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Today’s Outline - February 26, 2015
• Example 8.2
• Connection formula
Homework Assignment #05:Chapter 8: 1,3,5,6,9,13due Thursday, March 05, 2015
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 1 / 15
Today’s Outline - February 26, 2015
• Example 8.2
• Connection formula
Homework Assignment #05:Chapter 8: 1,3,5,6,9,13due Thursday, March 05, 2015
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 1 / 15
Today’s Outline - February 26, 2015
• Example 8.2
• Connection formula
Homework Assignment #05:Chapter 8: 1,3,5,6,9,13due Thursday, March 05, 2015
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 1 / 15
Today’s Outline - February 26, 2015
• Example 8.2
• Connection formula
Homework Assignment #05:Chapter 8: 1,3,5,6,9,13due Thursday, March 05, 2015
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 1 / 15
Example 8.2
The WKB approximation can be used to compute the spontaneous decayof an nucleus by alpha emission.
The alpha particle (2 protons and 2 neutrons) is positively charged andwill be repelled by the remaining nucleus once it escapes but there is apotential barrier that can be significantly larger than the kinetic energy ofthe alpha particle.
Approximate the nuclear potentialby a square well of width r1 joinedto a Coulomb repulsion. E is theenergy of the alpha particle. Theouter turning point is defined by
1
4πε0
2Ze2
r2= E
E
V(r)
r
-V0
r1 r2
Coulomb repulsion
Nuclear binding
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 2 / 15
Example 8.2
The WKB approximation can be used to compute the spontaneous decayof an nucleus by alpha emission.
The alpha particle (2 protons and 2 neutrons) is positively charged andwill be repelled by the remaining nucleus once it escapes but there is apotential barrier that can be significantly larger than the kinetic energy ofthe alpha particle.
Approximate the nuclear potentialby a square well of width r1 joinedto a Coulomb repulsion. E is theenergy of the alpha particle. Theouter turning point is defined by
1
4πε0
2Ze2
r2= E
E
V(r)
r
-V0
r1 r2
Coulomb repulsion
Nuclear binding
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 2 / 15
Example 8.2
The WKB approximation can be used to compute the spontaneous decayof an nucleus by alpha emission.
The alpha particle (2 protons and 2 neutrons) is positively charged andwill be repelled by the remaining nucleus once it escapes but there is apotential barrier that can be significantly larger than the kinetic energy ofthe alpha particle.
Approximate the nuclear potentialby a square well of width r1 joinedto a Coulomb repulsion. E is theenergy of the alpha particle.
Theouter turning point is defined by
1
4πε0
2Ze2
r2= E
E
V(r)
r
-V0
r1 r2
Coulomb repulsion
Nuclear binding
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 2 / 15
Example 8.2
The WKB approximation can be used to compute the spontaneous decayof an nucleus by alpha emission.
The alpha particle (2 protons and 2 neutrons) is positively charged andwill be repelled by the remaining nucleus once it escapes but there is apotential barrier that can be significantly larger than the kinetic energy ofthe alpha particle.
Approximate the nuclear potentialby a square well of width r1 joinedto a Coulomb repulsion. E is theenergy of the alpha particle. Theouter turning point is defined by
1
4πε0
2Ze2
r2= E
E
V(r)
r
-V0
r1 r2
Coulomb repulsion
Nuclear binding
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 2 / 15
Example 8.2
The WKB approximation can be used to compute the spontaneous decayof an nucleus by alpha emission.
The alpha particle (2 protons and 2 neutrons) is positively charged andwill be repelled by the remaining nucleus once it escapes but there is apotential barrier that can be significantly larger than the kinetic energy ofthe alpha particle.
