wkb approximation: particle decay department of physics...
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1
WKB approximation: particle decay Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: October 06, 2015)
WKB approximation
This method is named after physicists Wentzel, Kramers, and Brillouin, who all developed it in 1926. In 1923, mathematician Harold Jeffreys had developed a general method of approximating solutions to linear, second-order differential equations, which includes the Schrödinger equation. But even though the Schrödinger equation was developed two years later, Wentzel, Kramers, and Brillouin were apparently unaware of this earlier work, so Jeffreys is often neglected credit. Early texts in quantum mechanics contain any number of combinations of their initials, including WBK, BWK, WKBJ, JWKB and BWKJ. The important contribution of Jeffreys, Wentzel, Kramers and Brillouin to the method was the inclusion of the treatment of turning points, connecting the evanescent and oscillatory solutions at either side of the turning point. For example, this may occur in the Schrödinger equation, due to a potential energy hill. (from http://en.wikipedia.org/wiki/WKB_approximation) ________________________________________________________________________ Gregor Wentzel (February 17, 1898 – August 12, 1978) was a German physicist known for development of quantum mechanics. Wentzel, Hendrik Kramers, and Léon Brillouin developed the Wentzel–Kramers–Brillouin approximation in 1926. In his early years, he contributed to X-ray spectroscopy, but then broadened out to make contributions to quantum mechanics, quantum electrodynamics, and meson theory. http://en.wikipedia.org/wiki/Gregor_Wentzel Hendrik Anthony (2 February 1894 – 24 April 1952) was a Dutch physicist who worked with Niels Bohr to understand how electromagnetic waves interact with matter. http://en.wikipedia.org/wiki/Hans_Kramers Léon Nicolas Brillouin (August 7, 1889 – October 4, 1969) was a French physicist. He made contributions to quantum mechanics, radio wave propagation in the atmosphere, solid state physics, and information theory. http://en.wikipedia.org/wiki/L%C3%A9on_Brillouin ________________________________________________________________________ 1. Classical limit
Change in the wavelength over the distance x
xdx
d .
When x
dx
d .
2
In the classical domain,
dx
d or 1
dx
d,
which is the criterion of the classical behavior. 2. WKB approximation
The quantum wavelength does not change appreciably over the distance of one wavelength. We start with the de Broglie wave length given by
h
p ,
)(2
1 2 xVpm
,
or
)]([22
2 xVmh
p
,
or
)((2 xVmp .
Then we get
])(
[22 32
dx
xdVm
dx
dh ,
or
dx
xdV
p
mh
dx
xdV
p
h
h
m
dx
xdV
h
m
dx
d )()()(3
3
23
2
.
When 1dx
d, we have
1)(
3
dx
xdV
p
mh (classical approximation)
3
If dV/dx is small, the momentum is large, or both, the above inequality is likely to be satisfied Around the turning point, p(x) = 0. |dV/dx| is very small when V(x) is a slowly changing function of x. Now we consider the WKB approximation,
)()(2
)(2
22
xxVxm
x
.
When 0V ,
ipxikx AeAex )( .
If the potential V is slowly varying function of x, we can assume that
)()(
xSi
Aex ,
.....)(!3
)(!2
)(!1
)()( 3
3
2
2
10 xSxSxSxSxS
.
____________________________________________________________________ ((Mathematica))
4
WKB approximation
eq1 —2
2 mDx, x, 2 Vx x ∂ x;
rule1 Exp—
S & ;
eq2 eq1 . rule1 Simplify
Sx
— 2 m ∂ 2 m Vx Sx2 — Sx2 m
rule2
S
S0 — S1 —2
2S2
—3
3S3 —4
4S4 & ;
eq3 2 ∂ m 2 m Vx Sx2 — Sx;
eq4 eq3 . rule2 Expand;
list1 Tablen, Coefficienteq4, —, n, n, 0, 6 Simplify;
TableForm
0 2 m ∂ 2 m Vx S0x2
1 2 S0x S1x S0x2 S1x2 S0x S2x S1x3 S1x S2x 1
3S0x S3x 1
2 S2x
4 112
3 S2x2 4 S1x S3x S0x S4x 2 S3x5 1
244 S2x S3x 2 S1x S4x S4x
6 172
2 S3x2 3 S2x S4x
_______________________________________________________________________ For each power of ħ, we have
0)]('[)(22 20 xSxmVm ,
5
)(")(')('2 010 xiSxSxS ,
)(")(')(')]('[ 120
21 xiSxSxSxS ,
..................................................................................................................................... (a) Derivation of )(0 xS
Suppose that )(xV . Then we have
)()]([2)]('[ 220 xpxVmxS ,
where
)]([2)(2 xVmxp , or
)()('0 xpxS ,
or
x
x
dxxpxS0
)()(0 .
