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Lecture 11: Nucleon-Nucleon Interaction •Basic properties •The deuteron •NN scattering • exchange model

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

• Interactions between / can be understood by only considering their 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)

≲ • is (generally) the same for all of a given element Electrons (and, by inference, perhaps other ) are immune from the strong force

• After correcting for Coulomb effects, protons are no more/less bound than The strong force must be mostly charge-independent 2 The simplest two-nucleon system: the deuteron K.S. Krane, Introductory Nuclear (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, = ( ) , inside & outside of the well 𝑢𝑢 𝑟𝑟 ( ) Ψ• 𝑟𝑟Inside: ⁄𝑟𝑟 = sin + cos , where = 0 2𝑚𝑚 𝑉𝑉 +𝐸𝐸 2 • Outside:𝑢𝑢 𝑟𝑟 =𝐴𝐴 𝑘𝑘𝑘𝑘 + 𝐵𝐵 , where𝑘𝑘𝑟𝑟 = 𝑘𝑘 �ћ

(note < 0 for bound states)−κ𝑟𝑟 κ𝑟𝑟 −2𝑚𝑚𝑚𝑚 2 𝑢𝑢 𝑟𝑟 𝐶𝐶𝑒𝑒 𝐷𝐷𝑒𝑒 κ ⁄ћ • To keep 𝐸𝐸 finite at 0 and , = = 0 𝑢𝑢 𝑟𝑟 Employing continuity in the wavefunction and it’s derivative at = , • 𝑟𝑟 𝑟𝑟 → 𝑟𝑟 → ∞ 𝐵𝐵 𝐷𝐷 cot = • Employing the measured binding of the ground state 𝑟𝑟= 2.𝑅𝑅 , 𝑘𝑘 𝑘𝑘𝑘𝑘 −κ we have a transcendental equation relating and It turns out, when solving the Schrödinger equation𝐸𝐸 for a224𝑀𝑀 3D square𝑀𝑀𝑀𝑀 -well, to have a • For the measured 2. , we find 35 : V ≥(π2ћ2)/(8M R2) 3 𝑉𝑉0 𝑅𝑅 0 red 𝑅𝑅 ≈ 14𝑓𝑓𝑓𝑓 𝑉𝑉0 ≈ 𝑀𝑀𝑀𝑀𝑀𝑀 The simplest two-nucleon system: the deuteron