Approximate the nuclear potentialby a square well of width r1 joinedto a Coulomb repulsion. E is theenergy of the alpha particle. Theouter turning point is defined by
1
4πε0
2Ze2
r2= E
E
V(r)
r
-V0
r1 r2
Coulomb repulsion
Nuclear binding
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 2 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du
=
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recall that for tunneling
in this case
but r2E = 2Ze2
4πε0
substituting
r ≡ r2 sin2 u
dr = 2r2 sin u cos u du
T ' e−2γ , γ ≡ 1
~
∫ a
0|p(x)| dx
γ =1
~
∫ r2
r1
√2m
(1
4πε0
2Ze2
r− E
)dr
=
√2mE
~
∫ r2
r1
√r2r− 1 dr
γ =
√2mE
~
∫ r2
r1
2r2
√1
sin2 u− 1 sin u cos u du
=
√2mE
~
∫ r2
r1
2r2√
1− sin2 u cos u du =
√2mE
~
∫ r2
r1
2r2 cos2 u du
=
√2mE
~r2[u + sin u cos u
∣∣∣r1r2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 3 / 15
Example 8.2
Recasting the limits, when r −→ r1,
u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2,
u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u cos u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]
=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]
=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]
since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
Recasting the limits, when r −→ r1, u −→ sin−1√
r1r2
and when r −→ r2, u −→ sin−1√
r2r2
= π/2
γ =
√2mE
~r2[u + sin u
√1− sin2 u
∣∣∣π/2sin−1√
r1/r2
=
√2mE
~r2
[π
2− sin−1
√r1r2
+ 0−√
r1r2
√1− r1
r2
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1r2
√r2(r2 − r1)
]=
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]since r2 � r1 this expression can be approximated
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 4 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]
'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~
= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~
= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]
=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~
= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~
= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~
= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~
= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~
= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~
= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~
= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~
= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~
= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~
= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
γ =
√2mE
~
[r2
(π
2− sin−1
√r1r2
)−√
r1(r2 − r1)
]'√
2mE
~
[r2
(π
2−√
r1r2
)−√r1r2
]=
√2mE
~
[r2π
2− 2√r1r2]
γ ' K1Z√E− K2
√Zr1
K1 ≡(
e2
4πε0
)π√
2m
~= 1.980 MeV1/2
K2 ≡(
e2
4πε0
)4√m
~= 1.485 fm−1/2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 5 / 15
Example 8.2
The lifetime of the nucleuscan be estimated by assum-ing that the alpha particlehas an average velocity v in-side the nucleus
this means that it will“strike” the barrier every2r1/v seconds
combining with the probabil-ity of decay gives lifetime
the figure shows how wellthis correlation with 1/
√E is
for a number of different nu-clei in the U and Th decaychains
τ =2r1v
e2γ
log10 τ = C
(K1
Z√E− K2
√Zr1
)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 6 / 15
Example 8.2
The lifetime of the nucleuscan be estimated by assum-ing that the alpha particlehas an average velocity v in-side the nucleus
this means that it will“strike” the barrier every2r1/v seconds
combining with the probabil-ity of decay gives lifetime
the figure shows how wellthis correlation with 1/
√E is
for a number of different nu-clei in the U and Th decaychains
τ =2r1v
e2γ
log10 τ = C
(K1
Z√E− K2
√Zr1
)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 6 / 15
Example 8.2
The lifetime of the nucleuscan be estimated by assum-ing that the alpha particlehas an average velocity v in-side the nucleus
this means that it will“strike” the barrier every2r1/v seconds
combining with the probabil-ity of decay gives lifetime
the figure shows how wellthis correlation with 1/
√E is
for a number of different nu-clei in the U and Th decaychains
τ =2r1v
e2γ
log10 τ = C
(K1
Z√E− K2
√Zr1
)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 6 / 15
Example 8.2
The lifetime of the nucleuscan be estimated by assum-ing that the alpha particlehas an average velocity v in-side the nucleus
this means that it will“strike” the barrier every2r1/v seconds
combining with the probabil-ity of decay gives lifetime
the figure shows how wellthis correlation with 1/
√E is
for a number of different nu-clei in the U and Th decaychains
τ =2r1v
e2γ
log10 τ = C
(K1
Z√E− K2
√Zr1
)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 6 / 15
Example 8.2
The lifetime of the nucleuscan be estimated by assum-ing that the alpha particlehas an average velocity v in-side the nucleus
this means that it will“strike” the barrier every2r1/v seconds
combining with the probabil-ity of decay gives lifetime
the figure shows how wellthis correlation with 1/
√E is
for a number of different nu-clei in the U and Th decaychains
τ =2r1v
e2γ
log10 τ = C
(K1
Z√E− K2
√Zr1
)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 6 / 15
Example 8.2
The lifetime of the nucleuscan be estimated by assum-ing that the alpha particlehas an average velocity v in-side the nucleus
this means that it will“strike” the barrier every2r1/v seconds
combining with the probabil-ity of decay gives lifetime
the figure shows how wellthis correlation with 1/
√E is
for a number of different nu-clei in the U and Th decaychains
τ =2r1v
e2γ
log10 τ = C
(K1
Z√E− K2
√Zr1
)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 6 / 15
Example 8.2
The lifetime of the nucleuscan be estimated by assum-ing that the alpha particlehas an average velocity v in-side the nucleus
this means that it will“strike” the barrier every2r1/v seconds
combining with the probabil-ity of decay gives lifetime
the figure shows how wellthis correlation with 1/
√E is
for a number of different nu-clei in the U and Th decaychains
τ =2r1v
e2γ
log10 τ = C
(K1
Z√E− K2
√Zr1
)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 6 / 15
Non-vertical walls
Suppose that the boundarywhich separates the “classi-cal” and “non-classical” re-gions is not vertical?