Since )()( xkxp ,
x
x
dxxkxS0
)()(0 .
(b) Derivation of )(1 xS
)(")(')('2 010 xiSxSxS ,
)('
)('
2)('2
)(")('
0
0
0
01 xS
xSdx
di
xS
xiSxS ,
which is independent of sign.
6
)](ln[2
)]('ln[2
)(')( 011 xki
xSi
dxxSxS ,
or
2/11 )](ln[)](ln[
2
1)( xkxkxiS ,
or
)(
1)(1
xke xiS
.
(c) Derivation of )(2 xS
)(")(')(')]('[ 1202
1 xiSxSxSxS ,
)('
)]('[)(")('
0
211
2 xS
xSxiSxS
.
Then the WKB solution is given by
...)](ln[2
)(
.....)(!3
)(!2
)(!1
)()(
0
3
3
2
2
10
xki
dxxk
xSxSxSxSxS
x
x
The wave function has the form
)]})(exp["])(exp['))]{(ln(2
1exp[)(
00
x
x
x
x
dxxkiBdxxkiAxkx ,
or
])(exp[)(
])(exp[)(
])(exp[)(
'])(exp[
)(
')(
00
00
x
x
x
x
x
x
x
x
dxxkixk
Bdxxki
xk
A
dxxkixk
Bdxxki
xk
Ax
.
or
7
])(sin[)(
])(cos[)(
)(00
11 x
x
x
x
dxxkxk
Bdxxk
xk
Ax for for )(xV ,
where we put
'AA ,
'BB , BAA 1 , )(1 BAiB
We now assume that Suppose that )(xV . )(xk is replaced by
)()( xixk where
])([2)( xVmx
Then we have the wave function
])(exp[)(
"])(exp[
)(
")(
00
x
x
x
x
dxxx
Bdxx
x
Ax
for )(xV .
where "A and "B are constants. 3. The probability current density
We now consider the case of B = 0.
])(exp[)(
)(0
x
x
dxxkixk
Ax .
The probability is obtained as
mv
A
xk
AxxxP
22
)()()(*)( ,
where mvxk )( . The probability current density is
22
2A
mmv
AvvJ
.
8
Fig. 2avdtJadt , or
2vJ
4. WKB approximation near the turning points
We consider the potential energy V(x) and the energy shown in the following figure. The inadequacy of the WKB approximation near the turning point is evident, since
0)( xk implies an unphysical divergence of )(x . (a) V(x): increasing function of x around the turning point x = a
Vx
xaO
E
(i) For x>>a where V(x)>,
])(exp[)(
])(exp[)(
)( x
a
x
a
I dxxx
Bdxx
x
Ax
,
where A1 and B1 are constants, and
)(21
)( xVmx
,
9
(ii) For x<a where V(x)<,
])(sin[)(
])(cos[)(
)( a
x
a
x
II dxxkxk
Ddxxk
xk
Cx ,
where
)(21
)( xVmxk
.
________________________________________________________________________ (b) V(x): decreasing function of x around the turning point
Vx
xbO
E
(i) For x<<b where V(x)>,
])(exp[)(
])(exp[)(
)( b
x
b
x
I dxxx
Bdxx
x
Ax
,
with
)(21
)( xVmx
,
(ii) For x>b where V(x)<,
)))(sin()(
))(cos()(
)( x
b
x
b
II dxxkxk
Ddxxk
xk
Cx
where
10
)(21
)( xVmxk
_______________________________________________________________________ 5. Exact solution of wave function around the turning point x= a
Vx
xaO
E
The Schrödinger equation is given by
)()(2 2
22
xxVdx
d
m
or
0])([2 2
22
xV
dx
d
m
where is the energy of a particle with a mass m. We assume that
)()( axgxV in the vicinity of x =a, where g>0. Then the Schrödinger equation is expressed by
0)(2
22
2
axg
m
dx
d
.