•Hold it right there, you low-budget Houdini, you assumed radial symmetry of the wavefunction, i.e. = 0, but how do you know that’s valid!? •From experiments, we know the deuteron has = 1 , which is created from the addition of 𝑙𝑙 the and angular momenta and𝜋𝜋 their+ respective orbital angular momentum, = + + and = ( 1) , where protons𝐽𝐽 and neutrons are ½ •The positive deuteron therefore𝑙𝑙 implies = + 𝐽𝐽⃗ 𝑠𝑠𝑛𝑛 𝑠𝑠𝑝𝑝 𝑙𝑙⃗ 𝜋𝜋 𝜋𝜋𝑛𝑛𝜋𝜋𝑝𝑝 − •Taking into account angular momentum addition, = 0 2 𝑙𝑙 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 •Now consider the implications of on the (See Lectures 2& 3) 𝑙𝑙 𝑜𝑜𝑜𝑜 = ( ) + ( ) •For an odd nucleon, 𝑙𝑙( ) ( ) 𝑗𝑗 𝑗𝑗+1 +𝑙𝑙 𝑙𝑙+1 −𝑠𝑠 𝑠𝑠+1 𝑗𝑗 𝑗𝑗+1 −𝑙𝑙 𝑙𝑙+1 +𝑠𝑠 𝑠𝑠+1 •If = 0, the total magnetic𝑔𝑔𝑗𝑗 moment2𝑗𝑗 𝑗𝑗+ of1 the deuteron𝑔𝑔𝑙𝑙 is just2𝑗𝑗 𝑗𝑗+1= +𝑔𝑔𝑠𝑠 0.88 •The experimental value is 0.86 𝑙𝑙 𝜇𝜇 𝜇𝜇𝑛𝑛 𝜇𝜇𝑝𝑝 ≈ 𝜇𝜇𝑁𝑁 •Therefore, the deuteron ground state𝑁𝑁 is mostly = 0, but there is some admixture with = 2 (To get the correct , = 𝜇𝜇 =≈0 + 𝜇𝜇 ( = 2), where = 0.96, = 0.04, i.e. 96% = 0) This jives with the fact that the deuteron has𝑙𝑙 a non2 -zero electric2 quadrupole moment 𝑙𝑙 4 𝜇𝜇 Ψ 𝑎𝑎0ψ 𝑙𝑙 𝑎𝑎2ψ 𝑙𝑙 𝑎𝑎0 𝑎𝑎2 𝑙𝑙 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 , it must depend on the direction • The only reference axes we have are the nucleon spins and 𝑟𝑟 𝑟𝑟⃗ • In quantum mechanics, the relative angles between and are provided by the dot products: 𝑠𝑠⃗𝑛𝑛 𝑠𝑠⃗𝑝𝑝 • , , and (which with fully describe the geometry) 𝑟𝑟⃗ 𝑠𝑠⃗𝑖𝑖 • The first was taken into account to find out = 0 and = 2 are possible 𝑠𝑠⃗𝑛𝑛 � 𝑠𝑠⃗𝑝𝑝 𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟⃗ 𝑠𝑠⃗𝑝𝑝 � 𝑟𝑟⃗ 𝑟𝑟 • 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 𝑙𝑙 𝑙𝑙 • 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 , and so that won’t get the job done 2 𝑟𝑟 • Therefore, something depending on the product is needed 5 𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟⃗ 𝑠𝑠⃗𝑝𝑝 � 𝑟𝑟⃗ The deuteron and the non-central (a.k.a. tensor) potential • We’ve established that a non-central force relying on is needed • The form adopted is ( ), where ( ) takes care of all of the radial dependence 𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟⃗ 𝑠𝑠⃗𝑝𝑝 � 𝑟𝑟⃗ • 𝑆𝑆12𝑉𝑉𝑇𝑇 𝑟𝑟 𝑉𝑉, 𝑇𝑇 𝑟𝑟 4 3 2 2 • 𝑆𝑆12 ≡ ћis to𝑟𝑟 remove𝑠𝑠⃗𝑛𝑛 � 𝑟𝑟⃗ the𝑠𝑠⃗𝑝𝑝 radial� 𝑟𝑟⃗ − dependence𝑠𝑠⃗𝑛𝑛 � 𝑠𝑠⃗𝑝𝑝 • −2 is to make = 0 when averaged over ∝ 𝑟𝑟 • The𝑛𝑛 end𝑝𝑝 result is that 12the tensor component of the potential deepens the overall interaction potential−𝑠𝑠⃗ � 𝑠𝑠⃗ 𝑆𝑆 𝑟𝑟⃗ L. van Dommelen, Quantum Mechanics for Engineers (2012)

(using Argonne ) = 0, central only 18 = 2, central only𝑣𝑣 Dineutron𝑙𝑙 (unbound!) 𝑙𝑙 = 0 + = 2, including non-central 6 𝑙𝑙 𝑙𝑙 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 (1988) plane wave and an outgoing spherical wave • To keep life simple, we’ll consider an = 0 (a.k.a. s-wave, if you’re apart of the secret society) plane wave, which practically means a𝑙𝑙 low energy ( ) beam to keep

≪ 20𝑀𝑀𝑀𝑀𝑀𝑀 𝜇𝜇𝑟𝑟𝑟𝑟𝑟𝑟𝑣𝑣𝑣𝑣 ≪ ћ

7 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, = sin( ), ( ) where = , but here is set by the center of 𝑖𝑖 energy for1 the scattering 0 𝑢𝑢 𝑟𝑟 𝐴𝐴 𝑘𝑘 𝑟𝑟 2𝑚𝑚 𝑉𝑉 +𝐸𝐸 2 • Outside𝑘𝑘 1the barrier, �ћ = C sin 𝐸𝐸 + D cos( ), where =

