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Second Physics ConferenceMay, 2007
A confined N – dimensional harmonic oscillator
Sami M. Al – JaberDepartment of physics
An- Najah National University,
Nablus , Palestine
Abstract
We compute the energy eigenvalues for the N- dimensional harmonic oscillator confined in an impenetrable spherical cavity. The results show their dependence on the size of the cavity and the space dimension N. The obtained results are compared with those for the free N- dimensional harmonic oscillator, and as a result, the notion of fractional dimensions is pointed out. Finally, we examine the correlation between eigenenergies for confined oscillators in different dimensions.
2- The N- dimensional harmonic oscillator
The radial part solution, , satisfies
)4(,2
1
2
11
222
22
22
rERrRrmmr
N
dr
d
r
N
dr
d
m
letting R (r)=u (r) / r(N-1)/2
and Looking for solutions of the
)6(,22
2
1
Veu
N
We get
And thus
)12(.,2
,422
1 2211
222
r
NNFeArrR
r
)16(.2
NnE
1F1(a,b,z)=
Where,
)17(,
!n
z
b
a n
n
n
1..................21 naaaaa n
Connection between a confined harmonic oscillator and a free one
The value of λ is
We consider the case a = -1and
)19(24 Na
.0
Table 1: a = -1 andNZ0SR0s/r0
2111.58110.6324
31.51.22471.73210.7071
421.41421.87080.7599
52.51.581120.7906
631.73212.12130.8165
73.51.87082.23610.8366
8422.34520.8528
94.52.12132.44950.8660
1052.23612.54950.8771
100507.07117.17640.9853
100050022.360722.23940.9985
.0
Table 2: The case a = -1 and
NZ0Sr0S/r0
221.41421.87080.7559
32.51.581120.7906
431.73212.12130.8165
53.51.87082.23610.8366
6422.34520.8528
74.52.12132.44950.8660
852.23612.54950.8771
95.52.34522.64580.8864
1062.44952.73860.8944
100517.14147.24570.9856
100050122.383022.41650.9985
1
For a = -2, 1F1(a,b,z) has two roots. and For the ground state we have
and b =N /2, and thus The two roots z1 and z2 yield two radii for the
cavity , S1 and S2, which correspond to the states = ( 4 ,0) and ( 4 , 2) for the
free harmonic oscillator
111 bb 112 bb
,0 8N
.2
4
NE
,n
Table 3: a = -2 and
NZ1S1=Z2S2
20.58580.76533.41421.8477
30.91880.95854.08112.0200
41.26791.12604.73212.1753
51.62921.27645.37082.3175
621.414262.4495
72.37871.54236.62132.5732
82.76401.66257.23612.6900
93.15481.77627.84522.8010
103.55051.88438.44952.9068
10043.85866.622658.14147.6251
1000478.617021.8773523.383022.8776
r0
2.1213
2.2361
2.3452
2.4495
2.5495
2.6457
2.7386
2.8284
2.9155
7.3144
22.4388
0
General Case
= even integer implies even or odd.
The n = even case yields mod 4,
a = negative integer, and this
energy corresponds to the states for the free harmonic oscillator with whose
number is n /2.
,cN .22
NcE
c nc
24N
.2
NnE
,n ,2,.....2,0 n
•n= odd case yields mod 4, a = negative half – integer, and the energy corresponds to the states
for the free harmonic oscillator with
whose number is (n+1)/2.
