wkb approximation: masatsugu sei suzuki department of physics, suny …bingweb.binghamton.edu ›...
TRANSCRIPT
1
WKB approximation: particle decay Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: January 13, 2012)
________________________________________________________________________ Gregor Wentzel (February 17, 1898, in Düsseldorf, Germany – August 12, 1978, in Ascona, Switzerland) 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 "Hans" Kramers (Rotterdam, February 2, 1894 – Oegstgeest, April 24, 1952) was a Dutch physicist. http://en.wikipedia.org/wiki/Hendrik_Anthony_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 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,
2
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) ________________________________________________________________________ 1. Classical limit
Change in the wavelength over the distance x
dxdx
d .
When x
dx
d .
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 .
3
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)
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 ,
ipx
ikx 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
)()]([2)]('[ 22
0 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.
)](ln[2
)]('ln[2
)(')( 011 xki
xSi
dxxSxS ,
or
6
2/1
1 )](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
.
where we put
'A
A , 'B
B
3. The probability current density
7
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
.
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
8
Vx
xaO
E
(i) For x>>a where V(x)>,
))(exp()(
))(exp()(
)( 11 x
a
x
a
I dxxx
Bdxx
x
Ax
,
where A1 and B1 are constants, and
)(21
)( xVmx
,
(ii) For x<a where V(x)<,
))(sin()(
))(cos()(
)( 22 a
x
a
x
II dxxkxk
Bdxxk
xk
Ax ,
where
)(21
)( xVmxk
.
________________________________________________________________________ (b) V(x): decreasing function of x around the turning point
9
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()(
2)(
x
b
x
b
II dxxkxk
Bdxxk
xk
Ax
where
)(21
)( xVmxk
_______________________________________________________________________ 5. Exact solution of wave function around the turning point x= a
10
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
)()( axgExV 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
.
Then we get
0)()(
2
2
zzdz
zd .
11
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
-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
12
ezzBi
4/12/1)( , for z>0
with
2/3||3
2z
-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
13
For z>0,
2/13/1
22||
2)(
2)( z
mgaxg
mx
we have
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
(i) Asymptotic form for z<0 (x<a) The asymptotic form of the wave function for z<0 can be expressed by
)]4
)(sin()(
1)
4)(cos(
)(
12[
2
)4
sin(||)4
cos(||2)()(2
21
6/1
22/1
4/12/12
4/12/1121
a
x
a
x
ii
dxxkxk
Cdxxkxk
Cmg
zCzCzBCzAC
where
14
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 can be expressed by
)])(exp()(
1))(exp(
)(
1[
2
)()(2
21
6/1
22/1
4/12/12
4/12/1121
x
a
x
a
ii
dxxx
Cdxxx
Cmg
ezCezCzBCzAC
where
2/3||3
2)( zdxx
x
a
, 2/13/1
2||
2)( z
mgx
.
The 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.
Vx
xaO
E
15
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
0)(2
22
2
bxg
m
dx
d
.
Here we put
)(2
3/1
2bx
mgz
.
Then we get
16
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
__________________________________________________________________ 8. Connection formula-II (downward)
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
17
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.
18
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
.
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
19
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
,
and
))(exp( b
a
dxxr ,
Next, the connection formula (I; upward) is applied to the boundary between the regions II and I.
20
)])(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
.
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
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
21
))(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.
22
-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
23
])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:
24
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))
25
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
26
9. Bound state
ab
I II III
Vx
E
For x<b (region I), the unnormalized wave function is
))(exp()(
1b
x
I dxxx
,
Using the connection rule (II; downward)
)]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
27
]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),
)])(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
28
))(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
,
where n = 0, 1, 2, ... 10. 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
ndxxpa
x
29
we have
)2
1(
0
n ,
or
)2
1( n
_______________________________________________________________________ APPENDIX Connection formula
)(21
)( xVmxk
)(21
)( xVmx