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1
Hydrogen atom Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: January 13, 2012)
Bohr model Schrödinger equation Hydrogen atom
Niels Henrik David Bohr (7 October 1885 – 18 November 1962) was a Danish physicist who made fundamental contributions to understanding atomic structure and quantum mechanics, for which he received the Nobel Prize in Physics in 1922. Bohr mentored and collaborated with many of the top physicists of the century at his institute in Copenhagen. He was part of a team of physicists working on the Manhattan Project. Bohr married Margrethe Nørlund in 1912, and one of their sons, Aage Bohr, grew up to be an important physicist who in 1975 also received the Nobel prize. Bohr has been described as one of the most influential scientists of the 20th century.
http://en.wikipedia.org/wiki/Niels_Bohr _______________________________________________________________________ Erwin Rudolf Josef Alexander Schrödinger (12 August 1887– 4 January 1961) was an Austrian theoretical physicist who was one of the fathers of quantum mechanics, and is famed for a number of important contributions to physics, especially the Schrödinger equation, for which he received the Nobel Prize in Physics in 1933. In 1935, after extensive correspondence with personal friend Albert Einstein, he proposed the Schrödinger's cat thought experiment.
2
http://en.wikipedia.org/wiki/Erwin_Schr%C3%B6dinger 1. Orbital angular momentum in quantum mechanics
The orbital angular momentum is defined as
zx
zyx
ppp
zyx
yˆˆˆ
ˆˆˆˆˆˆeee
prL
yzx pzpyL ˆˆˆˆˆ
zxy pxpzL ˆˆˆˆˆ
xyz pypxL ˆˆˆˆˆ
zyx
zyxz
zyxz
zxyzyx
Lipxpyi
xpzppzpy
pxpzpzpy
pxpzpzpyLL
ˆ)ˆˆˆˆ(
ˆ]ˆ,ˆ[ˆˆ]ˆ,ˆ[ˆ
]ˆˆ,ˆˆ[]ˆˆ,ˆˆ[
]ˆˆˆˆ,ˆˆˆˆ[]ˆ,ˆ[
or
zyx LiLL ˆ]ˆ,ˆ[ ,
3
Similarly,
xzy LiLL ˆ]ˆ,ˆ[ , yxz LiLL ˆ]ˆ,ˆ[
2L is defined by
2222 ˆˆˆˆ
zyx LLLL
We have
0]ˆ,ˆ[]ˆ,ˆ[
]ˆ,ˆ[]ˆ,ˆ[]ˆ,ˆˆˆ[]ˆ,ˆ[22
222222
yzxz
zyzxzzyxz
LLLL
LLLLLLLLLL
using the relations
)ˆˆˆˆ(]ˆ,ˆ[ˆˆ]ˆ,ˆ[]ˆ,ˆ[ 2yxxyxzxxxzxz LLLLiLLLLLLLL
)ˆˆˆˆ(]ˆ,ˆ[ˆˆ]ˆ,ˆ[]ˆ,ˆ[ 2yxxyyzyyyzyz LLLLiLLLLLLLL
Similarly
0]ˆ,ˆ[ 2 xLL , 0]ˆ,ˆ[ 2 yLL 0]ˆ,ˆ[ 2 zLL
2. Quantum mechanical orbital angular momentum: spherical coordinates
4
x
y
z
q
dq
f df
r
drr cosq
r sinq
r sinq df
rdq
er
ef
eq
The orbital angular momentum in the quantum mechanics is defined by
)( riprL using the expression
sin
11
rrrr eee
in the spherical coordinate. Then we have
)sin
1(
)sin
11()(
ee
eeeerL
i
rrrrii rr
.
The angular momentum Lx, Ly, and Lz (Cartesian components) can be described by
5
]sin
1)sinsincoscos(cos)cossin([
zyyi eeeeeL xx.
or
)coscot(sin
iLx
)sincotcos(
iLy
iLz
We define L+ and L- as
)cot(
ieiiLLL iyx
and
)cot(
ieiiLLL i
yx
We note that the operator can be expressed using the operator L as
2r
i
rr
Lre
The proof of this equation is given as follows.
