A student in Chemistry StackExchange asked the effect of radio waves on matter, which led to an interesting set of arguments by some users. It was pointed out that in a proton nuclear magnetic resonance (or any other nuclei for that matter), there is no absorption of the photon corresponding to the radio waves. The following reference was cited NMR signal reception: Virtual photons and coherent spontaneous emission which says (in the abstract)

In portions of the magnetic resonance community, there is a misunderstanding of the process of nuclear magnetic resonance (NMR) signal generation and reception, and even in accepted texts, it is frequently described in terms of absorption and emission of radio waves, or radiation, by a two‐level quantum system. While this explanation can be refuted, for those who do understand that the NMR free induction decay signal is easily explained by Faraday's law of induction, reconciling the presence of an induced electromotive force with an apparent absence of transitions between nuclear energy levels causes conceptual problems.

I had never heard of emission of (electromagnetic radio waves) by the precessing nuclei in an NMR experiment, but a Google search of NMR + "absorption of radio waves" shows the concept of radio wave absorption is very common.

This leads to my main question: What does the RF transmitter do in the NMR experiment and what is practically happening in the RF coil?

(a) My understanding was that the RF transmitter, emits (electromagnetic) radio waves. If the RF coil is being energized with AC current of MHz frequencies, isn't the coil emitting electromagnetic radiation in MHz range? The magnetic field component of the radio waves interacts with the nuclei magnetic dipole.

(b) Another user stated "The "transmitter" powers the pulsed radio frequency B1 field. It never decouples from the probe coil to become electromagnetic radiation, i.e. actual photons. " Although this word "decouple" is more of a play of words, can an isolated oscillating B exist without a corresponding oscillating electric field?

I thought to learn from physical chemists and engineers who will have a better picture of the RF transmitters used in the NMR than us analytical chemists. My key question is related the Rf transmitter, not FID.


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    $\begingroup$ Note that it is closely related to how caesium atomic clocks work. In NMR case, radio photons flips proton spins, as there is spin energy difference in magnetic field. In the clocks case, there is Cs kernel spin energy difference due kernel electron spin coupling leading to hyperfine spectrum. Effectively, the unpaired electron acts as atomic embedded magnet, creating magnetic field, instead of external magnets. $\endgroup$
    – Poutnik
    Commented Sep 10, 2019 at 16:15
  • $\begingroup$ I think this is $the$ key point, is there an absorption of radio waves (as you call it radio photon) in an NMR experiment? I disagree with point (b). $\endgroup$
    – ACR
    Commented Sep 10, 2019 at 16:28
  • $\begingroup$ I agree with (a) energy is absorbed by the sample and spins are moved from one energy levels to another. (b) implies that the coil and sample are coupled, it is just a matter of how you look at this, (as you write a play on words). Of course they are coupled, energy is absorbed. The quote you give mixes up quantum and classical ideas. NMR is a funny mix of individual spin changes studied in a quantum way and then objects such as the magnetisation which are more easily understood in a semi-classical way. $\endgroup$
    – porphyrin
    Commented Sep 10, 2019 at 16:56
  • $\begingroup$ See physics.stackexchange.com/questions/334826/… or other similar NMR questions on Physics SE. $\endgroup$
    – Jon Custer
    Commented Sep 10, 2019 at 17:24
  • $\begingroup$ NMR requires a QM description to make sense of more advanced concepts, even if some of the basics can be explained using a classical description (Bloch equations). I have to admit that I don't really understand the quoted paragraph. Perhaps I belong to the portion of the NMR community that misunderstands NMR signal generation/reception ;-) $\endgroup$
    – Buck Thorn
    Commented Sep 10, 2019 at 18:46

2 Answers 2


I finally asked an NMR spectroscopist who was the editor of the Journal of Magnetic Resonance and knows the inside out of NMR spectrometers. He sent a scanned copy of Hoult's paper from 1989. The title explains it all "The magnetic resonance myth of radio waves ." Surprisingly very low citations (15). It is amazing how much nonsense we have been fed with radio waves absorption. Sometimes innocent student questions on these forums can lead to introspection.

