Exercise 1: Evaluate the
integral I_{(0)} =
+∞∫0
e^{−γ rk} dr
and discuss the two cases of k = 1 (e.g., 1s Slater-type
orbital) and k = 2 (e.g., 1s Gaussian-type orbital).
By using the method of u-substitution,
[N5]
let u = γr^{k} so that
r =
(u/γ)^{1⁄k
} and dr =
(γ)^{−1⁄k}
(u)^{(1⁄k)−1}/k
du.
Now substitute into the integrand, replacing all forms of r:
Case k = 1: 1s-STO.
For positive integers n, the gamma function is related to the factorial
function: Γ(n) = (n
− 1)!, so that for k = 1 the gamma function is
Γ(2) = 1! = 1 and I_{(0)} =
1/γ. The same result can be obtained by direct integration, after
u-substitution (i.e. u = −
γ r, from which r =
− ^{u}⁄_{γ} and
dr = −
^{1}⁄_{γ} du):
The result can be proven by direct integration, after
u-substitution (i.e. u^{2}
= γ r^{2}, from which
r =
√ 1 /γu and
dr =
√ 1 /γ du):
I_{(0)} =
+∞∫0
e^{−r2} dr
=
√ 1 /γ+∞∫0
e^{−u2} du
(E1.4)
To derive the value for I_{(0)}, the following steps are used.
First, the value of I_{(0)} is squared. Second, the squared
value is rewritten as a double integral. Third, the double integral is
evaluated by transforming to polar coordinates. Fourth, the
I_{(0)} is explicitly solved for. The first two steps yield
I_{(0)}^{2} =
1/γ+∞∫0
e^{−x2} dx ×
+∞∫0
e^{−y2} dy I_{(0)}^{2}
=
1/γ+∞∫0+∞∫0
e^{−(x2 + y2)} dx
dy
(E1.5)
where it has been taken into account that the integration variable u
is silent, in the sense that it can be chosen freely and, if necessary,
replaced with a symbol more appropriate to the context.
The region which defines the first quadrant, is the region of integration for
the integral in eq. (E1.5).
The bivariate transformation
x = r cos θ,
y = r sin θ
will transform the integral problem from cartesian coordinates,
(x, y), to polar coordinates, (r, θ).
These new variables will range from 0 ≤ r
≤ ∞ and 0 ≤ θ
≤ ^{π}⁄_{2} for the first quadrant.
The Jacobian determinant of the transformation is
where the substitution u =
−r^{2}, du
= −2rdr, and
dr = −
1/
2 r
du has been applied to solve the radial integral.
Hence,
I_{(0)} =
1/2√π/γ
(E1.9)
as it was anticipated.
Solution. The solution
of this exercise is summarized by the following equation
I_{(0)} =
+∞∫0
e^{−γ rk} dr
=
(γ)^{−1⁄k}
Γ
(1 +
1/k)
=
{1/γ
for k = 1 (STO), /1/
2√π/γ
for k = 2 (GTO),
(E1.10)
Exercise 2: Evaluate the
integral I_{(n)}
=
+∞∫0r^{n} e^{−γ rk} dr
and discuss the two cases of k = 1 (e.g., Slater-type
orbital) and k = 2 (e.g., Gaussian-type orbital).
∂^{n}/
∂γ^{n}+∞∫0
e^{−γ rk} dr
=
+∞∫0∂^{n}/
∂γ^{n}
e^{−γ rk} dr
(k = 1, 2)
(E2.1)
The n-th derivative of I_{(0)} in eq. (E1.10) equals
the product of the gamma function Γ
(1 +
1/k) times the n-th derivative of
(γ)^{−1⁄k}
with respect to γ:
I_{(n) =
∂n
I(0)
/
∂γn
I(n)2
= (−1)n
1
/
k
(1 +
1
/
k
)
(2 +
1
/
k
) …
(n − 1 +
1
/
k
)
(γ)−1⁄k−n
Γ
(1 +
1
/
k
)
I(n)2
= (−1)n
1⋅(k + 1)⋅(2k +
1)⋅(3k + 1)⋅…⋅[(n−1)
k + 1]
/
kn γ
(nk+1)/k
Γ
(1 +
1
/
k
)
I(n)2
= (−1)n
[kn − (k − 1)]!(k)
/
kn+1 γ
(nk+1)/k
Γ
(
1
/
k
)
I(n)2
= (−1)n
1
/
k γ
(nk+1)/k
Γ
(n +
1
/
k
)
}
In the eqs (E2.5-7) the identities
Γ(n + 1) = n!,
Γ
(n +
1/2) =
(2n − 1)!!/
2^{n}√π, and
Γ
(n + 1/2) =
(n − 1)!!/
2^{n/2}√π
have been used.
Exercise 3: Evaluate the
normalization constant of a Slater-type orbital (STO, k = 1) and
a Gaussian-type orbital (GTO, k = 2),
expressed in spherical coordinates χ_{n,ℓ,m}(η,k; r,θ,φ)
= R_{n}(η,k; r)
Y_{ℓ,m}(θ,φ), where the
Y_{ℓ,m}(θ,φ) are
orthonormalized spherical harmonics, and R_{n}(η,k; r) =
N_{n}(η,k) r^{n−1}
e^{−η rk} is the
radial function, with N_{n}(η,k)
being the normalization constant to be determined.