)sin
1()
sin
1(
)(
eeeeeLr
rri r
or
rrrri rr
eeeLr
sin
11)(2
or
6
2
)(
r
i
rr Lr
e
From 2222zyx LLL L , we have
)](sinsin
1
sin
1[
2
2
222
L
Using
)( 2222
2
rr
rr
L
we can also prove that
L
i
rrr )1(2
((Note))
2
22
22
2
22
22
22
222
22
222
2
1
)(11
)(11
)(11
rp
r
rrrr
rr
rrr
rr
rrr
L
L
L
L
where the definition of pr is given below. This expression can be rewritten as
2
222
2
222 r
pr
L
______________________________________________________________________ 3. Radial momentum operator pr in the quantum mechanics (a) In classical mechanics, the radial momentum of the radius r is defined by
)(1
pr r
prc
7
(b) In quantum mechanics, this definition becomes ambiguous since the component of p and r do not commute. Since pr should be Hermitian operator, we need to define as the radial momentum of the radius r is defined by
)(2
1
rrprq
rpp
r
This symmetric expression is indeed the canonical conjugate of r.
irprp rqrq
Note that
rrr
irr
iprq
1
)()1
)((
This can be proved as follows.
)1
(
)1
(
)2
2)((2
1
)](1
)[(2
1
)]},,([),,(.){(2
1),,(
22
rrr
i
rrri
rrri
rrrr
i
rrirp rrrq
ee
where
sin
11
rrrr eee
A
rA
rAr
rr r
sin
1)(sin
sin
1)(
1 22
A
For convenience we use
req pp
then we have
8
rrr
rrr
irrr
ipr 2
222 1
)1
)(1
(
________________________________________________________________________ 4. Central field problem
Free particle wave function satisfies the Schrödinger equation
ErV )](2
[ 22
,
where is the reduced mass of particle, E is the energy eigenvalue of the system. The wavefunction can be expressed by
),,( rmk
),,(),,()]([2
12
22
rErrV
rp mnmnr
L
(separation variables),
),()(),,( mnmn YrRr
with
),()1(),( 22 mm YY L
),(),( mmz mYYL
Note that ),( mY is the spherical harmonics.
)()()()(])1(
[2
12
22 rERrRrVrR
r
llp nnnr
Since rrri
pr
1
, we have
)]([1
)()1
(1
)(2
222 rrR
rrrRr
rrir
rrirRp nnnr
or
9
)(2
)()]1(1
)(2
[)]([1
2222
2
rERrRr
rVrrRrr nnn
or
0)()]1(1
))((2
[)]([1
222
2
rRr
rVErrRrr nn
.
Note that for a fixed l, the energy eigenvalue is independent of m, and is at least (2l+1)-fold degenerate. We assume that
r
rurRnl
)()( , = -1
u"(r) [
l(l 1)
r 2 2(1 V (r)
2 ]u(r) 0
We further assume a Coulomb potential given by
r
ZekrV
2
)(
with
04
1
k
Then
0)(])(2)1(
[)("2
2
1
2
rur
Zek
r
llru
We now introduce a new variable
222
28
8 2
2
22
4221
Z
an
kZe
n
n
eZk
r
where
10
2
22
22
422
1 22 n
Z
a
ke
n
eZk
, 1nE
2
2
eka
(a = a0: Bohr radius when = m (electron mass))
a
ke
2
2
(Rydberg energy when a = a0)
na
Z
Since
rna
rZ 22
we get
d
dr
ddr
d
d 2
d
d
d2
dr 2 2d
d(2
d
d) 4 2 d2
d2
4 2u"() [4 2 l(l 1)
2 2 (1 4n
)]u( ) 0
where
)4
1()(2 2
2
12 n
r
Zek
Then we have
u"() [l(l 1)
2 (1
4
n
)]u() 0
or
[d2
d2 l(l 1)
2 n
1
4]u( ) 0 (1)
11
5. Series expansion method Solution of radial part of the hydrogen atom (we need to show that = n; positive integer)
[d2
d2 l(l 1)
2
1
4]u() 0 (1)
with
1
2
2
kZe
, which corresponds to the eigenvalue.