Here is the article Radio Waves Myth in NMR. Karl will like it.

The conclusion is that there is no involvement of radio waves, neither in absorption nor in emission (of course). The emission process is easier to understand but the transmission process seems to have confused even the experienced spectroscopists and more MRI people.

This is from another paper of Hoult. "The Origins and Present Status of the Radio Wave Controversy in NMR"

The origins, history, and present status of the controversy surrounding a quantum description of the NMR signal as being due to radio waves are traced. With the Principle of Relativity and Coulomb’s Law as formal starting points and the minimum of mathematics needed for understanding, the derivation of a classical electromagnetic theory of signal reception is first given. The agreement between that classical theory and a recent NMR experiment is then presented, leading to proof that, except for the highest field imaging experiments, there is no significant contribution of radio waves to the signal.


An energy exchange (which you could call absorption) happens during the preparatory $B_1$ pulse, it is irrelevant during the measurement, i.e. the acquisition of an FID. Longitudinal relaxation (energy dissipation to the lattice) is also obviously not necessary to record the FID.

Hoult explains in his paper (cited in the question) that the recording of the FID (that is the presence of a probe coil and preamp with their finite resistance, in which a current must flow) draws a very small amount of power out of the spin ensemble of the sample, via the exchange of virtual photons, but this energy is far smaller than the one pumped into the sample in a ninety degree pulse, and usually effects no measureable additional decay on the magnetisation.

An effect is there with conductive samples (usually dubbed "radiation damping"), but lets just say that is a complicated matter. Hoult also elaborates on that.

(A "virtual" photon is one that is only present in the near field of a dipole oscillator. It cannot travel as an electromagnetic wave unless it becomes a "real" photon. As long as it is virtual, it can be created and destroyed without effort, just like the virtual particles that make up the vacuum in our universe. We only know it is there because it can, in its brief existence, interact with actual electrons, or, in NMR, magnetic nuclei.)

The energy transfer during the $B_1$ pulse (your original question) happens in much the same way as that energy transfer during acquisition, only at far higher power levels. The outside world sees radio EM waves/photons from the probe, but that is just due to the inevitable (and unwanted) outside $B_1$ field of the probe coil. The photons that are produced inside the probe must stay virtual, because otherwise how would they vanish again when reaching the other side of the sample chamber. A standing EM wave in a "resonator" a tenth of the size of the wavelenght does not make much sense.

The oscillating magnetic field in the probe coil is of course accompanied by an electric field, but electromagnetic they can only become when they have enough space, like in the resonator of your EPR, or like in your microwave oven, where you have cold spots at the knots of your EM wave pattern.

  • $\begingroup$ I might also add that the matter is somewhat more complicated in high field MRI with humans, where the sample dimensions reach a similar dimension than the wave lenght of photons with that energy level. $\endgroup$
    – Karl
    Commented Sep 10, 2019 at 19:56
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    $\begingroup$ That last paragraph is incredibly cryptic. $\endgroup$
    – Buck Thorn
    Commented Sep 10, 2019 at 20:19
  • $\begingroup$ @BuckThorn Ja. ;) It d be great if anyone could improve on that. The last sentence reads like a "stop bugging me you idiots, I havent understood that myself". $\endgroup$
    – Karl
    Commented Sep 10, 2019 at 20:27
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    $\begingroup$ @BuckThorn On second thought, I think the statement is still true and to the point. Imagine putting a powerful dipole emitter inside the probe. Photons (EM radiation) coming out of it would not be absorbed by the probe coil, at least not strongly. Ergo, the probe also cannot produce much photons on its inside. $\endgroup$
    – Karl
    Commented Sep 10, 2019 at 21:07
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    $\begingroup$ @M.Farooq The word "transmitter" is just jargon. Its a hf amplifier that could also be used to power a radio station. The actual probe circuit (coil+capacitors) is not part of the box labelled as transmitter. In between, there is usually a T junction with crossed diodes, where the "receiver" (the preamp, and mixer, ADC etc. behind it) are connected. $\endgroup$
    – Karl
    Commented Sep 11, 2019 at 7:39

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