3.2 Hydrogen Atom
Nicholas BrownECE 6451
Introduction to the Theory of MicroelectronicsFall 2005
3.2 Hydrogen Atom
22
21 mdd
=Φ
Φ−
φ
In section 3.1, we solved the two differential equations below where Eq. 3.2.1 is dependent on phi and Eq. 3.2.2 which is dependent on theta
θλ
θθ
θθ 2
2
sinsin
sin1 Θ
=Θ+
Θ m
dd
dd
In section 3.2, we will solve the radial dependent differential equation and analyze its results in the context of the Hydrogen atom.
(3.2.1)
(3.2.2)
3.2 Hydrogen Atom
ERRrVRmr
Rr
rrmr
=++
∂∂
∂∂
− )(22 2
22
2
2 hh λ
We will relate our equation to the Hydrogen atom by considering the form of V(r).
The differential equation below is the radial dependent component of the 3-D Schrödinger Equation in spherical coordinates
(3.2.3)
The potential of the Hydrogen atom, V(r) is inversely proportional to rr
rV 1~)(
3.2 Hydrogen Atom
+=
rdrd
ipr
1h
We will define an operator called the radial-momentum operator
Using this operator we can redefine the differential term of the kinetic energy operator in spherical coordinates as
mpr2
2
(3.2.4)
(3.2.5)
For the skeptics out there, the next slide will illustrate the equivalence of Eq. 3.2.5 and the differential term of the kinetic energy operator.
3.2 Hydrogen Atom
+−=
−
drd
rdrd
drdr
drd
r21
2
222
22 hh
Proof that the radial-momentum operator is equivalent to the differential term of the kinetic energy operator:
differential term of the kinetic energy operator
+−=
++−+−=
+
+−=
drd
rdrd
rdrd
rrdrd
rdrd
rdrd
rdrdpr
2
1111
11
2
22
222
22
22
h
h
h
radial-momentum operator same as the differential term of the kinetic energy operator
3.2 Hydrogen Atom
2
22
22 mrL
mpT r +=
)()()(2
)1(2
)()()(22
)()(
2
22
2
22
rErrVmrll
mp
rErrVmrL
mp
rErH
r
r
Ψ=Ψ
+
++
Ψ=Ψ
++
Ψ=Ψ
h
Ψ=Ψ λ2L
So rewriting the kinetic energy operator in terms of pr results in
Using Eq. 3.2.6, Schrödinger's Equation is given by
where )1( += llλ
(3.2.7)
(3.2.6)
3.2 Hydrogen Atom
rrur )()( =ΨLet where u(r) is an polynomial of r
Consider
2
2
22
2
2
2
22
22
1111
1111
)(1)(111
)(11)()(
drud
rdrdu
rdrud
rdrdu
r
drdu
rdrdurdr
ddrdurrdr
d
rur
rurdr
durrdr
drru
rdrd
rdrd
rruprp rr
=
++−=
+
=
+=
+−
+=
+
+==Ψ
(3.2.8)
3.2 Hydrogen Atomrrur )()( =Ψ
EuurVumrll
drud
m
ruE
rurV
ru
mrll
drud
mr
=++
+−
=++
+−
)(2
)1(2
)(2
)1(2
2
2
2
22
2
2
2
2 2
hh
hh
Substituting and Eq. 3.2.8 into Eq. 3.2.7
(3.2.9)
Define V(r) (in cgs units) we have
rZqrV
2
)( −=
where Z is the atomic number, q is the charge of an electron and ris the distance between force center and particle under observation.
(3.2.10)
3.2 Hydrogen Atom
ρoar = εoEE =
We can take some steps to make Eq. 3.2.9 dimensionless to simplify our analysis. Defining
no units
Substituting Eq. 3.2.11 and Eq. 3.2.10 into Eq. 3.2.9 results in
(3.2.11)
uEuaZqu
mall
dud
am oooo
ερρρ
=−+
+− 2
22
2
2
2
2
2
2)1(1
2hh
(3.2.12)
and
3.2 Hydrogen Atom
ooo
Emaa
Zq== 2
22 h Zmqao 2
2h=
2
42
h
mqZEo =
Defining ao and Eo from Eq. 3.2.11 so that Eq. 3.2.12 is dimensionless
(3.2.13)
Substituting Eq. 3.2.13 and Eq. 3.2.14 into Eq. 3.2.12 we get the dimensionless expression of the Schrödinger's Equation
uulldd ε
ρρρ=
−
++−
12
)1(21
22
2
(3.2.15)
(3.2.14)
3.2 Hydrogen Atom
m1cm100 =
J1erg10/scmg1erg1
7
22
=
⋅=
esu103C1 9×=
ao is known as the Bohr radius which is considered to be the effective radius of a Hydrogen atom. To calculate the Bohr radius we must consider the cgs units for energy, length and electric charge.
The cgs unit for length is the centimeter.
The cgs unit for energy is the erg.
The cgs units for electric charge is “electrostatic unit” or esu.
Conversion between SI and cgs.