Here we put
)(2
3/1
2ax
mgz
.
11
((Note))
dz
dmg
dz
d
dx
dz
dx
d3/1
2
2
2
23/2
22
2 2
dz
dmg
dx
d
Then we get
0)()(
2
2
zzdz
zd .
The solution of this equation is given by
)()(2)( 21 zBCzACz ii
where we use 2C1 instead of C1. The asymptotic form of the Airy function Ai(z) for large |z| is given by
)4
cos(||)( 4/12/1 zzAi , for z<0
and
ezzAi
4/12/1
2
1)( , for z>0
where
2/3||3
2z
12
-10 -8 -6 -4 -2 2 4z
0.5
1.0
Aiz
Fig. Plot of the Ai(z) (red) and its asymptotic form (blue) as a function of z for z<0. The asymptotic form of the Airy function Bi(z) for large |z|,
)4
sin(||)( 4/12/1 zzBi , for z<0
ezzBi
4/12/1)( , for z>0
with
2/3||3
2z
13
-10 -8 -6 -4 -2 2z
-0.5
0.5
1.0
1.5
2.0
Biz
Fig. Plot of the Bi(z) (red) and its asymptotic form (blue) as a function of z for z<0. Here we note that For z<0,
2/13/1
22||
2)(
2)( z
mgxag
mxk
.
Then we have
2/3
2/32/1
2
2/1
2
||3
2
)(2
3
2
2)(
z
xamg
dxxamg
dxxka
x
a
x
For z>0,
2/13/1
22||
2)(
2)( z
mgaxg
mx
,
we have
14
2/3
2/32/1
2
2/1
2
||3
2
)(2
3
2
2)(
z
axmg
dxaxmg
dxxx
a
x
a
______________________________________________________________________ 6. Connection formula (I; upward)
Vx
xaO
E
15
Fig. Connection formula (I; upward). The condition 1dx
d for the WKB
approximation is not satisfied in the vicinity of x = a. We need the asymptotic form the exact solution of the Schrödinger equation. Note that
)()( axax IIII and )()( axax IIII . The forms of )( bxII
and )( bxII are determined, depending on the nature of travelling waves, and the convergence of wave function at the infinity.
(i) Asymptotic form for z<0 (x<a) The asymptotic form of the wave function for z<0 (x<a) can be expressed by
)]4
)(sin()(
1
)4
)(cos()(
12[
2
)4
sin(||)4
cos(||2)()(2
2
1
6/1
22/1
4/12/12
4/12/1121
a
x
a
x
ii
dxxkxk
C
dxxkxk
Cmg
zCzCzBCzAC
(1)
where
16
2/3||3
2)( zdxxk
a
x
, 2/13/1
2||
2)( z
mgxk
.
(ii) The asymptotic form for z>0;
The asymptotic form of the wave function for z>0 (x>a) can be expressed by
)])(exp()(
1
))(exp()(
1[
2
)()(2
2
1
6/1
22/1
4/12/12
4/12/1121
x
a
x
a
ii
dxxx
C
dxxx
Cmg
ezCezCzBCzAC
(2)
where
2/3||3
2)( zdxx
x
a
, 2/13/1
2||
2)( z
mgx
.
From Eqs.(1) and (2) we have the connection rule (I; upward) as follows.
])(exp[)(
])(exp[)(
]4
)(sin[)(
]4
)(cos[)(
2
x
a
x
a
a
x
a
x
dxxx
Bdxx
x
A
dxxkxk
Bdxxk
xk
A
(I; upward)
at the boundary of x = a.
17
Vx
xaO
E
where C1 = A and C2 = B. ______________________________________________________________________ 7. Exact solution of wave function around the turning point x= b
Vx
xbO
E
The Schrödinger equation is given by
)()(2 2
22
xxVdx
d
m
,
or
0])([2 2
22
xV
dx
d
m
,
where is the energy of a particle with a mass m. We assume that
)()( bxgxV , in the vicinity of x =b, where g>0. The Schrödinger equation is expressed by
18
0)(2
22
2
bxg
m
dx
d
.
