Exact Solution of the Energy Shift in each Quantum Mechanical Energy Levels in a One Dimensional Symmetrical Linear Harmonic Oscillator

7 Pages·0.11 MB·English

by  Hendry I. Elim

Exact Solution of the Energy Shift in each Quantum Mechanical Energy Levels in a
One Dimensional Symmetrical Linear Harmonic Oscillator
Hendry Izaac Elim∗)1)
1) Department of Physics,Pattimura University, Ambon, Indonesia
January 7, 1999
9
9 Abstract
9 Anexactsolutionoftheenergyshiftineachquantummechanicalenergylevelsinaonedimensional
1 symmetricallinearharmonicoscillatorhasbeeninvestigated. Thesolutionwehaveusedhereisfirstly
derivedbymanipulatingSchr¨odingerdifferentialequationtobeconfluenthypergeometricdifferential
n
equation. Thefinalexactnumericalresultsoftheenergyshiftsarethenfoundbycalculatingthefinal
a
J analyticalsolutionoftheconfluenthypergeometricequationwiththeuseofasoftware(MathcadPlus
6.0)oraprogramprogrammedbyusingTurboPascal7.0. Wefindthattheresultsoftheenergyshift
6
2 in our exact solution method are almost thesame as that in Barton et. al. approximation method.
Thus, the approximation constants appeared in Barton et. al. method can also be calculated by

4 using theresults of theexact method.
v
9
0
1. Introduction
0
1
0 In1990,Bartonet. al. hadalreadypublishedanapproximationmethodtocalculatetheenergyshiftineachquantum
9 mechanical energy levels in a one dimensional symmetrical linear harmonic oscillator. They found a formula of the
9 energy shifts either for the ground state or for the excited states calculated in a finite interval L. The explanation
h/ about the use of their approximation formula is for bound states in one dimensional symmetrical linear harmonic
p oscillator with a potential V (x)which is proportional to x2.[1]However promising their method, the exact energy
- shifts in each energy levels have not been solved until now yet. In this paper We use another method to solve
t
n the problem. Instead of using the Barton et. al. approximation method, We apply a new method by changing
a Schr¨odinger differential equation to a confluent hypergeometric differential equation. After finding the confluent
u
hypergeometric differential equation, We solve exactly the equation and then use a certain program or software to
q
calculate the final exact energy shifts.
:
v This paper is organized as follows. In section 2, we will perform the exact solution method of the energy shift
i in each quantum mechanical energy levels in a one dimensional symmetrical linear harmonic oscillator and will also
X
compare our results with the results in ref.[1]. Section 3 is devoted for discussions and conclusions.
r
a
2. Exact solution method of the energy shift in each quantum mechanical energy levels in a one dimensional
symmetrical linear harmonic oscillator
Aparticlewithmassmmovinginaonedimensionalspacefrom to withtheinfluenceofapotentialV (x)obeys
the following Schr¨odinger differential equation[1] −∞ ∞
¯h2 d2
+V (x) ψ (x) = E ψ (x), < x < . (2.1)
−2mdx2 0 0 0 −∞ ∞
(cid:18) (cid:19)
Ontheotherhand,ifthemotionoftheparticleisrestrictedbyafiniteinterval L<x< L inthesamepotential,
−2 2
∗
email: [email protected]
1
the wave function ψ(x) and the energy E can be obtained by solving the following time independent Schr¨odinger
differential equation
¯h2 d2 L L
+V (x) ψ(x) = Eψ(x), < x < . (2.2)
−2mdx2 −2 2
(cid:18) (cid:19)
Eventhoughtheenergyshift∆E E E issmallforthelargeL,thiscaseisnotincludedinaclassicalperturbation
0
≡ −
theory.