herethe WKB approximation for-mally breaks down
first move x-axis to put theright hand boundary (for apotential well) at x = 0
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
ψ(x) =
1√p(x)
[Be
i~∫ x0 p(x ′) dx ′ + Ce−
i~∫ x0 p(x ′) dx ′
], x < 0
1√|p(x)|
De−1~∫ x0 |p(x
′)| dx ′ , x > 0
if E < V for all x > 0 then F ≡ 0
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 7 / 15
Non-vertical walls
Suppose that the boundarywhich separates the “classi-cal” and “non-classical” re-gions is not vertical? herethe WKB approximation for-mally breaks down
first move x-axis to put theright hand boundary (for apotential well) at x = 0
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
ψ(x) =
1√p(x)
[Be
i~∫ x0 p(x ′) dx ′ + Ce−
i~∫ x0 p(x ′) dx ′
], x < 0
1√|p(x)|
De−1~∫ x0 |p(x
′)| dx ′ , x > 0
if E < V for all x > 0 then F ≡ 0
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 7 / 15
Non-vertical walls
Suppose that the boundarywhich separates the “classi-cal” and “non-classical” re-gions is not vertical? herethe WKB approximation for-mally breaks down
first move x-axis to put theright hand boundary (for apotential well) at x = 0
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
ψ(x) =
1√p(x)
[Be
i~∫ x0 p(x ′) dx ′ + Ce−
i~∫ x0 p(x ′) dx ′
], x < 0
1√|p(x)|
De−1~∫ x0 |p(x
′)| dx ′ , x > 0
if E < V for all x > 0 then F ≡ 0
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 7 / 15
Non-vertical walls
Suppose that the boundarywhich separates the “classi-cal” and “non-classical” re-gions is not vertical? herethe WKB approximation for-mally breaks down
first move x-axis to put theright hand boundary (for apotential well) at x = 0
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
ψ(x) =
1√p(x)
[Be
i~∫ x0 p(x ′) dx ′ + Ce−
i~∫ x0 p(x ′) dx ′
], x < 0
1√|p(x)|
De−1~∫ x0 |p(x
′)| dx ′ , x > 0
if E < V for all x > 0 then F ≡ 0
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 7 / 15
Non-vertical walls
Suppose that the boundarywhich separates the “classi-cal” and “non-classical” re-gions is not vertical? herethe WKB approximation for-mally breaks down
first move x-axis to put theright hand boundary (for apotential well) at x = 0
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
ψ(x) =
1√p(x)
[Be
i~∫ x0 p(x ′) dx ′ + Ce−
i~∫ x0 p(x ′) dx ′
], x < 0
1√|p(x)|
De−1~∫ x0 |p(x
′)| dx ′ , x > 0
if E < V for all x > 0 then F ≡ 0
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 7 / 15
Non-vertical walls
Suppose that the boundarywhich separates the “classi-cal” and “non-classical” re-gions is not vertical? herethe WKB approximation for-mally breaks down
first move x-axis to put theright hand boundary (for apotential well) at x = 0
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
ψ(x) =
1√p(x)
[Be
i~∫ x0 p(x ′) dx ′ + Ce−
i~∫ x0 p(x ′) dx ′
], x < 0
1√|p(x)|
De−1~∫ x0 |p(x
′)| dx ′ , x > 0
if E < V for all x > 0 then F ≡ 0C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 7 / 15
Turning point solution
The idea is to match the “classical” and “non-classical” solutions butwhen p(x) = 0, WKB breaks down (ψ →∞)
use a patching function toconnect the two sides and as-sume the potential has a lin-ear dependence in the patch-ing region
V (x) ' E +dV
dx
∣∣∣∣0
x
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 8 / 15
Turning point solution
The idea is to match the “classical” and “non-classical” solutions butwhen p(x) = 0, WKB breaks down (ψ →∞)
use a patching function toconnect the two sides
and as-sume the potential has a lin-ear dependence in the patch-ing region
V (x) ' E +dV
dx
∣∣∣∣0
x
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 8 / 15
Turning point solution
The idea is to match the “classical” and “non-classical” solutions butwhen p(x) = 0, WKB breaks down (ψ →∞)
use a patching function toconnect the two sides and as-sume the potential has a lin-ear dependence in the patch-ing region
V (x) ' E +dV
dx
∣∣∣∣0
x
E
x0Classical
region
Nonclassical
region
Linearized
potential
Turning
point
V(x)
Patching
region
