Today’s Outline - February 26, 2015segre/phys406/15S/lecture_11.pdfToday’s Outline - February 26, 2015 Example 8.2 Connection formula Homework Assignment #05: Chapter 8: 1,3,5,6,9,13

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





• Example 8.2





• Example 8.2




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


Example 8.2




1

4πε0

2Ze2

r2= E

E

V(r)

r

-V0

r1 r2

Coulomb repulsion

Nuclear binding


Example 8.2



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


Example 8.2




1

4πε0

2Ze2

r2= E

E

V(r)

r

-V0

r1 r2

Coulomb repulsion

Nuclear binding


Example 8.2




1

4πε0

2Ze2

r2= E

E

V(r)

r

-V0

r1 r2

Coulomb repulsion

Nuclear binding


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


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√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


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


Example 8.2


in this case

but r2E = 2Ze2

4πε0

substituting

r ≡ r2 sin2 u


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


=

√2mE

~

∫ r2

r1

2r2√


√2mE

~

∫ r2

r1

2r2 cos2 u du

=

√2mE

~r2[u + sin u cos u

∣∣∣r1r2


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


Example 8.2

Recasting the limits, when r −→ r1, u −→ sin−1√

r1r2


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)



Example 8.2


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)



Example 8.2


r1r2


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)



Example 8.2


r1r2


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)



Example 8.2


r1r2


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)



Example 8.2


r1r2


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)



Example 8.2


r1r2


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)



Example 8.2


r1r2


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


Example 8.2


r1r2


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)



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


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


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


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


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


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


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


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


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

)


Example 8.2





√E is


τ =2r1v

e2γ

log10 τ = C

(K1

Z√E− K2

√Zr1

)


Example 8.2





√E is


τ =2r1v

e2γ

log10 τ = C

(K1

Z√E− K2

√Zr1

)


Example 8.2





√E is


τ =2r1v

e2γ

log10 τ = C

(K1

Z√E− K2

√Zr1

)


Example 8.2





√E is


τ =2r1v

e2γ

log10 τ = C

(K1

Z√E− K2

√Zr1

)


Example 8.2





√E is


τ =2r1v

e2γ

log10 τ = C

(K1

Z√E− K2

√Zr1

)


Example 8.2





√E is


τ =2r1v

e2γ

log10 τ = C

(K1

Z√E− K2

√Zr1

)


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


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


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



Non-vertical walls



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



Non-vertical walls



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



Non-vertical walls



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



Non-vertical walls



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




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




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


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)


Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp



d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)



Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp

canceling the E terms

and gather-ing the constants


d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)



Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp

canceling the E terms

and gather-ing the constants


d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)



Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp



d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)



Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp



d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)



Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp



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)


Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp



d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)

]1/3



Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp



d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)

]1/3



Patching function


− ~2

2m

d2ψp

dx2+ [E + V ′(0)x ]ψp = Eψp



d2ψp

dz2= zψp

~2

2m

d2ψp

dx2= V ′(0)xψp

d2ψp

dx2= α3xψp

α ≡[

2m

~2V ′(0)



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


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

]



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

]



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

]



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







p(x) ∼=√

2m(E − E − V ′(0)x) =2

3~(αx)3/2








p(x) ∼=√

2m(E − E − V ′(0)x) =2

3~(αx)3/2








p(x) ∼=√

2m(E − E − V ′(0)x) =2

3~(αx)3/2








p(x) ∼=√

2m(E − E − V ′(0)x)

=2

3~(αx)3/2








p(x) ∼=√

2m(E − E − V ′(0)x) =2

3~(αx)3/2








p(x) ∼=√

2m(E − E − V ′(0)x) =2

3~(αx)3/2



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



∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′

=2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2


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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D

, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




∫ x

0|p(x ′)| dx ′ ∼= ~α3/2

∫ x

0

√x ′ dx ′ =

2

3~(αx)3/2



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


a =

√4π

α~D, b = 0




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





∫ 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


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,


C√~α





∫ 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


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,


C√~α





∫ 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,


C√~α





∫ 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


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,


C√~α





∫ 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


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,


C√~α





∫ 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


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,


C√~α





∫ 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


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





∫ 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


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,


C√~α


Fully patched WKB solution

a =

√4π

α~D,

a

2i√πe iπ/4 =

B√~α,


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



a =

√4π

α~D,

a

2i√πe iπ/4 =

B√~α,


C√~α


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



a =

√4π

α~D,

a

2i√πe iπ/4 =

B√~α,


C√~α


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


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Today’s Outline - February 26, 2015segre/phys406/15S/lecture_11.pdfToday’s Outline - February 26, 2015 Example 8.2 Connection formula Homework Assignment #05: Chapter 8: 1,3,5,6,9,13