′ 2𝑚𝑚𝑚𝑚 2 • This is generally rewritten𝑢𝑢𝑜𝑜 𝑟𝑟 as =𝑘𝑘2𝑟𝑟sin ′ +𝑘𝑘2𝑟𝑟, so = 𝑘𝑘cos2 ( ) and⁄ ћ = sin( ), where is the “phase shift” we’ll care about later in the semester′ when we discuss′ scattering 𝑜𝑜 2 • Matching ( ) and it’s derivative𝑢𝑢 𝑟𝑟 at 𝐶𝐶and doing𝑘𝑘 𝑟𝑟 some𝛿𝛿 algebra𝐶𝐶 𝐶𝐶results𝛿𝛿 in the𝐷𝐷 transcendental𝐷𝐷 𝛿𝛿 equation𝛿𝛿 cot + = cot( ), relating and for a given 𝑢𝑢 𝑟𝑟 𝑅𝑅 ( ) 2 2 1 1 0 • Defining 𝑘𝑘 𝑘𝑘cot𝑅𝑅 ( 𝛿𝛿 ) and𝑘𝑘 flexing𝑘𝑘 𝑅𝑅 your trigonometry𝑉𝑉 𝑅𝑅skills, sin 𝐸𝐸= 𝛼𝛼 cos 𝑘𝑘2𝑅𝑅 + 𝑘𝑘/ sin 𝑘𝑘2𝑅𝑅 2 2 By inspection, the1 amplitude1 of the term of the outgoing wavefunction is going to tell2 2 us • 𝛼𝛼 ≡ −𝑘𝑘 𝑘𝑘 𝑅𝑅 𝛿𝛿 1+𝛼𝛼 𝑘𝑘2 what the total scattering was, since the term describes a plane wave like the incident one, so sin ( ) …so we can predict𝐷𝐷� the overall scattering cross section 2 𝐶𝐶� 𝜎𝜎𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 ∝ 𝛿𝛿 8 Nucleon-nucleon scattering • For = (from the deuteron) and (e.g. 10’s of keV), 0.92 , 0.016 ( ) −1 −1 • For reasons0 we’ll get into in the scattering lecture,0 = 1 2 𝑉𝑉 35𝑀𝑀𝑀𝑀𝑀𝑀 𝐸𝐸 ≪ 𝑉𝑉 𝑘𝑘 ≈ 2 𝑓𝑓𝑓𝑓 𝑘𝑘 ≲ 𝑓𝑓𝑓𝑓 4𝜋𝜋𝑠𝑠𝑠𝑠𝑠𝑠 𝛿𝛿 Then, using from the deuteron (which means,𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 for our 2 , 0.2 ), • 𝜎𝜎 𝑘𝑘 1 + = 4.6 −1 1 4𝜋𝜋 𝑅𝑅 ≈ 2𝑓𝑓𝑓𝑓 𝑘𝑘 𝛼𝛼 ≈ 𝑓𝑓𝑓𝑓 2 20.4 • Comparing𝛼𝛼 to data, at low energy, 𝜎𝜎 ≈ 𝛼𝛼𝛼𝛼 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏Maybe our 1st grade teacher was • Where did we go wrong!? right and finger-painting was our true calling.𝜎𝜎 ≈ 𝑏𝑏 • 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 = 1 and mostly = 0, is in the = 1 triplet state (because it has 3 spin projections), but nucleons can also𝜋𝜋 be anti+ -aligned in the = 0 (one spin projection) • Good old degeneracy𝐽𝐽 dictates that = 1𝑙𝑙 will contribute𝑆𝑆 3× as much as the = 0, so = + 𝑆𝑆 3 1 𝑆𝑆 The upshot𝑆𝑆 is that the central • Our calculation4 +4 data then imply, = 4. , = 67.8 potential is spin-dependent 𝜎𝜎 𝜎𝜎𝑆𝑆=1 𝜎𝜎𝑆𝑆=0 9 𝜎𝜎𝑆𝑆=1 6𝑏𝑏 𝜎𝜎𝑆𝑆=0 𝑏𝑏 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 ( ) • Then, for a given , whether an incident𝑠𝑠 nucleon𝑠𝑠 ⃗ ⃗ is spin-up or spin-down will change the𝑉𝑉 sign𝑟𝑟 of𝑙𝑙 � 𝑆𝑆 the interaction 𝑙𝑙

• 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 ( ) = ( ) is non-zero, in particular for larger , where𝑁𝑁 higher↑ −𝑁𝑁 ↓ play a role ( = 0 would not yield polarization) 𝑃𝑃 𝑁𝑁 ↑ +𝑁𝑁 ↓ • This means polarized beams can be formed just using scattering 𝑙𝑙 𝑙𝑙

10 The nucleon-nucleon potential • Bringing it all together, the contributions to the nucleon-nucleon potential are, • Central component of strong force [which is spin-dependent] 𝑁𝑁𝑁𝑁 • Coulomb potential 𝑉𝑉 • Non-central (tensor), spin-spin potential • Spin-orbit potential • = + + + ( )

𝑉𝑉𝑁𝑁𝑁𝑁 𝑟𝑟⃗ 𝑉𝑉𝑛𝑛𝑛𝑛 𝑟𝑟 𝑉𝑉𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 𝑟𝑟 𝑉𝑉𝑠𝑠𝑠𝑠 𝑟𝑟⃗ 𝑉𝑉𝑠𝑠𝑠𝑠 𝑟𝑟⃗ • 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!