Basedonthissimplephysicalphenomena,Bartonet. al.derivedgenerallyanapproximationmethodforcalculating
the energy changes or the energy shifts in each quantum mechanical energy levels including the systems in linear
harmonic oscillator and hydrogen atom. They got the following formulas
−1
L/2
¯h2 dx
1
∆E = ,for ground state, (2.3a)
m  ψ2(x )
Z 0 1
0
 
 
and
−1
L/2
¯h2 dx L L
1
∆E (0 < a < , andx ), (2.3b)
≃ m  ψ2(x ) 2 1 ∼ 2
Z 0 1
a
 
 
forexcitedstates. However,forthe calculationofthe energyshiftineachquantummechanicalenergylevelsinaone
dimensional symmetrical linear harmonic oscillator with a potential V (x)=mω2x2/2, they found
∆E(0) 2 L L2 l2
= exp 1+O , for ground state (2.4a)
E π1/2 l −4l2 L2
0 (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:18) (cid:19)(cid:19)
where ψ (x) = 1 exp x2 ,l = h¯ 1/2, and the initial energyis E =1¯hω. And for the excited states,
0 π1/4l1/2 −2l2 mω 0 2
they got (cid:0) (cid:1) (cid:16) (cid:17) (cid:0) (cid:1)
∆E(n) 2 L 2n+1 L2 l2
= exp 1+O , (2.4b)
E(n) (2n+1)π1/2n!2n l −4l2 L2
0 (cid:18) (cid:19) (cid:18) (cid:19)(cid:18) (cid:18) (cid:19)(cid:19)
where ψ(n)(x)= π1/2l22n! −1/2H x exp x2 andE(n)= 1 +n ¯hω, n=1,2,3,.... Forthe largex, H x
0 n l −2l2 0 2 n l ∼
2n x n. (cid:0) (cid:1) (cid:0) (cid:1) (cid:16) (cid:17) (cid:0) (cid:1) (cid:0) (cid:1)
l
Now,wearegoingtochangetheSchr¨odingerdifferentialequationtobecomeaconfluenthypergeometricdifferential
(cid:0) (cid:1)
equation. We provide eigen function Hψ =Eψfor a one dimensional symmetrical linear harmonic oscillator with a
potential V (x)=mω2x2/2 into two areas :[2]
¯h2 d2 1
+ mω2x2 ψ (x) = Eψ (x), forx > 0, (2.5a)
−2mdx2 2 + + +
(cid:18) (cid:19)
and
¯h2 d2 1
+ mω−2x2 ψ−(x) = Eψ−(x), forx < 0. (2.5b)
−2mdx2 2
(cid:18) (cid:19)
By substituting parameters 2mE =λ,and mω =αorα= 1, we find
h¯2 h¯ l2
d2ψ± λ α2x2 ψ± = 0, for x > 0. (2.6)
dx2 − − ±
(cid:2) (cid:3)
2
We then introduce parameter λ=α(2n′+1) and a free variable z =x√α to eq.(2.6), we get
d2ψdnz′2(z) − 2n′+1−z2 ψn′(z) = 0, for±z > 0, (2.7)
(cid:2) (cid:3)
and
1
′
En′ = h¯ω +n . (2.8)
2
(cid:18) (cid:19)
It means that
ω− ω 2n′+1
= = = r = 1, (2.9)
ω ω 2n′+1
+
which states that our solution is for a one dimensional symmetrical linear harmonic oscillator.
If ψn′(z)is the solution of eq.(2.7), then ψn′( z)is also the solution. On the other hand, if the Wronskian
−
W (ψn′(x√α),ψn′( x√α))is not the same as zero, then both solutions can be used as a basic solution for imple-
menting boundary c−onditions whether in x L or in x=0. To solve eq.(2.7), we firstly substitute the following
→±2
function
1
ψn′(z) = exp z2 Hn′(z), (2.10)
−2
(cid:18) (cid:19)
we then get
d2 d
′
dz2 −2z dz −2n Hn′(z) = 0, (2.11)
(cid:20) (cid:18) (cid:19) (cid:21)
whereHn′(z)isaHermitefunction. Byintroducinganewvariablet=z2,wefindaconfluenthypergeometric(CH)
differential equation
d2 1 d 1
′
tdz2 − 2 −t dt − 2n Hn′(t) = 0, (2.12)
