Basic Concepts in Nuclear Physics

Paolo Finelli

Literature/Bibliography

Some useful texts are available at the Library:

Wong, Nuclear Physics

Krane, Introductory Nuclear Physics

Basdevant, Rich and Spiro, Fundamentals in Nuclear Physics

Bertulani, Nuclear Physics in a Nutshell

Introduction

Purpose of these introductory notes is recollecting few basic notions of Nuclear Physics. For more details, the reader is referred to the literature.

Binding energy and Liquid Drop Model

Nuclear dimensions

Saturation of nuclear forces

Fermi gas

Shell model

Isospin

Several arguments will not be covered but, of course, are extremely important: pairing, deformations, single and collective excitations,α decay, β decay, γ decay, fusion process, fission process,...

The Nuclear LandscapeThe scope of nuclear physics is

Improve the knowledge of all nuclei Understand the stellar nucleosynthesis

© Basdevant, Rich and Spiro

e−5

e−6

e−7

≥ e−4

Stellar Nucleosynthesis

Dynamical r-process calculation assuming an expansion with an initial density of 0.029e4 g/cm3, an initial temperature of 1.5 GK and an expansion timescale of 0.83 s.

The r-process is responsible for the origin of about half of the elements heavier than iron that are found in nature, including elements such as gold or uranium. Shown is the result of a model calculation for this process that might occur in a supernova explosion. Iron is bombarded with a huge flux of neutrons and a sequence of neutron captures and beta decays is then creating heavy elements.

The evolution of the nuclear abundances. Each square is a nucleus. The colors indicate the abundance of the nucleus:

©JINA

mNc2 = mAc2 − Zmec2 +

Z

i=1

Bi mAc2 − Zmec2

B = (Zmp + Nmn) c2 −mNc2 [Zmp + Nmn − (mA − Zme)] c2

B =Zm( 1

H) + Nmn −m( AX)

c2

Binding energy

Electrons Mass (~Z)

Atomic Mass Electrons Binding Energies (negligible)

© Basdevant, Rich and Spiro

E/A

(Bin

ding

Ene

rgy

per n

ucle

on)

A (Mass Number)

Average mass of fission fragments is 118

Fe Nuclear Fission Energy

Nuclear Fusion Energy

235U

© Gianluca Usai

The most bound isotopes

Binding energy

Binding energy and Liquid Drop Model

© Basdevant, Rich and Spiro

Volume term, proportional to R3 (or A): saturation

Surface term, proportional to R2 (or A2/3)

Coulomb term, proportional to Z2/A1/3

Pairing term, nucleon pairs coupled to JΠ=0+ are favored

Asymmetry term, neutron-rich nuclei are favored

Binding energy and Liquid Drop Model

© Gianluca Usai

Contributions to B/A as function of A

Comparison with empirical data

Nuclear Dimensions

Ground state

Excited States (~eV)

© Gianluca Usai

Ground state

Ground state

Excited States (~ MeV)

Excited States (~ GeV)

Nuclear Dimensions: energy scales

ρ(r) =ρ(0)

1 + e(r−R)/s

R : 1/2 density radiuss : skin thickness

Nuclear Dimensions

© Basdevant, Rich and Spiro

Fermi distribution

Nuclear forces saturation

An old (but still good) definition:

© E. Fermi, Nuclear Physics

Mean potential method: Fermi gas modelIn this model, nuclei are considered to be composed of two fermion gases, a neutron gas and a proton gas. The particles do not interact, but they are confined in a sphere which has the dimension of the nucleus. The interaction appear implicitly through the assumption that the nucleons are confined in the sphere. If the liquid drop model is based on the saturation of nuclear forces, on the other hand the Fermi model is based on the quantum statistics effects.

The Fermi model could provide a way to calculate the basic constants in the Bethe-Weizsäcker formula

H =A

i=1

Ti ≡A

i=1

− 22M

∇2i

Hψ(r1,r2, . . .) = Eψ(r1,r2, . . .)

ψ(r1,r2, . . .) = φ1(r1)φ2(r2) . . .