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 8 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms
and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms
and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3
The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3
The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3
The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
Patching function
Need to solve the approximate Schrodinger equation
− ~2
2m
d2ψp
dx2+ [E + V ′(0)x ]ψp = Eψp
canceling the E terms and gather-ing the constants
making a change of variables to z ≡αx , dz = αdx results in the Airyequation
d2ψp
dz2= zψp
~2
2m
d2ψp
dx2= V ′(0)xψp
d2ψp
dx2= α3xψp
α ≡[
2m
~2V ′(0)
]1/3The solutions are two linearly independent Airy functions, Ai(z) and Bi(z)
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 9 / 15
The Airy functions
Ai(z) =1
π
∫ ∞0
cos
(s3
3+ sz
)ds
Bi(z) =1
π
∫ ∞0
[e−
s3
3+sz + sin
(s3
3+ sz
)]ds
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 10 / 15
Asymptotic forms of Airy functions
z � 0
Ai(z) ∼ 1
2√πz1/4
e−23z3/2
Bi(z) ∼ 1√πz1/4
e23z3/2
z � 0
Ai(z) ∼ 1√π(−z)1/4
sin
[2
3(−z)3/2 +
π
4
]Bi(z) ∼ 1
√π(−z)1/4
cos
[2
3(−z)3/2 +
π
4
]
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 11 / 15
Asymptotic forms of Airy functions
z � 0
Ai(z) ∼ 1
2√πz1/4
e−23z3/2
Bi(z) ∼ 1√πz1/4
e23z3/2
z � 0
Ai(z) ∼ 1√π(−z)1/4
sin
[2
3(−z)3/2 +
π
4
]Bi(z) ∼ 1
√π(−z)1/4
cos
[2
3(−z)3/2 +
π
4
]
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 11 / 15
Asymptotic forms of Airy functions
z � 0
Ai(z) ∼ 1
2√πz1/4
e−23z3/2
Bi(z) ∼ 1√πz1/4
e23z3/2
z � 0
Ai(z) ∼ 1√π(−z)1/4
sin
[2
3(−z)3/2 +
π
4
]
Bi(z) ∼ 1√π(−z)1/4
cos
[2
3(−z)3/2 +
π
4
]
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 11 / 15
Asymptotic forms of Airy functions
z � 0
Ai(z) ∼ 1
2√πz1/4
e−23z3/2
Bi(z) ∼ 1√πz1/4
e23z3/2
z � 0
Ai(z) ∼ 1√π(−z)1/4
sin
[2
3(−z)3/2 +
π
4
]Bi(z) ∼ 1
√π(−z)1/4
cos
[2
3(−z)3/2 +
π
4
]C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 11 / 15
Using the patching function
The patching wave func-tion is thus a linear com-bination of the Airy func-tions
in order to be able tosolve the entire poten-tial, this function must bematched to the WKB so-lutions in the overlap re-gions on either side
ψp(x) = aAi(αx) + bBi(αx)
in these regions, the potential is assumed to be linear but in a regimewhere the WKB approximation still holds
p(x) ∼=√
2m(E − E − V ′(0)x) =2
3~(αx)3/2
let’s look specifically at overlap region 2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 12 / 15
Using the patching function
The patching wave func-tion is thus a linear com-bination of the Airy func-tions
in order to be able tosolve the entire poten-tial, this function must bematched to the WKB so-lutions in the overlap re-gions on either side
ψp(x) = aAi(αx) + bBi(αx)
in these regions, the potential is assumed to be linear but in a regimewhere the WKB approximation still holds
p(x) ∼=√
2m(E − E − V ′(0)x) =2
3~(αx)3/2
let’s look specifically at overlap region 2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 12 / 15
Using the patching function
The patching wave func-tion is thus a linear com-bination of the Airy func-tions
in order to be able tosolve the entire poten-tial, this function must bematched to the WKB so-lutions in the overlap re-gions on either side
ψp(x) = aAi(αx) + bBi(αx)
in these regions, the potential is assumed to be linear but in a regimewhere the WKB approximation still holds
p(x) ∼=√
2m(E − E − V ′(0)x) =2
3~(αx)3/2
let’s look specifically at overlap region 2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 12 / 15
Using the patching function
The patching wave func-tion is thus a linear com-bination of the Airy func-tions
in order to be able tosolve the entire poten-tial, this function must bematched to the WKB so-lutions in the overlap re-gions on either side
ψp(x) = aAi(αx) + bBi(αx)
in these regions, the potential is assumed to be linear but in a regimewhere the WKB approximation still holds
p(x) ∼=√
2m(E − E − V ′(0)x) =2
