The WKB approximation - UNIOS · The WKB approximation. ContentsGeneral remarksThe \classical" regionTunnelingThe connection formulasLiterature Contents 1 General remarks 2 The \classical"

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Contents General remarks The “classical” region Tunneling The connection formulas Literature

The WKB approximationQuantum mechanics 2 - Lecture 4

Igor Lukacevic

UJJS, Dept. of Physics, Osijek

12. studenog 2013.

Igor Lukacevic UJJS, Dept. of Physics, Osijek

The WKB approximation


1 General remarks

2 The “classical” region

3 Tunneling

4 The connection formulas

5 Literature




Contents

1 General remarks


3 Tunneling


5 Literature




WKB = Wentzel, Kramers, Brillouin

in Holland it’s KWB

in France it’s BKW

in England it’s JWKB (for Jeffreys)

























Basic idea:

1 particle Epotential V (x) constant

if E > V ⇒ ψ(x) = Ae±ikx , k =

√2m(E − V )

~




Basic idea:



√2m(E − V )

~

A question

What’s the character of A and λ = 2π/k here?




Basic idea:



√2m(E − V )

~2 suppose V (x) not constant, but varies slowly wrt λ




Basic idea:



√2m(E − V )


A question

What can we say about ψ, A and λ now?




Basic idea:



√2m(E − V )


A question

What can we say about ψ, A and λ now?

We still have oscillating ψ, but with slowly changable A and λ.




Basic idea:



√2m(E − V )


3 if E < V , the reasoning is analogous




Basic idea:



√2m(E − V )



A question

What if E ≈ V ?




Basic idea:



√2m(E − V )



A question

What if E ≈ V ? Turning points




Contents

1 General remarks


3 Tunneling


5 Literature




S.E.

− ~2

2m∆ψ + V (x)ψ = Eψ ⇐⇒ ∆ψ = −p2

~2ψ , p(x) =

√2m [E − V (x)]




S.E.

− ~2

2m∆ψ + V (x)ψ = Eψ ⇐⇒ ∆ψ = −p2

~2ψ , p(x) =

√2m [E − V (x)]

“Classical” region

99K E > V (x) , p real




S.E.

− ~2

2m∆ψ + V (x)ψ = Eψ ⇐⇒ ∆ψ = −p2

~2ψ , p(x) =

√2m [E − V (x)]

“Classical” region

99K E > V (x) , p real

99K ψ(x) = A(x)e iφ(x)

A(x) and φ(x) real




Putting ψ(x) into S.E. gives two equations:

A′′ = A

[(φ′)2 − p2

~2

](1)(

A2φ′)′

= 0 (2)





A′′ = A

[(φ′)2 − p2

~2

](1)(

A2φ′)′

= 0 (2)

Solve (2)

A =C√φ′, C ∈ R





A′′ = A

[(φ′)2 − p2

~2

](1)(

A2φ′)′

= 0 (2)

Solve (2)

A =C√φ′, C ∈ R

Solve (1)

Assumption: A varies slowly

⇒ A′′ ≈ 0

φ(x) = ±1

~

∫p(x)dx




Solve (2)

A =C√φ′, C ∈ R

Solve (1)


⇒ A′′ ≈ 0

φ(x) = ±1

~

∫p(x)dx

Resulting wavefunction

ψ(x) ≈ C√p(x)

e±i~

∫p(x)dx

Note: general solution is a linear combination of these.




Solve (1)

A =C√φ′, C ∈ R

Solve (2)


⇒ A′′ ≈ 0

φ(x) = ±1

~

∫p(x)dx

Resulting wavefunction

ψ(x) ≈ C√p(x)

e±i~

∫p(x)dx

Note: general solution is a linear combination of these.

Probability of finding a particle at x

|ψ(x)|2 ≈ |C |2

p(x)




Example: potential well with two vertical walls

V (x) =

{some function , 0 < x < a∞ , otherwise




Example: potential well with two vertical walls

V (x) =

{some function , 0 < x < a∞ , otherwise

Again, assume E > V (x) =⇒

ψ(x) ≈ 1√p(x)

[C+e

iφ(x) + C−e−iφ(x)

]=

1√p(x)

[C1 sinφ(x) + C2 cosφ(x)]

where

φ(x) =1

~

∫ x

0

p(x ′)dx ′




Example (cont.)

