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