5.73 Lecture #2
2-1
free particle V(x)=V0
Last Time:
general solution
ψ = Aeikx + Be–ikx
A,B are complex constants, determined by “boundary conditions”
k=
p
h
(from e ikx , eigenfunction of p,√ and the real number, p, is the eigenvalue)
1/2

2m 
k = (E − V0 ) 2 

h 
probability
for E ≥ V0
P( x ) = ψ * ψ = |1
A4
|2 2
+4
| B4
|2 + 21Re
( A * B) cos 2 kx2
+ 4444444
2 Im( A * B)sin 23
kx
4
3
4444444
const.
distribution
wiggly
only get wiggly stuff when 2 or
more different values of k are
superimposed. In this special
case we had +k and –k.
TODAY
and
1. infinite box
2. δ(x) well
3. δ(x) barrier
and
5.73 Lecture #2
2-2
What do we know about ψ(x) for physically realistic V(x)?
ψ ( ±∞) = ?
ψ * ( x)ψ ( x) for all x?
∫
∞
−∞
ψ * ( x)ψ ( x)dx ?
Continuity of ψ and dψ /dx ?
Computationally convenient potentials have steps and flat regions.
infinite step
finite step
infinitely high but infinitely thin step,“ δ-function”
ψ continuous
dψ d 2 ψ
,
not continuous for infinite step, and not for δ-function
dx dx 2
dψ
is continuous for finite step
dx
More warm up exercises
1.
Infinite box
V(x)
0
L
ψ ( x) = Ae ikx + Be − ikx = C cos kx + D sin kx
[C=A+B, D=iA – iB]
ψ (0) = 0 ⇒ C = 0
ψ (L) = 0 ⇒ kL = nπ
n = 1, 2 , …
( why not n
= 0?)
x
5.73 Lecture #2
recall
2-3
2m n 2 π 2
= 2
V0 = 0
L
h2
Insert kL = nπ boundary condition.
k 2 = ( E − V0 )
En = n2
2

h2 π2
2 h
2
2 = n 
mL
2 mL
8


here.
En is integer multiple
of common factor, E1.
Important for
∞ # of bound levels wavepackets!
n = 0 would be
empty box
E1
normalization (P=1 for 1 particle in well)
L
⇒ |D|= (2 / L)1/2
1 =| D |2 ∫ dx sin 2 ( nπx)
0
ψ n ( x) = (2 / L )1/ 2 sin( nπx)
because
L
2
∫0 sin ( nπx)dx = L / 2
iα
D = (2 / L)1/2 e{
arbitrary
phase
factor
cartoons of ψn(x): what happens to {ψn} and {En} if
we move well:
left or right in x?
up or down in E?
Infinite well was easy: 2 boundary conditions plus normalization requirement.
Generalize to stepwise constant potentials: in each V(x)=constant region,
need to know 2 complex coefficients and, if the particle is confined within a
finite range of x, there is quantization of energy.
* boundary and joining conditions
* normalization
* overall phase arbitrariness
So next step is to deal with case where boundary conditions are not so
obvious. δ(x) well and barrier.
5.73 Lecture #2
2-4
0
x
V(x)
V(x) = –a δ(x)
a has units Energy x Length
(because, as we will see, δ(x) has
units of reciprocal length)
a>0
= 0 everywhere except V(0) = –a “∞”
“strength” of the δ-function well
Schrödinger
Equation
Integrate:

 2m
d2ψ
(3
E4
+2
aδ4
x) 2 ψ
2 = –  (1
dx
 E − V( x )  h
 +ε

2 mE
2 ma

 dx
dx
x
x
x
lim
=
−
lim
ψ
(
)
+
δ
(
)
ψ
(
)

 h2

ε →0
ε →0 
dx2
h2
−ε
 −ε

dψ
dψ
=
LHS =
±
size of discontinuity
dx x =+ε dx x =−ε
+ε
∫
d2ψ
∫
in
dψ
at x = 0
dx

RHS = 0

–
because
2mE
ψ(0)
h2
is finite and integral
over region of length
2ε ♦ 0.
2ma
−
h2

ψ (0) 