Note that according to the Bohr model, = n (positive integer) since
2
22
22
422
1 22 n
Z
a
ke
n
eZk
We solve the differential equation to determine the eigenvalue and eigenfunction. In the limit of 0, we assume that it behaves at the origin like
u s
[s(s 1) l(l 1)] s 2 s 1 1
4 s 0
Note that the s-2 term dominates for small .
s(s 1) l(l 1) 0 , or
(s 1)(s l) 0
s l 1 or s = - l. We must discard those solutions that behave as l . So we get the form around = 0:
1)( lu In the limit of ,
12
(d2
d 2 1
4)u() 0
The solution for this equation is
u() Ae / 2 Be / 2 The constant B should be equal to zero ( ):
u() e / 2 Thus we can attempt to find a solution of the form
u() l1e / 2 F() With this substitution, the differential equation (1) becomes
d2 F()
d2 (2l 2
1)
dF()
d (
l 1
)F( ) 0
We assume that
F() Ckk
k 0
with C0 ≠ 0.
k(k 1)Ck}k2
k2
(2l 2)kCkk 2
k1
[k (l 1)]Ck}k1
k 0
0
or
{k(k 1) (2l 2)(k 1)]Ck1 [k (l 1)]Ck}k1
k0
0
leading to
Ck1
Ck
k l 1
(k 1)(k 2l 2)
((Note))
13
Coefficient of 0
0)22()1( 10 ClCl
Coefficient of 1
0)23(2)2( 21 ClCl Coefficient of 2
0)24(3)3( 32 ClCl
Coefficient of 3
0)25(4)4( 43 ClCl
Coefficient of 4
0)26(5)5( 54 ClCl
Coefficient of 5
0)27(6)6( 65 ClCl
............................................................................................ Note that
Ck1
Ck
1
k
which is the same asymptotic behavior as e. Thus, unless the series terminate, u() will grow exponentially like e.
To avoid this, we must have For k = nr,
nr l 1 0 or
14
nr l 1 Then we have
Cnr1 Cnr 2 ... 0
For nr = 0, l 1
F() C0 For nr = 1, l 2
F() C0 C1 For nr = 2, l 3
F() C0 C1 C22
The function F will thus be a polynomial of degree of nr, known as an associated Laguerre polynomial.
rnlE
Zek 1
2
2
or
22
422
)1(2 rnl
eZkE
Since l = 0, 1, 2, 3,..., nr = 0, 1, 2,..., we introduced a principal quantum number n, defined by
n = l + 1 + nr with n = 1, 2, 3, .... Thus, in terms of n,
2
22
22
422
22 an
Zke
n
eZkEn
_____________________________________________________________________
n = 1, nr = 0, l = 0 (1s) ______________________________________________________________________
n = 2, nr = 0, l = 1 (2p)
15
n = 2, nr = 1, l = 0 (2s)
______________________________________________________________________ n = 3, nr = 0, l = 2 (3d)
n = 3, nr = 1, l = 1 (3p)
n = 3, nr = 2, l = 0 (3s)
______________________________________________________________________ n = 4, nr = 0, l = 3 (4f)
n = 4, nr = 1, l = 2 (4d)
n = 4, nr = 2, l = 1 (4p)
n = 4, nr = 3, l = 0 (4s)
______________________________________________________________________
n = 5, nr = 0, l = 4 (5g)
n = 5, nr = 1, l = 3 (5f)
n = 5, nr = 2, l = 2 (5d)
n = 5, nr = 3, l = 1 (5p)
n = 5, nr = 4, l = 0 (5s) _______________________________________________________________________ These nlm states have the same energy which is only dependent on n. 6. Spherical harmonics ),( lmY
),(),(),(
lmlmlmz YmYi
YL
(1)
The and dependence of ),( m
lY is given by
),()1(
),()](sinsin
1
sin
1[),(
2
2
2
222
ml
ml
ml
Yll
YY
L (2)
Equation (1) shows that
16
imm
lm
l eY )(),(
where ),( m
lY is normalized as
),(),(sin),(),(*'
'
*''',', m
lm
lm
lm
lmmll YYddYYd
where is a solid angle and ddd sin . We must require that the eigenfunction be single valued
eim eim ( 2 ) which means that m = 0, ±1, ±2, (integers). Equation (2) can be rewritten as
0)()]1(sin
)(sinsin
1[
2
2
mlll
m
d
d
d
d
The result for m≥0 is
lml
ml
mim
l
lm
l d
de
ml
mll
lY 2)(sin
)(cossin
1
)!(
)!(
4
)12(
!2
)1(),(
and we define Yl
m(,) by
Ylm(,) (1)m[Yl
m( ,)]* or
),()1()],([ * ml
mml YY
7. Quantum numbers
n: the principal quantum number. l : the azimuthal quantum number m: the magnetic quantum number
For the fixed n (=1, 2, 3, 4, ...),
l = n-1, n-2,......., 1, and 0.