Here we put
)(2
3/1
2bx
mgz
.
Then we get
0)()(
2
2
zzdz
zd .
The solution of this equation is given by
)()(2)( 21 zBCzACz ii .
We note the following. (i) For z<0 (x>b) k(x) is expressed by
2/13/1
22||
2)(
2)( z
mgbxg
mxk
,
2/32/3
2/1
2
2/1
2||
3
2)(
2
3
22)( zbx
mgdxbx
mgdxxk
x
b
x
b .
(ii) For z>0 (x<b), where >V(x) (x) is expressed by
2/13/1
2
22
2
)(2
)]([2
)(
zmg
xbgm
xVm
x
2/32/32/1
2
2/1
2 3
2)(
2
3
22)( zxb
mgdxxb
mgdxx
b
x
b
x
19
__________________________________________________________________ 8. Connection formula-II (downward)
Fig. Connection formula (II; downward). The condition 1dx
d for the WKB
approximation is not satisfied in the vicinity of x = b. We need the asymptotic form the exact solution of the Schrödinger equation. Note that
)()( bxbx IIII and )()( bxbx IIII . The forms of )( bxII
and )( bxII are determined, depending on the nature of travelling waves, and the convergence of wave function at the infinity.
The asymptotic form for z<0;
)]4
)(sin()(
1
)4
)(cos()(
12[
2
)4
sin(||)4
cos(||2)()(2
2
1
6/1
22/1
4/12/12
4/12/1121
x
b
x
b
ii
dxxkxk
C
dxxkxk
Cmg
zCzCzBCzAC
20
The asymptotic form for z>0;
)])(exp()(
1
))(exp()(
1[
2
)()(2
2
1
6/1
22/1
4/12/12
4/12/1121
b
x
b
x
ii
dxxx
C
dxxx
Cmg
ezCezCzBCzAC
Then we have the connection formula (II; downward) as
]4
)(sin[)(
]4
)(cos[)(
2
])(exp[)(
])(exp[)(
x
b
x
b
b
x
b
x
dxxkxk
Bdxxk
xk
A
dxxx
Bdxx
x
A
(II, downward)
with
Vx
xbO
E
where C1 = A and C2 = B. 9. Tunneling probability
We apply the connection formula to find the tunneling probability. In order that the WKB approximation apply within a barrier, it is necessary that the potential V(x) does not change so rapidly. Suppose that a particle (energy and mass m) penetrates into a barrier shown in the figure. There are three regions, I, II, and III.
21
ba
I II III
Vx
E
Fig. The connection formula I (upward) is used at x = a and the connection formula II
(downward) is used at x = b. For x>b, (region III)
]4
)(sin[)(
]4
)(cos[)(
)]4
)((exp[)(
1
1
1
1
1
1
x
b
x
b
x
b
III
dxxkxk
iAdxxk
xk
A
dxxkixk
A
(we consider on the wave propagating along the positive x axis), where
))((2
)(21 xVm
xk
for x>b
The connection formula (II, downward) is applied to the boundary between the regions III and II.
]4
)(sin[)(2
]4
)(cos[)(
])(exp[)(2
])(exp[)(2
1
1
1
1
x
b
x
b
b
x
b
x
dxxkxk
Bdxxk
xk
A
dxxx
Bdxx
x
A
(II, downward)
Here we get
22
iAB 2 .
Then we get the wave function of the region II,
]4
)(sin[)(
]4
)(cos[)(
])(exp[)(
])(exp[)(2
1
1
1
1
x
b
x
b
III
b
x
b
x
II
dxxkxk
iAdxxk
xk
A
dxxx
iAdxx
x
A
or
])(exp[2)(
])(exp[1
)(
])()(exp[)(
])()(exp[)(2
x
a
x
a
x
a
b
a
b
a
x
a
II
dxxr
x
Adxx
rx
iA
dxxdxxx
iAdxxdxx
x
A
where
])([2
)(2
xVm
x
, for a<x<b
and
])(exp[ b
a
dxxr ,
Next, the connection formula (I; upward) is applied to the boundary between the regions II and I.
