spherical potential well

5/25/2018 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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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


5/16

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 (, )

Lecture 25 4


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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.

Lecture 25 5


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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

Lecture 25 6


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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.

Lecture 25 7


9/16

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

Lecture 25 8


10/16

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

Lecture 25 9


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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

Lecture 25 10


12/16

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

Lecture 25 11


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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)

Lecture 25 12


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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.

Lecture 25 13


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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.

Lecture 25 14


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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

spherical potential well

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