fermi gas model heisenberg uncertainty principle particle in dx will have a minimum uncertainty in p...
Post on 19-Dec-2015
237 views
TRANSCRIPT
Heisenberg Uncertainty Principle
€
dpx ⋅dx ≥ h → dpx =h
dx
€
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
€
dx ⋅dpx = h
Heisenberg Uncertainty PrincipleIdentical conditions apply for the y, py, and z, pz --
€
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
€
dN =dVps
h3
In some dVps the maximum number dN of unique quantum states (fermions) is
px
pz
py
p
€
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--
€
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
€
V = V (x) + V (y) + V (z)
€
−h2
2M∇2ψ + Vψ = Eψ
€
ψ =ϕ x (x) ⋅ϕ y (y) ⋅ϕ z (z)
€
E = E x + E y + Ez
€
ϕ x (x) =2
Lsin
nxL
⎛
⎝ ⎜
⎞
⎠ ⎟
€
E x =hπ( )
2
2ML2nx
2
FGM for the nucleus
€
E = E x + E y + Ez
€
E =hπ( )
2
2ML2nx
2 + ny2 + nz
2( ) ni =1,2,3,⋅⋅⋅
Total energy eigenvalue
€
ψ =ϕ x (x) ⋅ϕ y (y) ⋅ϕ z (z)
€
ϕ x (x) =2
Lsin
nxL
⎛
⎝ ⎜
⎞
⎠ ⎟
€
ψnxnynzx,y,z( )
€
E x =hπ( )
2
2ML2nx
2
€
E x =px
2
2M → px =
hπ
Lnx
unique states
degenerate eigenvalues
FGM for the nucleus
€
px =hπ
Lnx → p =
hπ
L
⎛
⎝ ⎜
⎞
⎠ ⎟ nx
2 + ny2 + nz
2 = p nxnynz( )
€
ψnxnynzx,y,z( )
unique states quantized momentum states
px
pz
py
p
€
px , px , px( ) =hπ
Lnx ,ny ,nz( )
€
px , px , px( )
€
dp3
€
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
€
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 --
€
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
€
N =1
8
⎛
⎝ ⎜
⎞
⎠ ⎟4
3
π pF3
2πh
L
⎛
⎝ ⎜
⎞
⎠ ⎟3
N =1
8
⎛
⎝ ⎜
⎞
⎠ ⎟
2
2πh( )3
4
3π pF
3V
€
N = πV
3
2MEF
h2π 2
⎡
⎣ ⎢ ⎤
⎦ ⎥
3/2
€
pF = 2MEF
€
L3 = V( )
€
EF =hπ( )
2
2M
3N
πV
⎡ ⎣ ⎢
⎤ ⎦ ⎥2 /3
Fermi energy (most energetic nucleon(s)
€
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
€
N =1
8
⎛
⎝ ⎜
⎞
⎠ ⎟4
3
π pF3
dp3
FGM for the nucleus
€
pF = h 3π 2( )
1/3 Z
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥1/3
Protons Neutrons
€
pF = h 3π 2( )
1/3 A − Z
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥
1/3
Assume Z = N
€
pF = h 3π 2( )
1/3 A /2
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥1/3
€
pF = h 3π 2( )
1/3 A /2
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥1/3
€
V =4
3π R3
R = RoA1/3
€
V =4
3π Ro
3A = 4.18Ro3A
€
pF =h
Ro
3π 2
2 ⋅4.18
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
1/3
FGM for the nucleus
€
pF = h 3π 2( )
1/3 Z
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥1/3
Protons Neutrons
€
pF = h 3π 2( )
1/3 A − Z
V
⎡ ⎣ ⎢
⎤ ⎦ ⎥
1/3
Assume Z = N
€
pF =197Mev
Roc
3π 2
2 ⋅4.18
⎛
⎝ ⎜ ⎜
⎞
⎠ ⎟ ⎟
1/3
=300
RoMeV /c
€
pF = 231 MeV /c (Ro =1.3F)
€
EF =pF( )
2
2M= 28 MeV