23
)])(exp()(
))(exp()(
)]4
)(sin()(
)4
)(cos()(
22
2
2
2
x
a
x
a
a
x
a
x
dxxx
Ddxx
x
C
dxxkxk
Ddxxk
xk
C
(I; upward)
Here we get
riAC
1 ,
rA
D2
.
)]([2
)(22 xVm
xk
for x<a
Then we have the wave function of the region I,
)]}4
)((exp[)]4
)(({exp[4)(
)]}4
)((exp[)]4
)(({exp[1
)(
]4
)(sin[)(2
]4
)(cos[1
)(
2
22
2
22
2
2
2
2
2
a
x
a
x
a
x
a
x
a
x
a
x
I
dxxkidxxkiir
xk
A
dxxkidxxkirxk
iA
dxxkrxk
A
dxxkrxk
iA
or
)]}4
)((exp[))4
1()]
4)((exp[)
4
1{(
)(
)]}4
)((exp[)1
4()]
4)((exp[)
1
4{(
)(
22
2
22
2
x
a
x
a
a
x
a
x
I
dxxkir
rdxxki
r
rxk
iA
dxxkir
rdxxki
r
r
xk
iA
24
The first term corresponds to that of the reflected wave and the second term corresponds to that of the incident wave. Then the tunneling probability is
))(2exp()
41
(
1 2
2
b
a
dxxrr
r
T
where
))(exp( b
a
dxxr
9. -particle decay: quantum tunneling
r1 r2
Coulomb repusion
Nuclear binding
Vr
rO
E
Fig. Gamov's model for the potential energy of an alpha particle in a radioactive
nucleus.
25
-2
0
2
4
6
8
10
r
nucl
ear
pote
ntia
lMeV
Re yar
Ea
Fig. The tunneling of a particle from the 238U (Z = 92). The kinetic energy 4.2 MeV.
http://demonstrations.wolfram.com/GamowModelForAlphaDecayTheGeigerNuttallLaw/
For r1<r<r2.
)(21
)( rVmr
At r = r2,
20
21
4
2
r
eZ
.
The tunneling probability is
])(2exp[1
2 r
r
drreP ,
where
26
])1([arccos2
])(arccos[2
12
)(2
)(
2
1
2
1
2
12
1212
12
22
1
2
1
2
1
r
r
r
r
r
rr
m
rrrr
rr
m
drr
rm
drrVm
drr
r
r
r
r
r
r
where m is the mass of -particle (= 4.001506179125 u). fm = 10-15 m (fermi).
The quantity P gives the probability that in one trial an particle will penetrate the barrier. The number of trials per second could estimated to be
12r
vN ,
if it were assumed that a particle is bouncing back and forth with velocity v inside the nucleus of diameter 2r1. Then the probability per second that nucleus will decay by emitting a particle, called the decay rate R, would be
2
12 e
r
vR .
((Example))
We consider the particle emission from 238U nucleus (Z = 92), which emits a K = 4.2 MeV particle. The a particle is contained inside the nuclear radius r1 = 7.0 fm (fm = 10-15 m). (i) The distance r2: From the relation
20
2
4
2
r
ZeK
,
we get
r2 = 63.08 fm. (ii) The velocity of a particle inside the nucleus, v:
27
From the relation
21 2
1vmK
where m is the mass of the a particle; m = 4.001506179 u, we get
v = 1.42318 x 107 m/s (iii) The value of :
])(arccos[2
1212
12 rrr
r
rr
mK
=51.8796.
(iv) The decay rate R:
2
12 e
r
vR = 8.813 x 10-25.
((Mathematica))
28
Clear"Global`";
rule1 u 1.660538782 1027, eV 1.602176487 1019,
qe 1.602176487 1019, c 2.99792458 108,
— 1.05457162853 1034, 0 8.854187817 1012,
MeV 1.602176487 1013, Ma 4.001506179125 u,
fm 1015, Z1 92, r1 7 fm , K1 4.2 MeV;
eq0 K1 2 Z1 qe2
4 0 r. rule1
6.729141013 4.245021026
r
eq01 Solveeq0, r; r2 r . eq0116.308421014
r2
fm. rule1
63.0842
eq1 1
2Ma v2 K1 . rule1; eq2 Solveeq1, v;
v1 v . eq221.42318107
2 Ma K1
—r2 ArcCos r1
r2 r1 r2 r1 .