The normalization constant of an atomic orbital χ(r)
can be computed by imposing the condition
∫
ℝ^{3}|χ(r)|^{2} d^{3}r
= 1
(E3.1)
which, for STOs and GTOs in spherical coordinates, can be written as
Exercise 4: Evaluate the
normalization constant N_{a,b,c}(α) of a
Gaussian-type orbital (GTO) expressed in cartesian coordinates χ_{a,b,c}(α;
x,y,z) = N_{a,b,c}(α)
x^{a}y^{b}z^{c}
e^{−αr2}, where a,
b, and c are integers, whose sum equals the azimuthal quantum
number ℓ = a + b +
c.
The normalization constant of an atomic orbital χ(r) is
computed by imposing the condition
∫
ℝ^{3}| χ_{a,b,c}(α; x,y,z)
|^{2} d^{3}r = 1
(E4.1)
Substituting χ(r) with its expression in Cartesian
coordinates and observing that r^{2} = x^{2} +
y^{2} + z^{2}, the previous
equation can be rearranged as follows:
Exercise 5: Evaluate the
normalization constant N_{a,b,c,k}(ζ) of a
Slater-type orbital (STO) expressed in cartesian coordinates χ_{a,b,c,k}(ζ;
x,y,z) = N_{a,b,c,k}(ζ)
x^{a}y^{b}z^{c}r^{k−1}
e^{−ζ r}, where a,
b, and c are integers, whose sum equals the azimuthal quantum
number ℓ = a + b +
c, and k is the difference between the principal
quantum number n and the azimuthal quantum number ℓ,
k = n −
ℓ = n − (a + b + c).
where Z is the atomic number. Use the variation method with the
Slater-type orbital χ_{n,ℓ,m}(ζ; r,θ,φ)
= R_{n}(ζ; r)
Y_{ℓ,m}(θ,φ)
as a trial function to show that the orbital energies are an upper bound to
the exact orbital energies (which are given by E_{n} = −
Z^{2}/
2 n^{2}). Compare the behavior of Slater-type orbitals against that of exact
hydrogenic orbitals, and find the relationship between hydrogenic and
Slater-type orbitals.
Left multiplying both sides of the Schrödinger equation by
⟨χ❘, yields the formal solution
E =⟨χ❘
−
1/2
∇^{2} −
Z/r❘χ⟩/
⟨χ❘χ⟩^{ }
(E6.1)
As shown in the statement of the Exercise, the Slater-type function,
χ_{n,ℓ,m}(ζ; r,θ,φ),
is the product of the orthonormalized real spherical harmonics,
Y_{ℓ,m}(θ,φ), with the radial function
R_{n}(ζ; r) =
N_{n}(ζ) r^{n−1}
e^{−ζ r}
(E6.2)
where N_{n}(ζ) is the normalization constant
(cf Exercise 3):
N_{n}(ζ) =
√
(2ζ)^{2n+1}/
(2n)!
(E6.3)
Substituting χ in eq (E6.1) with its explicit expression, yields
In the last equality of eq (E6.4), the normalization conditions of the functions
χ_{n,ℓ,m}(ζ; r,θ,φ),
R_{n}(ζ; r), and
Y_{ℓ,m}(θ,φ)
have been taken into account.
Substituting the Laplace operator ∇^{2} in eq. (E6.4) with
its expression in spherical coordinates yields
Differentiating eq (E6.9) with respect to ζ and equating to zero
yields
d E_{n}/
d ζ =
n ζ −
(2n − 1) Z/n (2n − 1) =
0 ⇒
ζ =
2n − 1/nZ
(E6.10)
so that the orbital energy has the expression
E_{n} = −
Z^{2}/
2 n^{2}
(2n − 1)
(E6.11)
which overestimates the exact value by a factor of (2n − 1)
and only in the case of n = 1 gives the exact ground state energy,
which is -0.5 hartree for the hydrogen atom (Z = 1).
where Y_{ℓ,m}(θ,φ) are orthonormalized real spherical harmonics and the R_{n,ℓ}(α; r)
are the eigenfunctions of the one-dimensional Schrödinger equation for the
radial distance r:
It's worth noting that the overall “size” of the hydrogenic
wavefunctions becomes larger with larger n.
The number of zeros in the wavefunction is n
− 1. The associated Laguerre polynomials in the radial
wavefunction have n − ℓ −
1 zeros, and the spherical harmonics have ℓ nodal
“circles”.
For every value of the principal quantum number n there is a degeneracy
equal to n^{2}, i.e. there are n^{2} different
atomic orbitals, all of which possess the same value for the energy,
E_{n}. These degenerate orbitals are characterized by the
n possible values of the azimuthal quantum number ℓ
(ℓ = 0, 1, …, n−1) and, for every value of
ℓ, by the (2ℓ + 1) possible values of the magnetic
quantum number m (m = −ℓ, …, −2,
−1, 0, 1, 2, …, ℓ).
The main difference between hydrogenic wavefunctions and Slater-type orbitals
is the presence of nodes in the former.
Unlike hydrogenic atomic orbitals, whose radial function depends on both the
principal and azimuthal quantum numbers, the radial function of Slater-type
orbitals depends only on the principal quantum number.
However, it is easy to show that the hydrogenic wavefunction can be written
as a linear combination of Slater-type orbitals.
In fact, eq (E6.13) can be rewritten as follows
Explicit expressions of the expansion coefficients
u_{ℓq}(n) are reported in Table E6.1 for the
lowest values of n. It's worth noting the special values
u_{0,0}(n) = 1 and
u_{n−1,0}(n)
= 1 for every value of n = 1, 2, ….
