Lecture 11: Nucleon-Nucleon Interaction

•Basic properties
•The deuteron
•NN scattering
•Meson exchange model

Lecture 11: Ohio University PHYS7501, Fall 2017, Z. Meisel (meisel@ohio.edu) Zach Weinersmith (SMBC)
Apparent properties of the strong force
Some basic observations provide us with general properties of the strong force:
• Nuclei exist with several protons within close proximity
 The strong force must be stronger than the Coulomb force at close distances

• Interactions between atoms/molecules can be understood by only considering their

electrons
 The strong force must be weaker than the Coulomb force at “long” distances (≳atomic size)

• Nuclei are larger than a single nucleon

 The strong force must be repulsive at very short distances (≲ nucleon size)

• Chemistry is (generally) the same for all isotopes of a given element

 Electrons (and, by inference, perhaps other particles) are immune from the strong force

• After correcting for Coulomb effects, protons are no more/less bound than neutrons
 The strong force must be mostly charge-independent
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The simplest two-nucleon system: the deuteron K.S. Krane, Introductory Nuclear Physics (1988)

• 2H is the only bound two nucleon system, so it’s a good place to start for
understanding properties of the nucleon-nucleon interaction
• Using the simplest possible potential, a 3D square well,
we can write down the radial form of the wavefunction,
Ψ 𝑟𝑟 = 𝑢𝑢(𝑟𝑟)⁄𝑟𝑟 , inside & outside of the well
2𝑚𝑚(𝑉𝑉0 +𝐸𝐸)
• Inside: 𝑢𝑢 𝑟𝑟 = 𝐴𝐴sin 𝑘𝑘𝑘𝑘 + 𝐵𝐵cos 𝑘𝑘𝑟𝑟 , where 𝑘𝑘 = �ћ2

