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(,21

211

222

22

22

rERrRrmmrN

drd

rN

drd

m

letting R (r)=u (r) / r(N-1)/2

and Looking for solutions of the

)6(,22

21

VeuN

We get

And thus

)12(.,2

,422

1 2211

222

r

NNFeArrR r

)16(.2

NnE

1F1(a,b,z)=

Where,

)17(,

!nz

ba 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

31.51.22471.73210.7071421.41421.87080.759952.51.581120.7906631.73212.12130.816573.51.87082.23610.83668422.34520.852894.52.12132.44950.8660

1052.23612.54950.8771100507.07117.17640.9853

100050022.360722.23940.9985

.0

Table 2: The case a = -1 and NZ0Sr0S/r0

221.41421.87080.7559

32.51.581120.7906431.73212.12130.816553.51.87082.23610.83666422.34520.852874.52.12132.44950.8660852.23612.54950.877195.52.34522.64580.88641062.44952.73860.8944

100517.14147.24570.9856100050122.383022.41650.9985

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 thecavity , S1 and S2, which correspond to the states = ( 4 ,0) and ( 4 , 2) for thefree harmonic oscillator

111 bb 112 bb

,0 8N

.2

4

NE

,n

Table 3: a = -2 and

NZ1S1=Z2S2

20.58580.76533.41421.8477

30.91880.95854.08112.020041.26791.12604.73212.175351.62921.27645.37082.3175621.414262.449572.37871.54236.62132.573282.76401.66257.23612.690093.15481.77627.84522.8010

103.55051.88438.44952.906810043.85866.622658.14147.62511000478.617021.8773523.383022.8776

r0

2.1213

2.2361

2.34522.44952.54952.64572.73862.82842.91557.3144

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

,2c

n

2NnE

2

12

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....,..........,

252,

232

4mod1.1..,..........,2

32,2

12

cfornn

cfornn

2

12

12 NnE

,2

12n

4mod3.1,............,

252,

212

4mod10.,..........,2

52,

212

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

nn

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 = 1NZ0Sr0

22.30131.51700.707133.13611.77091.2247

43.91291.97811.414254.65362.15721.581165.36952.31721.732176.06672.46311.870886.74942.5980297.42032.72402.1213108.08142.84282.23612014.37063.79093.16235031.90145.64815

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

2nc

.24

NnE

22

1

41 Nn

E

,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

SrZNNFerAr nSrz

nimn

Table 6: computed energy levels

NE10E21E32E10E21E32

21.12226.825817.57741.00084.16228.546231.76488.279519.69741.50284.75849.418042.47189.582821.90632.00705.383610.330053.246911.093424.20282.51566.040411.278264.092612.689426.58623.03126.728412.261075.010114.370029.05503.55147.448413.278686.000016.133631.60924.08868.200014.330497.062917.979234.24764.63688.983215.4156

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

42.000054.003790.002500.0948552.500144.508320.005760.1848863.000375.016050.012200.3210073.500845.528700.024000.5218084.001786.048000.044500.8000094.503496.576000.077601.16923

105.006447.114000.128801.628572010.2800013.420002.800011.00005033.6680042.2900034.67256.6300

%100%1002

220

2

10 xEEE

EEE

N

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,

4221

rNN

,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.