5.73 Lecture #6
6-1
Last time: Normalization of eigenfunctions which belong to continuously
variable eigenvalues.
1.
2.
identities
ψ δk , ψ δp , ψ δE , ψ box
3.
trick using box normalization
 # states   # particles 
 δθ  

δx
∝L
4.
1.
∝ 1/L
for box normalization
dn
often needed - alternate method via JWKB next lecture
dE
V(x) = αx linear potential
solve in momentum representation, φ(p), and take F.T. to ψ(x) → Airy functions
2.
Semi-classical (JWKB) approx. for ψ(x)
*
p( x) = [(E − V( x))2 m ]
*
ψ ( x) = p( x)
1/2
−1/2
envelope
 i
exp ±
 h
∫
x
c

p( x′ )dx′ 

variable
* visualize y(x) as plane wave with x-dependent wave vector
* useful for evaluating stationary phase integrals (localization, causality)
**** splicing across classical (E > V)|| forbidden (E < V) ] Next lecture
WKB Quantization Condition
h
x (E)
∫x _+(E) p(x ′)dx ′ = 2 (n + 1 2) n = 0,1,…
updated 9/12/02 11:39 AM
5.73 Lecture #6
6-2
Linear Potential. V(x) = αx
p√2
√
H=
+ αx√
2m
coordinate representation
momentum representation
x√ → x
p√ → p
∂
h ∂
p√ →
x√ → ih
i ∂x
∂p
√ p√] = ih in both
 note [ x,



 representations - prove this?
2
d2
p2
d
h
√
√
+ αx
+ ihα
H=
H=−
2
2 m dx
2m
dp
2nd order
1st order - easier!
Solve in momentum representation (a sometimes useful trick)
Schr. Eq.
solution
(
)
dφ(p)
i
=−
E − p 2 2m φ(p)
dp
hα
φ(p) = Ne ap+bp
3
gives p 2 times φ ( p )
gives constant times φ ( p )
iE
hα
i
b=
6hα m
plug into Schr. Eq. and identify, term-by-term, to get a = –
(
)
 i

φ(p) = N exp −
Ep − p 3 / 6m 
 hα

φ * (p) φ (p) = 1!
easy?
∴ N = 1!
updated 9/12/02 11:39 AM
5.73 Lecture #6
6-3
Now p is an observable, so it must be real. Thus φ(p) is defined for all (real) p
and is oscillatory in p for all p. NEVER exponentially increasing or decreasing!
IT IS STRANGE THAT φ(p) does not distinguish between classically allowed
and forbidden regions. IS THIS REALLY STRANGE? If we allow p to be
imaginary in order to deal with classically forbidden regions, φ(p) becomes an
increasing or decreasing exponential.
If we insist on working in the ψ (x) picture, we must perform a
Fourier Transform
ψ (x) = N′
ψ (x) = N′
e
iθ
∫
∞
∫
∞
−∞
−∞
e ipx / h φ (p) dp

i
exp
 hα





3
(
)
p
α
x
−
E
p
/
m
+
6
 144424443   dp
 odd function of p: O(p)  
∫
= cos
θ + isin
θ
{
{
even
ψ(x) =
sin O (p) dp = 0
since O(p) is odd wrt p → –p.
 (αx − E)p + p 3 / 6m 
cos 
dp
hα


Ai ( z) = π −1/2
Surprise!
−∞
odd
∞
N ′ ∫−∞
∞
∫
∞
0
(
)
cos s3 3 + sz ds
This is a named (Airy) and tabulated integral
* numerical tables for x near turning point
i.e., x ≈ E/ α
∗ analytic “a s y m p t o t i c” functions for x far from turning point
useful for deriving f(q.n.) and for matching across boundaries.
* zeroes of Airy functions [Ai(zi)=0] and of derivatives of Airy
functions [Ai′(z′i)=0] are tabulated. (Useful for matching across
center of potentials with definite even or odd symmetry.) [Two
kinds of Airy functions, Ai and Bi.]
updated 9/12/02 11:39 AM
5.73 Lecture #6
6-4
 s3