because, by the definition of a δ–fn
∫ δ(x)ψ(x)dx = ψ(0)
or, more generally
∞
∫±∞ δ(x ± a)ψ(x)dx = ψ(a)
5.73 Lecture #2
2-5
Since the potential has even symmetry wrt
x → –x, ψ(x) must be even or odd (not a
mixture) with respect to x → – x, thus ψ(x) = ±ψ(–x). If ψ(x) is even, there must be a
cusp in ψ(x) at x = 0
ψ(x) is
ψ(x)
continuous
0
OR
BUT NOT
ψ(x) is not
continuous
at x = 0
So what happens
when ψ(x)
is an odd function?
dψ(+) dψ(±)
2ma
±
= ± 2 ψ(0)
dx
dx
h
The new
boundary condition
since there is + reflection symmetry for an even ψ(x)
dψ(+)
dψ(±)
=±
dx
dx
ma
dψ(±)
= m 2 ψ(0)
dx
h
Now find the eigenfunctions and eigenvalues. Standard procedure: divide
space into regions and match ψ and dψ/dx across boundaries.
5.73 Lecture #2
2-6
Region I
Region II
0
x
ψ L = ψ I = A L e + ρx + BL e − ρx
(8 unknowns, because A and B can
be complex numbers)
V(x)
|x|>0
E= –|E|
Let E < 0
ψ R = ψ II = A R e + ρx + BR e − ρx
1/ 2
| E | 2m 
ρ = 
 h 2 
(THIS IS WHAT WE DO WHEN k
WOULD BE IMAGINARY)
ψ(+∞
∞)=0
AR = 0
unknowns
determined
(2)
ψ(–∞
∞)=0
BL = 0
(2)
ψL(–εε)=ψ
ψR(+εε)0
AL = BR ≡ A
(2)
arbitrary phase
(1)
normalization
(1)
(8)
dψ R ( + )
− ma
= − ρAe −0 = 2 ψA(0)
h
dx
∴ ρ=
ma
h2
dψ L (–)
+ ma
= + ρAe +0 = 2 ψA
(0)
h
dx
again
ma
ρ= 2
h
ψ L = Ae ρx
ψ R = Ae −ρx
Done!
required discontinuity in dψ/dx at
x = 0.
5.73 Lecture #2
2-7
Only one acceptable value of ρ → one value of E < 0
2 2
2
ma
ρ
ma
h
ρ = 2 |E|=
=
= ±E
2
h
2m
2h
E=±
ma
2h 2
Actually, the above solution was specifically for an even ψ(x). What
about odd ψ(x)? No calculation is needed. Why?
Normalization of ψ
∞
1 = ∫ | ψ |2 dx
−∞
ψ R = Ae − max/ h
∞
2
1 = 2 ∫ | A |2 e
(
0
ma
A = ± 2 
h 
)
− 2 ma h 2 x
 h2 
dx = 2 | A |2 

 2 ma 
see Gaussian
Handout
1/ 2
 ma 
ψ δ = ± 2 
h 
1/2
e −ma|x|/h
2
only one bound
level, regardless
of magnitude of a
large a, narrower and taller ψ
There is a continuum of ψ’s possible for E > 0. Since the particle
is free for E > 0, specific form of ψ must reflect specific problem:
e.g., particle probability incident from x < 0 region. It is even
more interesting to turn this into the simplest of all barrier
scattering problems. See Non-Lecture pp. 2-8, 9, 10.
5.73 Lecture #2
2-8
Nonlecture
Consider instead scattering off V(x) = + αδ(x)
a>0
V(x) = +αδ(x)
x
0
ψ L = A L e ikx + BL e −ikx
 2mE 
k= 2 
 h 
ψ R = A R e ikx + BR e −ikx
1/2
In this problem we have flux entering exclusively from left.
The entering probability flux is |AL|2.
Two things can happen:
1.
transmit through barrier
∝ |AR|2
2.
reflect at barrier
∝ |BL|2
2
There is no way that BR can become different from 0. Why?
2
2
Our goal is to determine A R and BL vs. E
ψL(0) = ψR(0)
continuity of ψ
A L + BL = AR + B R
but BR = 0
A L + BL = AR
2ma
 dψ R (+0) dψ L (±0) 
±
= + 2 ψ(0)
 dx

dx 
h
2ma
ψR(0)
ikA R ± (ikA L − ikBL ) = 2 A R
h
AR = AL + BL
2ma
ik( A L + BL ) − ik(A L − BL ) = 2 ( A L + BL )
h
ψL(0)
5.73 Lecture #2
2-9
2ma
( A L + BL )
h2
2ma  2ma

BL  2ik − 2  = 2 A L

h 
h
2ikBL =
h2 
2ma  ikh 2
AL
=
−1 ≡ α
 2ik − 2  =
BL 2ma 
ma
h 
α +1 =
ikh 2
ma
B
A R = A L + BL = A L L + BL = α BL + BL = BL (α + 1)
BL
α = AL/BL
 ikh 2 
A R = BL 

 ma 
Transmission is
T=
AR
AL
Reflection is
R=
BL
AL
2
2
2
2
What is T(E), R(E)?
AR
2
= BL
2
k 2h 4
2 2
m a
= BL
2
2mE h 4
h
2
2 2
m a
= BL
2
2h 2 E
ma 2
*
 ikh 2   ikh 2 
 AL   AL 
− 1  −
− 1


 =
 BL   BL 
 ma
  ma

2
k 2h4
2h 2 E + ma 2
2 = 2 2 +1 =
m a
ma 2
BL
AL
 2h 2 E 
ma 2
R(E) = 2
=
+ 1

2h E + ma 2  ma 2

−1
decreasing to zero as E increases
−1
 ma 2

2h 2 E
T(E) = 2
=
+
1

 .
2h E + ma 2  2h 2 E 
R(E) + T(E) = 1
increasing to one as E increases
5.73 Lecture #2
Note that:
2 - 10
R(E) starts at 1 at E = 0 and goes to 0 at E → ∞
T(E) starts at 0 and increases monotonically to 1 as E increases.
Note also that, at E = −
ma 2
2h 2
R → ∞ as E approaches –ma2/2h2 from above and
then changes sign as E passes through –ma2/2h2 !
This is the energy of the bound state in the δ(x)-function well
problem.
See CTDL Chapter 1 Problem #3b (page 87) for a
related problem