Coauthors:
M. Gajda, M. ZaĿuska-Kotur, J. Mostowski
Pauli Crystals
hidden geometric structures
of the quantum statistics
Tomasz Sowiński
Institute of Physics of the Polish Academy of Sciences ArXiv:1511.01036
instead of the
introduction
(r
1, . . . , r
N) = Det
2
6 6
6 4
1
(r
1)
2(r
1) . . .
N(r
1)
1
(r
2)
2(r
2) . . .
N(r
2)
.. . .. . .. . .. .
1
(r
N)
2(r
N) . . .
N(r
N)
3
7 7
7 5
ground-state of spinless fermions
V (r)
{ i(r)}
~
22m r
2+ V (r)
i(r) = E
i i(r)
• We consider spinless fermions
confined in some external potential
• In principle the corresponding Schrödinger equation can be solved
• The set of eigenstates is known
• Wave function of the noninteracting many-body ground-state
(r
1, . . . , r
N) = Perm
2
6 6
6 4
1
(r
1)
2(r
1) . . .
N(r
1)
1
(r
2)
2(r
2) . . .
N(r
2)
.. . .. . .. . .. .
1
(r
N)
2(r
N) . . .
N(r
N)
3
7 7
7 5
(r
1, . . . , r
N) =
1(r
1)
2(r
2) · . . . ·
N(r
N)
small digression
• Fundamentally distinguishable particles
• Bosons
1
(r) =
2(r) = . . . =
N(r)
Ground-state is obtained for
-3 -2 -1 0 1 2 3 0.2
0.4 0.6
0.5 1.5 2.5 3.5
• two distinguishable particles
˜
x
1= 0 x ˜
2= ±1
• the most probable configuration
density profiles
⇢(x
1, x
2) = | (x
1, x
2) |
2⇢(˜ x
1, ˜ x
2) = max [⇢(x
1, x
2)]
first observation
(x
1, x
2) = '
0(x
1)'
1(x
2)
first observation
0.5 1.5 2.5 3.5
• two identical bosons
density profiles
-3 -2 -1 0 1 2 3
0.2 0.4 0.6
(x
1, x
2) = 1
p 2 ['
0(x
1)'
1(x
2) + '
1(x
1)'
0(x
2)]
˜
x
1= ± 1
p 2 x ˜
2= ˜ x
1• the most probable configuration
boson
enhancement
-3 -2 -1 0 1 2 3 0.2
0.4 0.6
first observation
0.5 1.5 2.5 3.5
• two identical bosons
density profiles
(x
1, x
2) = 1
p 2 ['
0(x
1)'
1(x
2) + '
1(x
1)'
0(x
2)]
˜
x
1= ± 1
p 2 x ˜
2= ˜ x
1• the most probable configuration
boson
enhancement
⇢(x) = Z
dx2 | (x, x2)|2
single-particle density
first observation
0.5 1.5 2.5 3.5
• two identical fermions
• the most probable configuration
density profiles
(x1, x2) = 1
p2 ['0(x1)'1(x2) '1(x1)'0(x2)]
˜
x
1= ± 1
p 2 x ˜
2= x ˜
1many-body
ground-state
(x
1,x
2) =
1 p 2
['
0(x
1)'
1(x
2) + '
1(x
1)'
0(x
2)]
-3 -2 -1 0 1 2 3
0.2 0.4 0.6
⇢(x) = Z
dx2 | (x, x2)|2
single-particle density
general scheme
(r
1, . . . , r
N) = Det
2
6 6
6 4
1
(r
1)
2(r
1) . . .
N(r
1)
1
(r
2)
2(r
2) . . .
N(r
2)
.. . .. . .. . .. .
1
(r
N)
2(r
N) . . .
N(r
N)
3
7 7
7 5
P(r
1, . . . , r
N) = | (r
1, . . . , r
N) |
2• probability density of finding given configuration
• we can find its maximum (the most probable configuration)
• we can simulate an experiment with given number of particles!
With Metropolis algorithm we can generate
an ensemble of configurations
according to given density distribution
N. Metropolis et al, J. Chem. Phys. 21, 1087 (1953)
two-dimensional
harmonic trap
Two-dimensional harmonic trap
nm
(x, y) = N
nmH
n(x)H
m(y)e
(x2+y2)/2,
V (r) = m⌦
22 r
2= m⌦
22 (x
2+ y
2)
• single-particle basis
Energy shell Energy (osc. units)
Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
• unique ground-state
N = 1, 3, 6, 10, 15, . . .
