2. Consider a2D parabolic quantum dot with single particle Energy levels of 2 HO E(n,m)=w0 (n+m+1) a)Draw energy spectrum, including s and p-shell, and add (1,1) state from the d-shell b)Assign...

Question 2 (a) The 2D quantum dot in the x-y plane in the Hamiltonian
equation is represented by
22
2 2 2
2 2
e o
dot
e e
m h k kh
H k w r p
m m
  

Exact diagonalization results for the ground-state energy, E0 and the
Magnetisation M = 0 for ground state.
The second-quantized formalism below, with the non interacting
Hamiltonian:
2
22
t
o j j j
j
H E C C
J
E hw




when
, 1j n mE E n m   

Pauli matrices , ,x y z refe
ing to the electron spin.
The operator
( )y x x yk k
P
k
 
 has the eigen value of + - 1.
Low-energy states have high positive helicity with
11
( )
2 ie

 
    
 

a U(1) degeneracy is realized, co
esponding
to a ring in momentum space.
The non interacting ground state is a Fermi sea with all states and the exact
diagonalisation results for the ground state energy of N = 2 and N =3
electrons in the dot for a = 10 and a = 15.
(b) The second quantisation provides the means to heavily condense the
epresentation. Thus, in contrast to the fe
omagnet, the spin–wave
excitations of the anti-fe
omagnet
exhibit a linear dispersion in the limit k → 0. Surprisingly, although
developed in the limit of large spin, experiment shows that even for S =
1/2 spin chains, the integrity of the linear dispersion is maintained.
The Hamiltonian assumes the diagonal form:
2 2 | sin | tk kH NJS JS     
number operato
1t
k k kUa U 
 one can obtain the ground state as
| . |G S U   .
2
3 1/20 1281 ( )
2 15 2
d
E n V
na
L
 
  
 

Note that string indices for reverse-lexical ordering are not necessarily the
same as indices for lexical ordering. For the string
1
2 3 4
t t t
a a a kUa U  

from the indices level.
The index is calculated as 0+0+0+1+1+0+1+1= 4.
Question 2 (c).
The evaluation of the numerator, requiring the calculation of one– and
two–body operator matrix elements, is more involved. The single particle
part is
Question 1.(a). The spectrum of exciton becomes very dense with the
increase in energy, like in a hydrogen atom. The electrons and holes
ehave like independent particles. The excitons are mobile but ca
y no
charge. The Hamiltonian equation for an exciton is given by:
When no magnetic energy is applied, there is a change in its discrete
energy levels. The particle motion is confined to two dimensional motion
Where m is the electron mass (effective), r is the displacement, z is the
angular momentum.
1
.
2
c
A B
and
eB
w
m



The quantum numbers (n,m) follows: n=0,1,2,3,... azimuthal momentum
quantum number m = −n, −n+2,...


 lkji
ijkl
ee
m
mmm ccccklVijccEH ''
'
||
2
1   

(b) The energy
2 ( ) 2 ( )
[ , ]
fE
kinE k d N
if
W W
  

 
 
 

2 2( ) (0)kin fE N E W n 
The gain is maximum for half filling n = 1/2 and...