The Shell Model Can we treat the nucleus in an analogous way to Atomic Physics? i.e. Electrons moving in a potential well. There is evidence for Magic Numbers, just as there are Magic Numbers in atomic physics. e.g. The Nobel Gasses appear at Z = 2, 10, 18, 36, 54, 80, 86
e.g. First ionization energies for the atoms. The Shell Model In Nuclei, the Magic Numbers are N or Z = 2, 8, 20, 28, 50, 82, 126 Evidence: • For Z = magic, there are more stable isotopes. • For N = magic, there are more stable isotones. The Shell Model In Nuclei, the Magic Numbers are N or Z = 2, 8, 20, 28, 50, 82, 126 Evidence: • Neutron separation energies are greater for N = magic. • Proton separation energies are greater for Z = magic. The Shell Model In Nuclei, the Magic Numbers are N or Z = 2, 8, 20, 28, 50, 82, 126 Evidence: • Nuclear binding energies are greater when either N or Z = magic. • They are even greater when both N and Z are magic. These are Doubly Magic nuclei.
• First excited state energies are greater for nuclei with either N or Z = magic.
The Shell Model
First Excited state energies The Shell Model
This suggests a model in which the nucleons move within a potential well. Unlike the atomic case there is no central potential. [Note: Potential = PE just as in QM, not PE/q as in E&M.] The short range force would imply that a nucleon, within a nucleus, would feel an approximately constant potential. So in this model, we can imagine each nucleon, moving in a potential well which is established by the presence of the other nucleons. The Shell Model
So each nucleon moves in a potential that looks like:
V(r) V(r)
0 r 0 r
For Neutrons For Protons The Shell Model
In the Atomic case – to first order – we ignore the interaction between the electrons. In the Nuclear case the interaction between the nucleons is strong, but this is taken into account by the potential. i.e. We can treat each nucleon as moving independently in the potential. This is known as the Independent Particle Model. We solve the Schrödinger equation for a single nucleon moving in the potential well. (We assume nucleons are non-relativistic. Can be justified using the uncertainty principle.) The Shell Model
If we assume the potential is spherically symmetric, we can write the wave function for a particular nucleon as
(r) R(r)Yl,m (,) where,
Y l , m ( , ) are the spherical harmonics, l = orbital angular momentum quantum number = 0, 1, 2,… m = magnetic quantum number. l m l 2 So 2 L l(l 1) Lz m
We will use the usual spectroscopic notation for l. i.e. l = 0, 1, 2, 3, … s p d f …
The Shell Model
The time-independent Schrödinger equation becomes, 2 d 2u(r) l(l 1)2 2 V (r) 2 u(r) Eu(r) 2M dr 2Mr with u(r) rR(r) V (r) (Effective potential) eff For nucleons bound within the potential V(r) the boundary conditions force us to have a discrete set of solutions
u ( r ) u n , l ( r ) and E En,l with n = principal quantum number. The full wave function is u (r) (r,,) R (r)Y (,) n,l Y (,) n,l,m n,l l,m r l,m The Shell Model
l(l 1)2 is the potential modified by the Veff (r) V (r) 2 2Mr switch from a 3-dim to a 1-dim problem. e.g. for a neutron.
