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8.1 The hydrogen atom solutions
Slides: Video 8.1.1 Separating for the radial equation
Text reference: Quantum Mechanics for Scientists and Engineers
Section 10.4 (up to “Solution of the hydrogen radial wavefunction”).
The hydrogen atom solutions
Separating for the radial equation
Quantum mechanics for scientists and engineers David Miller
Internal states of the hydrogen atom
We start with the equation for the relative motion of electron and proton
We use the spherical symmetry of this equation
and change to spherical polar coordinates From now on, we drop the subscript r in the
operator
2
2
2 HV U E U
r r r r
2
Internal states of the hydrogen atom
In spherical polar coordinates, we have
where the term in square brackets is the operator we introduced
in discussing angular momentum Knowing the solutions to the angular momentum problem
we propose the separation
22 2
2 2 2 2
1 1 1 1sinsin sin
rr r r r
,U R r Y r
2 2 2,
ˆ /L
Internal states of the hydrogen atom
The mathematics is simpler using the form
where, obviously
This choice gives a convenient simplification of the radial derivatives
1 ,U r Yr r
r rR r
22
2 2
1 1r rr
r r r r r r
Internal states of the hydrogen atom
Hence the Schrödinger equation becomes
Dividing byand rearranging, we have
222
2 3
1 1 ˆ, , ,2 2
1 ,H
r r rY L Y Y V r
r r r r
E r Yr
2 3, / 2r Y r
222 2
2 2 2
2 1 1 ˆ ,,H
rr r E V r L Yr r Y
Internal states of the hydrogen atom
In
in the usual manner for a separation argument the left hand side depends only on rand the right hand side depends only on and
so both sides must be equal to a constantWe already know what that constant is explicitly
i.e., we already know that so that the constant is
222 2
2 2 2
2 1 1 ˆ ,,H
rr r E V r L Yr r Y
2 2ˆ , 1 ,lm lmL Y l l Y 1l l
1l l
Internal states of the hydrogen atom
Hence, in addition to the eigenequationwhich we had already solved
from our separation above, we also have
or, rearranging
which we can write as an ordinary differential equationAll the functions and derivatives are in one variable, r
2L̂
222
2 2
2 1H
rr r E V r l lr r
22 2
2 2
12 2 H
d r l lV r r E r
dr r
Internal states of the hydrogen atom
Hence we have mathematical equation
for this radial part of the wavefunctionwhich looks like a Schrödinger wave equation
with an additional effective potential energy term of the form
22 2
2 2
12 2 H
d r l lV r r E r
dr r
2
2
12
l lr
Central potentials
Note incidentally thatthough here we have a specific form for
in our assumed Coulomb potential
the above separation works for any potential that is only a function of r
sometimes known as a central potential
V r
2
4e po e p
eV
r rr r
Central potentials
The precise form of the equation
will be different for different central potentialsbut the separation remains
We can still separate out the angular momentum eigenequation
with the spherical harmonic solutions
22 2
2 2
12 2 H
d r l lV r r E r
dr r
2L̂
Central potentials
Since a reasonable first approximation for more complicated atoms
is to say that the potential is still approximately “central”
approximately independent of anglewe can continue to use the spherical harmonics
as the first approximation to the angular form of the orbitals
and use the “hydrogen atom” labels for theme.g., s, p, d, f, etc.
8.1 The hydrogen atom solutions
Slides: Video 8.1.3 – 8.1.4 Radial equation solutions
Text reference: Quantum Mechanics for Scientists and Engineers
Sections 10.4 starting with “Solution of the hydrogen radial wavefunction”, and 10.5
Note: The section Text reference: Quantum Mechanics for Scientists and Engineers Section 10.4 contains the complete mathematical details for solving the radial equation in the hydrogen atom problem. For this course, not all those details are required and they are consequently not all covered in the online lectures, so the additional detail, in particular on power series solutions in section Text reference: Quantum Mechanics for Scientists and Engineers Section 10.4, is optional for the student.