Table E6.1 -
Expansion coefficients of hydrogenic wavefunctions into Slater-type orbitals.
For each value of n, the functions
u_{ℓq}(n,α) of Eq (E6.20) define an
n×n upper triangular matrix U_{n}, whose
elements U_{n,ij}(i = ℓ + 1,j = ℓ + q + 1, ℓ = 0, 1, …,
n−1, and
q = 0, 1, …,
n−ℓ−1) are equal to 0 for
i > j, and have the expression of
eq (E6.20), U_{n,ij} ≡
U_{n;ℓ+1,ℓ+q+1} =
u_{ℓq}(n,α), for
i ≤ j.
Therefore, if y_{n}(α; r) = |
R_{n,0}(α; r),
R_{n,1}(α; r), …,
R_{n,n−1}(α; r)
|^{†} and x_{n}(α;
r) = | R_{1}(α; r),
R_{2}(α; r), …,
R_{n}(α; r) |^{†},
eq (E6.18) can be written in matrix notation as
y_{n}(α;
r) = U_{n}x_{n}(α; r).
Low order (n = 1, …, 5) hydrogen radial wavefunctions are compared
in Table E6.2 with the corresponding Slater-type orbitals with three different
exponent values (ζ = α,
(2n−1)α, Z). For every value of n, the
two functions coincide only if ℓ = n
− 1 (in addition to sharing the same exponent, α).
Table E6.2 -
Comparison of various low order radial wavefunctions of hydrogenic atoms
with the corresponding STO approximations.
√15 Z^{5}/
192r(4 − Z r +
1/
20Z^{2}r^{2})
e^{−Z r/4}
R_{4,2} ≡ R_{4d}
√5 Z^{7}/
3840r^{2}
(12 − Z r)
e^{−Z r/4}
R_{4,3} ≡ R_{4f}
√35 Z^{9}/
26880r^{3}
e^{−Z r/4}
The exercise is completed by the
graphs of Figure E6.1, illustrating the behavior of the radial wavefunctions
(amplitudes,[N0] R(r)) shown in Table E6.2, and by the graphs of
Figures E6.2 and E6.3, which show how the associated squared amplitudes,
|R(r)|^{2}, and radial densities,
ρ(r) = r^{2}
|R(r)|^{2}, vary as the radius increases.
The radial functions are computed over about 200 points, whose coordinates
along a line are expressed in atomic units of length. An ad hoc
fortran program provides a human-readable block-data file for each value of
n. In each file, the data are structured in (n + 4) columns,
the first of which contains the values of the radial coordinate, r.
The next n columns contain the values of the hydrogen radial
eigenfunctions for ℓ = 0, 1, 2, …, n−1.
The last three columns contain the STO values for
ζ =
Z/n, ζ =
(2n−1) Z/n, and ζ =
Z, respectively.
C ------------------------------------------------------------------
Program Hydrogen
C ------------------------------------------------------------------
C
C Compare the radial functions of hydrogenic and Slater-type
C orbitals
C
C ------------------------------------------------------------------
Implicit Double Precision (a-h,o-z)
Parameter (nmax=5)
Dimension u(nmax), y1(nmax)
C-tst Dimension y5(nmax)
Data Zero /0.0d0/, One /1.0d0/, Two /2.0d0/
Data RMax /15.0d0/, dr /0.075d0/
C
C Table E6.1
C
Do n = 1,nmax
Write(*,'(/,1x,79("-"),/," Table E6.1 for n = ",i2,/,
$ 1x,79("-"))') n
Do l = 0,n-1
Call STO_ULK (n,l,u)
If (l+1.eq.2) then
Write(*,'(10x," l = ",i2,15x,5f12.5)') l,(u(i),i=1,n-l)
else if (l+1.eq.3) then
Write(*,'(10x," l = ",i2,27x,5f12.5)') l,(u(i),i=1,n-l)
else if (l+1.eq.4) then
Write(*,'(10x," l = ",i2,39x,5f12.5)') l,(u(i),i=1,n-l)
else if (l+1.eq.5) then
Write(*,'(10x," l = ",i2,51x,5f12.5)') l,(u(i),i=1,n-l)
else
Write(*,'(10x," l = ",i2,3x,5f12.5)') l,(u(i),i=1,n-l)
end If
end Do
end Do
write(*,*)
C
C Profiles
C
iop = 1
Z = one ! atomic number
C
iu = 10
ju = 20
ku = 30
Do n = 1,nmax
iu = iu + 1
ju = ju + 1
ku = ku + 1
write(*,'(" Profiles of Hydrogenic and Slater-type",
$ " Orbitals for n =",i2," are written on unit",i3)') n, iu
zeta = z/float(n) ! orbital exponent
zetamod = zeta * float(2*n-1)
C
r = - dr
do while (r.lt.rmax)
r = r + dr
C
C Hydrogenic Atomic Orbitals
Do l = 0, n-1
y1(l+1) = HAO (n,l,z,r,iop)
end Do
C
C Slater-type Orbital, with zeta = Z/n
y2 = STO (n,zeta,r,iop)
C
C Slater-type Orbital, with zeta = Z*(2n-1)/n
y3 = STO (n,zetamod,r,iop)
C
C Slater-type Orbital, with zeta = Z
y4 = STO (n,z,r,iop)
C
C HAO as a linear combination of STOs
C (values to be tested against y1)
C-tst Do l = 0, n-1
C-tst Call STO_ULK (n,l,u)
C-tst y5(l+1) = 0.0d0
C-tst do k=1,n-l
C-tst y5(l+1) = y5(l+1) + u(k) * STO(l+k,zeta,r,iop)
C-tst end do
C-tst end Do
C
C Write out radial amplitude profiles
C
write(iu,'(14f12.8)') r, (y1(i),i=1,n), y2, y3, y4
C-tst$ , (y5(i),i=1,n)
C
C Write out squared radial amplitude profiles
C
write(ju,'(14f12.8)') r, (y1(i)**2,i=1,n), y2**2,
$ y3**2, y4**2
C
C Write out radial density profiles
C
r2 = r*r
Do l = 0, n-1
y1(l+1) = r2 * y1(l+1)**2
end Do
y2 = r2 * y2**2
y3 = r2 * y3**2
y4 = r2 * y4**2
write(ku,'(14f12.8)') r, (y1(i),i=1,n), y2, y3, y4
C
end do
end Do
End
C ------------------------------------------------------------------
Double Precision Function STO (n,z,r,iop)
C ------------------------------------------------------------------
C
C Return the radial part of a Slater-type orbital
C
C n ..... principal quantum number (0, 1, 2, ...)