• Outside: 𝑢𝑢 𝑟𝑟 = 𝐶𝐶𝑒𝑒 −κ𝑟𝑟 + 𝐷𝐷𝑒𝑒 κ𝑟𝑟 , where κ = −2𝑚𝑚𝑚𝑚⁄

ћ2
(note 𝐸𝐸 < 0 for bound states)
𝑢𝑢 𝑟𝑟
• To keep finite at 𝑟𝑟 → 0 and 𝑟𝑟 → ∞, 𝐵𝐵 = 𝐷𝐷 = 0
𝑟𝑟
• Employing continuity in the wavefunction and it’s derivative at 𝑟𝑟 = 𝑅𝑅,
𝑘𝑘cot 𝑘𝑘𝑘𝑘 = −κ
• Employing the measured binding energy of the ground state 𝐸𝐸 = 2.224𝑀𝑀𝑀𝑀𝑀𝑀,
we have a transcendental equation relating 𝑉𝑉0 and 𝑅𝑅 It turns out, when solving the Schrödinger
equation for a 3D square-well, to have a
• For the measured 𝑅𝑅 ≈ 2.14𝑓𝑓𝑓𝑓, we find 𝑉𝑉0 ≈ 35𝑀𝑀𝑀𝑀𝑀𝑀 bound state: V0≥(π2ћ2)/(8MredR2) 3
The simplest two-nucleon system: the deuteron
•Hold it right there, you low-budget Houdini, you assumed radial symmetry of the wavefunction,
i.e. 𝑙𝑙 = 0, but how do you know that’s valid!?
•From experiments, we know the deuteron has 𝐽𝐽𝜋𝜋 = 1+ , which is created from the addition of
the neutron and proton spin angular momenta and their respective orbital angular momentum,
𝐽𝐽⃗ = 𝑠𝑠𝑛𝑛 + 𝑠𝑠𝑝𝑝 + 𝑙𝑙⃗ and 𝜋𝜋 = 𝜋𝜋𝑛𝑛 𝜋𝜋𝑝𝑝 (−1)𝑙𝑙 , where protons and neutrons are ½+
•The positive deuteron parity therefore implies 𝑙𝑙 = 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
•Taking into account angular momentum addition, 𝑙𝑙 = 0 𝑜𝑜𝑜𝑜 2
•Now consider the implications of 𝑙𝑙 on the magnetic moment (See Lectures 2& 3)
•For an odd nucleon, 𝑔𝑔𝑗𝑗 = 𝑗𝑗 𝑗𝑗+1 +𝑙𝑙 𝑙𝑙+1 −𝑠𝑠(𝑠𝑠+1)
2𝑗𝑗(𝑗𝑗+1)
𝑔𝑔𝑙𝑙 + 𝑗𝑗 𝑗𝑗+1 −𝑙𝑙 𝑙𝑙+1 +𝑠𝑠(𝑠𝑠+1)
2𝑗𝑗(𝑗𝑗+1)
𝑔𝑔𝑠𝑠
•If 𝑙𝑙 = 0, the total magnetic moment of the deuteron is just 𝜇𝜇 = 𝜇𝜇𝑛𝑛 + 𝜇𝜇𝑝𝑝 ≈ 0.88𝜇𝜇𝑁𝑁
•The experimental value is 𝜇𝜇 ≈ 0.86𝜇𝜇𝑁𝑁
•Therefore, the deuteron ground state is mostly 𝑙𝑙 = 0, but there is some admixture with 𝑙𝑙 = 2
(To get the correct 𝜇𝜇, Ψ = 𝑎𝑎0 ψ 𝑙𝑙 = 0 + 𝑎𝑎2 ψ(𝑙𝑙 = 2), where 𝑎𝑎02 = 0.96, 𝑎𝑎22 = 0.04, i.e. 96% 𝑙𝑙 = 0)
This jives with the fact that the deuteron has a non-zero electric quadrupole moment 4
The deuteron and the non-central (a.k.a. tensor) potential
• The non-spherical nature of the deuteron tells us something pretty interesting:
the internucleon potential can’t just depend on the radius 𝑟𝑟, it must depend on the direction 𝑟𝑟⃗
• The only reference axes we have are the nucleon spins 𝑠𝑠⃗𝑛𝑛 and 𝑠𝑠⃗𝑝𝑝
• In quantum mechanics, the relative angles between 𝑟𝑟⃗ and 𝑠𝑠⃗𝑖𝑖 are provided by the dot products:
⃗ and 𝑠𝑠⃗𝑝𝑝 � 𝑟𝑟⃗ (which with 𝑟𝑟 fully describe the geometry)
• 𝑠𝑠⃗𝑛𝑛 � 𝑠𝑠⃗𝑝𝑝 , 𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟,
• The first was taken into account to find out 𝑙𝑙 = 0 and 𝑙𝑙 = 2 are possible
• The second two need to be treated together, since a rotation (e.g. 180° from a parity change)
will affect them in the same way
• Furthermore, it must be some 2nd order combination of the two because the strong force
conserves parity and a 1st order dependence would mean this non-central contribution would
change sign under a parity change
• Squaring only the 2nd or 3rd term would just result in some integer multiple of 𝑟𝑟 2 , and so that
won’t get the job done
• Therefore, something depending on the product 𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟⃗ 𝑠𝑠⃗𝑝𝑝 � 𝑟𝑟⃗ is needed
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The deuteron and the non-central (a.k.a. tensor) potential
• We’ve established that a non-central force relying on 𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟⃗ 𝑠𝑠⃗𝑝𝑝 � 𝑟𝑟⃗ is needed
• The form adopted is 𝑆𝑆12 𝑉𝑉𝑇𝑇 (𝑟𝑟), where 𝑉𝑉𝑇𝑇 (𝑟𝑟) takes care of all of the radial dependence
4 3
• 𝑆𝑆12 ≡ 𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟⃗ 𝑠𝑠⃗𝑝𝑝 � 𝑟𝑟⃗ − 𝑠𝑠⃗𝑛𝑛 � 𝑠𝑠⃗𝑝𝑝 ,
ћ2 𝑟𝑟 2
• ∝ 𝑟𝑟 −2 is to remove the radial dependence
• −𝑠𝑠⃗𝑛𝑛 � 𝑠𝑠⃗𝑝𝑝 is to make 𝑆𝑆12 = 0 when averaged over 𝑟𝑟⃗
• The end result is that the tensor component of the potential deepens the overall interaction
potential
L. van Dommelen, Quantum Mechanics for Engineers (2012)

(using Argonne 𝑣𝑣18 )

𝑙𝑙 = 0, central only
𝑙𝑙 = 2, central only
Dineutron (unbound!)
𝑙𝑙 = 0 + 𝑙𝑙 = 2, including non-central
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Nucleon-nucleon scattering
• To move beyond the single naturally-occurring scenario the deuteron provides us,
we need to move to nucleon-nucleon (NN) scattering to learn more about the NN interaction
• To get the cleanest observables, the optimum target is a single nucleon, namely hydrogen
Why not a neutron target? Because they decay and so it would be really hard to make!
• Like with the deuteron, we’ll stick to a square well,
but now we’re considering an incoming K.S. Krane, Introductory Nuclear Physics (1988)

plane wave and an outgoing spherical wave

• To keep life simple, we’ll consider an 𝑙𝑙 = 0
(a.k.a. s-wave, if you’re apart of the secret society)
plane wave, which practically means a
low energy (≪ 20𝑀𝑀𝑀𝑀𝑀𝑀) beam to keep
𝜇𝜇𝑟𝑟𝑟𝑟𝑟𝑟 𝑣𝑣𝑣𝑣 ≪ ћ