∞
Ai(z) = π −1/2 ∫0 cos + sz ds
3

s ≡ p(2mha)−1/3
(if α > 0)
for our specific problem (αx − E)
1/3
2mα h 2
z≡
α
Turning point
[
]
V(x) = α x
E
α >0
x+(E)
x
At turning point E = V(x + ) = αx + ∴ x + ( E ) = E / α
Problems with linear potentials:
V(x)
αx
+αx
–αx
boundary conditions
When there is symmetry (or 1/2 symmetry) we need to know where the zeroes of ψ(x)
{
and
d
ψ /4
dx
12
4
3
for odd functions
or ∞ barrier
are located.
for even functions
tables of zeroes of
Ai(z) and Ai ′(z)
"z n "
See Handout.
" z ′n "
When there is no symmetry, must match Ai (or, more precisely, a linear
combination of Ai and Bi) and Ai′ across boundaries, but we do not have to actually
look at the Airy function itself near the joining point.
updated 9/12/02 11:39 AM
5.73 Lecture #6
E
6-5
This is not as bad as it seems because we are
usually far from turning point at internal
joining point and can use analytic asymptotic
x
expressions for Ai(z).
2 linear potentials of
different |slope|.
For α > 0 there are 2 cases (classical and nonclassical)
αx
(i)
z << 0 , E > V(x) classically allowed region




2
−1/ 4
3/2
)
sin
(
−z
)
+
π
/
4
Ai(z) → π-1/2 ( −z

{
{
{
positive
 3 x is in here
phase 

shift 

asymptotic form for z << 0.
* oscillatory, but wave vector, k, varies with x
* Ai vanishes as x → ± ∞ because of (-z) -1/4 factor
* Bi is needed for case where Airy function must vanish as x → +∞ in classical region
(ii) z >> 0 , E < V( x) forbidden region

-1/2
1/4 −(2/3 ) z 3/2
/2 1
z −2
Ai(z) → π
3 e1424
3
positive decreasing

exponential
well

behaved * not oscillatory, monotonic
* Ai vanishes as x → + ∞

* Bi vanishes as x → ± ∞ in forbidden region
(
)
asymptotic form for z >> 0.
Cartoon
need numerical tables
to define ψ(x) in this region
|
|
The two asymptotic forms of Ai(z) are not normalized,
but their amplitudes (& phases) are matched. Links B.C.
at x → +∞ to B. C. at x → –∞.
updated 9/12/02 11:39 AM
5.73 Lecture #6
6-6
NonLecture
OTHER CASE: α < 0 → z ≡ -
(|α|x + E) 2m|α| 1/3

h 2 
|α|
for this case, need Bi(z) instead of Ai(z)
 2 3/2 
Bi ( z) → π −1/2 2 |z|−1/4 exp − z 
 3

 2 3/2 π 
Bi ( z) → π −1/2 |z|−1/4 cos  z
+ 
4
3
(
)
E
α<0
x
( forbidden region, z << 0.)
(allowed region, z >> 0.)
What is so great about V(x) =αx? ψ(x) is ugly — need lookup tables, complicated solutions!
Ai(z) turns out to be key to generalization of quantization of all (well behaved) V(x)!
There are semi-classical JWKB ψ(x)’s — These blow up near turning points (i.e. on both
sides). The Ai(z)’s permit matching of JWKB ψ(x)’s across the large gap where ψJWKB is
invalid, ill-defined.
(JEFFREYS)
WENTZEL
KRAMERS
BRILLOUIN
JWKB provides a way to get ψn(x) and En without solving
differential equations or performing a FT.
But actually, the differential equations are easy to solve numerically. The reason
we care about JWKB is that it provides a basis for:
* physical interpretation (semi-classical)
* RKR inversion from EvJ ♦ VJ(R).
* semi-classical quantization.
* the link to classical mechanics is essential to wavepacket pictures.
updated 9/12/02 11:39 AM
5.73 Lecture #6
6-7
(generalize on eikx for free particle by letting k =p(x)/h depend explicitly on x (why does this
not violate [x,p]=ih ?)
 i x
 k(x) and p(x) are classical mechanical
−1/2
ψ JWKB = p(x)
p( x ′ )dx ′  functions of x, not QM operators.
exp  ±
1424
3
 h c

∫
classical envelope
p(x) = [2m(E – V(x)]1/2
phase factor: adjustable to
satisfy boundary conditions
p(x) is pure real (classically allowed) or pure imaginary (classically forbidden). p(x) is not
Q.M. momentum. It is a classically motivated function of x which has the form of the
classical mechanical momentum and has the property that the λ = h varies with x in a
p
reasonable way.
*
p( x)
−1/2
is probability amplitude envelope because
probability ∝
*
*
*
i
exp− 
h
∫
x
c
1
so amplitude ∝
v
1
v

p( x′ )dx′  is generalization of e ipx/h to non - constant V(x).