C z ..... orbital exponent
C r ..... radial distance
C iop ... 0 unnormalized
C 1 normalized
C
C ------------------------------------------------------------------
Implicit Double Precision (a-h,o-z)
Data One /1.0d0/, Two /2.0d0/
C
STO = (r**(n-1)) * exp(-z*r)
C
C Normalization constant
C
if (iop.eq.1) then
nn = 2*n
STO = STO * sqrt( ((Two*z)**(nn+1)) / Fact(nn) )
end If
C
C
Return
End
C ------------------------------------------------------------------
Double Precision Function HAO (n,l,z,r,iop)
C ------------------------------------------------------------------
C
C Return the radial part of a hydrogenic atomic orbital
C
C n ..... principal quantum number (0, 1, 2, ...)
C l ..... azimuthal quantum number (0, 1, 2, ..., n-1)
C z ..... atomic number
C r ..... radial distance
C iop ... 0 unnormalized
C 1 normalized
C
C ------------------------------------------------------------------
Implicit Double Precision (a-h,o-z)
Data Zero /0.0d0/, One /1.0d0/, Two /2.0d0/
C
alpha = z / float(n)
ar = alpha*r
tar = two*ar
s = zero
do k = 0, n-l-1
s = s + (tar**(k+l)) * ((-one)**k)/
$ ( fact(n-l-k-1) * fact(l+l+k+1) * fact(k) )
end do
HAO = s * fact(n+l) * exp( -ar )
C
C Normalization constant
C
if (iop.eq.1) then
CN = ((two*alpha)**3) * fact(n-l-1) /
$ ( float(2*n) * fact(n+l) )
HAO = HAO * sqrt( CN )
end If
C
Return
End
C ------------------------------------------------------------------
Subroutine STO_ULK (n,l,u)
C ------------------------------------------------------------------
C
C Return the coefficients required to express the radial part of an
C hydrogenic wavefunction as a linear combination of the radial part
C of Slater-type orbitals
C
C n ... principal quantum number (0, 1, 2, ...)
C l ... azimuthal quantum number (0, 1, 2, ..., n-1)
C z ... orbital exponent
C u ... coefficients
C
C ------------------------------------------------------------------
Implicit Double Precision (a-h,o-z)
Data One /1.0d0/, Two /2.0d0/
Dimension u(*)
C
UA = float(2*n) * fact(n+l) * fact(n-l-1)
C
j = 0
Do i = 0, n-l-1
j = j + 1
UB = sqrt( UA * fact(2*l + 2*i + 2) )
UD = fact(n-l-i-1) * fact(2*l+i+1) * fact(i) * float(2*n)
u(j) = (-One)**i * UB/UD
end Do
C
Return
End
C ------------------------------------------------------------------
Double Precision Function Fact (N)
C ------------------------------------------------------------------
C
C The Factorial of a non-negative integer N, denoted by N!, is the
C product of all positive integers less than or equal to N.
C ------------------------------------------------------------------
Implicit Double Precision (A-H,O-Z)
Data One /1.0D0/
C
Fact = One
If (N.lt.1) return
Do I=1,N
Fact = Fact * Float(I)
end Do
C
Return
End
The last step is to use GRACE [LL1"Grace (plotting tool)", Wikipedia, The Free
Encyclopedia, LL2GRACE
website] to read the block-data files generated by the
fortran code and to arrange the data interactively until the most satisfactory
representation is obtained.
The graphs of Figure E6.1 show that each time the quantum number n
increases, an additional node is created in the radial wavefunction.
This is made more clear in the upper graph of Figure E6.2, which shows the
ns radial wavefunctions R_{n,s}(r) of the
hydrogen atom. They differ from the
ns orbitals ψ_{n,s}(r,θ,φ) by
a constant angular factor, the spherical harmonic Y_{00} =
1/√4π. Therefore the ns orbitals
are all spherically symmetrical and, like the radial functions, they have
n − 1 radial nodes (spherical shells on which the wavefunction
equals zero).
Each time the quantum number ℓ is increased (lower graph of Figure
E6.2), one of the spherical nodes disappears again. It is replaced by a planar
node that goes through the nucleus.
In the ground state of the hydrogen atom, n = 1, the radial wavefunction
is nodeless (i.e. all positive).
