Asymptotic formula for quantum harmonic oscillator tunneling probabilities

0.27 MB·English

by  Arkadiusz Jadczyk

Asymptotic formula for quantum harmonic
5
1 oscillator tunneling probabilities
0
2
r Arkadiusz Jadczyk
a
M Laboratoire de Physique Th´eorique, Universit´e de Toulouse III & CNRS
118 route de Narbonne, 31062 Toulouse, France

3
E-mail: [email protected]

]
h March 4, 2015
p
-
t
n
Abstract
a
u
Using simple methods of asymptotic analysisit is shown that for a
q
quantum harmonic oscillator in n-th energy eigenstate the probability
[
of tunneling into the classically forbidden region obeys an unexpected

6 but simple asymptotic formula: the leading term is inversely propor-
v
tional to the cube root of n.
3
8
Keywords: quantum harmonic oscillator, turning points, tunneling proba-
4
7 bilities, asymptotic, Hermite polynomials, Airy function
0
.
1
1 Introduction
0
5
1 Consider the quantum harmonic oscillator in dimensionless variables, where
:
v the Hamiltonian, in terms of position and momentum operators xˆ,pˆ, satis-
Xi fying [xˆ,pˆ]= i, is given by the formula
r 1
a Hˆ = (pˆ2+xˆ2).
2
Its normalized eigenstates, when represented as square integrable functions
of the position variable x are
ψ (x) = π 1/4(2nn!) 1/2H (x)e x2/2,
n − − n −
H are Hermite’s polynomials: H (x) = ( 1)nex2 dn e x2. Thecorrespond-
n n − dxn −
ing probability densities P (x) are then given by ψ (x)2, i.e.
n n
| |
2
P (x) = a H (x)e x2/2 ,
n n n −
(cid:16) (cid:17)
1
where
1
a = , (1)
n
π1/22nn!
so that the normalization condition holds:
∞
P (x)dx = 1.
n
Z
−∞
The functions P (x) are symmetric with respect to the origin x = 0. The
n
classical turning points are at x = √2n+1, thus the probability of quan-
±
tum tunneling into the classically forbidden region are given by the formula
∞
P = 2 P (x)dx = 2a Q , (2)
n,tun n n n
Z
√2n+1
where
2
Q = ∞ H (x)e x2/2 dx. (3)
n n −
Z√2n+1(cid:16) (cid:17)
The probability distributions P (x) are usually represented (see, for exam-
n
ple, [1, p. 168]) as in Fig. 1
The tunneling probabilities P are rarely discussed in textbooks on
n,tun
quantum mechanics. Sometimes they are discussed in quantum chemistry
textbooks. For instance in [7, p. 92] the first three are calculated (though
the second one with an error: it should be 0.1116 instead of 0.116). In [8, p.
66-67; p. 98-99 in 2nd ed (2014)] P is calculated with a comment: “The
0,tun
probability of being found in classically forbidden regions decreases quickly
with increasing n, and vanishes entirely as n approaches infinity.” Yet the
question of “how quickly” is left open there. To answer this question we will
derive the asymptotic behavior, first by using a heuristic method based on
a simple formula by Szego¨, then, in a rigorous way, using another formula
due to Olver. Both methods lead to the same final result. The problem
that we solve is closely related to a more general problem of large quantum
numbers and the correspondence principle. As can be seen from Refs. [2–5],
even for such a simple quantum system as the harmonic oscillator, there are
surprises (a more general discussion can be found in the monograph [6]).
Our result showing a very slow approach to zero of P for large n may
n,tun
be considered as another surprise.
Insection 2wederivetheleadingtermformulausingheuristicreasoning.
In section 3 we use mathematically rigorous arguments based on dominated
convergence theorem, while controlling the order of terms. In section 4 we
derive the second order correction to the leading term. The proof of the
main lemma used in this paper is given in the Appendix. Our theoretical
results are also verified numerically.
2
n = 40
2.5
Tunneling probability
0.16
2.0
0.14
0.12
0.1
1.5 0.08
0.06
0.04
1.0 0.02
0 50 100
0.5
0.0
-1.25 -1.0 -0.75 -0.5 -0.25 0.0 0.25 0.5 0.75 1.0 1.25