Boundary conditions: ψ(0) = 0, ψ(a) = 0⇒ φ(a) = nπ , n = 1, 2, 3, . . .⇒∫ a

0

p(x)dx = nπ~




Example (cont.)


0

p(x)dx = nπ~

Take, for example, V (x) = 0⇒

En =n2π2~2

2ma2

We got an exact result...is this strange?




Example (cont.)


0

p(x)dx = nπ~

Take, for example, V (x) = 0⇒

En =n2π2~2

2ma2

We got an exact result...is this strange? No, since A =√

2/a = const.




Contents

1 General remarks


3 Tunneling


5 Literature




Now, assume E < V :

ψ(x) ≈ C√|p(x)|

e±1~

∫|p(x)|dx

where p(x) is imaginary.




Now, assume E < V :

ψ(x) ≈ C√|p(x)|

e±1~

∫|p(x)|dx

where p(x) is imaginary.

Consider the potential:

V (x) =

{some function , 0 < x < a0 , otherwise




x < 0

ψ(x) = Ae ikx + Be−ikx




x < 0 x > a

ψ(x) = Ae ikx + Be−ikx ψ(x) = Fe ikx

Transmission probability: T =|F |2

|A|2




x < 0 0 ≤ x ≤ a x > a

ψ(x) = Ae ikx + Be−ikx ψ(x) ≈ C√|p(x)|

e1~

∫ x0 |p(x

′)|dx′ ψ(x) = Fe ikx

+ D√|p(x)|

e−1~

∫ x0 |p(x

′)|dx′

Transmission probability: T =|F |2

|A|2




x < 0 0 ≤ x ≤ a x > a


e1~

∫ x0 |p(x


+ D√|p(x)|

e−1~

∫ x0 |p(x

′)|dx′

Transmission probability:

T =|F |2

|A|2High, broad barrier 1st termgoes to 0Why?




x < 0 0 ≤ x ≤ a x > a


e1~

∫ x0 |p(x


+ D√|p(x)|

e−1~

∫ x0 |p(x

′)|dx′

Transmission probability:

T =|F |2

|A|2 ∼ e−2~

∫ a0 |p(x

′)|dx′

High, broad barrier 1st termgoes to 0Why?

T ≈ e−2γ , γ =1

~

∫ a

0

|p(x)|dx




Example: Gamow’s theory of alpha decay

first time that quantummechanics had beenapplied to nuclearphysics




Example: Gamow’s theory of alpha decay (cont.)

first time that quantummechanics had beenapplied to nuclearphysics

turning points:1 r1 7−→ nucleus radius

(6.63 fm for U238)

2 r2 7−→1

4πε0

2Ze2

r2= E





γ =1

~

∫ r2

r1

√2m

(1

4πε0

2Ze2

r2− E

)dr =

√2mE

~

∫ r2

r1

√r2r− 1dr





γ =1

~

∫ r2

r1

√2m

(1

4πε0

2Ze2

r2− E

)dr =

√2mE

~

∫ r2

r1

√r2r− 1dr

Substituting r = r2 sin2 u gives

γ =

√2mE

~

[r2

(π

2− sin−1

√r1r2

)−√

r1(r2 − r1)

]





γ =1

~

∫ r2

r1

√2m

(1

4πε0

2Ze2

r2− E

)dr =

√2mE

~

∫ r2

r1

√r2r− 1dr


γ =

√2mE

~

{r2

[π

2− sin−1

√r1r2︸︷︷︸

r1�r2−−−→√r1/r2

]

︸︷︷︸π2r2−2√r1r2

−√

r1(r2 − r1)︸︷︷︸√

r1r2−r21

r1�r2−−−→√r1r2

}






γ =

√2mE

~

{r2

[π

2− sin−1

√r1r2︸︷︷︸

r1�r2−−−→√r1/r2

]

︸︷︷︸π2r2−2√r1r2

−√

r1(r2 − r1)︸︷︷︸√

r1r2−r21

r1�r2−−−→√r1r2

}

γ ≈√

2mE

~

[π2r2 − 2

√r1r2]

= K1Z√E− K2

√Zr1

where

K1 =

(e2

4πε0

)π√

2m

~= 1.980MeV1/2

K2 =

(e2

4πε0

)1/24√m

~= 1.485 fm−1/2





v average velocity

2r1/v average timebetween “collisions”with the nucleuspotential “wall”

v/2r1 averagefrequancy of “collisions”

e−2γ “escape”probability

(v/2r1)e−2γ “escape” probability perunit time

Lifetime:

τ =2r1v

e2γ





v average velocity

2r1/v average timebetween “collisions”with the nucleuspotential “wall”

v/2r1 averagefrequancy of “collisions”

e−2γ “escape”probability

(v/2r1)e−2γ “escape” probability perunit time

Lifetime:

τ =2r1v

e2γ ⇒ ln τ ∼ 1√E




HW

Solve Problem 8.3 from Ref. [2].