N=3
0 1 2
-1 -2
0 1 2 -1
-2
N=3
P(r
1, . . . , r
N) = | (r
1, . . . , r
N) |
2• the most probable configuration
Energy shell Energy (osc. units)
Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
0
1
2
-1
-2
0 1 2
-1
-2
N=3
Energy shell Energy(osc. units)
Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
⇢(r) =
Z
. . .
Z
dr
2. . . dr
N| (r, r
2, . . . , r
N) |
2Averaged positions 106 elements
0 1 2
-1 -2
0 1 2 -1
-2
P (r
1,. .., r
N)= | (r
1,. .., r
N)|
20 1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
N=3
⇢(r) = Z
. . . Z
dr2 . . . drN| (r, r2, . . . , rN)|2
Energy shell Energy (osc. units)
Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
P(r
1, . . . , r
N) = | (r
1, . . . , r
N) |
2N=6
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2
N=6
Energy shell Energy(osc. units)
Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
N=10
N=6
N=10
Energy shell Energy(osc. units)
Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2
Energy shell Energy (osc. units)
Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
N=10
N=15
0 1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
0 1 2
-1 -2
0 1 2 -1
-2 0
1 2
-1 -2
0 1 2 -1
-2
N=3 N=6
N=10 N=15instead of
conclusions
RIDDLE!
N=5
Energy shell Energy (osc. units)Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
N=5
Energy shell Energy (osc. units)Degeneracy Excitation structure (n,m)
s 1 1 (0,0)
p 2 2 (1,0) (0,1)
d 3 3 (2,0) (1,1) (0,2)
f 4 4 (3,0) (2,1) (1,2) (0,3)
g 5 5 (4,0) (3,1) (2,2) (1,3) (0,4)
⋮ ⋮ ⋮ ⋮
Appendix:
numerical
recipes
Towards the most probable config.
P({R
i}) > P({r
i}
t) ) {r
i}
t+1= {R
i}
P({R
i}) P({r
i}
t) ) {r
i}
t+1= {r
i}
tP({R
i}) and P({r
i}
t)
{r
i}
0= RAND
{R
i} = {r
i}
t+ RAND( {
i})
1. Start with some random configuration
2. Shift the configuration randomly
3. Calculate corresponding probabilities
4. Conditionally accept a new configuration
5. Go to 2.
Ensemble of configurations
Towards the most probable config.
P({R
i}) > P({r
i}
t) ) {r
i}
t+1= {R
i}
P({R
i}) P({r
i}
t) ) {r
i}
t+1= {r
i}
tP({R
i}) and P({r
i}
t)
{r
i}
0= RAND
{R
i} = {r
i}
t+ RAND( {
i})
1. Start with some random configuration
2. Shift the configuration randomly
3. Calculate corresponding probabilities
4. Conditionally accept a new configuration
5. Go to 2.
Ensemble of configurations
THE MOST PROBABLE CONFIGURATION
Ensemble of configurations
P({R
i}) and P({r
i}
t)
{r
i}
0= RAND
{R
i} = {r
i}
t+ RAND( {
i})
1. Start with some random configuration
2. Shift the configuration randomly
3. Calculate corresponding probabilities
5. Go to 2.
P({R
i}) > P({r
i}
t) ) {r
i}
t+1= {R
i}
P({R
i}) P({r
i}
t) ) {r
i}
t+1= RAND ( {r
i}
t, {R
i})
4. Conditionally accept a new configuration
N. Metropolis et al, J. Chem. Phys. 21, 1087 (1953)
Ensemble of configurations
P({R
i}) and P({r
i}
t)
{r
i}
0= RAND
{R
i} = {r
i}
t+ RAND( {
i})
1. Start with some random configuration
2. Shift the configuration randomly
3. Calculate corresponding probabilities
5. Go to 2.
P({R
i}) > P({r
i}
t) ) {r
i}
t+1= {R
i}
P({R
i}) P({r
i}
t) ) {r
i}
t+1= RAND ( {r
i}
t, {R
i})
4. Conditionally accept a new configuration P({rP({Ri})
i}t)
1 P({Ri}) P({ri}t)
N. Metropolis et al, J. Chem. Phys. 21, 1087 (1953)