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Nucleon-nucleon scattering
• To solve for properties of our system, we’ll again do the wavefunction matching gymnastics we
did for the deuteron
• The wavefunction inside the barrier is like for the deuteron, 𝑢𝑢𝑖𝑖 𝑟𝑟 = 𝐴𝐴sin(𝑘𝑘1 𝑟𝑟),
2𝑚𝑚(𝑉𝑉0 +𝐸𝐸)
where 𝑘𝑘1 = �ћ2 , but here 𝐸𝐸 is set by the center of mass energy for the scattering

• Outside the barrier, 𝑢𝑢𝑜𝑜 𝑟𝑟 = C ′ sin 𝑘𝑘2 𝑟𝑟 + D′cos(𝑘𝑘2 𝑟𝑟), where 𝑘𝑘2 = 2𝑚𝑚𝑚𝑚⁄
ћ2

• This is generally rewritten as 𝑢𝑢𝑜𝑜 𝑟𝑟 = 𝐶𝐶sin 𝑘𝑘2 𝑟𝑟 + 𝛿𝛿 , so 𝐶𝐶 ′ = 𝐶𝐶cos(𝛿𝛿) and 𝐷𝐷′ = 𝐷𝐷sin(𝛿𝛿),
where 𝛿𝛿 is the “phase shift” we’ll care about later in the semester when we discuss scattering
• Matching 𝑢𝑢(𝑟𝑟) and it’s derivative at 𝑅𝑅 and doing some algebra results in the transcendental
equation 𝑘𝑘2 cot 𝑘𝑘2 𝑅𝑅 + 𝛿𝛿 = 𝑘𝑘1 cot(𝑘𝑘1 𝑅𝑅), relating 𝑉𝑉0 and 𝑅𝑅 for a given 𝐸𝐸
𝛼𝛼
cos 𝑘𝑘2 𝑅𝑅 + sin(𝑘𝑘2 𝑅𝑅)
𝑘𝑘2
• Defining 𝛼𝛼 ≡ −𝑘𝑘1 cot(𝑘𝑘1 𝑅𝑅) and flexing your trigonometry skills, sin2 𝛿𝛿 =
1+𝛼𝛼2 /𝑘𝑘22
• By inspection, the amplitude of the 𝐷𝐷𝐷 term of the outgoing wavefunction is going to tell us
what the total scattering was, since the 𝐶𝐶𝐶 term describes a plane wave like the incident one,
so 𝜎𝜎𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∝ sin2 (𝛿𝛿) …so we can predict the overall scattering cross section
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Nucleon-nucleon scattering
• For 𝑉𝑉0 = 35𝑀𝑀𝑀𝑀𝑀𝑀 (from the deuteron) and 𝐸𝐸 ≪ 𝑉𝑉0 (e.g. 10’s of keV), 𝑘𝑘1 ≈ 0.92𝑓𝑓𝑓𝑓−1 , 𝑘𝑘2 ≲ 0.016𝑓𝑓𝑓𝑓−1
4𝜋𝜋𝑠𝑠𝑠𝑠𝑠𝑠2 (𝛿𝛿)
• For reasons we’ll get into in the scattering lecture, 𝜎𝜎𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
𝑘𝑘 2
• Then, using 𝑅𝑅 ≈ 2𝑓𝑓𝑓𝑓 from the deuteron (which means, for our 𝑘𝑘1 , 𝛼𝛼 ≈ 0.2𝑓𝑓𝑓𝑓−1 ),
4𝜋𝜋
𝜎𝜎 ≈ 2 1 + 𝛼𝛼𝛼𝛼 = 4.6 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
𝛼𝛼
• Comparing to data, at low energy, 𝜎𝜎 ≈ 20.4𝑏𝑏
Maybe our 1st grade teacher was
• Where did we go wrong!? right and finger-painting was our
true calling.
• We didn’t consider the fact Anyhow, it’s too late for that.
that the nucleon-nucleon
K.S. Krane, Introductory Nuclear Physics (1988)

interaction could be spin-dependent!