l = 0 sharp (s) m = 0
17
l = 1 principal (p)
m = 1, 0, -1
l = 2 diffuse (d) m = 2, 1, 0, -1, -2
l = 3 fundamental (f)
m = 3, 2, 1, 0, -1, -2, -3
l = 4 (g) m = 4, 3, 2, 1, 0, -1, -2, -3, -4
There are (2l+1) solutions to the Schrodinger equation corresponding to the same energy eigenvalue En.
Degeneracy of En = 21
0 2
)1(2)12( nn
nnl
n
l
.
((Note)) Including spin degeneracy of states is 2n2.
n l m ms 1s 1 0 0 2/1 2s 2 0 0 2/1 2p 2 1 0, ±1 2/1 3s 3 0 0 2/1 3p 3 1 0, ±1 2/1 3d 3 2 0, ±1, , ±2 2/1 4s 4 0 0 2/1 4p 4 1 0, ±1 2/1 4d 4 2 0, ±1, , ±2 2/1 4f 4 3 0, ±1, , ±2, ±3 2/1
8. Vector model of the orbital angular momentum
We consider a case which l is some fixed number (l = 1, 2, 3,...). Then the total angular momentum may be represented by a vector of length
)1( ll
The component m in the z direction is
m = l, l - 1, l - 2, ........, -l+ 1, -l The vector J should be thought of as covering a cone, with vector angle given by
18
)1(cos
ll
mm
where m is the angle between the z axis and L. (a)
l = 1 m = 1, 0, -1
1 1+ 1
m=1
m=0
m=-1
(b)
l= 3 m = 3, 2, 1, 0, -1, -2, -3
19
3 3+ 1
m=3
m=2
m=1
m=0
m=-1
m=-2
m=-3
(c)
l = 6 m = 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, -5, -6
6 6+ 1
m=6
m=5
m=4
m=3
m=2
m=1
m=0
m=-1
m=-2
m=-3
m=-4
m=-5
m=-6
9. SphericalPlot3D of 2
),( mlY
We make a SphericalPlot3D of the spherical harmonics. ________________________________________________________________________ (i)
l=0 m=0 ),(00 lY
20
0 0 12
l = 0, m = 0
_______________________________________________________________________ (ii) m = -1, 0, 1
l=1 m ),(1 mlY
1 1 12 3
2 Sin
1 0 12
3
Cos
1 1 12 3
2 Sin
l = 1, m = ±1 l = 1, m = 0 ________________________________________________________________________ (ii) m = -2, -1, 0, 1, 2
l=2 m ),(2 mlY
21
2 2 142 15
2 Sin2
2 1 12 15
2 Cos Sin
2 0 18
5
1 3 Cos2
2 1 12 15
2 Cos Sin
2 2 142 15
2 Sin2
l = 2, m = ±2 l = 2, m = ±1 l = 2, m = 0
_______________________________________________________________________ (iv) m = -3, -2, -1, 0, 1, 2, 3
l=3 m ),(3 mlY
22
3 3 183 35
Sin3
3 2 142 105
2 Cos Sin2
3 1 116
21
3 5 Cos2 Sin
3 0 116
7
3 Cos 5 Cos3
3 1 116
21
3 5 Cos2 Sin
3 2 142 105
2 Cos Sin2
3 3 183 35
Sin3
l = 3, m = ±3 l = 3, m = ±2 l = 3, m = ±1 l = 3, m = 0 10. SphericalPlot of 2p orbitals
23
p-orbitals
ypx= 1
2-Y1
1(q,f)+Y1-1(q,f)]
ypy=Â 1
2Y1
1(q,f)+Y1-1(q,f)]
ypz=Y10(q,f)
px orbit
24
py orbit
Out[53]=
pz orbit
11. SphericalPlot of 3d orbitals
25
de-orbitals
yxy=-Â 1
2[Y2
2(q,f)-Y2-2(q,f)]
yyz=-Â 1
2[Y2
2(q,f)+Y2-2(q,f)]
yzx=- 1
2[Y2
1(q,f)- Y2-1(q,f)]
dg-orbitals
yx2-y2= 1
2[Y2
2(q,f)+Y2-2(q,f)]
y3 z2-r2=Y2
0(q,f)
d(xy) orbit
26
d (yz) orbit
d (zx) orbit
27
d(x2-y2) orbit
d(3z2-r2) orbit ________________________________________________________________________ 12. The form of radial wave function Rnl(r)
The radial wave function Rnl(r) is given by
rRnl (r) unl (r) Ae / 2 l1Ln l12l1 () ,
28
where
2r . Then we have
Rnl (r) unl (r)
r
A
re / 2 l1Ln l1
2l1 () 2Ae / 2 lLn l12l1 ( )
A is determined from the condition of normalization.