Its maximum is at r = 0, i.e. the point in space with the highest (joint)
probability density of finding the electron is actually inside the nucleus!
This is true for all the ns orbitals, whose maximum value is at the
origin of the radial coordinate. Therefore, only ℓ = 0 electrons
have a finite (joint) probability density at the nucleus.
At this point it is appropriate to spend a few words trying to overcome
some misunderstanding about the position of the electron in the hydrogen atom.
The statement that “the probability density of the electron's
position in a ns orbital is maximal at the origin” refers to the
Born rule, [LL5Born, M. “The
statistical interpretation of quantum mechanics”, Nobel lecture, December
11, 1954.] which postulates that the probability density of
finding a particle in some region is proportional to the square of the modulus
of the wave function, e.g.
This is a joint probability density, where the word “joint”
stresses the fact that it is linked to the occurrence of a triple of values,
one for each of the three variables r, θ, φ.
When a joint probability density is multiplied by an infinitesimal volume
element (which in polar spherical coordinates is dV =
r^{2} sinθ dr
dθ dφ) gives the probability that the values
of the three variables r, θ, φ fall in that
infinitesimal volume element, i.e. the probability of finding the electron
there. This probability is infinitesimal, even if the volume element dV
is positioned at the origin, where the probability density is maximal.
Integrating the probability over a finite volume V yields the
electronic charge contained in V and, if V is extended to all
space, the total charge that amounts to 1e is retrieved, so satisfying
the normalization condition.
Thus, the probability of finding the electron in an infinitesimal volume
element can be written as
ρ_{n,s}(r,θ,φ) r^{2}
sinθ dr dθ dφ =
|ψ_{n,s}(r,θ,φ)|^{2}r^{2} sinθ dr dθ dφ
(E6.22)
The use of spherical-polar coordinates is convenient because people are
used to think of an atom as some sort of fuzzy ball of electrons around a
small, central, heavy nucleus, and extend such a mental picture even to
bonded atoms in molecules and crystals. In this case, the ns
orbitals of the hydrogen atom are spherically-symmetrical (i.e. do not
depend on θ and φ), so that integration of eq. (E6.22)
over θ and φ gives 4π, the total solid
angle of a sphere:
ρ_{n,s}(r) dr =
|ψ_{n,s}(r)|^{2}r^{2}
drπ∫
0 sinθ dθ2π∫
0 dφ =
4π
|ψ_{n,s}(r)|^{2}r^{2}
dr
(E6.23)
Eq. (E6.23) introduces a “marginal” probability density,
namely the “radial” probability density:
Eq. (E6.24) can be generalized for all the permissible values of the
quantum numbers n and ℓ, thanks to the form of the wave
functions (ψ_{n,ℓ,m} or χ_{n,ℓ,m})
that are defined as the product of the orthonormalized real spherical harmonics,
Y_{ℓ,m}(θ,φ), with a radial function,
R_{n,ℓ}(α; r) or
R_{n}(ζ; r):
where the double integral expresses the normalization condition of
the real spherical harmonics and is equal to one.
Rather than considering the probability of finding the electron in one
particular small element of space, the radial density is associated to the
probability of finding the electron in a spherical shell of space of
infinitesimal thickness, 4π r^{2}
dr (cf eq. E6.23).[N4Volume of
a spherical shell of infinitesimal thickness]
The plot of radial probability passes through zero at r = 0 since
the surface area, 4π r^{2},
the surface area of a sphere of zero radius is zero. As the radius of the
sphere is increased, the surface area of the spherical shells increases.
However, it increases more rapidly with increasing r than the joint
electron probability density decreases, so that the radial probability has
a maximum at r_{max} = a_{0}, which is the Bohr
radius for the n = 1 orbit (cf first graph in Figure E6.3). Thus more of
the electronic charge is present at a distance a_{0}, out from
the nucleus than at any other value of r.
Where is the electron in a hydrogen atom?
http://spiff.rit.edu/classes/phys314/lectures/hyd_probs/hyd_probs.html
Most probable point for finding an electron in the 1s orbital of a Hydrogen atom
https://chemistry.stackexchange.com/questions/84726/most-probable-point-for-finding-an-electron-in-the-1s-orbital-of-a-hydrogen-atom
http://www.falstad.com/
Exercise 7: find the average
distance of the electron from the nucleus in each state of the hydrogen atom.
The average distance of the electron from the nucleus is found by weighting
each possible distance with the probability that the electron will be found
at that distance
<r>_{n.ℓ} =
+∞∫
0r ρ_{n,ℓ}(r) dr
(E7.1)
where ρ_{n,ℓ}(r)
= r^{2}
|R_{n,ℓ}(r)|^{2}
is the radial probability density. Eq. (E7.1) can be easily
generalized to compute the m-th moment of the radial distribution
<r^{m}>_{n.ℓ} =
+∞∫
0r^{m} ρ_{n,ℓ}(r) dr
(E7.2)
Using eqs. (E6.13) and (E6.14), eq. (E7.2) becomes
Introducing the scaled distance s =
2αr, from which r^{m+2} =
s^{m+2}/
(2α)^{m+2}
and dr =
ds/
2α, and rewritting
the associated Laguerre polynomials
L^{2ℓ+1}_{n−ℓ−1}(2αr) of eq. (E6.15)
as
where the identity α =
Z/n has been used.