n = odd integer : using
n =half odd integer. This energy could be written in two ways:
The first: which is the same
as that of the states of the free harmonic oscillator in (N-1) dimensions. The values of are given by
2N
2/NnE ,n
............,3,1 n
,2
cn
2
NnE
2
1
2
12 NnE
,2
12n
The second: which is the same as that of the states
of the free harmonic oscillator in (N+1) dimensions. The values of are given by
4mod30....,..........,
2
52,
2
32
4mod1.1..,..........,2
32,
2
12
cfornn
cfornn
2
1
2
12 NnE
,2
12n
4mod3.1,............,
2
52,
2
12
4mod10.,..........,2
52,
2
12
cfornn
cfornn
defined
C(n+ , )in (N-1)dim(n- , )in (N+1)dim
1(1,1)(0,0)
3(2,0)(1,1)
5(3,1), (3,3)(2,0), (2,2)
7(4,0), (4,2)(3,1), (3,3)
9(5,1), (5,3), (5,5)(4,0), (4,2), (4,4)
11(6,0), (6,2), (6,4)(5,1), (5,3), (5,5)
13(7,1), (7,3), (7,5), (7,7)(6,0), (6,2), (6,4), (6,6)
15(8,0), (8,2), (8,4), (8,6)(7,1), (7,3), (7,5), (7,7)
,2
12
n
n
For computational purposes, we choose c=1 and
and it is just the energy of the
state ( 0 ,0) or ( 1 , 1) for the free harmonic
oscillator in ( N+1) or ( N-1) dimensions respectively,
1N .25.04/ Na
,22
1
NE
Table 5. c = 1
NZ0Sr0
22.30131.51700.7071
33.13611.77091.2247
43.91291.97811.4142
54.65362.15721.5811
65.36952.31721.7321
76.06672.46311.8708
86.74942.59802
97.42032.72402.1213
108.08142.84282.2361
2014.37063.79093.1623
5031.90145.64815
10059.73947.72917.0711
c=half odd integer: For example if c= , then the energy of the confined oscillator becomes
This energy could be written as which is just the ground – state energy of the state
(n' , ) (0 ,0 ) for the free harmonic oscillator in the fractional dimension
if we let with n being odd, then the energy of the confined harmonic oscillator becomes
This energy could be written as which is exactly the energy of the state (n' , )= for the free harmonic oscillator in the fractional
dimension
21
.24
1
NE
22
10
NE
2
1N
2
nc
.24
NnE
22
1
4
1 NnE
,4
1n
21N
4- Energy levels and eigenfunctions for a confined harmonic oscillator
Letting the value of, for a given at which 1F1 has its zero at
We numerically compute E10, E21, and E32 for two cavities.
na thn thn22SZ
24 Nann
.2
22
NaE nn
)22(./,2
,422
1 2211
2/,
22
SrZ
NNFerAr nSrz
nimn
Table 6: computed energy levels
NE10E21E32E10E21E32
21.12226.825817.57741.00084.16228.5462
31.76488.279519.69741.50284.75849.4180
42.47189.582821.90632.00705.383610.3300
53.246911.093424.20282.51566.040411.2782
64.092612.689426.58623.03126.728412.2610
75.010114.370029.05503.55147.448413.2786
86.000016.133631.60924.08868.200014.3304
97.062917.979234.24764.63688.983215.4156
108.198719.906336.96965.20089.797616.5342
ASforEn 21
ASforEn 102
Table 7:Results for E10 & E20 for S=4A0 .
N E10 E20
21.0000033.000640.000150.02133
31.5000153.501700.001000.04857
42.000054.003790.002500.09485
52.500144.508320.005760.18488
63.000375.016050.012200.32100
73.500845.528700.024000.52180
84.001786.048000.044500.80000
94.503496.576000.077601.16923
105.006447.114000.128801.62857
2010.2800013.420002.800011.0000
5033.6680042.2900034.67256.6300
%100%1002
220
2
10 xE
EE
E
EEN
N
N
N
•It is observed that E10 corresponds to the state (n , ) = (0 , 0) for the free N –dimensional harmonic oscillator.E20 corresponds to the state ( 2, 0) with energy
One also observes that the effect of the boundary is relatively larger on the energy level E20 compared to its effect on E10.
Mapping of energy eigenvalues
For example, the energies for in dimension N =5 are identical to the three- dimensional solutions for the case
0
,1
•For N = odd and angular momentum the energies correspond to solutions for the roots of 1F1
These correspond exactly to those either for dimension (N-2) and angular momentum
or they correspond to the three dimensional case with
For N = even and angular momentum the energies correspond to those in dimension (N-2) with angular momentum or they correspond to those for two- dimensional case (N=2) with
,
22,
2,
422
1r
NN
,1
2/3 N,
,1
.2/2' N
Conclusion
The energy eigenfunctions were computed numerically using mathematica.
We pointed out the connection between solutions for the confined oscillator and the free one.
The effect of the boundaries and the dimension N were discussed.
The notion of fractional dimensions was explored. Mapping between energy eigenvalues in different
dimensions was examined.
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