Veff(r) l = 0 Veff(r) l = 1 0 r 0 r
Sometimes called the Angular Momentum barrier. The Shell Model We can choose any convenient form for V(r). We know analytical solutions for potentials such as the simple harmonic oscillator or a square well. However these simple models do not reproduce the magic numbers. The best results come from assuming a potential shape that mimics the nuclear density. i.e. V(r) V a V (r) 0 ra r 1 e d for neutrons a 1.2A1/3 fm, d 0.75 fm
The Schrödinger equation must be solved numerically. The Shell Model The result is a single-particle energy level scheme that looks something like… Nucleons are fermions so the 1g 18 Pauli Exclusion principle applies. 40 # m values = 2l + 1 2p l 6 nucleon spin s = ½ 1f 14 2 fermions in each ml (ms = + ½, ½) 20 Prediction for magic numbers. 2s 2 But, the magic numbers are E 1d 10 n,l actually 2, 8, 20 28, 50… 8
1p 6 Predicting the excited states 2 of nuclei with these energy 1s 2 levels did not work well. n(l) # neutrons (or protons) The Shell Model What we have neglected is a contribution to the potential due to the spin of the neutron and protons. In the Atom, the spin ½ electron has a magnetic moment , which interacts with the magnetic field generated by its orbital motion. The result is the Spin-Orbit interaction that gives rise to the Fine-Structure splitting in atomic states. This is a very small effect in atoms. Magnetic Moments S Electrons g S Spin angular momentum s B S 2 2 1 z 1 With S s(s 1) s z gs B S 2 z 2 e B = Bohr Magneton 2me
gs 2.00232 = Landé g factor for the electron. So z B opposite to S
Note: This is what is quoted in tables of magnetic moments for nuclei, rather than , because this is what is measured (in a Stern-Gerlach type experiment, for example). Spin-Orbit Effect The Spin-Orbit Effect in atoms If an electron is in a state with non-zero orbital angular momentum, it has a net motion around the nucleus. In the frame of reference of the electron it would appear that the nucleus is moving around the electron. That is like a current going around B the electron so there would be magnetic field at the position of nucleus e the electron. +Ze
The magnetic moment of the electron will tend to “align” with this magnetic field. The Spin-Orbit Effect in atoms B B There are two possible “alignments” of the magnetic moment with the magnetic field. z The system is in a lower energy state z when “aligned.” “Aligned” “Anti-Aligned” The potential energy is U B
z B
We will attempt to calculate this energy shift U for a given hydrogen-like state of the electron. To do that we will calculate U . First we need to find the magnetic field. The Spin-Orbit Effect in atoms B The magnetic field in the centre nucleus of a current loop is e r +Ze I where is the permeability of B 0 0 2r free space. Ze The current I with period T T We can relate T to the angular momentum of the electron. 2 2 For the electron going around the nucleus: L (mer ) Ze T B 0 L So 2 2r 2 mer 1 Using c and recognizing that B 0 0 is in the same direction as L 1 Ze keZe B 2 3 L 2 3 L 4 0 mec r mec r The Spin-Orbit Effect in atoms k Ze The energy shift U B B e L m c2r 3 S e And we had g s B e B keZe So U g S L and since B s 2 3 2m mec r e 2 keZe U gs 2 2 3 S L 2me c r But actually our analysis is not entirely correct because we did this in the non-inertial frame of reference of the electron. The correct result is: 2 keZe U (gs 1) 2 2 3 S L 2me c r The Spin-Orbit Effect in atoms First we look at: L S J 2 J J (L S)(L S) L2 S 2 2LS 1 2 2 2 So LS 2 (J L S ) 2 Therefore LS (n,l, j,m ) j( j 1) l(l 1) s(s 1) (n,l, j,m ) j 2 j Then U *UdV k Ze2 1 (g 1) e * L SdV s 2 2 3 2me c r 2 2 keZe 1 (gs 1) 2 2 j( j 1) l(l 1) s(s 1) 3 2me c 2 r 1 Z 3 for We can show that 3 1 3 3 l 0 r l(l 2)(l 1)n a0 The Spin-Orbit Effect in atoms Combining these we get an estimate for the energy shift: Z 2 j( j 1) l(l 1) 3 2 2 2 Z E0 U En 1 Where En 2 2n l(l 2 )(l 1) n k e2 1 and e a dimensionless quantity. c 137 is called the Fine Structure Constant
e.g. For the 2p states in Hydrogen
2 1 1 3 1 1 1 13.6 eV 2 ( 2 1) 1(11) 2 5 2 p 1 U 3.010 eV 2 state 2 1 2 2 137 2 1(1 2 )(11) 2 3 3 3 1 1 1 13.6 eV 2 ( 2 1) 1(11) 2 5 2 p3 U 1.510 eV 2 state 2 1 2 2 137 2 1(1 2 )(11) So the energy difference between these states is 5105 eV The Spin-Orbit Effect in atoms This introduces a splitting of the spectral lines which can be seen with a high resolution spectrometer.