• The deuteron, with 𝐽𝐽𝜋𝜋 = 1+ and mostly 𝑙𝑙 = 0, is in the 𝑆𝑆 = 1 triplet state (because it has 3 spin projections),
but nucleons can also be anti-aligned in the 𝑆𝑆 = 0 singlet state (one spin projection)
• Good old degeneracy dictates that 𝑆𝑆 = 1 will contribute 3× as much as the 𝑆𝑆 = 0,
so 𝜎𝜎 = 34𝜎𝜎𝑆𝑆=1 + 14𝜎𝜎𝑆𝑆=0
The upshot is that the central
• Our calculation + data then imply, 𝜎𝜎𝑆𝑆=1 = 4.6𝑏𝑏, 𝜎𝜎𝑆𝑆=0 = 67.8𝑏𝑏 potential is spin-dependent
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More nucleon-nucleon scattering K.S. Krane, Introductory Nuclear Physics (1988)

• Since the nucleon-nucleon potential is spin-dependent,

let’s keep an open mind and assume spin-orbit
interactions can play a role too
• We can consider a spin orbit potential 𝑉𝑉𝑠𝑠𝑠𝑠 (𝑟𝑟)𝑙𝑙⃗ � 𝑆𝑆⃗
• Then, for a given 𝑙𝑙, whether an incident nucleon
is spin-up or spin-down will change the sign of
the interaction
• A spin-up nucleon would be scattered one way, R. Wilson, The Nucleon-Nucleon Interaction (1963)
while a spin-down would be scattered another
• Indeed, such an effect is seen, where the polarization
𝑁𝑁 ↑ −𝑁𝑁(↓)
𝑃𝑃 = is non-zero, in particular for larger energies,
𝑁𝑁 ↑ +𝑁𝑁(↓)
where higher 𝑙𝑙 play a role (𝑙𝑙 = 0 would not yield polarization)
• This means polarized beams can be formed just using scattering

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The nucleon-nucleon potential
• Bringing it all together, the contributions to the nucleon-nucleon potential 𝑉𝑉𝑁𝑁𝑁𝑁 are,
• Central component of strong force [which is spin-dependent]
• Coulomb potential
• Non-central (tensor), spin-spin potential
• Spin-orbit potential
• 𝑉𝑉𝑁𝑁𝑁𝑁 𝑟𝑟⃗ = 𝑉𝑉𝑛𝑛𝑛𝑛 𝑟𝑟 + 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑟𝑟 + 𝑉𝑉𝑠𝑠𝑠𝑠 𝑟𝑟⃗ + 𝑉𝑉𝑠𝑠𝑠𝑠 (𝑟𝑟)
⃗

• Coming up with an adequate form that describes observables from nuclei is an outstanding
problem in nuclear physics
• A further complication which the data seems to require is an inclusion of higher-order
interactions, i.e. so-called 3-body interactions
• Practically speaking, this means we’re a far way off from describing observables for a large
range of nuclei starting purely from a theoretical description of the NN interaction
Hurray for job security!

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The NN interaction as a derivative of meson exchange
• Like all forces, the nucleon-nucleon interaction is R. Wilson, The Nucleon-Nucleon Interaction (1963)