For ℓ = n − 1, only the term with q = 1,
i.e. t_{1}^{2} (2ℓ + 2 + m)! =
(2n + m)!, survives, so that you have
The standard deviation, i.e. the square root of the variance, of the
electron's distribution in the radial direction,
σ_{r;n,ℓ}, has the following expression:
The summation involves only the first term for ℓ = n − 1,
which allows to derive
<r^{m}>_{10},
<r^{m}>_{21},
<r^{m}>_{32},
<r^{m}>_{43}, etc. For ℓ =
n − 2, the summation extends to the first two terms, allowing to
recover
<r^{m}>_{20},
<r^{m}>_{31},
<r^{m}>_{42}, and so on for any pair
(n,ℓ).
Using Slater-type orbitals. If eq. (E7.2) is
written in terms of STOs, it yields
where the last equality is justified by eq. (E2.5). Eq. (E7.16)
can be rewritten explicitly for cases where m = 1 or 2:
<r>_{n} =
2n + 1/
2ζ
(E7.17)
<r^{2}>_{n} =
(2n + 1) (2n + 2)/
4 ζ^{2}
(E7.18)
The standard deviation, i.e. the square root of the variance, of the
electron's distribution in the radial direction,
σ_{r;n}, has the following expression:
From eqs (E7.17) and (E7.19) it follows that the relative error on the
average distance of the electron from the nucleus is
σ_{r;n}/<r>_{n} =
1/√
2n + 1 , which for n = 1 is equal to 57.7%
The numerical values of first and second moments with the associated standard
deviation, calculated by means of eqs (E7.13-15) and (E7.17-19), are
compared in Table E7.1.
Table E7.1 -
First and second moments of the electron's distribution in the radial
direction, and associated standard deviation for hydrogenic orbitals and
Slater-type orbitals (ζ = Z).
n,ℓ
exact hydrogenic orbitals
Slater-type orbitals
Z <r>_{n,ℓ}
Z^{2} <r^{2}>_{n,ℓ}
Zσ_{r;n,ℓ}
ζ <r>_{n}
ζ^{2} <r^{2}>_{n}
ζσ_{r;n}
1,0
1.5
3
0.87
1.5
3
0.87
2,0
6
42
2.45
2.5
7.5
1.12
2,1
5
30
2.24
3,0
13.5
207
4.97
3.5
14
1.32
3,1
12.5
180
4.87
3,2
10.5
126
3.97
4,0
24
648
8.49
4.5
22.5
1.50
4,1
23
600
8.43
4,2
21
504
7.94
4,3
18
360
6.00
Exercise 8: compare the most
probable distances of the electron from the proton in the hydrogen 1s, 2s and
2p states with the corresponding average radius.
The most probable distance of the electron from the nucleus is the value of
r which maximizes the radial probability density
ρ_{n,ℓ}(r)
= r^{2}
|R_{n,ℓ}(r)|^{2}.
Since ρ_{n,ℓ}(r) is largest where
rR_{n,ℓ}(r) reaches its maximum, we can look
for the most probable r_{max} distance by setting
d [rR_{n,ℓ}(r)]/
dr = 0
(E8.1)
using the functions R_{n,ℓ}(r) from Teble E6.2.
For the 1s state of hydrogen, the condition for a maximum is
d [rR_{1s}(r)]/
dr =
d/
dr(
2 √Z^{3}
e^{−Z r})
=
2 √Z^{3}
e^{−Z r} (1 − Z r) = 0
⇒
r_{max} =
1/Z
(E8.2)
The average or "expectation value" of the radius for the electron in the ground
state of hydrogen was obtained in the Exercise 7 and is <r> =
3/
2Z. It may seem a bit surprising that the average value of r
is 1.5 times the most probable value. The extended tail of the probability
density accounts for the average being greater than the most probable value.
For the 2s state of hydrogen, the condition for a maximum is
d [rR_{2s}(r)]/
dr =
d/
dr(√2 Z^{3}/4
(2r − Z r^{2}) e^{−Z r/2}) =
√2 Z^{3}/8
(Z^{2}r^{2} − 6 Z r + 4)
e^{−Z r/2} = 0
⇒
r_{max} =
3 ± √5/Z
(E8.3)
The two possible solutions indicate the presence of two humps in the radial
probability density for the 2s orbital (cf the upper-right graph of Figure
E6.3).
For the 2p state of hydrogen, a similar analysis yields,
d [rR_{2s}(r)]/
dr =
d/
dr(√6 Z^{5}/
12r^{2} e^{−Z r/2}) =
√6 Z^{5}/
24
(4 r − Z r^{2}) e^{−Z r/2} = 0
(E8.4)
with the obvious roots r_{min} = 0 and r_{max} =
4/Z.
Thus, the most probable distances of the electron from the nucleus for the
1s and 2p states are the same predicted by the simple Bohr model for the
radius of the first and second orbits. As in the case of the ground state,
the most probable value of r in the 2s and 2p states is lower than
the average value.
As usual, it is worth to compare the results for the hydrogenic orbitals
with those that can be obtained using Slater-type orbitals:
with the obvious roots r_{min} = 0 and r_{max} =
n/ζ
(cf the graph on the right in Figure E6.4).
Again the value of r_{max} is lower than the average value
and the following relationship holds:
r_{max,n} =
2 n/
2 n + 1 <r>_{n}
(E8.6)
Exercise: Find the moment generating function of an exponential random
variable r with parameter ζ, e.g. the radial density function
of the gound state of the hydrogen atom
(click to view/hide details).