− 22M

∇2iφ(ri) = Eφ(ri)

E = E1 + E2 + E3 + . . . =A

i=1

Ei

k2i ≡ (k2ix + k2iy + k2iz) =2MEi

2 > 0

φi(r) ≡ φi(x, y, z) = N sin(kixx) sin(kiyy) sin(kizz)

d2φi(x)

dx2= −k2ixφi(x)

Fermi gas model (I)

Hamiltonian

Wavefunction factorization

Boundary conditions

Separable equations

Gasiorowicz, p.58

φi(x) = B sin(kixx)

1 =

L

0dx|φi(x)|2 = B2

L

0dx sin2(kixx) = B2L

2B =

2

L

φi(r) =

2

L

3/2

sin(kixx) sin(kiyy) sin(kizz)

kix =π

Ln1i , kiy =

π

Ln2i , kiz =

π

Ln3i

(n1i, n2i, n3i = positive integers)

Ei =2k2i2M

=22M

(k2ix + k2iy + k2iz)

Ei(n1i, n2i, n3i) =2π2

2ML2(n2

1i + n22i + n2

3i)

Fermi gas model (II)

Solution

Normalization

∆kx,y,z =π

L(n1,2,3 + 1− n1,2,3) =

π

L

dn(k) =1

84πk2dk

1

(π/L)3≡ Ω

(2π)3dk

n(k) =

k

0dn(k) =

Ω

(2π)34π

3k3

A = 4

kF

0dn(k) =

Ω

(2π)344π

3k3 = Ω

2k3F3π2

ρ0 =2k3F3π2

Ω ≡ L3

ρ0 = A/Ω

Fermi gas model (III)

Density of states

Numberof particles

Densityof particles

spin-isospin

Fermi momentum

θ(kF − k)

Fermi gas model (IV)

Fermi gas distribution:N(k)

kkF

1

0

Step function

filled empty

ρ0 = 0.17 fm−3 kF = 1.36 fm−1

F =2k2F2M

= 38.35 MeV

T = 23 MeV

4dn(k)N(k) = 4Ω

(2π)3θ(kF − k)dk

T = Ω2

π2

2k22M

k2dkθ(kF − k) = Ω2k3F3π2

3

5

2k2F2M

= A3

5F

(BE)vol = −bvolA (bvol = 15.56 MeV)

< U >= −15.56− < T > −39 MeV

Fermi gas model (V)

The fermi level is the last level occupied

Evidences of Shell Structure in Nuclei

© Basdevant, Rich and Spiro

En = (n + 3/2)ω

H = Vls(r)l · s/2

l·s2 = j(j+1)−l(l+1)−s(s+1)

2= l/2 j = l + 1/2= −(l + 1)/2 j = l − 1/2

Mean potential method: Shell modelThe shell model, in its most simple version, is composed of a mean field potential (maybe a harmonic oscillator) plus a spin-orbit potential in order to reproduce the empirical evidences of shell structure in nuclei

© Basdevant, Rich and Spiro

Mean potential method: Shell model

Hi =1

2mp2i +

1

2Mω

20r

2i − V0

p2

2M+

1

2Mω2

0r2

ψ(r) = (E + V0)ψ(r)

ψ(r) = Rnl(r)Ylm(θ,φ)

Rnl(r) = (−1)n

2

(l + 1/2)!

l + n+ 1/2

n

rle−λr2/21F1

−n, l +

3

2,λr2

Shell model (I)

H =A

i=1

Hi

1F1 (−n, µ+ 1, z) =Γ(n+ 1)Γ(µ+ 1)

Γ(n+ µ+ 1)Lµn(z)

EN =

N +

3

2

ω0 N = 2n+ l

d = 2N

l=0

(2l + 1) = 2

[N/2]

n=0

(2(N − 2n) + 1) =

= 2(2N + 1)

N

2+ 1

− 8

[N/2]

n=0

d = (N + 1)(N + 2)

Shell model (II)

Degeneracy

Shell model (III)

Shell model (IV)

Shell model (V)

Shell model (V)

Isospin

In 1932, Heisenberg suggested that the proton and the neutron could be seen as two charge states of a single particle.