2p 2 p3 2
2 p 1 2 One photon energy Two, slightly different, photon energies
1s 1s 1 2 No Spin-Orbit Effect With Spin-Orbit Effect The spectral line splitting is known as the Fine Structure Splitting. The Spin-Orbit effect is also known as L-S Coupling. Because it is the result of the L S term. The Spin-Orbit Effect in atoms
Actually the Spin-Orbit effect is not the only contribution to the Fine Structure splitting. There is also a correction to the energy levels due to relativity. (Remember that the Schrödinger Equation is a non-relativistic equation.) 1 k e2 In the Bohr model v e v 2106 m/s ~1% of c n n 1 Corrections for relativity are of the order
2 2 2 2 v kee 2 1 5 510 c c 137 So corrections for relativity will be small, but are of the same order of magnitude as the Spin-Orbit effect. The Spin-Orbit Effect in atoms Dirac’s relativistic treatment of the Hydrogen atom, which naturally includes the spin-orbit effect, results in a set of energy levels given by: 2 1 3 where the E are the E E 1 n n, j n 1 Bohr model energies. n j 2 4n Notice that in the Dirac treatment the energies do not depend on l, only on n and j. In 1947 Lamb and Retherford showed that there is a slight shift depending on l, of the order 106 eV. The shift, known as the Lamb Shift, is because the potential has a slight departure from Coulomb’s Law. In effect the nucleus does not act like a point charge. The effect is accurately predicted by Quantum Electrodynamics (QED). Magnetic Moments S Electrons g S Spin angular momentum s B S With S 2 s(s 1)2 s 1 g z 2 z s B e B = Bohr Magneton 2me
gs 2.00232 = Landé g factor for the electron. So z B opposite to S e = Nuclear Magneton Protons 2.79N with N 2mp
Mass of proton Neutrons 1.91N Therefore N << B The Shell Model
The potential energy of a magnetic moment in a magnetic field is U B and for the spin-orbit effect the magnetic field is related to the orbital angular momentum.
Therefore, since N << B, we would expect that the effect of the spin-orbit interaction to be much smaller than in the Atomic case. i.e. it would be negligible. Nevertheless, there is a significant spin-orbit interaction in nuclei. The origin is not due the interaction of the magnetic moment with the magnetic field due to the orbital motion. i.e. it is not an electromagnetic effect. The origin is that it is part of the more fundamental nucleon- nucleon interaction (which we will discuss later). The Shell Model The orbital angular momentum relevant in N-N interaction is the relative angular momentum between nucleons. Inside the nucleus this would cancel, or average to zero, for a particular nucleon.
But near the edges of the nucleus this cancellation is not complete. Therefore we can model the average effect of this N-N interaction spin orbit effect by a term in the shell model potential that has a radial dependence. V(r) V(r) W(r)LS W(r) is the same for neutrons and protons (does not depend on magnetic moments). Known as the spin orbit coupling term.
W(r) would only be significantly non-zero near r = RNucleus. 1 dV (r) In fact, often W(r) r dr The Shell Model
Before including the spin-orbit term, the single nucleon wave functions are (r) with energies E n,l,ml n,l (The energies are degenerate with respect to m .) l The total angular momentum is J L S
with J z Lz Sz With the spin-orbit term included, the “good” quantum numbers are n, j, mj, l, s. From the rules of angular momentum addition l s j l s so with s = ½ for protons or neutrons, we have, 1 1 j l 2 or j l 2 Reminder General Rules for the Addition of Angular Momentum Suppose the state of a system is composed of two elements with angular momenta, J with quantum numbers j and m . 1 1 1 J 2 with quantum numbers j2 and m2. The total angular momentum of the system is with quantum numbers and . J J1 J2 j m The allowed values of j are found from j j j j j and the allowed 1 2 1 2 values step by 1. i.e. the allowed values are:
j j1 j2 , j1 j2 1, j1 j2 2, ... j1 j2 For each j there are 2j+1 m values. m j, j 1, ..., j 1, j The Shell Model