rule1
51.8796
R1 v1
2 r1Exp2 . rule1
8.812821025
29
10. Schottky barrier
We consider the Schottky barrier which exists on the junction between a metal surface and a n-typed semiconductor surface. The form of the potential energy V(x) for an electron can be determined from a Poisson equation with appropriate boundary condition. The parabolic form for V(x) is expected. For simplicity, here, we assume that V(x) has a triangular form given by
)1()( 0 w
xVxV
where w is the width of the depletion layer in the Schottky barrier (see a Fig. below)
30
31
Fig. Electronic band scheme of a metal/semiconductor (n-doped) junction: pinning of
the Fermi-level FE in interface states near the neutrality level causes the
formation of a Schottky-barrier SBe and a depletion space charge layer within the
semiconductor. VD is the built-in diffusion voltage. (a) In thermal equilibrium, (b) under external bias U.
When )(xV , we have the value of x0
)1(0
0 Vwxx
where is the energy of electron with a mass m. The transition probability can be expressed by
32
])(2exp[0
0x
dxxT
where
])([2
)(2
xVm
x
for 00 xx .
Using the Mathematica we calculate the integral as
wV
Vm
dxw
xVV
m
dxxVm
dxx
x
xx
0
2/30
2
0
002
02
0
3
)(22
2
])([2
)(
0
00
where )1()( 0 w
xVxV . Then we have
]3
)(42exp[)(
0
2/30
2w
V
VmT
Suppose that =0.
]2
3
4exp[)0(
20 w
mVT
or
)]()(83089.6exp[)0( 0 nmweVVT
when V0 = 0.7 eV,
33
w nm
T
0.2 0.4 0.6 0.8 1.0
0.2
0.2
0.4
0.6
0.8
1.0
Fig. Transition probability T as a function of the width of the depletion layer. V0 = 0.7
eV. 1nm = 10 Å 9. Bound state: Bohr-Sommerfeld condition.
ab
I II III
Vx
E
For x<b (region I), the un-normalized wave function is
])(exp[)(
1b
x
I dxxx
,
Using the connection rule (II; downward)
34
]4
)(sin[)(2
]4
)(cos[)(
])(exp[)(2
])(exp[)(2
x
b
x
b
II
b
x
b
x
dxxkxk
Bdxxk
xk
A
dxxx
Bdxx
x
AI
(II; downward)
we get
2A , B = 0 Then we have
)]4
)(cos[)(
2 x
b
II dxxkxk
for b<x<a
This may also be written as
]4
)(sin[])(cos[)(
2
]4
)(cos[])(sin[)(
2
]4
)()(sin[)(
2
]24
)()(cos[)(
2
]4
)()(cos[)(
2]
4)(cos[
)(
2
a
x
a
b
a
x
a
b
a
x
a
b
a
x
a
b
a
x
a
b
x
b
II
dxxkdxxkxk
dxxkdxxkxk
dxxkdxxkxk
dxxkdxxkxk
dxxkdxxkxk
dxxkxk
Here we use the connection rule (I, upward),
35
])(exp[)(
])(exp[)(
]4
)(sin[)(
]4
)(cos[)(
2
x
a
x
a
a
x
a
x
dxxx
Bdxx
x
A
dxxkxk
Bdxxk
xk
A
(I; upward)
From this we have
]4
))(sin[])(cos[)(
2
]4
))(cos[])(sin[)(
2
a
x
a
b
a
x
a
b
II
dxxkdxxkxk
dxxkdxxkxk
with
])(sin[a
b
dxxkA , ])(cos[2 a
b
dxxkB .
Since III should have such a form
])(exp[)(
x
a
III dxxx
A
for x>a. Then we need the condition that
0])(cos[2 a
b
dxxkB ,
or
)2
1()( ndxxk
a
b
or
)2
1()( ndxxp
a
b
,
36
where n = 0, 1, 2, ... (Bohr-Sommerfeld condition).
hnndxxp )2
1(2)
2
1()( .
is an area inside the trajectory of the particle in the phase space.
10. Eample-1 (Simple harmonics):
We consider a simple harmonics,
2200
22 2)2
1(2)]([2)( xxmxmmxVmxp
where
20
0
2
mx .