939.6 MeV938.3 MeV

EM ≠ 0 EM = 0n

pN

Protons and neutrons have almost identical mass

Low energy np scattering and pp scattering below E = 5 MeV, aftercorrecting for Coulomb effects, is equal within a few percent

Energy spectra of “mirror” nuclei, (N,Z) and (Z,N), are almost identical

ψN (r, σ, τ) =

ψp(r, σ, 12 ) proton

ψn(r, σ,− 12 ) neutron

η 12 , 1

2= |π =

10

η 1

2 ,− 12

= |ν =

01

Isospin is an internal variable that determines the nucleon state

One could introduce a (2d) vector space that is mathematical copy of the usual spin space

proton state neutron state

Isospin (II)

τ3|π = |πτ3|ν = −|π

ψN = a|π + b|ν =

ab

[ti, tj ] = iijktk

Pp = 1+τ32 = Q

ePn = 1−τ3

2

τ1, τ2, τ3

ti =12τi

t+|ν = |πt−|π = |νt+|π = 0t−|ν = 0

t± = t1 ± it2

Isospin

eigenstates of the third component of isospin

In general

The isospin generators

Projectors Raising and lowering operators

Pauli matrices

neutron to proton proton to

neutron

Fundamental representations

T = t1 + t2 T = 0, 1

T = 0 η0,0 = 1√2(π1ν2 − ν1π2)

T = 1

η1,1 = π1π2

η1,−1 = ν1ν2

η1,0 = 1√2(π1ν2 + π2ν1)

Isospin for 2 nucleons

|T = 1, Tz = 1 = |pp

|T = 1, Tz = −1 = |nn1√2

[|T = 1, Tz = 0+ |T = 0, Tz = 0] = |pn

Proton-proton state

Neutron-neutron state

Proton-neutron state

Isospin for 2 nucleons

ψ(1, 2) = ψpp(r1, σ1, r2, σ2)η1,1 + ψnn(r1, σ1, r2, σ2)η1,−1 + ψanp(r1, σ1, r2, σ2)η1,0 + ψs

np(r1, σ1, r2, σ2)η0,0

PT=0 =1− τ (1)τ2

4PT=1ν=1 =

1 + τ (1)3

21 + τ (2)

3

2

PT=1ν=0 =

14(1 + τ (1)τ (2) − 2τ (1)

3 τ (2)3 )

η0,0η1,1

PT=1ν=−1 =

1− τ (1)3

21− τ (2)

3

2 η1,−1 η1,0

antisymmetric symmetric

Wavefunction

Ψ(r,s1,s2,t1,t2) = φ(r)fσ(s1,s2)fτ (t1,t2)

(−)L+S+T = (−)

Symmetry for two nucleon states

the overall wavefunction must be antisymmetric

L=0, S=1 T=0 3S1isospin singlet

Sistema di 2nucleoni identici

(pp,nn)

Sistema di 2nucleoni distinti

(pn)

ISOSPIN SPAZIO SPIN

Tz = ±1

Tz = 0

Funzione simmetrica(tripletto T=1)

Funzione antisimmetrica(singoletto T=0)

L dispari

L dispari

L pari

L dispari

ψ(x)

ψ(x)

ψ(x)

ψ(x)

antisimmetrica

antisimmetrica

simmetrica

simmetrica

S=1

simmetrica(no onda S)

S=0

ψ(σ)

ψ(σ) antisimmetrica

1S0

L pari

L pari

ψ(x) simmetrica

ψ(x) antisimmetrica

S=1

S=1

simmetricaψ(σ)

simmetricaψ(σ)

S=0

S=0

ψ(σ) antisimmetrica

ψ(σ) antisimmetrica

(no onda S)

1S0

(no onda S)

3S1Tz = 0

Funzione simmetrica(tripletto T=1)

pp np nn

0.0

60 eV

-2.23 MeV3S1 (T=0)

1S0 (T=1)1S0 (T=1)1S0 (T=1)

Coulomb

Additional slides

...many open questions

v(r − r) = −v0δ(r − r)

V (r) =

dr v(r − r)ρ(r)

dr v(r) ∼ 200 MeV fm3

V (r) =V0

1 + e(r−R)/R

Mean potential methodThe concept of mean potential (or mean field) strongly relies on the basic assumptionof independent particle motion, i.e. even if we know that the “real” nuclear potentialis complicated and nucleons are strongly correlated, some basic properties can be adequately described assuming individual nucleons moving in an average potential (it means that all the nucleons experience the same field).

a rough approximation could be

where v0 can be phenomenologically estimated to be

Then one can use a simple guess for V: harmonic oscillator, square well, Woods-Saxon shape...