To first order, we can estimate the energy shift caused by the addition of the spin-orbit term. (First order perturbation.) E W(r)LS Using the unperturbed wave functions n,l, j to evaluate the expectation value. * n,l,m (r)W(r)L S (r)dV l n,l,ml V Now J 2 L2 S 2 2LS 1 2 2 2 LS 2 (J L S ) 2 LS j( j 1) l(l 1) s(s 1) 2 The Shell Model Putting in s = ½ . 2 1 L S 2 l for j l 2 2 1 2 (l 1) for j l 2
2 2 Therefore En,l,( jl 1 ) 2 l W (r) (r) dV 2 V
2 2 En,l,( jl 1 ) 2 (l 1) W (r) (r) dV 2 V
To agree with data we must make W(r) negative so that E (j = l + ½ ) < 0 and E (j = l ½ ) > 0 [This is opposite to the Atomic case!] So E (j = l + ½ ) = l (constant) E (j = l ½ ) = +(l+1) (constant)
The Shell Model Single-particle State labeling: S ml m Without spin-orbit coupling s L n, l, ml, ms
degenerate S With spin-orbit coupling the mj energy depends on the L relative orientation of L and S J n, l, j, mj
degenerate The Shell Model magic Result: Label states with n(l)j numbers. 1g7/2 8
1g 50 1g9/2 10 2p1/2 2 2p 1f5/2 6 2p 4 1f 3/2 28 1f7/2 8 20 2s 1d3/2 4 2s 2 1d 1/2 1d5/2 6 8 1p1/2 2 1p 1p3/2 4 2 2 1s 1s1/2 No LS term With LS term # neutrons Print this page as a handy reference. or protons, 2j+1 The Shell Model
Labeling of nuclear states. States are usually labeled by simply quoting just the total spin quantum number, j, and the parity of the state. These are the things than can be deduced easily from experiment. The Shell Model Parity. Recall the total space wave function for a single particle (r) R (r)Y (,) n,l,ml n,l l,ml The Parity operator: Pˆ makes r r The wave functions for a spherically symmetric potential have symmetry with respect to parity. 2 2 i.e . (r) (r) or Pˆ (r) 1 (r) (r ) i.e. 1 From the properties of spherical harmonics. PˆY (,) (1)l Y (,) l,ml l,ml Pˆ (r) (1)l (r) i.e. (1)l so n,l,ml n,l,ml The Shell Model
Therefore the parity of a single-particle wave function in a nucleus is either even (+) or odd (). The parity of the whole nucleus is the product of the parities for all of the single particle wave functions. l i.e. (1) i i1,A Aside – Multi-Particle Systems
E.g. To simplify: a 1-dimensional system with two non- interacting particles.
The potential energy of the system can be written
V(x1, x2 ) V1(x1) V2 (x2 ) VInteraction (x1, x2 ) 2 2 2 2 TDSE: 2 2 V1(x1) V2 (x2 ) i 2m1 x1 2m2 x2 t
Note: is the wave function for the whole (x1, x2 ,t) system.
Because we have non-interacting particles, we can use separation of variables. i.e. We can write (x1, x2 ,t) 1(x1) 2 (x2 ) f (t) Aside – Multi-Particle Systems
Inserting (x1, x2 ,t) 1(x1) 2 (x2 ) f (t) into the TDSE we get 2 2 1(x1) 1 2 V1(x1) 1(x1) 2m1 x1 1(x1) 2 2 2 (x2 ) 1 2 V2 (x2 ) 2 (x2 ) 2m2 x2 2 (x2 ) f (t) 1 i t f (t)
Since LHS = RHS for all x1, x2, t = constant = E
As well, since LHS1(x1) + LHS2(x2) = constant for all x1, x2, t
Then LHS1(x1) = constant = E1 so E1 E2 E and LHS2(x2) = constant = E2 Aside – Multi-Particle Systems
2 2 d 1(x1) So we can write: 2 V1(x1)1(x1) E11(x1) …(1) 2m1 dx1 2 d 2 (x ) 2 2 …(2) 2 V2 (x2 ) 2 (x2 ) E2 2 (x2 ) 2m2 dx2 f (t) i Ef (t) …(3) t We can solve (1) and (2) independently to find
E1 and 1(x1) and E2 and 2(x2) . E i t E E E From (3): i t with 1 2 f (t) e e So the total wave function is
i t i t (x1, x2 ,t) 1(x1) 2 (x2 )e (x1, x2 )e i.e. the product of single particle wave functions. The Shell Model
Therefore the parity of a single-particle wave function in a nucleus is either even (+) or odd (). The parity of the whole nucleus is the product of the parities for all of the single particle wave functions. l i.e. (1) i i1,A If a single particle level is filled with neutrons (or protons). one arrow = 1 nucleon e.g. 1d5/2 There is always an even number of nucleons in it. Therefore: The parity of that filled state is even (+). Therefore: Any even-even nucleus has a ground state parity that is even. The Shell Model For odd A nuclei (even-odd or odd-even) n p n p There is always one unpaired nucleon in the ground state. Therefore, the parity of the ground state of an odd A nucleus is that of the single particle state of the unpaired nucleon.
