# Why is non-relativistic quantum mechanics used in nuclear physics?

I am a little bit surprised to see people doing nuclear physics use the Schrödinger equation. I expected the Dirac equation. The energy scale of nuclear physics is on the order of MeV, right? Why is the non-relativistic quantum mechanics relevant for such high energies?

• The rest mass of nucleons is about 940 MeV. This makes the binding energy of nucleons small compared to the rest mass leading some (but not all) to use nonrelativistic methods. – Lewis Miller Mar 11 at 1:11
• I know this is not directly relevant to the specific equations you are talking about, but the Schrödinger equation sweeps a lot of nuance under the rug with the Hamiltonian. It is completely compatible with the mathematics to pick a Hamiltonian with relativistic mechanics. The basic wave equation is compatible with special relativity. – Tim Seguine Mar 12 at 13:28

When you have MeV-scale energies but GeV-scale particles, you can get a lot of mileage from working in the non-relativistic limit.

• And to clarify, "GeV-scale particles" refers to the masses of the particles times $c^2$. – Ruslan Mar 11 at 17:32

The force which drives apart atomic fission products is their electrostatic repulsion. Their departure speed is determined by their mass and that force. While the resulting speed of the fragments is high (translating into a very high temperature of the resulting fireball), it is still not high enough to require any relativistic corrections.

Nuclear physics is done in a non-relativistic limit, but it is not mere quantum mechanics, since the main forces are strong and weak interactions (more first than second) rather than the electromagnetic forces exclusively treated in the QM. In simpler terms: one can get a lot of mileage by replacing Coulomb potential with the Yukava one.

The reasonf or this are the energy scales, as it has already been pointed out by @rob: the products of nuclear reactions move at sub-relativistic velocities and the transformation of particles, which are the domain of the HEP, are not needed to be taken into account.

I recommend the introductory chapter of Fundamentals in nuclear physics by Besdevant for the excellent discussion of this approximations.

• You can get a lot of mileage from an even simpler potential: treating the nucleus as a finite square well. A famous example is Gamow's alpha-tunneling decay rate estimates. There are dozens of other examples in the theory of thermal (and colder) neutrons interacting with matter. – rob Mar 11 at 16:24
• @rob Yes, this is correct. – Vadim Mar 11 at 16:45

There is a nice answer by rob, I would just add a few interesting notes. Your question is an interesting one, because nuclear reactions are extremely fast, and there are neutrinos involved, which do travel at relativistic speeds. However, when we say that we do not need relativity in the calculations, what we mean is that nucleons (neutrons or protons) and fragments travel at slower speeds.

Now I have found basically official information only for two cases that we consider:

1. slower or thermal nucleons, that travel at speeds around 600 km/s

Let us compare this to a characteristic proton-proton collision rate, npσvth, where n ≈ 6 × 1025 cm−3 is the central number density of protons and vth = (3kBTc/mp)1/2 ≈ 600 km s−1 is their thermal velocity

https://www.astro.princeton.edu/~burrows/classes/403/nucl.masses.fusion.pdf

1. faster nucleons that travel at speeds around 20000 km/s, so less then 10% of the speed of light

Nuclear fission produces neutrons with a mean energy of 2 MeV (200 TJ/kg, i.e. 20,000 km/s), which qualifies as "fast". However the range of neutrons from fission follows a Maxwell–Boltzmann distribution from 0 to about 14 MeV in the center of momentum frame of the disintegration, and the mode of the energy is only 0.75 MeV, meaning that fewer than half of fission neutrons qualify as "fast" even by the 1 MeV criterion.[5]

https://en.wikipedia.org/wiki/Neutron_temperature#Fast

So the ultimate answer to your question is that even these fast nucleons travel only at speeds less then 10% of the speed of light and thus do not require relativistic calculations.

• So, in principle, one would still need the Dirac or Majorana equations to describe emitted neutrinos? – iSeeker Mar 16 at 20:47

Lets take the Energy to be 1000Mev (1Gev). Then the rest mass of proton is 940 Mev.(Not exactly)

You will apply relativity when Total Energy >> Rest Mass Energy.

In nuclear physics, we don't deal with relativistic particles, the calculations I deed is just to convert energy scale to velocity scale so as to make it more apparent that relativity is not needed.

Lets take 1000 Mev as the kinetic energy.$$m_0c^2(\frac{1}{\sqrt{1-\beta^2}})= 1000\ Mev$$

put $$m_0c^2=940Mev$$ $$(\frac{1}{\sqrt{1-\beta^2}})=1.06382$$

$$\frac{1}{1-\beta^2}=1.13173$$ $$1-\frac{v^2}{c^2}=0.8836$$ $$\frac{v^2}{c^2}=0.1164$$ $$v=0.36c$$

again, These calculations are just an analogy. when you are taking energy in GeV the Relativistic effect starts rising up. While in nuclear physics, The order of energy is few Mev. So relativistic effects are negligible(Not here) in nuclear problems we are dealing with

• You write "you will apply relativity when total energy $\gg$ rest energy." This suggests, incorrectly, that you can ignore relativity in the semi-relativistic case. I start keeping track of relativistic issues when the kinetic energy is more than about 1% of the mass; your example here is above my cutoff. – rob Mar 11 at 16:14