Spherical Potential Well
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Transcript of Spherical Potential Well
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Lecture 25
Spherical Potential Well
December 6, 2010
Lecture 25
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Review: Spherical Harmonics
For spherically symmetric potential: V(r)
r|Elm = Elm(r,,) = REl(r)Yml (, )
Ym
l (, ) = |lm
Yml (, ) = (1)m 1
2ll!
(2l + 1)!(l + m)!
4(2l)!(l m)!
1/2eim(sin )
m
dlm
d(cos)lm(sin )2l
= (1)m
(2l + 1)(l m)!4(2l)!(l + m)!
1/2eimPml (cos )
Z Ym
l (, )Ym
l (, ) d = ll mm
Lecture 25 1
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Review: Spherical Harmonics
|
= ZZ |r
r
|
r
2dr
d
| =X
l
Xm
|rlmrlm| r2dr
r| = (r,,) = ZXlXm r|r
lmrlm| r2dr
=X
l
Xm
Yml (, )rlm| =X
l
Xm
Yml (, )Clm(r)
rlm| = Clm(r) = ZZrlm|rr| r2dr d
=
Z Y
m
l (, )r| d =
Z Y
m
l (, )(r, , ) d
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Probabilities
Prob(r,,) =
|r
|
|2 = X
lX
m |Yml (, )
|2
|Clm(r)
|2
Prob(, ) =
ZXl
Xm
|Yml (, )|2|Clm(r)|2 r2
dr
=
Xl Xm|Yml (, )|2clm
Prob(r,l,m) = |rlm||2 = |Clm(r)|2
=
ZZ |Yml (, )|2|(r,,)|2 d
Prob(lm) =
Z |Clm(r)|2 r2 dr
Lecture 25 3
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Schrodinger Equation for a Spherically Symmetric
Potential
"h
2
2m2 + V(r) E#Elm(r,,) = 0
with Elm(r,,) = REl(r)Ym
l (, )
Take the expression for the Laplacian in spherical coordinates that
you derived in homework 6.
2 = 2r2
+ 2r
r
+ 1r2
2
2 + cot
r2
+ 1
r2 sin2 2
2
It is not difficult to show that this can be written as:
2 = 1
r2
rr2
r+
1
r2 "
1
sin
sin
+
1
sin
2
2
2
But we know that: r , ,|L2|E , l , m
= h2"
1
sin
sin
+
1
sin2
2
2
#Elm(r,,)
= h2"
1
sin
sin
+
1
sin2
2
2
#REl(r)Y
ml (, )
= h2l(l + 1)REl(r)Ym
l (, )
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The Radial Solutions
" h2
2m
1
r2
d
dr
r
2 d
dr
+
l(l + 1)h2
2mr2 + V(r) E#Elm(r) =The angular dependent part of the energy eigenstates for any
spherically symmetric potential, V(r) is given by the sphericalharmonics. So that part of the work is already done for us. The
remaing task for a given V(r) is to solve for REl(r).
" h
2
2m
1
r2d
dr
r
2 d
dr
+
l(l + 1)h2
2mr2 + V(r) E
#REl(r) =
It will help to make a change of function:
REl(r)
El
(r)
r" h
2
2m
d2
dr2 +
l(l + 1)h2
2mr2 + V(r) E
#
El(r) = 0
This is a one-dimensional Schrodinger equation with an angular
momentum barrier.
Lets look at the asymptotic solutions.
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Asymptotic Solution for Small r
Assume that r2V(r) 0 as r 0
" h
2
2m
d2
dr2 +
l(l + 1)h2
2mr2
#
El(r) = 0
El (r) = l(l + 1)
r2 El (r)
Try the trial solution: El
(r) r
( 1) = l(l + 1)
= l (irregular solution) or =l+1 (regular solution
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l= 0 Case
El (r) rl REL (r) r(l+1)
|REL
(r)|2 1r(2l+2)
This is not normalizable.
Z0
|REL
(r)|2r2 dr Z
0
r2l dr = r2l+1
0=
only El
(r) rl+1 is allowed
REL
(r) rl REL
(r) = 0 if l = 0
This is the effect of the angular momentum barrier term.