(cid:20) (cid:18) (cid:19)(cid:18) (cid:19) (cid:21)
which has a general solution in a linear combination of the following two CH functions[3]
n′ 1 1 n′ 3
A F ; ;t=z2 +B √t F − ; ;t=z2 . (2.13)
1 1 1 1
− 2 2 2 2
(cid:18) (cid:19) (cid:16) (cid:17) (cid:18) (cid:19)
In eq.(2.10), the Hermite function Hn′(z)is just a special function of the following equation
2n′Γ 1 n′ 1 2n′Γ 1 1 n′ 3
Hn′(z)= Γ 1(1 2n′) 1F1 − 2 ;2;z2 + Γ −n′2 (z)1F1 −2 ;2;z2 . (2.14)
2 −(cid:0) (cid:1) (cid:18) (cid:19) −(cid:0)2 (cid:1) (cid:18) (cid:19)
Here, we have ch(cid:0)osen const(cid:1)ants A and B as follows[4](cid:0) (cid:1)
2n′Γ 1
A = 2 , (2.15a)
Γ 1(1 n′)
2 −(cid:0) (cid:1)
and (cid:0) (cid:1)
2n′Γ 1
B = −2 . (2.15b)
Γ n′
−(cid:0)2 (cid:1)
The boundar(cid:0)y con(cid:1)ditions in z are
3
2n′Γ 1
Hn′(0) = Γ 1(1 2n′) , (2.16a)
2 −(cid:0) (cid:1)
and (cid:0) (cid:1)
dHn′(0) = 2n′Γ −21 = −2n′+1√π. (2.16b)
dz Γ n′ Γ n′
−(cid:0)2 (cid:1) − 2
We can also show th(cid:0)at[3](cid:1) (cid:0) (cid:1)
2n′√π
W ψn′ x√α ,ψn′ −x√α = Γ( n′)exp z2 . (2.17)
−
(cid:0) (cid:0) (cid:1) (cid:0) (cid:1)(cid:1) (cid:0) (cid:1)
Hence, if n′is not exactly the same as n′ =1,2,3,...,then both of the solutions Hn′( z) have a linear independent
±
property. If this is work, then both of the solutions can be chosen as a basic solution of eq.(2.7), we get the general
solution as
1
ψn′(z) = [AHn′(z)+BHn′( z)]exp z2 . (2.18)
− −2
(cid:18) (cid:19)
Here, the boundary condition for the eigen function in z = L√α is
± ±2
L
ψn′( z) = ψn′ √α = 0. (2.19)
± ±2
(cid:18) (cid:19)
By substituting the boundary condition in eq.(2.19) into eq.(2.18), we find
L L L L2
ψn′ √α = AHn′ + √α +BHn′ √α exp α = 0. (2.20)
±2 2 −2 − 8
(cid:18) (cid:19) (cid:20) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21) (cid:18) (cid:19)
Eq.(2.20) can be divided into two equations with two conditions as follows
(1). For the condition with even parity:
L L L L2
ψn′ √α = AHn′ + √α +BHn′ √α exp α = 0,
2 2 −2 − 8
(cid:18) (cid:19) (cid:20) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21) (cid:18) (cid:19)
or
2n′Γ 1 n′ 1 L2
2 2 F ; ; α = 0. (2.21a)
"Γ 12(1−(cid:0) n(cid:1)′) 1 1(cid:18)− 2 2 4 (cid:19)#
(2). Fo(cid:0)r the condi(cid:1)tion with odd parity:
L L L L2
ψn′ √α = AHn′ + √α BHn′ √α exp α = 0,
2 2 − −2 − 8
(cid:18) (cid:19) (cid:20) (cid:18) (cid:19) (cid:18) (cid:19)(cid:21) (cid:18) (cid:19)
or
2n′Γ 1 L 1 n′ 3 L2
2 −2 √α F − ; ; α = 0. (2.21b)
" Γ −(cid:0)n2′ (cid:1)(cid:18)2 (cid:19) 1 1(cid:18) 2 2 4 (cid:19)#
In both(cid:0)eq.(2(cid:1).21a) and (2.21b), hermite functions Hn′ ±L2√α are[5]
L 2n′Γ 1 n′ 1 L2 (cid:0) (cid:1)
Hn′(cid:18)+2√α(cid:19)="Γ 12(1−(cid:0)2n(cid:1)′) 1F1(cid:18)− 2 ;2; 4 α(cid:19)#
(cid:0)2n′Γ 1 (cid:1) L 1 n′ 3 L2
+ −2 √α F − ; ; α , (2.22a)
" Γ −(cid:0)n2′ (cid:1)(cid:18)2 (cid:19) 1 1(cid:18) 2 2 4 (cid:19)#
(cid:0) (cid:1)
4
and
L 2n′Γ 1 n′ 1 L2
Hn′(cid:18)−2√α(cid:19)="Γ 12(1−(cid:0)2n(cid:1)′) 1F1(cid:18)− 2 ;2; 4 α(cid:19)#
