University of Connecticut
DigitalCommons@UConn
Chemistry Education Materials
Department of Chemistry
6-14-2006
Introduction to Atomic Units, Normalization and
Orthogonalization (Part 2 of a Series)
Carl W. David
University of Connecticut, [email protected]
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Introduction to Atomic Units, Normalization and Orthogonalization (Part 2 of a
Series)
C. W. David
Department of Chemistry
University of Connecticut
Storrs, Connecticut 06269-3060
(Dated: June 14, 2006)
I.
involves choosing forms which force probabilities of finding a particle somewhere in its domain, the values of the
normalization constants change according to the nature
of the coördinate system being employed. This means
that care must be taken to use a consistent “space” when
considering quantum chemistry manipulations.
Consider normalization of the 1s orbital,
ABSTRACT
Normalization in physical X, Y, and Z space, where
dimensions are measured in centimeters (for instance),
leads to different values for normalization constants than
when dimensionless spaces such as that employed in
atomic units are employed.
ψ1s (x, y, z) = ce−Ar
II.
in atomic units (where A is the atomic number of the
nucleus, we often set A = 1 for illustrative purposes).
We are interested in evaluating
The use of atomic units comes with a non-obvious
penalty worth mentioning. Where normalizing of orbitals
Z
∞
Z
∞
dx
−∞
Z
∞
dy
dz
−∞
2
ψ1s
Z
∞
=
−∞
Z
∞
dx
−∞
Z
∞
dz c2 e−2Ar
dy
−∞
2
− Ame
h̄2
Ψ1s (X, Y, Z) = Ce
R
III.
− aA
0
→ Ce
which works out to be centimeters. The normalization
integral for this wavefunction is
Z ∞
Z ∞
Z ∞
dX
dY
dZΨ21s → 1
∞
−∞
Z
∞
dX
−∞
∞
dY
−∞
2
R
2 −2 Ame
h̄2
dZC e
→1
−∞
where we are using the “real” form of wavefunctions (and
therefore avoiding complex conjugates). This value is
chosen to force the probability of finding the electron
somewhere we it is known that the electron inhabits this
particular orbital, should be certainty. We know that we
will choose the normalization constant, C, so that this
integral has the value 1.
Typeset by REVTEX
What are the units of C?
The integrand
C 2 e−2
Ame2
h̄2
R
dXdY dZ
is supposed to be the probability of finding the electron
within dX of X, i.e., from X to X + dX, within dY
of Y and within dZ of Z. Probabilities are dimension1
less, so C 2 must have the units cm
3 to offset the units of
dXdYdZ.
IV.
−∞
Z
UNITS
R
where the mass, “m”, is in grams, the electron’s charge
“e” is in stat-coulomb, and h̄ is in erg-sec, “R” is in cm
h̄2
and a0 is the Bohr radius whose value is me
2 and which
has units
erg 2 sec2
gram(dyne1/2 cm)2
−∞
(2)
−∞
Converting back to “real” units, we have
which is
Z
(1)
INTRODUCTION
DOING THE INTEGRAL
Oddly enough, in doing the integral, we usually convert
back to a form which looks suprisingly enough like the
original dimensionless form we started with (but it isn’t).
We write
Z ∞Z ∞Z ∞
Zme2
C 2 e−2 h̄2 R dXdY dZ
−∞
−∞
−∞
as
C2
Z
∞
−∞
Z
∞
−∞
Z
∞
−∞
e−γR dXdY dZ
2
2
= 2 aA0 . Now, we can convert to spherwhere γ = 2 Zme
h̄2
ical polar coördinates, and obtain
Z 2π
Z π
Z ∞
C2
dΦ
sin ΘdΘ
dRe−γR R2
0
0
which yields
∞
Z
2
dR (R2 )e−γR = 3
γ
0
0
which is almost Equation 3.
which is
4πC 2
∞
Z
dRR2 e−γR → 1
(3)
0
A.