h
p(x)
gives easily identifiable stationary phase region for many wiggly integrands.
(Both ψ ’s have same λ at x s.p. )
node spacing λ (x) =
Long Nonlecture derivation/motivation.
updated 9/12/02 11:39 AM
5.73 Lecture #6
6-8
i x
Try ψ(x) = N(x) exp  ± ∫c p( x ′ )dx ′ 
h


plug into Schr. Eq. and get a new differential equation that N(x) must satisfy
d 2 ψ 2m
+
( E − V(x))ψ = 0
dx 2 h 2
d2ψ 1
2
2 + 2 p(x) ψ = 0
dx
h
* derived
in box
below
**

2ip( x)
ip′ ( x) 
 i
0 = N ′′ ±
N′ ±
N  exp ±
h
h


 h
∫
x
c

p( x′ ) dx′ 

This is a new Schr. Eq. for N(x). Now make an approximation, to be tested later, that N″ is
negligible everywhere. This is based on the hope that a slowly varying V(x) will lead to a slowly
varying N(x).
*
i
i x
dψ 
= N ′ ± p(x) exp  ± ∫c p( x ′ )dx ′ 
h
dx 
 h


d2ψ 
i
ip 
i
i
  exp  ± i x p( x ′ )dx ′ 
=
N
±
N
p
±
Np
N
p
±
N
±
′′
′
′
′
 h ∫c

h
h
h
h  
dx 2 

2i
ip′
= N ′′ ± N ′p ±
N−
h
h

p2 
 ± i x p( x ′ )dx ′ 
N
exp

 h ∫c

h 2 
d 2 ψ p2
i x
0 = 2 + 2 N exp  ± ∫c p( x ′ )dx ′ 
 h

h
dx
2ip(x)
i x
ip′ 

0 = N ′′ ±
N  exp  ± ∫c p( x ′ )dx ′ 
N′ ±
h
h 
 h


updated 9/12/02 11:39 AM
5.73 Lecture #6
6-9
so, if we neglect N ′′, we get
2pN ′ + p′N = 0
1/2  1/2
if p ≠ 0, then 2p
(
d Np1/2
p
) = N′p1/2 + 1 p−1/2 p′N

dx
∴

(
d Np1/2
dx
1 −1/2
N′ + p
p′N  = 0
2

)=0
2

N(x)p1/2 (x) = constant
∴ N(x) = cp(x)−1/2
OK, now we have a form for N(x) which we can use to tell us what conditions must be satisfied
for N″(x) to be negligible everywhere.
N = cp −1/2
1
dp −1/2
dp
= − p −3/2
2
dx
dx
dp  dV  −1
= −
p m
dx  dx 
p(x) = [2m( E − V(x))]
1/2
m dV
dp −1/2
= p −5/2
∴
2 dx
dx
2
d 2 p −1/2 m dV  5  −7/2  m dV 
−5/2 m d V
+p
=
− p
−


2


2 dx
2
2 {
dx
dx 2
 p dx 
ignore
updated 9/12/02 11:39 AM
5.73 Lecture #6
6 - 10
dV  2
5
∴ N ′′ = c m 2 p −9/2 
 dx 
4
But we have made several assumptions about N″
* N ′′ <<
2ip
icm −3/2 dV
p
N′ = +
h
h
dx
* N ′′ <<
ip′
icm −3/2 dV
N=−
p
h
h
dx
p2
c
* N ′′ << 2 N = 2 p +3/2
h
h
all of this is satisfied if
5 mh  dV  −3
p << 1
4 i  dx 
Is this the JWKB validity condition?
Spirit of JWKB: if initial JWKB approximation is not sufficiently accurate, iterate:
p(x) → ψ 0 (x)
(ordinary JWKB)
ψ 0 (x) → p1 (x)
p1 (x) → ψ1 (x)
(first order JWKB)
1/2
 h2 d2ψ0 
e .g.
+ 2 ψ 0 = 0 → p1 ( x) =  −
2 
h
dx2
 ψ 0 ( x) dx 
 i x

−1/2
ψ1 ( x) = p1 ( x)
exp ±
p1 ( x′ ) dx′ 
 h c

d2ψ0
p12
∫
see ** Eq.
on p. 6-8
iterative improvement
of accuracy
p1(x) is not smaller than p0(x). It has more of the correct wiggles in it.
END OF NONLECTURE
updated 9/12/02 11:39 AM
5.73 Lecture #6
6 - 11
Resume Lecture
ψ(x) ≈ p(x)
{
−1/ 2
envelope
provided that
 i
exp ±
 h
∫
x
c