1 [Rnl (r)]2 r2 dr0
A2
2e 2l2[Lnl1
2l1 ()]2 d0
, (2)
Here we use the formula:
e q1 Lpq ()Lp
q ()d0
( p q)!
p!(2 p q 1)
Note that
p n l 1, q 2l 1, p q n l , and 2 p q 1 2n Then we have
e 2l2 [Lnl12l1 ()]2 d
0
(n l)!
(n l 1)!(2n)
Using this formula, Eq.(2) can be rewritten as
1 A2
2(n l)!
(n l 1)!(2n)
or
A Z 1/ 2
na1/ 2
(n l 1)!
(n l)!
Thus we get
Rnl (r) Rnl() Anle / 2lLn l1
2l1 () with
29
)!(
)!1(22/32
2/3
ln
ln
an
ZAnl
((Note)) The final form of )(rRnl is as follows.
)2
()exp(2)!(
)!1(
)2
(2
)exp()!(
)!1(2)(
121
22/32/31
1212/32
2/3
rna
ZL
na
ZrrnZa
ln
ln
rna
ZLr
na
Z
na
Zr
ln
ln
an
ZrR
lln
lllll
lln
l
nl
since
rna
Z2
This function satisfies the differential equation
[d2
d2 2
d
d (
1
4
n
l(l 1)
2 )]Rnl ( ) 0 .
We also have
)()!(
)!1()()( 12
112/
2/1
2/1
lln
lnlnl Le
ln
ln
na
Zuru
13. Radial probability
The wave function )(rnlm is normalized as
drrdnlm2)(1 r
where is the solid angle and ddd sin
),()()( mlnlnlm YrRr
and
22222 )(),()(1 rRdrrYdrRdrr nlm
lnl
where
30
1),(2 m
lYd
We define Prdr as
drrRrdrP nlr22 |)(|
Then the average sr is defined by
0
22
0
22 )]([)]([ rRdrrrrRdrrr nlss
nls
where
an
kZeE
eka
n 2
2
2
2
2
The average sr is obtained as
]2/3)1(2)[1)(2/1(
)]1(3[45
244
lllllan
llnZr
)1)(2/1(4
33
llaln
Zr
)2/1(23
22
lan
Zr
an
Zr
21
10 r
)]1(3[2
2 llnZ
ar
)]1(315[2
222
22 llnn
Z
ar
)]5)1(6(5)2)(1()1(335[8
2423
33 llnllllnn
Z
ar
14. Form of the wave function
We use the radial wave function as
31
)()!(
)!1(2)()( 12
12/
2/32
2/3
lln
lnlnl Le
ln
ln
an
ZRrR
where
na
Zrr
22
________________________________________________________________________ (a) Expression of )(rRnl
n =1
)exp(2)(2/3
10 a
Zr
a
ZrR
n = 2
)2
exp()2
1(2
1)(
2/3
20 a
Zr
a
Zr
a
ZrR
)2
exp(62
1)(
2/3
21 a
Zr
a
Zr
a
ZrR
n=3
)3
exp()27
2
3
21(
33
2)(
2
222/3
30 a
Zr
a
rZ
a
Zr
a
ZrR
)3
exp()6
1(627
8)(
2/3
31 a
Zr
a
Zr
a
Zr
a
ZrR
)3
exp(3081
4)(
2
222/3
32 a
Zr
a
rZ
a
ZrR
(b) Expression of )(nlR
n =1
32
)2
exp(2)(2/3
10
a
ZR
n = 2
)2
exp()2(22
1)(
2/3
20
a
ZR
)2
exp(62
1)(
2/3
21
a
ZrR
n=3
)2
exp()66(39
1)( 2
2/3
30
a
ZrR
)2
exp()4(69
1)(
2/3
31
a
ZrR
)2
exp(309
1)( 2
2/3
32
a
ZrR
________________________________________________________________________ 15. Plot of the probability of the wave function and the average radius (i) 22 )]([ rRr nl vs r/a, where a = 1 and Z =1.