The hydrogen atom solutions
Radial equation solutions
Quantum mechanics for scientists and engineers David Miller
Radial equation solutions
Using a separation of the hydrogen atom wavefunctionsolutions into radial and angular parts
and rewriting the radial part using
we obtained the radial equation
where we know l is 0 or any positive integer
,U R r Y r
r rR r
22 2 2
2 2
12 4 2 H
o
d r l le r E rdr r r
Radial equation solutions
We now choose to write our energies in the form
where n for now is just an arbitrary real number We define a new distance unit
where the parameter is
2HRyEn
s r
2
2 22 Ho
Ena
Radial equation solutions
We therefore obtain an equation
Then we write
so we get
2
2 2
1 1 04
l ld nds s s
1 exp / 2ls s L s s
2
2 2 1 1 0d L dLs s l n l Lds ds
Radial equation solutions
The technique to solve this equation
is to propose a power series in sThe power series will go on forever
and hence the function will grow arbitrarilyunless it “terminates” at some finite power
which requires thatn is an integer, and
2
2 2 1 1 0d L dLs s l n l Lds ds
1n l
Radial equation solutions
The normalizable solutions of
then become the finite power seriesknown as the associated Laguerre polynomials
or equivalently
2
2 2 1 1 0d L dLs s l n l Lds ds
12 1
10
!1
1 ! 2 1 !
n lql q
n lq
n lL s s
n l q q l
0
!1
! ! !
pqj q
pq
p jL s s
p q j q q
Radial equation solutions
Now we can work back to construct the whole solutionIn our definition
we now insert the associated Laguerre polynomials
where Since our radial solution was
we now have
1 exp / 2ls s L s s
1 2 11 exp / 2l l
n ls s L s s
r rR r (2 / )os na r
1 2 11
1/ 2 exp / 2l lo n lR r na s s L s s
r
2 11 exp / 2l l
n ls L s s
Radial equation solutions - normalization
We formally introduce a normalization coefficient A so
The full normalization integral of the wavefunction
would be
but we have already normalized the spherical harmonicsso we are left with the radial normalization
2 11
1/ 2 exp / 2l lo n lR r na s s L s s
A
,U R r Y r
2
2 2
0 0 0
1 , sinr
R r Y r d d dr
Radial equation solutions - normalization
Radial normalization would be
We could show
so the normalized radial wavefunction becomes
2 2
0
1 R r r dr
22 2 1 21
0
2 !exp
1 !l l
n l
n n ls L s s s ds
n l
1/232 1
1
1 ! 2 2 2 exp2 !
ll
n lo o o o
n l r r rR r Ln n l na na na na
Hydrogen atom radial wavefunctions
We write the wavefunctions using the Bohr radius ao as the unit of radial distance
so we have a dimensionless radial distance
and we introduce the subscriptsn - the principal quantum number, andl - the angular momentum quantum number
to index the various functions Rn,l
/ or a
Radial wavefunctions - n = 1
Principal quantum numbern = 1
Angular momentum quantum number l = 0
0 5 10 15
0.5
1
1.5
2
Radius
R
1,0 2expR
1,0R
0 5 10 150.2
0.2
0.4
0.6
0.8
Radial wavefunctions - n = 2
l = 0
l = 1
Radius
R 2,02 2 exp / 2
4R
2,0R
2,1R
2,16 exp / 2
12R
0 5 10 150.1
0.1
0.2
0.3
0.4
Radial wavefunctions - n = 3
l = 0
l = 1
l = 2Radius
R
3,0
22 3 23 2 exp / 327 9
R
3,0R
3,1R
23,2
2 30 exp / 31215
R
3,16 24 exp / 3
81 3R
3,2R
1.2 MMZl
zx
1s
Hydrogen orbital probability density
n = 1l = 0m = 0
logarithmic intensity scale
x - zcross-section
at y = 0
1.2 MMZl
zx
Hydrogen orbital probability density
n = 2l = 0m = 0
logarithmic intensity scale
2s
x - zcross-section
at y = 0
1.2 MMZl
zx
Hydrogen orbital probability density
n = 3l = 0m = 0
logarithmic intensity scale
3s
x - zcross-section
at y = 0
1.5 MMZl
zx
Hydrogen orbital probability density
n = 3l = 1m = 0
logarithmic intensity scale
3p
x - zcross-section
at y = 0
Behavior of the complete hydrogen solutions
(i) The overall “size” of the wavefunctions becomes larger with larger n
(ii) The number of zeros in the wavefunction is n – 1The radial wavefunctions have n – l – 1 zeros
and the spherical harmonics have l nodal “circles” The radial wavefunctions appear to have an additional
zero at r = 0 for all , but this is already countedbecause the spherical harmonics have at least one nodal “circle” for all
which already gives a zero as in these cases1l
0r
1l
Behavior of the complete hydrogen solutions
In summary of the quantum numbers for the so-called principal quantum number
and We already deduced that l is a positive or zero integerWe also now know the eigenenergies
Given the possible values for n
Note the energy does not depend on l (or m)
1,2,3,n 1l n
2HRyEn
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