# What does the phrase "$s$-wave resonance" mean?

I saw the phrase "$$s$$-wave resonance" in various research articles:

1. "Comparison of average s-wave resonance spacings from proton and neutron resonances" by Vonach H, Uhl M, Strohmaier B, Smith BW, Bilpuch EG, Mitchell GE (Phys. Rev. C $$\bf{38}$$, $$2541$$)

2. "S-wave resonances in positron scattering by He$$^+$$" by Akinori Igarashi and Isao Shimamura (Phys. Rev. A $$\bf{56}$$,$$~~4733$$)

What does this phrase mean?

• Here is half an answer: When you represent a wave as a linear combination of spherical harmonics $Y_{\ell m}(\theta, \phi)$, the “s-wave” is the spherically-symmetric part with $\ell=0$. The terminology comes from the $s,p,d,f,...$ orbitals in atoms with $\ell=0,1,2,3,...$. Someone else can explain “resonance”. Jan 24 '20 at 5:24
• This was already asked on the chemistry site: chemistry.stackexchange.com/q/55209. Jan 24 '20 at 6:10
• Sorry @DanielSank, Your given link doesn't comply my query a little bit. Jan 24 '20 at 6:51
• I know about the terminology $~s,~p,~d,~f~$. Here I just want to know about 𝑠 -wave resonance@G.Smith & @DanielSank Jan 24 '20 at 6:54
• I am providing the references @sammygerbil Jan 26 '20 at 15:20

A "resonance" state is one that lives long enough to have a discretely identifiable nature, but which eventually decays. At its most basic, a resonance is any state that is a local (but not global) minimum of the potential (of field) energy. If a system is located at such a local minimum, it will generally take a quantum tunneling process to escape through the potential. This means that the resonance will live for an indeterminate amount of time; the mean lifetime is determined by the height and width of the potential barrier through which the decay products need to tunnel, but the actually length of time a resonance lives is stochastic, with an exponential probability density.

So after a certain amount of time, a resonance will eventually decay. The decay may be rapid or slow. What "rapid" and "slow" means here actually depends a great deal on the context. The $$\Delta$$ resonances in particle physics consist of three quarks (in up or down flavors) with their spins aligned; these resonances decay extremely quickly—with lifetimes of $$\sim10^{-24}$$ s, because the decay is mediated entirely by the strong interactions, and those are typical strong-interaction time scales. The $$\Lambda$$ hyperon is also made up of three quarks, but one of them is a strange quark, and strange quark decays are only mediated by the weak interaction. The weakness of the interaction slows down the decay rate to about $$\sim10^{-10}$$ s.

Sometimes, the resonance nature of a state is crucially important to its behavior, and sometimes the resonance character is not so important. If actuality, every unstable isotope is really a resonance state. However, many isotopes have mean lifetimes that are very, very long compared with the processes that are used to create them. The highly radioactive isotope $$^{235}$$U, which is radioactive enough that it is capable of sustaining nuclear fission chain reactions in sufficiently enriched bulk samples, still has a mean life of hundreds of millions of years. Many theoretically unstable (i.e. resonance nuclides) have expected lifetimes many times longer than the age of universe; however, since the decays of these isotopes essentially never occur in practice, it is not typically useful to treat them explicitly as resonances.

The "s-wave" part of "s-wave resonance" just indicates that the resonance has total angular momentum $$\ell=0$$. This property is completely unrelated to whether or not a state is a resonance. What makes s-wave state important is typically that they can be excited by quite low-energy collisions. In order to have a nonzero angular momentum $$\ell\hbar$$, an incoming particle must have a certain amount of linear momentum. That means that low-energy incident particles typically cannot change the angular momentum state when they collide with a target. If the initial state is a $$\ell=0$$ s state, then the final state of a low-energy nuclear or particle collision will also be an s-state resonance.

• How $~\ell=\color{red} -~0~$ in the last line ? @Buzz Jan 27 '20 at 4:38
• @nmasanta Just a typo. I fixed it.
– Buzz
Jan 27 '20 at 14:30

As has already been pointed out in the comments, the s-wave corresponds to $$l=0$$ , when the radial Schrödinger equation (see e.g. this question) takes on the particularly simple form

$$-\frac{\hbar^2}{2m}\left[\frac{d^2}{dr^2} +\frac{2}{r}\frac{d}{dr}\right]R + E_p(r)R = ER$$,

where $$R=R(r)$$ is the radial part of the wavefunction.

The “s” is a left over from the early days of spectroscopy, and stands for “sharp” (if I recall my student days correctly, they are actually the sharpest spectral lines you will see in a basic spectrometer with the naked eye).

Resonances correspond to peaks in scattering data at certain value of energy (or equivalently, momenta). These resonances typically correspond to unstable states of particles, e.g. excited states or something similar.

Finding such resonances in the scattering cross-section is one of the main ways new particles are detected (the experimental literature is full of plots of the scattering cross-sections of various processes versus energy). Theoretically, these correspond to poles of the system’s S-matrix, which is a representation of an operator which connects states “far in the past” with states “far in the future”.