Odd-odd nuclei have to be treated on a case-by-case basis.
Space wave function parity is the only type of parity we need to consider when talking about nuclei. More about parity later. The Shell Model
Angular Momentum The total angular momentum of a nuclear state is the vector sum of all the orbital and spin angular momenta of all the nucleons. Using the shell model it is possible to predict the angular momentum and parity for a particular configuration of neutrons and protons in the single particle states. That prediction is not always unambiguous. We will look at several examples and see if we can develop some general rules. The Shell Model
A single particle level, specified by n, l, & j, which is filled, has an even number of nucleons. A filled state has 2j + 1 nucleons in it.
There are also 2j + 1 mj values, and each nucleon must have a different mj value due to the Pauli Exclusion principle. Therefore all possible orientations of J are included, so the total angular momentum will be zero. e.g. filled 1p3/2 state 3 3 1 1 3 For each nucleon: j 2 mj 2 , 2 , 2 , 2 Total j = 0 Generalization: Closed shell nuclei have ground state j = 0 The Shell Model
12 2 4 2 4 e.g. g.s. of C n(1s1/ 2 ) (1p3/2 ) p(1s1/ 2 ) (1p3/ 2 )
Configuration: 1p3/2
(Arrow orientation 1s1/2 indicates spin n p projection wrt 퐿.)
For either neutrons or protons, each filled shell has j = 0. Therefore total angular momentum j = 0. Parity of each filled shell is even. Therefore the total parity is even.
Therefore: j = 0+ The Shell Model 13 2 4 1 2 4 e.g. g.s. of C n(1s1/ 2 ) (1p3/2 ) (1p1/ 2 ) p(1s1/2 ) (1p3/ 2 ) 1p Configuration: 1/2 1p3/2
1s1/2 n p Since the 12C “core” has j = 0, the total angular momentum of the nucleus must be the angular momentum of the unpaired nucleon. Thus j = ½ 1 one nucleon in state The parity is (1)(1) 1 p parity of pared nucleons parity of p state = (1)l = (1)1
1 Therefore: j 2 The Shell Model 11 2 3 2 4 e.g. g.s. of C n(1s1/ 2 ) (1p3/ 2 ) p(1s1/ 2 ) (1p3/2 )
Configuration: 1p3/2
1s1/2 n p There are two ways to think about this situation. Method 1. One can think of the missing neutron as a “hole” that acts like an unpaired nucleon. i.e. j of added nucleon j of nucleus = 0 = j of filled level + needed to fill hole
3 These must be equal j 2 The Shell Model e.g. g.s. of 11C
Configuration: 1p3/2
1s1/2 n p Method 2. We can assume that paired nucleons within a single particle level always couple to j = 0.