11 The NN interaction as a derivative of meson exchange • Like all , the nucleon-nucleon interaction is R. Wilson, The Nucleon-Nucleon Interaction (1963) a result (a.k.a. derivative) of some underlying exchange, which itself is described by some interaction • Evidence for this fact is provided by the A bit of secret-handshake trivia: The backward peak is angular distribution of scattering referred to as the “saturation of nuclear forces” • The average deflection angle should be, 𝑛𝑛𝑛𝑛

sin = = = 𝑑𝑑𝑑𝑑 𝑉𝑉0 𝑅𝑅 = ∆𝑝𝑝 𝐹𝐹∆𝑡𝑡 𝑑𝑑𝑑𝑑 ∆𝑡𝑡 𝑅𝑅 𝑣𝑣 𝑉𝑉0 • 𝜃𝜃For≈ 𝜃𝜃 𝑝𝑝 , 𝑝𝑝 10𝑝𝑝° …so≈ a backward𝑝𝑝 peak2𝐾𝐾𝐾𝐾 is unexpected • The interpretation is that some force swaps the neutron & proton positions, meaning𝐾𝐾𝐾𝐾 ≳ a 100𝑀𝑀forward𝑀𝑀𝑀𝑀-moving𝜃𝜃 ≲ neutron is converted to a forward moving proton • At the most fundamental level, exchange between within the nucleons is responsible for the . This is described by (QCD), which is complex enough as to not be all that useful at the moment. • Since quarks and 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 , subject to the uncertainty principle • The furthest range a force can be mediated by a particle is, since it ∆must𝑡𝑡 have , ∆𝐸𝐸∆𝑡𝑡 ≥ ћ and minimum energy = , is = 2 ћ𝑐𝑐 197𝑀𝑀𝑀𝑀𝑀𝑀 𝑣𝑣 ≤ 𝑐𝑐 So, to mediate a force across the nucleon ( ~ 2 ), 2 ~ • 𝐸𝐸 𝑚𝑚𝑐𝑐 𝑅𝑅 ≤ 𝑡𝑡𝑡𝑡 𝑀𝑀𝑐𝑐 ≈ 𝑀𝑀𝑐𝑐 𝑓𝑓𝑓𝑓 • Since this is between the and nucleon , 2 You expect originality from the meson 𝑅𝑅 1𝑓𝑓𝑓𝑓 𝑀𝑀𝑐𝑐 200𝑀𝑀people 𝑀𝑀who𝑀𝑀 only ever wear it was coined , for the Greek “meso” for “middle” checkered-patterned shirts!? • These turn-out to be pairs, the lightest (and most important) of which is the pi-meson (), which is a quark-antiquark pair in the ground state • There are three types of , with charge +1, -1, or 0: ±(~ ) and (~135 )

0 13 𝜋𝜋 140𝑀𝑀𝑀𝑀𝑀𝑀 𝜋𝜋 𝑀𝑀𝑀𝑀𝑀𝑀 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, and so we expect 1 • However, the finite𝑚𝑚 mass𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 of the force-mediating particle means the finite range must be taken 𝑉𝑉 ∝ 𝑟𝑟 into account with a penalty for interaction distances beyond the particle range, thus / • = = , where is the internucleon distance, is fit to data, −𝑟𝑟 𝑅𝑅 and 1. is the pion𝑒𝑒 range 𝑉𝑉𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑉𝑉𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌𝑌 −𝐶𝐶 𝑟𝑟 𝑟𝑟 𝐶𝐶 • So, even though protons in the nucleus are all capable of attracting each other by the strong force, the𝑅𝑅 strong≈ 4𝑓𝑓𝑓𝑓 force range is very limited, while the Coulomb force is not. Hence, Coulomb wins and too many protons splits a nucleus apart.

14 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 = and the nucleon has = +, so, since the nucleon field is generating the pion, a pion of any old can’t be generated • Also, charged𝜋𝜋 pions−exist and charge is conserved,𝜋𝜋 so if, e.g. a proton generates a , it must 𝑙𝑙 change into a neutron + 𝜋𝜋 • A lot of hard (but logical) work (See L van Dommelen Appendix 41) leads to the OPEP:

σ and τ are spin and (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 Physics (B.A. Brown) • Chapter 10: The (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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