Figure 1: Probability distribution P (x), with the classical turning points
40
scaledtox = 1,andthecurveP fornfrom0to40.TheblackU-shaped
n,tun
±
curve represents classical probability distribution P (x) = 1/(π√1 x2)
class
−
for an oscillator with known energy and unknown phase. Shadowed tails
beyond the classical turning points are responsible for non-zero quantum
tunneling probabilities.
2 Derivation of the asymptotic tunneling proba-
bility formula
We are interested in the approximate behavior of P , as a function of n,
n,tun
for “large” n.To this end we firstusethe asymptotic formula by Szego¨ [9,p.
201, (8.22.14)] ( valid in the transition regions )
x2 3312n2+41(n!)12 2
e− 2 Hn(x) = 3 1 A(t) 1+O(n−3) , (4)
π4n12 (cid:16) (cid:17)
x = (2n+1)1/2 2 1/23 1/3n 1/6t.
− − −
−
The function A(x) used by Szego¨ relates to the Airy functions Ai(x) by the
following formula (cf. [10, p.194]):
A(t) = 3 1/3πAi( 3 1/3t).
− −
−
3
Thus Eq. 4 can be written as1
e x2/2H (x) π1/42n/2+1/4(n!)1/2n 1/12Ai(s), (5)
− n −
≈
where
s = 21/2n1/6(x √2n+1).
−
The above asymptotic formula is proven to be valid for t bounded. In the
calculation below I will use this result for all s > 0, because the main
contributions to integral come from a very small neighborhood of s = 0, in
particular when n is large in (5).
Since dx = 2 1/2n 1/6, we can calculate now Q defined in Eq. 3 as
ds − − n
2
Q = ∞ H (x)e x2/2 dx. (6)
n n −
Z√2n+1(cid:16) (cid:17)
We get
Q π1/22nn!n 1/3 ∞Ai2(s)ds.
n −
≈ Z
0
The integral of the squared Airy function is known (cf. e.g. [12, Eq. (3.75),
p. 50]):
1
∞Ai2(s)ds = .
Z 32/3Γ2[1/3]
0
Therefore (cf. Eqs. (2),(1))
P = 2a Q (7)
n,tun n n
1 1
2 π1/22nn!n 1/3
−
≈ π1/22nn! 32/3Γ2[1/3]
2
= n 1/3. (8)
−
32/3Γ2[1/3]
Thus, numerically,
0.133975
P . (9)
n,tun
≈ n1/3
As it can be seen from Fig. 2, the asymptotic formula (9) seems to agree
withthenumericalcalculations fornupto612. Yetinits derivation wehave
assumed the validity of Szego¨’s asymptotic formula in the unbounded inter-
val. Can such an extension be mathematically justified? We will address
this question in the next section.
1AnotheruseoftheAiryfunctioninadiscussionoftheharmonicoscillatorwavefunc-
tionscanbefound,forinstance,in[11,Fig. 5-23],inthesectiononWKBapproximation.
4
0.08
0.07
0.06
0.05
n
u
t 0.04
n,
P
0.03
0.02
0.01
0.0
0 100 200 300 400 500 600
n
Figure 2: Comparison of the tunneling probabilities calculated from the
exact formula with the curve given by Eq. (9), starting from n = 5.
3 Using Olver’s formula
Olver [13, p. 403] gives the following formula valid for x 1:
≥
1
ν2x2 ζ 4 4
e− 2 Hn(νx) = cn(cid:18)x2 1(cid:19) Ai(ν3ζ)+ǫ(x) , (10)
(cid:16) (cid:17)
−
where
ν = √2n+1, cn = (2π)12e−ν42ν(3ν26−1), (11)
2/3
3 3
ζ = x(x2 1)1/2 cosh 1(x) , (12)
−
(cid:18)4 − − 4 (cid:19)
and
ǫ(x) 1.36 exp(0.09ν 2) 1 Ai(ν4/3ζ). (13)
−
| | ≤ −
(cid:0) (cid:1)
Remark 1. : The function cosh 1(x) in the above formula denotes the posi-
−
tive branch of the inverse function of cosh. For general aspects of asymptotic
methods for integrals see [14], where, in particular, Section 23.4 provides
a complete description of the Airy-type expansion of Hermite polynomials.
Notice also that ν 2 = 1/(2n+1) = (1/(2n))(1 1/(2n))+O(n 3).
− −
−
5
We will need Eq. (10) squared. Using Eqs. (10)-(13) we write it in the
form
2 1/2
ν2x2 1 ζ
(cid:18)e− 2 Hn(νx)(cid:19) = dn(cid:18)1+O(n)(cid:19)(cid:18)x2 1(cid:19) Ai2(ν4/3ζ),
−
where
d = 2πe ν2/2ν(3ν2 1)/3.
n − −
In order to use this formula for calculating P , we use Eq. (3) changing
n,tun
the integration variable to get
2
P = 2a ν ∞ H (νx)e ν2x2/2 dx
n,tun n n −
Z1 (cid:16) (cid:17)
1
= 4a πνe ν2/2ν(3ν2 1)/3(1+O( ))I ,
n − − n
n
where
1/2
ζ
I = ∞ Ai2(ν4/3ζ)dx.
n Z (cid:18)x2 1(cid:19)
1
−
An easy power expansion leads to
1
4a πνe ν2/2ν(3ν2 1)/3 = n1/3 27/3+O( ) ,
n − −
(cid:18) n (cid:19)