(ii) )]1(3[2
2 llnZ
ar , where a = 1 and Z = 1.
For the 1s state,
33
n=1
{=0
<r>a
0 1 2 3 4 5 6 7ra0.0
0.1
0.2
0.3
0.4
0.5
Pr
Fig. 1s (n = 1, l = 0). The straight line denotes the average value (<r>/a). For the 2s, 2p states
n=2
{=0
{=1
0 2 4 6 8 10 12 14ra0.00
0.05
0.10
0.15
0.20
Pr
Fig. 2s (n = 2, l = 0). 2p (n = 2, l = 1). The straight lines denote the average value
(<r>/a). For the 3s, sp, and 3d states,
34
n=3
{=0
{=1
{=2
0 5 10 15 20ra0.00
0.02
0.04
0.06
0.08
0.10
0.12
0.14
Pr
Fig. 3s (n = 3, l = 0). 3p (n = 3, l = 1). 3d (n = 3, l = 2). The straight lines denote the
average value (<r>/a). For the 4s, 4p, 4d, and 4f states,
n=4
{=0{=1{=2
{=3
0 5 10 15 20 25ra0.00
0.02
0.04
0.06
0.08
0.10
Pr
Fig. 4s (n = 4, l = 0). 4p (n = 4, l = 1). 4d (n = 4, l = 2). 4f (n = 4, l = 3). The straight
lines denote the average value (<r>/a). 16. ContourPlot of the wavefunctions (Mathematica)
The probability function 2
),,( rnlm is expressed using the spherical co-ordinate (r,
, ).
35
22),()(),,( m
lnlnlm YrRr
where
)2
()exp(2)!(
)!1()( 12
122/32/31 r
na
ZL
na
ZrrnZa
ln
lnrR l
lnlllll
nl
Using the relations
222 zyxr ,
222arccos
zyx
z , )arctan(x
y
the function can be expressed by the Cartesian co-ordinate. Here we assume that
Z = 1, = m (mass of electron), and a = a0 (Bohr radius). (a) d (zx) orbit
-15 -10 -5 0 5 10 15
-15
-10
-5
0
5
10
15
n Ø 3, l Ø 2, yzx, x, 0, z
(b) d (3z2-r2) orbit
36
-15 -10 -5 0 5 10 15
-15
-10
-5
0
5
10
15
n Ø 3, l Ø 2, y3z2 - r2, x, y, 0
_______________________________________________________________________ APPENDIX Vector analysis in the Spherical co-ordinates: (a) The gradient is given by
sin
11
rrrr eee
where is a scalar function of r, , and . (b) A The divergence is given by
A
rA
rAr
rr r
sin
1)(sin
sin
1)(
1 22
A
(c) A
37
A is given by
ArrAA
r
rr
rAhAhAh
r
hhh
hhhr
r
rr
rr
r
sin
sin
sin
112
eeeeee
A
(d) Laplacian
)]sin
1()(sin)sin([
sin
1
)]()()([1
22
2
rr
rr
h
hh
h
hh
rh
hh
rhhhrr
rr
or
2
2
2222
22
sin
1)(sin
sin
1)(
1
rrr
rrr
We can rewrite the first term of the right hand side as
)(1
)(1
22
2r
rrrr
rr
which can be useful in shortening calculations. Note that we also use the expression for the operator
}sin
1)(sin
sin
1{
1)(
1
sin
1)(sin
sin
1)(
1
2
2
2222
2
2
2
2222
22
rrr
rr
rrrr
rr