i.e. j( of pair) 0 3 Therefore, the total j is the j of the unpaired nucleon. j 2 The parity is the parity of the unpaired nucleon. i.e. (1)(1)1 1 3 Therefore: j 2 The Shell Model
The “Rule” that paired nucleons in a single particle level always couple to j = 0, might be better described as “a reasonable hypothesis that appears to work.” To be considered “paired” the two nucleons – must be of the same type – must have opposite projections of their spins. Using this rule the shell model predicts that the ground state spins and parities of – odd A nuclei is equal to the j of the unpaired nucleon. – even-even nuclei are all j = 0+. – odd-odd nuclei there is no clear rule. Often the shell model does not predict a unique result. Predicting the spins and parities of excited states is, in general, more difficult. Often there is not a unique result. The Shell Model e.g. First Excited state of 12C 1p1/2
Ground state 1p3/2 configuration: 1s1/2 n p Let us assume that the first excited state is a neutron
promoted from the 1p3/2 state to the 1p1/2. 2 3 1 2 4 n(1s1/ 2 ) (1p3/2 ) (1p1/2 ) p(1s1/ 2 ) (1p3/2 )
The total angular 1p1/2 momentum will be 1p3/2 the sum of the “hole” and the 1s1/2 promoted neutron. n p The Shell Model J Jhole Jneutron 1p1/2 3 1 jhole 2 jneuton 2 1p3/2
jhole jneutron j jhole jneutron 1s1/2 3 1 3 1 Sometimes we write: 2 2 j 2 2 n j 3 1 1, 2 1 j 2 2 2 j 1 or 2 “Coupled with” ( p ) ( p ) Parity: 3/ 2 1/ 2 Therefore: j 1 or 2 (1)(1) 1
Actual first excited state is j 2 at E = 4.44 MeV The Shell Model
12 Energy level diagram for C From NNDC Chart of the Nuclides
j (keV) The Shell Model
Many of the lowest energy excited states in nuclei can be described by Single-Particle excitations. This is where, compared to the ground state, only one nucleon is moved to a higher energy single particle state. In these cases the shell model does a reasonable job of predicting the j of low-lying excited states. However, in some cases, a unique prediction is not possible. e.g. the 12C case we looked at. For two or more particle excitations unique predictions are generally not possible. The Shell Model The order in which nucleons are added to the single particle states are as we gave in the energy level diagram up to magic number 50. For larger nuclei, the order of filling states is slightly different for neutrons and protons due to the slight difference in the potential that they experience.
For protons: magic numbers 2 4 2 6 2 4 8 4 (1s1/2) 2 (1p3/2) (1p1/2) 8 (1d5/2) (2sa/2) (1d3/2) 20 (1f7/2) 28 (2p3/2) 6 2 10 8 6 12 4 2 (1f5/2) (2p1/2) (1g9/2) 50 (1g7/2) (2d5/2) (1h11/2) (2d3/2) (3s1/2) 82 10 4 4 (1h9/2) (2f7/2) (3p3/2)
For neutrons: magic numbers 2 4 2 6 2 4 8 4 (1s1/2) 2 (1p3/2) (1p1/2) 8 (1d5/2) (2sa/2) (1d3/2) 20 (1f7/2) 28 (2p3/2) 6 2 10 6 8 2 4 12 (1f5/2) (2p1/2) (1g9/2) 50 (2d5/2) (1g7/2) (3s1/2) (2d3/2) (1h11/2) 82 8 10 4 6 2 14 10 6 (2f7/2) (1h9/2) (3p3/2) (2f5/2) (3p1/2) (1i13/2) (2g9/2) (3d5/2) 126 12 8 (1i11/2) (2g7/2)
The Shell Model
The shell model does quite a good job of predicting the ground state spins and parities of the ground states of nuclei. However the sequence given is not always strictly followed. e.g. There is a tendency for pairs of nucleons to go into higher orbital angular momentum states when the competing states are close together in energy. This is because of the pairing energy we discussed in connection with the liquid drop model. The shell model does not explain the pairing energy.
Mirror Nuclei Mirror nuclei are pairs of nuclei that have the same A, but the values of N and Z are interchanged. 7 7 11 11 13 13 e.g. 3 Li 4 and 4Be3 5 B6 and 6C5 6 C7 and 7N6 If we ignore the effects of the Coulomb interaction, then the shell model would predict identical properties for mirror nuclei. Thus experimentally observed differences between mirror nuclei indicate the importance of the Coulomb part of the potential. Mirror Nuclei
13 13 E.g. 6 C7 and 7N6 Ground states have been lined up, so the binding energy difference, due to the Coulomb interaction, has been accounted for.
Similarity indicates neutrons and protons behave similarly with respect to the nuclear force. But they clearly are not exactly the same. The Shell Model This is just a first approximation to the energy levels. • As the number of nucleons increases – the shape of the potential well changes. – the importance of the spin-orbit coupling term changes. – These effects change the spacing of the energy levels. • This simple model ignores the fact that there should be a different potential for neutrons and protons.