If the radial density function of the ground state of
the hydrogen atom is ρ(r) =
γ^{3}/2r^{2} e^{−γ r}, where
γ = 2 ζ, the moment generating function is by
definition:
M_{r}(s) =
E(e^{s r}) =
+∞∫0e^{s r} ρ(r) dr =
γ^{3}/2+∞∫0e^{−(γ−s)r}r^{2} dr =
γ^{3}/
(γ − s)^{3}
provided that s <
γ. Note that M_{r}(s)❘_{s=0}
= 1 is the normalization condition of the radial density
function ρ(r). It's now easy to verify that the n-th
derivative of M_{r}(s) with respect to s
(n times),
d^{n}/
d s^{n}M_{r}(s) =
d^{n}/
d s^{n}
[γ (γ − s)^{−1}]^{3}
=
n!/
2 (γ − s)^{n}
calculated at s = 0, equals the n-th moment of ρ(r):
d^{n}/
d s^{n}M_{r}(s)❘_{s=0}
=
n!/
2 γ^{n}.
Indeed, the n-th moment of ρ(r) is by definition:
⟨r^{n}⟩ ≡
E(r^{n}) =
+∞∫0r^{n} ρ(r) dr =
γ^{3}/2+∞∫0e^{−γ r}r^{n+2} dr =
n!/
2 γ^{n}.
Exercise: Use an 1s-GTO, as a trial function, to find an upper bound to
the exact ground state energy of the hydrogen atom
(click to view/hide details).
An 1s-GTO has the form
❘Φ⟩ =
(2α/π)^{3⁄4}
e^{−α r2}
(t)
and satisfies the normalization condition,
⟨Φ❘Φ⟩ =
1, i.e.
Exercise: Write the wave function files (.wfn) for the hydrogen atom
using both Slater and Gaussian basis sets
(click to view/hide details).
1) Slater basis set (a single 1s orbital, according to eq.16):
Name H Run Type SinglePoint Method ROHF Basis Set Slater
SLATER 1 MOL ORBITALS 1 PRIMITIVES 1 NUCLEI
H (CENTRE 1) 0.00000000 0.00000000 0.00000000 CHARGE = 1.0
CENTRE ASSIGNMENTS 1
TYPE ASSIGNMENTS 1
EXPONENTS 1.0000000E+00
MO 1 OCC NO = 1.00000000 ORB. ENERGY = -0.5
0.56418958E+00
END DATA
THE SCF ENERGY = -0.500000000000 THE VIRIAL(-V/T)= 2.00000000
2) Gaussian basis set (a single 1s orbital, according to eq.D?):
Name H Run Type SinglePoint Method ROHF Basis Set STO-1G
GAUSSIAN 1 MOL ORBITALS 1 PRIMITIVES 1 NUCLEI
H (CENTRE 1) 0.00000000 0.00000000 0.00000000 CHARGE = 1.0
CENTRE ASSIGNMENTS 1
TYPE ASSIGNMENTS 1
EXPONENTS 0.2829425E+00
MO 1 OCC NO = 1.00000000 ORB. ENERGY = -0.42441369
0.27649271E+00
END DATA
THE SCF ENERGY = -0.424413690000 THE VIRIAL(-V/T)= 2.00000000
References
“Exact Gaussian Expansions of Slater-Type Atomic
Orbitals” Gomes, A. S. P.; Custodio, R. J. Comput.
Chem.2002, 23, 1007-1012.
Szabo, A.; Ostlund, N. S. “Modern Quantum Chemistry
− Introduction to Advanced Electronic Structure Theory”
Macmillan Publishing Co., Inc., New York, 1982, p. 33.
ISBN 0-02-949710-8.
“A Concise Handbook of Mathematics, Physics, and
Engineering Sciences”; Polyanin, A. D., Chernoutsan, A. I., Eds.;
Taylor & Francis Group: New York, 2011.
A. V. Arecchi, T. Messadi, and R. J. Koshel, “Field Guide
to Illumination”, SPIE Press, Bellingham, WA, 2007, p. 2.
ISBN: 9780819467683.
Leonard I. Schiff, “Quantum Mechanics”.
McGraw-Hill, 1968, Chapter 16, pp. 88-99. ISBN-13: 9780070856431.
ISBN-10: 0070856435.
Paul A. Tipler; Gene Mosca; “Physics for Scientists and
Engineers: Extended Version”. W. H. Freeman, 2003,
Section 36-4, p. 1181. ISBN-13: 9780716743897. ISBN-10: 0716743892.
David A. B. Miller, “Quantum Mechanics for Scientists and
Engineers”. Cambridge University Press, 2008, Chapter 10,
pp. 259-278. ISBN-13: 9780521897839. ISBN-10: 0521897831.
Milton Abramowitz and Irene A. Stegun, eds.
“Handbook of Mathematical Functions with Formulas, Graphs, and
Mathematical Tables”. United States Government Printing Office,
1972, § 22.3, p. 775. ISBN-13: 9780318117300.
ISBN-10: 0318117304.
“Some Integrals of the Products of Laguerre
Polynomials” Lee, P.-A.; Ong, S.-H.; Srivastava, H. M.
Intern. J. Computer Math.2001, 78, 303-321,
equation 18. DOI: 10.1080/00207160108805112
Einstein, A.; Infeld, L. “The Evolution of
Physics” Cambridge University Press, 1971, p. 106,
ISBN 0-521-09687-1.
Internet
Archive: accessed February 24, 2019.
Notes
Probability amplitude.
