Pairing Evidence for pairing, what is pairing, why pairing exists, consequences of pairing – pairing gap, quasi-particles, etc. For now, until we see what pairing is, we will think of it as something that makes even-A nuclei different from odd-A nuclei, or as special features of even-A nuclei.

Odd-even staggering in two neutron separation energies: Even-even nuclei are more bound Evidence for pairing

Mean square charge radii δ (fm 2 ) Volume Deformation Dynamical effects Odd-even staggering (pairing) Isomer shifts (smaller than ground state in some cases!) Even-even nuclei have (relatively) larger radii

Pairing gap in Sn

Sn – Magic: no valence p-n interactions Both valence protons and neutrons

Independent particle model: magic numbers, shell structure, valence nucleons.  V ij r UiUi r = |r i - r j | Nucleon-nucleon force – very complex One-body potential – very simple: Particle in a box ~ Nucleons orbit the nucleus in states with good principal quantum number, n, and good angular momentum, j. This extreme approximation cannot be the full story. Will need “residual” interactions. But it works surprisingly well in special cases.

Pauli Principle Two fermions, like two protons or two neutrons, can NOT be in the same place at the same time: can NOT occupy the same orbit. Orbit with total Ang Mom, j, has 2j + 1 substates, hence can only contain 2j + 1 neutrons or protons. This, plus the clustering of levels in simple potentials, gives nuclear SHELL STRUCTURE.

Clusters of levels + Pauli Principle  magic numbers, inert cores, valence nucleons Key to structure. Many-body  few- body: each body counts. (Addition of 2 neutrons in a nucleus with 150 can drastically alter structure)

Independent Particle Model Some great successes (for nuclei that are “doubly magic” and “doubly magic plus 1”). Clearly inapplicable for nuclei with more than one particle outside a doubly magic “core”. In fact, in such nuclei, it is not even defined. Thus, as is, it is applicable to only a couple % of nuclei. Residual interactions and angular momentum coupling to the rescue.

Can we obtain such simple results by considering residual interactions? Typical spectra of singly-magic nuclei with two valence particles of the other type

How can we predict these energies? Need only two ingredients: –Nuclear force is short range and attractive –Pauli Principle Lets assume a  force that acts ONLY when the two particles are in contact. Consider ONLY the ANGLE between the orbital planes of the two nucleons

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This is the most important slide: understand this and all the key ideas about residual interactions will be clear !!!!!

Spectrum of  force (left) and the data (right): 0 + state lowest. Higher spin levels get closer and closer.

“Magic plus 2”: Characteristic spectra ~ 1.3 -ish Yaaaaaaaaaaaaaaaaaaay !!!! This, ultimately, is why all e-e nuclei have 0 + ground states !!!!!!!

Pairing: what it is and what it does Short range force between identical nucleons. Pairing is just a simplified approximation to a  force. As with any short range force, it favors coupling two nucleons in identical orbits to J = 0. Pairing force drives nuclei towards spherical shapes – J = 0 has no preferred direction in space. This is a “diagonal” effect on the energies. But there are also strong mixing effects that are extraordinarily important. -- They produce the well-known “pairing gap” in e-e nuclei.

Excited 0 + States in the Shell Model Occ Pairing couples pairs to J = 0. What is the energy of the first excited 0 + state? Suppose on average the single particle levels are separated by ~ keV (typ. for heavy nuclei) Then one would expect a 0+ state at ~ 500 keV. But even-even nuclei show a clear gap between the ground state and simple excited states. That gap is ~ 2 MeV. WHY ?? Occupation Shell model levels

The pairing interaction – maybe not what you always thought (or more than you thought). Suppose all j’s are the same. Then this expresses the attractive interaction of two identical particles in the same orbit when their angular momenta are coupled to J = 0  well known lowering of 0 + states that we have talked about. On the order of 800 keV. BUT, this interaction ALSO exists if j 1 = j 2 are different than j 3 = j 4. This represents a scattering (transfer) of two particles (together, coupled to J = 0) from one orbit to another. But what is that effect? It is mixing !! The final wave function of the ground state will have a mixture of two particles in j 1 and two in j 3. (This will lead to the idea of quasi-particles.)

Effects of Pairing – partial occupancies Concept of quasi-particles Occ. ~1.8 ~1.6 ~1.0 ~0.8 ~0.2 ~0.6 Now, the occupation of levels is spread out over several levels. Occupation Shell model levels

Effects of mixing in a simple toy model Notice the production of a gap! This is the basic reason the mixing gives a pairing gap. V

How can we describe this mathematically? Use a formalism called BCS, borrowed from theory of superconductivity in condensed matter systems I will not derive the results – they are in all standard textbooks – but will show the resulting formulas and how they work.

Single particle energies are replaced with quasi-particle energies involving a “gap” parameter related to the strength of the pairing matrix element. In even-even nuclei the ground state is lowered producing a “pairing” gap. An excited 0 + state means creating two quasi-particles so E > 2  > 2  In odd-even nuclei the excited states are produced by substituting one quasi-particle for another, so they can have very low energies ! So: e-e nuclei – pairing gap: o-e nuclei – energy compression !

The “gap” equation Standard parameter values The “occupation” amplitudes

Shell model levels

Energy compression near the ground state due to pairing in odd-A nuclei Neutron number A Excitation energy for particle(s) in orbit A Without pairing With pairing Shell model levels Nuclear levels

Three excitations within 75 keV !

Backup slides

Coupling of two angular momenta j 1 + j 2 All values from: j 1 – j 2 to j 1 + j 2 (j 1 = j 2 ) Example: j 1 = 3, j 2 = 5: J = 2, 3, 4, 5, 6, 7, 8 BUT: For j 1 = j 2 : J = 0, 2, 4, 6, … ( 2j – 1) (Why these?) /

How can we know which total J values are obtained for the coupling of two identical nucleons in the same orbit with total angular momentum j? Several methods: easiest is the “m-scheme”.

“pairing gap” g.s. ~1800 keV Collective states Pairing Gap in even-even nuclei

How to use the Independent Particle Model Put nucleons (protons and neutrons separately) into orbits. Special case: Put 2j + 1 identical nucleons (fermions) in an orbit with angular momentum j. Each one MUST go into a different magnetic substate. Remember, angular momenta add vectorially but projections (m values) add algebraically. So, total M is sum of m’s M = j + (j – 1) + (j – 2) /2 + (-1/2) [ - (j – 2)] + [ - (j – 1)] + (-j) M = 0. If the only possible M is 0, then J= 0 Thus, a full j- shell, and hence a full major shell of nucleons, always has total angular momentum 0. This simplifies things enormously !!! It allows us to often consider only the valence nucleons!