a result (a.k.a. derivative) of some underlying particle exchange,

which itself is described by some interaction
• Evidence for this fact is provided by the Atrivia:bit of secret-handshake
The backward peak is
angular distribution of 𝑛𝑛𝑛𝑛 scattering referred to as the
“saturation of nuclear forces”
• The average deflection angle should be,
𝑑𝑑𝑑𝑑 𝑉𝑉0 𝑅𝑅
∆𝑝𝑝 𝐹𝐹∆𝑡𝑡 𝑑𝑑𝑑𝑑
∆𝑡𝑡 𝑅𝑅 𝑣𝑣 𝑉𝑉0
𝜃𝜃 ≈ sin 𝜃𝜃 = = = ≈ =
𝑝𝑝 𝑝𝑝 𝑝𝑝 𝑝𝑝 2𝐾𝐾𝐾𝐾
• For 𝐾𝐾𝐾𝐾 ≳ 100𝑀𝑀𝑀𝑀𝑀𝑀, 𝜃𝜃 ≲ 10° …so a backward peak is unexpected
• The interpretation is that some force swaps the neutron & proton positions,
meaning a forward-moving neutron is converted to a forward moving proton
• At the most fundamental level, gluon exchange between quarks within the nucleons is
responsible for the nuclear force. This is described by quantum chromodynamics (QCD), which
is complex enough as to not be all that useful at the moment.
• Since quarks and gluons aren’t found in isolation anyways, a net force description is going to be
used, where groups of quarks mediate interactions between other groups of quarks 12
Why meson exchange?
• Considering the fact that particles are waves, waves transmit fields (e.g. E&M), and quantum
theory require small-scale phenomena to be quantized, quantum field theory postulates that
nucleons (or at least the quarks within them) interact by exchanging quanta of the nuclear field
• The mass of the quanta can be surmised by considering the fact that quanta will be
“virtual particles”, which are only allowed to exist for a finite time ∆𝑡𝑡, subject to the
uncertainty principle ∆𝐸𝐸∆𝑡𝑡 ≥ ћ
• The furthest range a force can be mediated by a particle is, since it must have 𝑣𝑣 ≤ 𝑐𝑐,
2 , is 𝑅𝑅 ≤ 𝑡𝑡𝑡𝑡 = ћ𝑐𝑐 ≈ 197𝑀𝑀𝑀𝑀𝑀𝑀 𝑓𝑓𝑓𝑓
and minimum energy 𝐸𝐸 = 𝑚𝑚𝑐𝑐 2 2
𝑀𝑀𝑐𝑐 𝑀𝑀𝑐𝑐
• So, to mediate a force across the nucleon (𝑅𝑅~1𝑓𝑓𝑓𝑓), 𝑀𝑀𝑐𝑐 2 ~200𝑀𝑀𝑀𝑀𝑀𝑀
You expect originality from the
• Since this is between the electron and nucleon masses, people who only ever wear
it was coined meson, for the Greek “meso” for “middle” checkered-patterned shirts!?
• These turn-out to be quark pairs, the lightest (and most important) of which is the pi-meson
(pion), which is a quark-antiquark pair in the ground state
• There are three types of pions, with charge +1, -1, or 0: 𝜋𝜋 ± (~140𝑀𝑀𝑀𝑀𝑀𝑀) and 𝜋𝜋 0 (~135𝑀𝑀𝑀𝑀𝑀𝑀)
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The Yukawa potential
• The preceding considerations lead us to the conclusion that a nucleon generates a pion field,
where the associated interaction is described by a pion potential
• A nearby nucleon will have its own pion field, which the first nucleon will interact with, and
which will interact with the pion field of the first nucleon
• Like the Coulomb potential, we have an interaction generated by some central force,
1
and so we expect 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 ∝
𝑟𝑟
• However, the finite mass of the force-mediating particle means the finite range must be taken
into account with a penalty for interaction distances beyond the particle range, thus
𝑒𝑒 −𝑟𝑟/𝑅𝑅
• 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑉𝑉𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 =
−𝐶𝐶 ,
where 𝑟𝑟 is the internucleon distance, 𝐶𝐶 is fit to data,
𝑟𝑟
and 𝑅𝑅 ≈ 1.4𝑓𝑓𝑓𝑓 is the pion range
• So, even though protons in the nucleus are all capable of attracting each other by the strong
force, the strong force range is very limited, while the Coulomb force is not.
Hence, Coulomb wins and too many protons splits a nucleus apart.

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One-Pion Exchange Potential (OPEP)
• Killjoys with an enthusiasm for rigor will notice a spin and charge dependent parts are required
to explain pion-nucleon interactions
• The pion has 𝜋𝜋 = − and the nucleon has 𝜋𝜋 = +, so, since the nucleon field is generating the
pion, a pion of any old 𝑙𝑙 can’t be generated
• Also, charged pions exist and charge is conserved, so if, e.g. a proton generates a 𝜋𝜋 + , it must
change into a neutron
• A lot of hard (but logical) work (See L van Dommelen Appendix 41) leads to the OPEP:
σ and τ are spin and isospin (here, charge)
flipping operators, respectively

15
Further Reading
• Chapter 5: Modern Nuclear Chemistry (Loveland, Morrissey, Seaborg)
• Appendices 40 & 41: Quantum Mechanics for Engineers (L. van Dommelen)
• Chapter 11: Lecture Notes in Nuclear Structure Physics (B.A. Brown)
• Chapter 10: The Atomic Nucleus (R.D. Evans)
• Chapter 4: Introductory Nuclear Physics (K.S. Krane)
• Chapter 11: Introduction to Special Relativity, Quantum Mechanics, and Nuclear Physics for
Nuclear Engineers (A. Bielajew)

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