclear clocks" once. Only by using multiple values of $t_1$, can one figure out the frequencies at which the nuclear clocks operate (i.e. the coupling constant) - or whether they are even operating in the first place (i.e. whether they are coupled).

What is it about applying a second pulse to partially relaxed protons, and then measuring the FID, that leads us to discover how the protons are correlated, and to the presence or absence of cross-peaks in our 2D spectrum?

I hope the above discussion was sufficient. And if you want more detail I'm afraid that you need to go into the maths.

  • $\begingroup$ When I wrote "noise", I was thinking in terms of "white noise" sound, which is an equal mixture of all audible frequencies. I thought that the pulse was similarly a mixture of all frequences across our frequency range of interest, so that it resonated with all protons at the same time in modern FT-NMR. Maybe based on your answer, this is not the case and this aspect is different. $\endgroup$
    – Mark R
    Commented Sep 17, 2017 at 20:54
  • $\begingroup$ I understand J-coupling in 1D NMR to be based on the possibilities of neighbouring protons being either aligned or anti-aligned with the externally applied magnetic field, and thus either slightly increasing or decreasing the effect of this field on our proton of interest, leading to peak splitting. It is not clear if this is the same mechanism that leads to "through-bond" coupling in COSY NMR. $\endgroup$
    – Mark R
    Commented Sep 17, 2017 at 20:54
  • $\begingroup$ Regardless of what the mechanism of coupling is, I am unclear on the reason it gives cross peaks, or rather I am unclear on why cross peaks do NOT occur when the protons are not coupled. When the transfer of magnetisation occurs from I to S, I assume S continues to precess at its own characteristic frequency. Is it something to do with I influencing S's phase that leads to the cross-peak? $\endgroup$
    – Mark R
    Commented Sep 17, 2017 at 20:54
  • $\begingroup$ Firstly regarding the noise bit, if you're interested you should ask a new question, otherwise this comment chain will become too long. I wasn't really correcting your understanding of pulses, it was just the term you used. Ok, as for the cross peaks. You are right, that S will always precess during $t_2$. However it is not simply any S-signal that becomes a cross peak. It has to be magnetisation that precesses at $ω_I$ during $t_1$, and then precesses at $ω_S$ during $t_2$. That's why it is critical that between $t_1$ and $t_2$, there has to be this magnetisation transfer. $\endgroup$ Commented Sep 18, 2017 at 3:45
  • $\begingroup$ Now if I and S are not coupled, all the magnetisation that precesses at $ω_I$ during $t_1$ will simply continue to precess at $ω_I$ during $t_2$. That gives rise to one diagonal peak. Likewise, any magnetisation that precesses at $ω_S$ during $t_1$ will simply continue to precesses at $ω_S$ during $t_2$. That's a second diagonal peak. But since there is no transfer, you cannot get the cross-peaks. / And lastly, through-bond coupling is the same as your J-coupling. It's just expressed this way to emphasise a difference from through-space coupling (important for NoE) $\endgroup$ Commented Sep 18, 2017 at 3:50

Two pulses are applied to your sample, with a distance $\tau$ between them, which gets incremented by $\Delta\tau$ in, lets say, 128 steps, starting at zero, or, more realistically (because you cant apply two pulses at the same time), at $\Delta\tau$.

There is an FID after each first pulse, which is discarded, and one after the second pulse, which gets recorded. In the end, we have 128 FIDs, each with e.g. 512 points.

Now for COSY, it is elementar that the FID is recorded in quadrature (all modern spectrometers do that), where each peak not only has an intensity, but also a phase, or, mathematically equivalent, a complex signal consisting of a "real" and an "imaginary" part. If you look at 1D NMR spectra, you usually only show the "real" part, and you have to phase the spectrum beforehand.

The absolute phase has no real physical meaning, but from a pulse on a totally relaxed sample, the phase of all peaks is (on an ideal spectrometer) the same, and we wilfully shift it to zero and call the real part "absorptive". Every additive part of the signal is a damped cosine, with slightly different frequencies, and so they all start at intensity $1$ x (number of protons)

Now if you apply the (second) pulse to a sample which is already showing an FID, each signal has a different phase, depending on how far the first FID had evolved yet. If you put all the 128 1D spectra behind each other, you see that the absorptive signal of each peak oscillates between full intensity, zero, negative, and back. (The "magnitude" signal of all 128 would look perfectly the same, just getting a bit weaker.)

They oscillate with the same frequency as in the 1D direction, so if you do the 2D FT, you get the diagonal peaks, and the projection in both directions gives you the ordinary 1D spectra.

Now if two signals (protons) are coupling, they each split into e.g. a doublett in 1D, and in 2D you get those notorious additional crosspeaks, because the magnetisation of each of them feels a coherent modulation with the lamor frequency of the other. The stronger the coupling, the stronger the crosspeaks, and no coupling, no crosspeaks.

If the modulation is close to the actual Lamor frequency, btw., you get those $AB$ spin systems. The modulation can now induce transitions, and so the two protons depopulate certain spin states in each other, depending on their own. In the end (no chemical shift difference), you get no visible coupling at all.


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