Contents

1 General remarks


3 Tunneling


5 Literature







Let us repeat:

ψ(x) ≈

1√p(x)

[Be

i~

∫ 0x p(x′)dx′ + Ce−

i~

∫ 0x p(x′)dx′

], if x < 0

1√|p(x)|

De−1~

∫ x0 |p(x

′)|dx′ , if x > 0




Let us repeat:

ψ(x) ≈

1√p(x)

[Be

i~

∫ 0x p(x′)dx′ + Ce−

i~

∫ 0x p(x′)dx′

], if x < 0

1√|p(x)|

De−1~

∫ x0 |p(x

′)|dx′ , if x > 0

Our mission: join these two solutions at the boundary.




Let us repeat:

ψ(x) ≈

1√p(x)

[Be

i~

∫ 0x p(x′)dx′ + Ce−

i~

∫ 0x p(x′)dx′

], if x < 0

1√|p(x)|

De−1~

∫ x0 |p(x

′)|dx′ , if x > 0


A problem

What happens with the w.f. whenE ≈ V ?




Let us repeat:

ψ(x) ≈

1√p(x)

[Be

i~

∫ 0x p(x′)dx′ + Ce−

i~

∫ 0x p(x′)dx′

], if x < 0

1√|p(x)|

De−1~

∫ x0 |p(x

′)|dx′ , if x > 0


A problem


E ≈ V ⇒ p(x)→ 0⇒ ψ →∞ !




Let us repeat:

ψ(x) ≈

1√p(x)

[Be

i~

∫ 0x p(x′)dx′ + Ce−

i~

∫ 0x p(x′)dx′

], if x < 0

1√|p(x)|

De−1~

∫ x0 |p(x

′)|dx′ , if x > 0


A problem


E ≈ V ⇒ p(x)→ 0⇒ ψ →∞ !

A solution

Construct a “patching”wavefunction ψp.




Approximation: we linearize the potential

V (x) ≈ E + V ′(0)x





V (x) ≈ E + V ′(0)x

From S.E. we get 99K

d2ψp

dz2= zψp , z = αx , α =

[2m

~2V ′(0)

] 13





V (x) ≈ E + V ′(0)x


d2ψp

dz2= zψp︸︷︷︸

Airy’s equation

, z = αx , α =

[2m

~2V ′(0)

] 13





V (x) ≈ E + V ′(0)x


d2ψp

dz2= zψp︸︷︷︸

Airy’s equation

, z = αx , α =

[2m

~2V ′(0)

] 13

ψp = a Ai(αx)︸︷︷︸Airy function

+b Bi(αx)︸︷︷︸Airy function







a delicate double constraint has to be satisfied




a delicate double constraint has to be satisfied

we need WKB w.f. and ψp for both overlap regions (OLR)




p(x) =√

2m(E − V ) ≈ ~α32√−x




p(x) ≈ ~α32√−x

OLR 2 (x > 0)∫ x

0

|p(x ′)|dx ′ ≈ 2

3~(αx)

32

ψWKB ≈D√

~α3/4x1/4e−

23(αx)3/2

ψz�0p ≈ a

2√π(αx)1/4

e−23(αx)3/2

+b√

π(αx)1/4e

23(αx)3/2

⇒ a = D

√4π

α~, b = 0




OLR 2 (x > 0)∫ x

0

|p(x ′)|dx ′ ≈ 2

3~(αx)

32

ψWKB ≈D√

~α3/4x1/4e−

23(αx)3/2

ψz�0p ≈ a

2√π(αx)1/4

e−23(αx)3/2

+b√

π(αx)1/4e

23(αx)3/2

⇒ a = D

√4π

α~, b = 0

OLR 1 (x < 0)∫ 0

x

p(x ′)dx ′ ≈ 2

3~(−αx)

32

ψWKB ≈1√

~α3/4(−x)1/4

[Be i

23(−αx)3/2

+Ce−i 23(−αx)3/2

]ψz�0

p ≈ 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




The connection formulas

B = −ie iπ/4 · D , C = ie−iπ/4 · D




The connection formulas

B = −ie iπ/4 · D , C = ie−iπ/4 · D

WKB w.f.