therefore
1
P = 27/3n1/3 1+O( ) I .
n,tun n
(cid:18) n (cid:19)
It is convenient now to introduce a new integration variable
t = ν4/3ζ. (14)
Fig. 3 shows the dependence of ζ on x. We find
dζ x2 1 1/2
= − ,
dx (cid:18) ζ (cid:19)
therefore
1/2
ζ
dx = ν 4/3 dt,
−
(cid:18)x2 1(cid:19)
−
and thus
I = ν 4/3 ∞f (t)Ai2(t)dt,
n − n
Z
0
6
ζ(x) = (3x(x2 1)1/2 3 cosh 1(x))2/3
−
4 − − 4
18
16
14
12
) 10
x
(
ζ 8
6
4
2
0
0 1 2 3 4 5 6 7 8 9 10
x
Figure 3: Plot of the function ζ(x) for x > 1.
where
ζ
f (t) =
n x2 1
−
4
is implicitly defined through ζ = ν−3t and Eq. (12) that allows us to find x
as a function of ζ. Since
1
ν 4/3 = (2n+1) 2/3 = 2 2/3n 2/3 1+O( ) ,
− − − −
(cid:18) n (cid:19)
for P we get
n,tun
1
P = 25/3n 1/3 1+O( ) ∞f (t)Ai2(t)dt. (15)
n,tun − n
(cid:18) n (cid:19)Z
0
Denote by F the integral in Eq. (15):
n
F = ∞f (t)Ai2(t)dt. (16)
n n
Z
0
We will show now that
1
F F =2 2/3 ∞Ai2(t)dt = , (17)
n ∼ ∞ − Z0 22/332/3Γ(13)2
7
and therefore
2
P n 1/3,
n,tun ∼ 32/3Γ(1)2 −
3
in agreement with Eq. (8). Fig. 4 shows numerically calculated ratios
1.03
1.025
1.02
n
F
/ 1.015
∞
F
1.01
1.005
1.0
0 50 100 150 200 250 300 350 400 450 500
n
Figure 4: Numerically calculated ratios F /F , 6 n 500.
n
∞ ≤ ≤
F /F for 6 n 500. In order to verify Eq. (17), we notice that (see the
n
∞ ≤ ≤
Appendix) the function
3x(x2 1)1/2 3 cosh 1(x) 2/3
f(x) = 4 − − 4 − , (x > 1) (18)
(cid:0) x2 1 (cid:1)
−
is monotonically decreasing, from f(1) = lim f(x) = 2 2/3 to 0 at
x 1+ −
→
x . It follows that the sequence f (t) is uniformly bounded. Moreover,
n
→ ∞
with fixed t > 0, and since ζ = (2n+1) 1/3t, when n , ζ 0, thus
−
→ ∞ →
x 1, therefore lim f (t)= f(1) = 2 2/3. Thus
n n −
→ →∞
lim F = ∞ lim f (t) Ai2(t)dt.
n n
n→∞ Z0 (cid:16)n→∞ (cid:17)
by the dominated convergence theorem.
8
4 Next order approximation
Using the method described in the previous section one can easily get the
next term in the asymptotic formula for F . The result is
n
2 2/3n 2/3 1
F = 2 2/3 ∞Ai2(t)dt − − ∞tAi2(t)dt+O( ), (19)
n −
Z − 5 Z n
0 0
which, after some simplifications leads to the formula
1 0.0122518
P = 0.133975 +O(n 4/3). (20)
n,tun n1/3 (cid:18) − n2/3 (cid:19) −
The second term in Eq. (19) can be derived from the second term in the
Taylor series expansion f (t) = f (0)+(df /dt) t+O(t2), if we use Eq.
n n n t=0
|
(14) and notice that
(df /dt) = ν 4/3(df /dζ) 2 2/3n 2/3( 1/5).
n t=0 − n ζ=0 − −
| | ∼ −
Fig. 5 shows the comparison of the exact data with the two curves. The
black curve, obtained from Eq. (20) is closer to the data than the grey
one, obtained from the simple approximation given by Eq. (8). Additional
correction terms have been obtained by Paris in [15] by different means.
5 Appendix
Fig. 6 shows the computer generated plot of the function ζ(x)/(x2 1) for
−
x > 1. While the behavior of this function can be easily guessed from the
plot, it is not that easy to prove rigorously that the function has the prop-
erties that are necessary for the application of the dominated convergence
theorem used in this paper.
Lemma 1. Let f(x) be as in Eq. (18). Then f is monotonically decreasing,
from f(1) = lim f(x)= 2 2/3 to 0 at x .
x 1+ −
→ → ∞
Proof. We have
2
3 3
f(x)= f˜(x) , (21)
(cid:18)4 (cid:19)
where
x(x2 1)1/2 cosh 1(x)
f˜(x) = − − − ,
(x2 1)3/2
−
9
0.0168
0.0166
0.0164
n
u
t
n, 0.0162
P
0.016
0.0158
520 530 540 550 560 570 580 590 600 610
n
Figure 5: Details of the comparison for n > 512 : exact data (dots), simple
formula P = 0.133975n 1/3 (upper curve), and the formula with the
n,tun −
first correction term P = 0.133975n 1/3 0.0122518n 1 (lower curve).
n,tun − −
−
0.6 lim ζ(x)/(x2 1) = 2
x→1+ 3
−
) 0.5
1
−
2
x 0.4
(
/
)
x
(
ζ 0.3
0.2
0 1 2 3 4 5 6 7 8 9 10
x
Figure 6: Plot of the function ζ(x)/(x2 1).
−
and it is enough to show that the statement in the lemma holds for f˜. Using
10