Integral Review, trickery
We know that
Z
∞
0
Alternative Integration using Integration by
Parts
Alternatively, we could write
∞
Z
2 −γR
R e
dR e−γR
e−γR
→
−γ
e
−γR
→ 1
dR e
→
−γ 0
γ
Further, we can take the derivative of this last equation
with respect to γ i.e.,
o
nR
∞
d 0 dRe−γR = γ1
d(uv)
dR =
dR
Z
and we can do this again, i.e.,
o
nR
∞
d 0 dR (−R)e−γR = − γ12
vdu +
udv
∞ Z ∞
e−γR e−γR
−
2R
dR
−γ 0
−γ
0
Z
dR =
0
0
∞
R2 e−γR dR =
0
Z
∞
2R
0
e−γR
dR
γ
which can be integrated again by parts, this time defining
u=R, to obtain
dγ
R e
Z
Since the first term on the r.h.s vanishes, we have
1
dR (−R)e−γR = − 2
γ
2 −γR
d(uv) =
R2 e−γR dR = R2
0
∞
Z
∞
Z
which is
Z
Z
We then obtain
dγ
0
udv
du
dv
d(uv)
=v
+u
dR
dR
dR
and multiplying by dR and integrating we would have
−γR ∞
∞
dR =
where u = R2 and dv = e−γR dR, employing the standard
Z
Z
Z
0
so the definite integral
Z
B.
∞
e−γR
2
2R
dR =
γ
γ
and again, the first term vanishes, yielding
∞ Z ∞
e−γR −γR
R
+
e
dR
γ 0
0
V.
CONTINUING
Equation 3 now becomes
Z
0
∞
2
R2 e−γR dR = 2
γ
∞ e−γR 2
= 3
−γ 0
γ
4πC 2
2
2
= 4πC 2 3 → 1
3
γ
2A
a0
so that, solving for C 2 we have
3
again.
C2 =
2A
a0
8π
(4)
3
making the integral (Equation 5)
so that
C=
v
u 3
u 2A
t a0
8π
=
v
u 3
u A
t a0
π
h̄2
Ame2
3 Z
∞
Z
∞
Z
dx
∞
dy
−∞
−∞
dzC 2 e−2r = 1
−∞
Then we have employing Equation 4’s definition of C 2 :
VI.
NORMALIZATION IN ATOMIC UNITS
Z ∞
Z ∞
a 3 A 3 1 Z ∞
0
dx
dy
dze−2r = 1
A
a0
π −∞
−∞
−∞
We now return to the original question, which was,
what is the normalization constant (c) when using atomic
units? The normalization integral was
Z ∞
Z ∞
Z ∞
Ame2
dX
dY
dZC 2 e−2 h̄2 R = 1
(5)
−∞
−∞
or, cancelling,
∞
Z
Z
∞
dx
−∞
Z
∞
dy
−∞
−∞
dze−2r = π
−∞
and we now define the coördinate transformation
which would be, in spherical polar coördinates:
Ame2
A
x=
2 X = a X
h̄
0
Z
with similar equations for Y and Z (r =
Then
Ame2
R
h̄2
=
A
a0 R).
(and, where γ = 2 now)
4
and therfore
dx
Ame2
h̄2
=
A
a0
VII.
h̄2
Ame2
3
dxdydz =
∞
Z
Z
a 3
0
A
∞
dx
−∞
2
=1
23
a tautology.
dX
so, dXdYdZ becomes
dXdY dZ =
R2 dRe−2r = π
0
Ame2
A
dx =
dX = dX
a0
h̄2
dX =
∞
4π
Z
dXdY dZ
∞
dy
−∞
which would be
−∞
2
dz ψ1s
=
RETURNING TO THE BEGINNING
Finally, we have from Equation 2
Z
∞
Z
∞
dx
−∞
Z
∞
dy
−∞
dz c2 e−2Ar
(6)
−∞
which says that
4πc2
Z
∞
r2 dre−2Ar
0
which is
4πc2
2
(2A)3
r
and if we force this to be 1, we have
4πc2
1
=1
4A3
A3
c=
π
which shows that the normalization constant’s form depends on the initial choice of coördinate system.