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l = 0 Case
El
(r)
r
0
a constant
REL
(r) 1r
so normalization is OK
But from electrostatics we know that: = q
ris a solution to:
2 = 4 with = q3(r)
R
EL(r)
1
r
would require a function potential at r =
El
(r) rl+1 for all l
REl
(r) rl for all l
El
( 0 ) = 0
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Asymptotic Solution for Large r
Assume that rV(r) 0 as r
Note that this excludes the case of a Coulomb potential. Wellcome back to this later when we discuss the hydrogen atom.
" h
2
2m
d2
dr2 E
#
El(r) = 0
ForE > 0,this is the equation for a one-dimensional free particle.
El
(r) = Aeikr + Beikr k =
2mE/h
It is subject to the boundary condition El
(0) = 0. Thisdetermines, f, the ratio ofB to A, and there is one free constant
left, A that is then fixed by the normalization.
REl
(r) = A
eikr + f eikr
r
!
For E < 0,
El
(r) = AeKr + BeKr K =q
2m|E|/h
Here we must require that B = 0 so that REl (r) does not blowup at r =. Again, there is one free constant left, A, that isfixed by the normalization
REl
(r) = AeKr
r
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Spherical Potential Well
V(x) =8 a (Region II)
Firtst solve for Region I.
h
2
2m2 E
!Elm(r , ,) = 0
2 + 2mE
h2
Elm(r , ,) = 0
2 + k2Elm(r , ,) = 0 with k =
s2mE
h2
This is the Helmhotz equation. The radial part is:
"d2
dr2 +
2
r
d
dr l(l + 1)
r2 + k2
#REl(r) = 0
Set = kr
"d2
d2 +
2
d
d l(l + 1)
2 + 1
#REl() = 0
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Spherical Bessel Functions
" d2
d2 +
2
d
d l(l + 1)
2 + 1# REl() = 0
This looks like the Bessel equation except that l(l+ 1) is not the
square of an integer. The solutions are the spherical Bessel functions
jl(kr) =
2kr
1/2
Jl+1/2(kr)
j0(kr) = sin kr
kr
j1(kr) = sin kr
(kr)2 cos kr
kr
j2(kr) =
3
(kr)3 1
kr
sin kr 3
(kr)2 cos kr
as r 0 jl(kr) (kr)l
(2l + 1)!!
as r jl(kr) sin(kr l/2)
kr
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Spherical Neumann Functions
The spherical Neumann functions are also solutions
nl(kr) = (1)l+1
2kr
1/2Jl1/2(kr)
n0(kr) = cos kr
kr
n1(kr) = cos kr
(kr)2 sin kr
kr
n2(kr) =
3
(kr)3 1
kr
cos kr 3
(kr)2 sin kr
as r 0 nl(kr) (2l 1)!!
(kr)l+1
as r nl(kr) cos(kr l/2)
kr
The general solution with quantum number l in region I is
Ajl(kr) + Bnl(kr)
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Overall Solution
For E > V0, the general solution in Region II with quantum
number l is
Cjl(kr) + Dnl(k
r) with k =
p2m(E V0)
h
The overall solution is then
Ajl(kr) + Bnl(kr) forr < a
Cjl(kr) + Dnl(k
r) forr > a
There are three boundary conditions
at the origin: REl(r) rl as r 0 B = 0
at r =a: continuity of the wavefunction and its derivative
There is also a normalization condition.
These together exactly use up the four available constants and thereis no constraint on the energy. The energy spectrum is continuous.
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Spherical Hankel Functions
Spherical Hankel functions
hl(kr) = ll(kr) + inl(kr)
hl (kr) = ll(kr) inl(kr)
The solution for r > a can then be written as
Ehl(kr) + F h
l (k
r)
Asr
hl(kr) e[ikr(l+1)/2]
kr hl (kr)
e[ikr(l+1)/2]
kr
These are outgoing and incoming spherical plane waves, respectively.
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Bound State Solutions
For E < V0 the solution is
Ajl(kr) forr < a
Chl(Kr) + Dhl (Kr) forr > a
K =
p2m(V0 E)
h
as r the asymptotic solution is
1Kr
CeKr + DeKr
The boundary condition at r = requires that D = 0
There are still two boundary conditions that must be satisfied.
Continuity of the wavefunction and its derivative at r = a as
well as the normalization condition. But there are only two freeconstants left. It is over constrained leading do a constraint on
the allowable energies. The energies, as for all bound states, are
quantized.
Lecture 25 15