(cid:0)2n′Γ 1 (cid:1) L 1 n′ 3 L2
−2 √α F − ; ; α , (2.22b)
−" Γ −(cid:0)n2′ (cid:1)(cid:18)2 (cid:19) 1 1(cid:18) 2 2 4 (cid:19)#
(cid:0) (cid:1)
where
F n′;1;L2α = Γ 21 ∞ Γ −n2′ +s L42α s
1 1 − 2 2 4 Γ n′ (cid:16) Γ 1 +(cid:17)s(cid:16)s! (cid:17)
(cid:18) (cid:19) −(cid:0) 2(cid:1) Xs=1 2
(cid:0) n′L(cid:1)2α n′2 (cid:0)2n′ L(cid:1)4α2
=1 + − +... , (2.23a)
− 4 96
(cid:0) (cid:1)
and
F 1−n′;3;L2α = Γ 32 ∞ Γ 1−2n′ +s L42α s
1 1(cid:18) 2 2 4 (cid:19) Γ (1(cid:0)−2n(cid:1)′) Xs=1 (cid:16) Γ 23 +(cid:17)s(cid:16)s! (cid:17)
(cid:16) (1 n(cid:17)′)L2α (1(cid:0) n′)((cid:1)3 n′)L4α2
=1+ − + − − +... . (2.23b)
12 480
From eq.(2.21b), we find a solution for the odd parity,
′ ′
n = 1+2m. (2.24a)
On the other hand, the even parity is found by calculating eq.(2.21a), we then get
′ ′
n = 2m. (2.24b)
Here (in eq.(2.24a)and (2.24b)), n′and m′are positive real values due to the boundary condition z = L√αwhere
±2
ψn′(±z) = 0vanishes in z = ±L2√α. n′in eq.(2.21a) and (2.21b) can exactly be calculated by using mathcad plus
6.0 or a programprogrammedby using Turbo Pascal7.0. The results of the calculation are provided in table 1 and
2.
The requirement in eq.(2.24b) produces the bound states quantum mechanical energy with even parity,
1
′ ′ ′
En′ = n + ¯hω, n = 2m , (2.25a)
2
(cid:18) (cid:19)
and the requirement in eq.(2.24a)produces the bound states quantum mechanicalenergy with odd parity as follows
1
′ ′ ′
En′ = n + ¯hω, n = 1+2m. (2.25b)
2
(cid:18) (cid:19)
On the other hand, for the calculation of the quantum mechanical energy in the boundary z = ,and the
±∞
boundary condition ψ ( z)=finite, we can use an asymtotic formula of H ( z) according to ref.[3] as follows
n n
± ±
3π
H ( z) ( 2z)n, for z or z if arg(z) < , (2.26a)
n
± ≃ ± → ±∞ | | → ∞ | | 4
and
H ( z) √π exp z2 (z )−n−1, for z . (2.26b)
n
± ≃ | | → ±∞
(cid:0) (cid:1) (cid:0) (cid:1)
5
Bysubstitutingtheconditionsineq.(2.26a)and(2.26b)intoeq.(2.18),wefindtheboundstatesquantummechanical
energy with even parity,
1
E = n+ ¯hω, n = 2m = 0,2,4,6... , (2.27a)
n
2
(cid:18) (cid:19)
and for the odd parity, we get
1
E = n+ ¯hω, n = 1+2m = 1,3,5,... . (2.27b)
n
2
(cid:18) (cid:19)
Hence, for a one dimensional symmetrical linear harmonic oscillator in the boundary z = , we get its wave
±∞
function as
1
ψ (x) = A H x√α exp αx2 , (2.28)
n n n
−2
(cid:18) (cid:19)
(cid:2) (cid:0) (cid:1)(cid:3)
where the constant parameter A can be found by a normalization.
n
Now,theenergyshiftineachquantummechanicalenergylevelsintheonedimensionalsymmetricallinearharmonic
oscillator can exactly be calculated by formulating a simple formula related to eq.(2.25) and eq.(2.27) as follows
∆E(n) =En′ En
−
′
=(n n)h¯ω, (2.29)
−
where n = 0,1,2,3,4,5,...(positive integer)and the values of parameter n′depend on L√α, for example : for
2
L√α=6,
2
n′ = 1.55x10−15, 1+1.082x10−13 , 2+3.671x10−12 , 3+0.805x10−10 ,etc.