p(x ′ ) dx ′ 

d2V
is negligible
dx 2
adjustable phase shift.
AND


dp
dλ
m dV
<< 1  same as λ(x)
< p( x) or
<< 1
3 dx
dx
dx


|p|
144244
3
h
required for N ′′(x )
to be negligible
Next need to work out connection of ψJWKB(x) across region
where JWKB approx. breaks down (at turning points!).
dλ
→ ∞ at turning point because p(x) → 0
dx
BUT ALL IS NOT LOST — near enough to a turning point
all potentials V(x) look like V(x)=αx!
Now our job is to show that asymptotic – AIRY and
JWKB are identical for a small region not too close and
not too far on both sides of each turning point.
THIS PERMITS ACCURATE SPLICING OF
ψ(x) ACROSS TURNING POINT REGION!
updated 9/12/02 11:39 AM
5.73 Lecture #6
6 - 12
I
V(x)
II
V(x) = αx
α>0
E
approx. linear V(x)
a-AIRY
a-AIRY
JWKB
JWKB
x
a
Region I
E > V(x) classical
2
I
ψ a−AIRY
~ π −1/2 (−z)−1/ 4 sin  (−z)3/2 + π
3
4 

1/3
z=
(αx − E)  2mα 
 h 2 
α
αx - E 
= (x − a)
at turning point E = V(a) = αa so 
 α 
 2mα 
z = (x − a) 2 
 h 
Region II
1/3
<< 0
when x << a
Region I/II splice
using a-Airy.
E < V(x) forbidden region, z >> 0
π
II
ψ a−AIRY
~
−1/2
2
z −1/ 4 e −2/3z
3/ 2
Now consider ψJWKB for a linear potential and show that it is identical to a-Airy!
ψ JWKB ~ c ± p(x)
−1/2
 i x

exp  ±
p( x ′ )dx ′ 
 h a

∫
both c+ and c– additive terms could be present
p(x) ≡ [2m( E − V(x))]
1/2
updated 9/12/02 11:39 AM
5.73 Lecture #6
x < a
x > a
6 - 13
p is real ,
ψ JWKM
p is imaginary , ψ JWKB
classical ,
forbidden ,
oscillates
is exponential
pretend V(x) looks linear near x = a
(l-JWKB )
p(x) = [2mα(a − x )]
linear
1/2
1/2
1/2 x
∫a (a − x′) dx′
x
∫a p( x′)dx′ = (2mα )
x
2
−  (a − x ′ )3/2
 3
a
1/2 
= (2mα )
2
= −(2mα )1/2 (a − x )3/2
3
Region I
ψ Il − JWKB (x) ~ p(x)
= p(x)
−1/ 2
−1/ 2
[Ae
iθ
+ Be − iθ
]
C sin(θ + φ)
Define the JWKB phase factor, θ (x):
θ=
1
h
∫
x
a
 2mα 
p(x ′ ) dx ′ = − 2 
 h 
1/ 2
2
( a − x) 3 / 2
3
Now compare θ (x) to z(x)
 2mα 
but, earlier, z = (x − a) 2 
 h 
1/3
2
∴ θ = − (−z)3/2
3
∴ p = (2mα h)1/3 ( −z)1/2
p
−1/2
= (2mα h)−1/6 ( −z)−1/ 4
for exponential factor
for pre-exponential factor
Thus, putting all of the pieces together
updated 9/12/02 11:39 AM
5.73 Lecture #6
6 - 14
−1/ 2
−θ4
−|p|
 647

8
6444
474444
8
2

I
ψ l−JWKB
= −(2mαh)−1/6 (−z)−1/ 4 C sin  (−z)3/2 − φ 
3



I
= ψ a−AIRY
If C = −(2mα h)1/6 π −1/2
φ = −π 4
I
I
ψ l−JWKB
exactly splices onto ψ a−AIRY
with a O/4 phase factor (shifted from what the argument of sine
would have been if one had started the phase integral at x = a
Similar result in Region II
−f (x)
ψ II
+ Be +f (x)
JWKB ~ Ae
at x → +∞
II
∴ ψ l−JWKB
f(x) → ∞
= A(2mα)
−1/ 4
∴B = 0
(x − a)
−1/ 4
  2mα  1/2 2

exp − 2 
(x − a)3/2 
3
  h 

II
if A = (2mαh)+1/6 π −1/2 2
which is equal to ψ a−AIRY
I
II
Final step: ψ IJWKB ↔ ψ a−AIRY
, ψ II
JWKB ↔ ψ a−AIRY
require A = –C/2
perfect match on opposite sides of turning point.
Ai(z) valid in region where ψJWKB is invalid.
updated 9/12/02 11:39 AM