ψ(x) ≈

2D√p(x)

sin

[1

~

∫ x2

x

p(x ′)dx ′ +π

4

], if x < x2

D√|p(x)|

exp

[−1

~

∫ x

x2

|p(x ′)|dx ′], if x > x2




Example: Potential well with one vertical wall




Example: Potential well with one vertical wall

Boundary condition: ψ(0) = 0, gives for ψWKB∫ x2

0

p(x)dx =

(n − 1

4

)π~ , n = 1, 2, 3, . . .




Example: Potential well with one vertical wall (cont.)

For instance, consider the “half-harmonic oscillator”:

V (x) =

1

2mω2x2 , x > 0

0 otherwise






V (x) =

1

2mω2x2 , x > 0

0 otherwise

Here we have

p(x) = mω√

x22 − x2






V (x) =

1

2mω2x2 , x > 0

0 otherwise

Here we have

p(x) = mω√

x22 − x2

So ∫ x2

0

p(x)dx =πE

2ω






V (x) =

1

2mω2x2 , x > 0

0 otherwise

Here we have

p(x) = mω√

x22 − x2

So ∫ x2

0

p(x)dx =πE

2ω

Comparisson now gives:

En =

(2n − 1

2

)~ω =

(3

2,

7

2,

11

2, . . .

)~ω






V (x) =

1

2mω2x2 , x > 0

0 otherwise

Here we have

p(x) = mω√

x22 − x2

So ∫ x2

0

p(x)dx =πE

2ω

Comparisson now gives:

En =

(2n − 1

2

)~ω =

(3

2,

7

2,

11

2, . . .

)~ω

Compare this result with an exact one.




Example: Potential well with no vertical walls





we have seen the connection formulas for upward potential slopes

for downward slopes (analogous):

ψ(x) ≈

D ′√|p(x)|

exp

[−1

~

∫ x1

x

|p(x ′)|dx ′], if x < x1

2D ′√p(x)

sin

[1

~

∫ x

x1

p(x ′)dx ′ +π

4

], if x > x1





we want the w.f. in the “well”, i.e. where x1 < x < x2:

ψ(x) ≈ 2D√p(x)

sin θ2(x) , θ2(x) =1

~

∫ x2

x

p(x ′)dx ′ +π

4

ψ(x) ≈ − 2D ′√p(x)

sin θ1(x) , θ1(x) = −1

~

∫ x

x1

p(x ′)dx ′ − π

4





sin θ1 = sin θ2 =⇒ θ2 = θ1 + nπ =⇒∫ x2

x1

p(x)dx =

(n − 1

2

)π~ , n = 1, 2, 3, . . .





sin θ1 = sin θ2 =⇒ θ2 = θ1 + nπ =⇒∫ x2

x1

p(x)dx =

(n − 1

2

)π~ , n = 1, 2, 3, . . .

0, two vertical walls 1/4, one vertical wall




Conclusions

WKB advantages

good for slowly changing w.f.

good for short wavelengths

best in the semi-classicalsystems (large n)

one doesn’t even have to solvethe S.E.

WKB disadvantages

bad for rapidly changing w.f.

bad for long wavelengths

inappropriate for lower states(small n)

constraint trade-off (sometimesnot possible)




Contents

1 General remarks


3 Tunneling


5 Literature




Literature

1 R. L. Liboff, Introductory Quantum Mechanics, Addison Wesley, SanFrancisco, 2003.

2 D. J. Griffiths, Introduction to Quantum Mechanics, 2nd ed., PearsonEducation, Inc., Upper Saddle River, NJ, 2005.

3 I. Supek, Teorijska fizika i struktura materije, II. dio, Skolska knjiga,Zagreb, 1989.

4 Y. Peleg, R. Pnini, E. Zaarur, Shaum’s Outline of Theory and Problems ofQuantum Mechanics, McGraw-Hill, 1998.



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The WKB approximation - UNIOS · The WKB approximation. ContentsGeneral remarksThe \classical" regionTunnelingThe connection formulasLiterature Contents 1 General remarks 2 The \classical"