Then we get
02
0
0
20
0
0
2200
2
2
1
422)(
00
0
mm
xmdxxxmdxxp
xx
x
When
)2
1()(
0
0
ndxxpx
x
we have
)2
1(
0
n ,
or
)2
1( n
11. Example-2: linear potential
We consider a particle moving in 1D potential of the form
37
xxV )(
)(2)( xmxp
0x .
)2
1()(2)(
00
00
odd
xx
x
ndxxpdxxp
or
2/30
0
0
0
0
0
3
222
22
)(22)(22
0
00
xm
dxxxm
dxxxmdxxm
x
xx
or
)2
1(
3
222 2/3
0 oddnxm .
because of the odd parity states. Then we get the energy as
3/22/12
])2
1()
2(
4
3[ oddn
m
When mg
3/122
3/23/1
3/23/2
)2
(
)()2(
1)
2
1()
4
3(
mgp
mgm
n
n
odd
We now calculate the value
3/23/23/23/2 )4
1()
2
3()
2
112()
4
3( nnpn ,
38
where nodd= 2n -1 (n = 1, 2, 3, 4,…), and
nz
for n = 1, 2, 3,….. Note that nz is the n-th zero points of the Airy function )(zAi with
odd parity. The value of zn can be obtained from the exact solution of the Schrödinger equation. The value pm is obtained from the WKB approximation. It is surprising that in spite of the approximation, the value of pn is so close to that of -zn
Ai z
z20 15 10 5 5
0.4
0.2
0.2
0.4
Fig. The Airyfunction Ai(z). The values of z at which Ai(z) becomes zero are denoted
by blue circles. z = zn. ____________________________________________________________________ Table
39
40
x0x0
V x
Fig. The potential energy xxV )( .
0x . Because of the symmetric potential,
the wave function should have either the even parity or the odd parity. ((Note)) Solution with the even parity
41
3/122
3/23/1
3/23/2
)2
(
)()2(
1)
2
1()
4
3(
mgq
mgm
n
n
even
where nneven 2 (n = 0, 1, 2, 3,…),
3/23/2 )
2
12()
4
3( nqn
Note that qn is nearly equal to -yn, where the derivative of the Airy function Ai(z) with respect to z, becomes zero at z = yn.
n qn -yn 0 1.11546 1.01879 1 3.26163 3,2482 2 4.82632 4.8201 3 6.16713 6.16331 4 7.37485 7.37218 5 8.49051 8.48849 6 9.53705 9.53545 7 10.529 10.5277 8 11.4762 11.4751 9 12.3857 12.3848 10 13.263 13.2622
Ai z
z20 15 10 5 5
0.4
0.2
0.2
0.4
Fig. The Airyfunction Ai(z). The values of z at which the derivative of Ai(z) with
respect to z becomes zero are denoted by blue circles. z = yn. _______________________________________________________________________
42
APPENDIX Connection formula
)(21
)( xVmxk
)(21
)( xVmx
(i) Connection formula at x = a (upward)
)])(exp()(
))(exp()(
)]4
)(sin()(
)4
)(cos()(
21
1
1
1
x
a
x
a
a
x
a
x
dxxx
Bdxx
x
A
dxxkxk
Bdxxk
xk
A
formula I (upward)
(ii) Connection formula at x = b (downward)
)]4
)(sin()(2
)4
)(cos()(
)])(exp()(2
))(exp()(2
2
2
2
2
x
b
x
b
b
x
b
x
dxxkxk
Ddxxk
xk
C
dxxx
Ddxx
x
C
formula II (downward)
43
(i) Connection formula at x = b
]4
)(sin[)(2
]4
)(cos[)(
])(exp[)(2
])(exp[)(2
1
1
1
1
x
b
x
b
b
x
b
x
dxxkxk
Ddxxk
xk
C
dxxx
Ddxx
x
C
formula II (downward,)
(ii) Connection formula at x = a
])(exp[)(
])(exp[)(
]4
)(sin[)(
]4
)(cos[)(
2
2
2
2
2
x
a
x
a
a
x
a
x
dxxx
Bdxx
x
A
dxxkxk
Bdxxk
xk
A
formula I (upward)