Some results of the en(cid:0)ergy shift are att(cid:1)ac(cid:0)hed in table 1 an(cid:1)d 2(cid:0). (cid:1)
Barton et. al. results (1990) Our results
k = L2√α see:eq.(2.4a), ∆E(0) ∆E(0) =E0′ −E0
(cid:0) (cid:1) (cid:0) (cid:1)
1 0.415 1+O 1 ¯hω 0.798h¯ω
4
3 4.177x10−4 1+O 1 ¯hω 3.911x10−4 ¯hω
(cid:0) (cid:0) (cid:1)(cid:1) 36
4 5.079x10−7 1+O 1 ¯hω 4.908x10−7 ¯hω
(cid:0) (cid:0)64(cid:1)(cid:1)
5 7.835x10−11 1+O 1 ¯hω 7.671x10−11¯hω
(cid:0) (cid:0) 10(cid:1)0(cid:1)
6 1.570x10−15 1+O 1 ¯hω 1.550x10−15¯hω
(cid:0) (cid:0)144(cid:1)(cid:1)
7 4.141x10−21 1+O 1 ¯hω 4.098x10−21¯hω
(cid:0) (cid:0)196(cid:1)(cid:1)
10 4.197x10−43 1+O 1 ¯hω 36.769x10−43¯hω
(cid:0) (cid:0)400(cid:1)(cid:1)
Table 1. Theenergy shift in the ground state of the one dimensional symmetrical linear harmonic oscillator.
(cid:0) (cid:0) (cid:1)(cid:1)
6
Barton et. al. results (1990) Our results
k = L2√α see:eq.(2.4b), ∆E(n) ∆E(n) =En′ −En
(cid:0) (cid:1) (cid:0) (cid:1)
3 ∆E(1) =7.52x10−3A¯hω ∆E(1) =6.081x10−3¯hω
∆E(2) =6.767x10−2A¯hω ∆E(2) = 4.119x10−2¯hω
where A= 1+O 1 we get A=0.606
36
(cid:0) (cid:0) (cid:1)(cid:1)
5 ∆E(1) =3.918x10−9B¯hω ∆E(1) =3.672x10−9¯hω
∆E(2) =9.794x10−8B¯hω ∆E(2) =8.402x10−8¯hω
∆E(3) =1.632x10−6B¯hω ∆E(3) =1.221x10−6¯hω
where B = 1+O 1 we get B =0.848
100
(cid:0) (cid:0) (cid:1)(cid:1)
6 ∆E(1) =1.13x10−13C¯hω ∆E(1) =1.08x10−13¯hω
∆E(2) =4.07x10−12C¯hω ∆E(2) =3.67x10−12¯hω
∆E(3) =0.98x10−10C¯hω ∆E(3) =0.80x10−10¯hω
where C = 1+O 1 we get C =0.892
144
Table 2. Theenergy shift in the excited states of the one dimensional symmetrical linear harmonic oscillator.
(cid:0) (cid:0) (cid:1)(cid:1)
3. Discussions and Conclusions
We have presented an exact solution method of the energy shift in each quantum mechanical energy levels in a one
dimensionalsymmetricallinearharmonicoscillatorby using the solutionofthe confluenthypergeometricdifferential
equation. We can conclude that the relationship of the energy shifts is as follows :
∆E(0) < ∆E(1) < ∆E(2) < ∆E(3) < ... < ∆E(n−1) < ∆E(n).
According to the comparison between our results and Barton et. al. results shown in table 1 and 2, we get that
their results are almost the same as our results. However, our results are the exact results. On the other hand,
our results will be more accurate than their method if in our calculation, we add the series of gamma function in
eq.(2.23a) and (2.23b).The longer gamma function series, the more accurate our results. Based on the comparison,
we also get the approximation constants (obliquity factors) appeared in Barton et. al. method.
In conclusion, our method is work without a detail calculation of the wave function in each energy levels.
Acknowledgements
We would like to thank G.Barton for useful discussions. We also would like to thank K.Esomar for his encour-
agement. The work of H.E. was supported by the PDM No.170/P2IPT/LITMUD/V/1997, Ditbinlitabnas-DIKTI,
Republic of Indonesia.
References
1. G. Barton, et. al., (1990), Am.J. Phys.58, p.p.751-753.
2. Muslim, (1989), Suatu Tinjauan Analitis Dinamika Kuantum Osilator Selaras Linier Taksimetri, Prosiding Simposium
Fisika Nasional; 4-5 Januari 1989, Yogyakarta, Indonesia, p.p.118-123.
3. N.N.Lebedev, (1972), Special Functions and Their Applications, DoverPubl., New York,p.p.269,286.
4. L.C. Andrews, (1992), Special Functions of Mathematical For Engineers, 2ndedition, McGraw-Hill, N.Y.,p.p.397-399.
5. S.Flu¨gge, (1971), Practical Quantum Mechanics, Berlin, Springer-Verlag, p.p. 303-306,310-312.
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