3~(αx)3/2
let’s look specifically at overlap region 2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 12 / 15
Using the patching function
The patching wave func-tion is thus a linear com-bination of the Airy func-tions
in order to be able tosolve the entire poten-tial, this function must bematched to the WKB so-lutions in the overlap re-gions on either side
ψp(x) = aAi(αx) + bBi(αx)
in these regions, the potential is assumed to be linear but in a regimewhere the WKB approximation still holds
p(x) ∼=√
2m(E − E − V ′(0)x)
=2
3~(αx)3/2
let’s look specifically at overlap region 2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 12 / 15
Using the patching function
The patching wave func-tion is thus a linear com-bination of the Airy func-tions
in order to be able tosolve the entire poten-tial, this function must bematched to the WKB so-lutions in the overlap re-gions on either side
ψp(x) = aAi(αx) + bBi(αx)
in these regions, the potential is assumed to be linear but in a regimewhere the WKB approximation still holds
p(x) ∼=√
2m(E − E − V ′(0)x) =2
3~(αx)3/2
let’s look specifically at overlap region 2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 12 / 15
Using the patching function
The patching wave func-tion is thus a linear com-bination of the Airy func-tions
in order to be able tosolve the entire poten-tial, this function must bematched to the WKB so-lutions in the overlap re-gions on either side
ψp(x) = aAi(αx) + bBi(αx)
in these regions, the potential is assumed to be linear but in a regimewhere the WKB approximation still holds
p(x) ∼=√
2m(E − E − V ′(0)x) =2
3~(αx)3/2
let’s look specifically at overlap region 2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 12 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′
∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′
=2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function is
ψ(x) ∼=D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D
, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 2 solution
∫ x
0|p(x ′)| dx ′ ∼= ~α3/2
∫ x
0
√x ′ dx ′ =
2
3~(αx)3/2
Thus the WKB wave function be-comes
and the patching function isψ(x) ∼=
D
2√π(αx)1/4
e−23(αx)3/2
ψp(x) ∼=a
2√π(αx)1/4
e−23(αx)3/2 +
b
2√π(αx)1/4
e23(αx)3/2
comparing the two, wehave
a =
√4π
α~D, b = 0
now for overlap region 1
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 13 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]
using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]
=a
√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]
a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α
,−a
2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Overlap region 1 solution
Now x ≤ 0, so the WKB exponentis
and the WKB wave function is
∫ 0
xp(x ′) dx ′ ∼=
2
3~(−αx)3/2
ψ(x) ∼=1√
~α3/4(−x)1/4
[Be i
23(−αx)3/2 + Ce−i
23(−αx)3/2
]using the asymptotic form of the Airy functions at large negative values ofx gives, the patching function
ψp(x) ∼=a
√π(−αx)1/4
sin
[2
3(−αx)3/2 +
π
4
]=
a√π(−αx)1/4
1
2i
[e iπ/4e i
23(−αx)3/2 − e−iπ/4e−i
23(−αx)3/2
]a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 14 / 15
Fully patched WKB solution
a =
√4π
α~D,
a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
B = −ie iπ/4D, C = ie−iπ/4
ψ(x) ∼=
2D√p(x)
sin[1~∫ x2x p(x ′) dx ′ + π
4
], x < x2
2D√p(x)
exp[−1
~∫ xx2|p(x ′)| dx ′
], x > x2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 15 / 15
Fully patched WKB solution
a =
√4π
α~D,
a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
B = −ie iπ/4D, C = ie−iπ/4
ψ(x) ∼=
2D√p(x)
sin[1~∫ x2x p(x ′) dx ′ + π
4
], x < x2
2D√p(x)
exp[−1
~∫ xx2|p(x ′)| dx ′
], x > x2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 15 / 15
Fully patched WKB solution
a =
√4π
α~D,
a
2i√πe iπ/4 =
B√~α,
−a2i√πe−iπ/4 =
C√~α
B = −ie iπ/4D, C = ie−iπ/4
ψ(x) ∼=
2D√p(x)
sin[1~∫ x2x p(x ′) dx ′ + π
4
], x < x2
2D√p(x)
exp[−1
~∫ xx2|p(x ′)| dx ′
], x > x2
C. Segre (IIT) PHYS 406 - Spring 2015 February 26, 2015 15 / 15
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