Fermi Gas Model

Heisenberg Uncertainty Principle

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dpx ⋅dx ≥ h → dpx =h

dx

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px = px + dpx = px +h

dx

Particle in dx will have a minimum uncertainty in px of dpx

dx

pxNext particle in dx will have a momentum px

Particles with px in dpx have minimum x-separation dx

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dx ⋅dpx = h

Heisenberg Uncertainty PrincipleIdentical conditions apply for the y, py, and z, pz --

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dVps ≡ dx ⋅dpx( )⋅ dy ⋅dpy( )⋅ dz ⋅dpz( )= h3

dVps = dV ⋅dp3

Therefore, in a fully degenerate system of fermions, (i.e., all fermions in their lowest energy state), we have 1 particle in each 6-dimensionl volume --

Phase space volume

Momentum volume

Spatial volume

=

Heisenberg Uncertainty Principle

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dN =dVps

h3

In some dVps the maximum number dN of unique quantum states (fermions) is

px

pz

py

p

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dN =dV ⋅4π p2dp

2π h( )3

Number of states in a shell in p-spacebetween p and p + dp

Only Heisenberg uncertainty principle; completely general

FGM for the nucleusTreat protons & neutrons separately

Consider a simple model for nucleus--

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V (x) = 0 ; 0 < x < L V (x) = ∞ ; x ≥ L

V (y) = 0 ; 0 < y < L V (y) = ∞ ; y ≥ L

V (z) = 0 ; 0 < z < L V (z) = ∞ ; z ≥ L

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V = V (x) + V (y) + V (z)

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−h2

2M∇2ψ + Vψ = Eψ

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ψ =ϕ x (x) ⋅ϕ y (y) ⋅ϕ z (z)

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E = E x + E y + Ez

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ϕ x (x) =2

Lsin

nxL

⎛

⎝ ⎜

⎞

⎠ ⎟

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E x =hπ( )

2

2ML2nx

2

FGM for the nucleus

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E = E x + E y + Ez

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E =hπ( )

2

2ML2nx

2 + ny2 + nz

2( ) ni =1,2,3,⋅⋅⋅

Total energy eigenvalue

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ψ =ϕ x (x) ⋅ϕ y (y) ⋅ϕ z (z)

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ϕ x (x) =2

Lsin

nxL

⎛

⎝ ⎜

⎞

⎠ ⎟

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ψnxnynzx,y,z( )

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E x =hπ( )

2

2ML2nx

2

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E x =px

2

2M → px =

hπ

Lnx

unique states

degenerate eigenvalues

FGM for the nucleus

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px =hπ

Lnx → p =

hπ

L

⎛

⎝ ⎜

⎞

⎠ ⎟ nx

2 + ny2 + nz

2 = p nxnynz( )

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ψnxnynzx,y,z( )

unique states quantized momentum states

px

pz

py

p

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px , px , px( ) =hπ

Lnx ,ny ,nz( )

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px , px , px( )

€

dp3

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dp3 =h

L

⎛

⎝ ⎜

⎞

⎠ ⎟3

=hπ

L

⎛

⎝ ⎜

⎞

⎠ ⎟3

from Heisenberg uncertainty relation

FGM for the nucleus

All momentum states up to pF are filled (occupied)

px

pz

py

p

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px , px , px( )

€

dp3

Assume extreme degeneracy all low levels filled up to a maximum -- called the Fermi level (EF)

We want to estimate EF and pF for nuclei --

The number N of momentum states within the momentum-sphere up to pF is --

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N =1

8

⎛

⎝ ⎜

⎞

⎠ ⎟4

3

π pF3

dp3one p-state per dp3

1/8 of sphere because nx, ny, nz > 0

FGM for the nucleus

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N =1

8

⎛

⎝ ⎜

⎞

⎠ ⎟4

3

π pF3

2πh

L

⎛

⎝ ⎜

⎞

⎠ ⎟3

N =1

8

⎛

⎝ ⎜

⎞

⎠ ⎟

2

2πh( )3

4

3π pF

3V

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N = πV

3

2MEF

h2π 2

⎡

⎣ ⎢ ⎤

⎦ ⎥

3/2

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pF = 2MEF

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L3 = V( )

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EF =hπ( )

2

2M

3N

πV

⎡ ⎣ ⎢

⎤ ⎦ ⎥2 /3

Fermi energy (most energetic nucleon(s)

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pF = h 3π 2( )

1/3 N

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1/3Fermi momentum

(most energetic nucleon(s)

protons N = Z

neutrons N = (A-Z)

2 spin states

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N =1

8

⎛

⎝ ⎜

⎞

⎠ ⎟4

3

π pF3

dp3

FGM for the nucleus

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pF = h 3π 2( )

1/3 Z

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥1/3

Protons Neutrons

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pF = h 3π 2( )

1/3 A − Z

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1/3

Assume Z = N

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pF = h 3π 2( )

1/3 A /2

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥1/3

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pF = h 3π 2( )

1/3 A /2

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥1/3

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V =4

3π R3

R = RoA1/3

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V =4

3π Ro

3A = 4.18Ro3A

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pF =h

Ro

3π 2

2 ⋅4.18

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

1/3

FGM for the nucleus

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pF = h 3π 2( )

1/3 Z

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥1/3

Protons Neutrons

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pF = h 3π 2( )

1/3 A − Z

V

⎡ ⎣ ⎢

⎤ ⎦ ⎥

1/3

Assume Z = N

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pF =197Mev

Roc

3π 2

2 ⋅4.18

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

1/3

=300

RoMeV /c

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pF = 231 MeV /c (Ro =1.3F)

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EF =pF( )

2

2M= 28 MeV

FGM potential

Test of FGM

not FGM