( )
m1029.5cm1029.5
)1(C1
esu103106.1kg1
g1000)kg1011.9(
J1erg10)sJ1005.1(
119
2921931
27234
2
2
−−
−−
−
×=×=
××
×
⋅×
==
o
o
a
CZmq
a h
3.2 Hydrogen Atom
......)( 11 ++++=
−−−nqn
qnnnn
o eAeAeAuρρρ
ρρρρρρ
Now let us assume a power series solution to Eq. 3.2.15 of the form
(3.2.16)
Applying the second derivative to the first term in Eq. 3.2.16,
nnnn
o
nn
no
nno
nno
nno
ennn
A
en
nddA
en
AenAddeA
dd
ρ
ρ
ρρρ
ρρρ
ρρρ
ρρρ
ρρ
−−−
−−
−−−−
−+−=
−=
−+=
212
1
12
2
)1(2
1
(3.2.17)
3.2 Hydrogen Atom
nnnn llnnn
ερρρρ=
++−−− −− 22
2 2)1()1(
21
21
nn
nερρ =− 2
121
221n
−=ε
Substituting this result into Eq. 3.2.15 and the first term of Eq. 3.2.16, we get
(3.2.18)
Equating the coefficients of like powers
(3.2.19)
Thus
22
42
22
42
221
nmqZ
nmqZEE o
hh−=
−== ε (3.2.20)
3.2 Hydrogen Atom
2
eV6.13n
E −=
2)1(
2)1( +=
− llnn
Substituting the appropriate values for the physical constants into Eq. 3.2.20 and setting Z=1 since we are considering the Hydrogen atom, we get
(3.2.21)
Equating the coefficients of ρn-2 leads to
(3.2.22b)
Solving for l in terms of n we get
10)1(2
−==−−+
nlnnll
(3.2.22a)
3.2 Hydrogen Atom
∑∞
=
−=
0q
nqq
s eAuρ
ρρ
Now, we have defined three quantum numbers – n, l, m – that are needed to specify the particles motion in a hydrogen atom.
What about the wave function of a particle in a hydrogen atom?
To accomplish this we will redefine the power series solution for Schrödinger's Equation
(3.2.23)
as
onar
ergru−
= )()( (3.2.24)
3.2 Hydrogen Atom
ooo
oo
nar
o
nar
o
nar
nar
o
nar
gena
egna
egdrud
gena
egdrdu
−−−
−−
+′−′′=
−′=
22
2
)(12
1
Also
(3.2.25)
Substituting the above result into Eq. 3.2.9 where V(r) is defined as Eq. 3.2.10 and E is defined as Eq. 3.2.20, dividing out the exponential terms and simplifying the result, we get
02
)1(22
2
2
22
=−+
+
′−′′− g
rZqg
mrllg
nag
m o
hh(3.2.26)
3.2 Hydrogen Atom
∑=q
s rArrg )(g(r) is defined as
(3.2.27)
So the first and second derivatives of g(r) are
...)1)((...
...)1()1()(
...)(...)1()(
2
11
2
11
1
+−+++
+++−=′′
++++++=′
−+
−−
−+−
qsq
ss
qsq
so
s
rAqsqs
rsAsrssrg
rAqsArsAsrrg(3.2.28)
We want to make Eq. 3.2.26 dimensionless so again we substitute Eq. 3.2.11 into Eq. 3.2.26
02)1(222
2
=++
−− ggllddg
ndgd
ρρρρ (3.2.29)
3.2 Hydrogen AtomSubstituting Eq. 3.2.11 into Eq. 3.2.27 and Eq. 3.2.28 and substituting that result into Eq. 3.2.29 gives us
[ ][ ][ ] 0......2
...)(...2
......)1(
...)1)((...)1(
11
11
22
22
=++++
+++++
++++−
+−++++−
−++−
−++−
−++−
−++−
qsqsoq
ssoo
qsqsoq
ssoo
qsqsoq
so
qsq
qso
so
so
aAaA
aAqsasAn
aAAll
AaqsqsAass
ρρ
ρρ
ρρ
ρρ
(3.2.30)
3.2 Hydrogen Atom
0)1()1( =+−− llss
Eq. 3.2.30 must equal zero for all ρ. Therefore the coefficients of theρs-2 must sum to zero.
(3.2.31)
The coefficients of the general term, ρs+q-1 are
02)1(
)(2))(1(
11
11
=++−
+−+++
++++
++++
qsoq
qsoq
qsoq
qsoq
aAaAll
qsaAn
aAqsqs(3.2.32)
3.2 Hydrogen Atom
1+= ls
Solving for s in terms of l using Eq. 3.2.31 we find
Substituting Eq. 3.2.33 into Eq. 3.2.32 and solving for Aq+1 we have
(3.2.33)
+−++++
−++=+ )1()1)(2(
2)1(2
1llqlql
qlnAA qq
(3.2.34)
Note: 1/ao factor is absorbed by the Aq term.
3.2 Hydrogen Atom
+++
−++=+ )22)(1(
2)1(2
1 lqq
qlnAA qq
Eq. 3.2.34 can be reexpressed as
(3.2.35)
Eq. 3.2.35 can be used to find the coefficients of the higher order terms of Eq. 3.2.23. The general wave function for the Hydrogen atom is defined by Eq. 3.2.36.
),()(),,( φθφθ lmYrRr =Ψ (3.2.36)
Eq. 3.2.35 and Eq. 3.2.23 are used to define R(r) in Eq. 3.2.36 and Ylm(θ,Φ) is defined in Section 3.1.
References• Brennan, Kevin F, “The Physics of Semiconductors with applications to
Optoelectronic Devices.”• http://www.physics.utoronto.ca/~jharlow/teaching/Units.htm• Halliday, D., Resnick, R., Walker, “Fundamentals of Physics”
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