In classical physics a wave describe "the motion of something which is not
matter, but energy propagated through matter".[last1Albert Einstein, Leopold Infeld, “The Evolution
of Physics” Cambridge University Press, 1971, p. 106.]
Not all waves require a medium, as electromagnetic waves and gravitational waves
can travel through a vacuum. Waves are characterized by properties like the
wavelength, the frequency, and the amplitude. The intensity of light or sound
is found to be proportional to the amplitude of the associated wave.
“When Schrödinger first discovered the correct laws of quantum
mechanics, he wrote an equation which described the amplitude to find a
particle in various places. This equation was very similar to the equations
that were already known to classical physicists − equations that they
had used in describing the motion of air in a sound wave, the transmission
of light, and so on.”[LL4“The
Feynman Lectures on Physics”, Online Edition, Vol. III, Ch. 3:
“Probability Amplitudes”]
Then Max Born postulated that the probability of finding a particle in
some region is proportional to the square of the modulus of the wave function
(Born rule).[LL5Born, M.
“The statistical interpretation of quantum mechanics”, Nobel
lecture, December 11, 1954.] So the term (probability) amplitude
can be used for the wave function.
Volume element in spherical coordinates.
Working in spherical coordinates, the diagram on the left enable us to
calculate the infinitesimal volume element as
dV = dr ⋅ r dθ ⋅
r sinθ dφ =
r^{2} sinθ dr dθ
dφ
The volume of a solid is always given by the integration of the infinitesimal
volume element dV over the entire solid. We cover a sphere of radius
R if we let r ∈ [0,R], θ ∈
[0,π], and φ ∈ [0,2π]. The volume of the
sphere is then given by
V =
∫sphere dV =
R∫0r^{2} drπ∫0
sinθ dθ2π∫0
dφ V =
[1/3r^{3}]^{R}_{0} ⋅
[−cosθ]^{π}_{0} ⋅
[φ]^{2π}_{0}
V =
4 π R^{3}/3
Surface area element in spherical coordinates.
The diagram of the previous note [N1] also enables us
to calculate the infinitesimal area elements when we integrate over only
two of the spherical coordinates. When we integrate over the surface of
a sphere, we only vary θ and φ by dθ
and dφ, respectively, and we do not vary r at all. The
corresponding area element is given in the diagram of the previous note as
dA = r^{2} sinθ dθ
dφ
The surface area of a sphere of radius R is obtained by direct
integration over the entire sphere, letting θ ∈
[0,π] and φ ∈ [0,2π] while using the
spherical coordinate area element R^{2} sinθ
dθ dφ,
A =
∫sphere dA =
R^{2}π∫0
sinθ dθ2π∫0
dφ A =
R^{2} ⋅
[−cosθ]^{π}_{0} ⋅
[φ]^{2π}_{0}
A =
4 π R^{2}
Solid angle. A solid angle is the 3 dimensional analog of
an ordinary angle. It is related to the surface area of a sphere in the same
way an ordinary angle is related to the circumference of a circle.
A plane angle, α, is made up of the lines that from two points A
and B meet at a vertex V: it is defined by the arc length (red
curve in the figure on the left) of a circle subtended by the lines and by the
radius R of that circle (dotted red lines),[4A. V. Arecchi, T. Messadi, and R. J. Koshel “Field Guide to Illumination”
SPIE Press, Bellingham, WA, 2007, p. 2.] as shown on the left. The
dimensionless unit of plane angle is the radian. The value in radians of a plane
angle α is the ratio
α =
ℓ/R
of the length ℓ of a circular arc to its radius R. It follows
that the plane angle subtended by the full circle measures 2π radians.
Si definisce angolo solido ciascuna delle due
regioni in cui viene suddiviso lo spazio dalla superficie formata dalle
semirette passanti per uno stesso punto (detto vertice dell'angolo solido) e
per i punti di una curva chiusa semplice tracciata su una superficie non
contenente il vertice. L'unità di misura dell'angolo solido Ω è lo
steradiante.
La misura in steradianti dell'angolo solido Ω è definita dal rapporto
Ω = A/R^{2} tra l'area A della porzione di
superficie sferica di raggio R vista sotto l'angolo Ω e il
quadrato del raggio, ed è indipendente dal particolare valore del raggio scelto.
L'angolo solido sotteso da una superficie generica rispetto a un punto P
è dunque equivalente a quello sotteso dalla proiezione della stessa superficie
su una sfera di raggio qualsiasi centrata in P.
Dalla precedente definizione consegue che l'angolo solido sotteso dall'intera
superficie sferica misura 4π. Per avere la misura in gradi quadrati si
moltiplica il valore in steradianti per (180/π)^{2}, ovvero per
3282,8 (circa). Quindi tutta la sfera corrisponde a circa 41253 gradi quadrati.
A solid angle is made up from all the lines that from a closed curve meet at
a vertex: it is defined by the surface area A of a sphere subtended by
the lines and by the radius R of that sphere,[4A. V. Arecchi, T. Messadi, and R. J. Koshel “Field Guide to Illumination”
SPIE Press, Bellingham, WA, 2007, p. 2.] as shown in the Figure on
the right. The dimensionless unit of solid angle is the steradian. The measure
in steradians of a solid angle Ω is the area A on the surface of
a sphere of radius R divided by the radius squared,
Ω =
A/R^{2},
and is independent of the particular value of the chosen radius. Because the
area A of the entire spherical surface is 4πr^{2},
the definition implies that a sphere subtends 4π steradians at its
center. By the same argument, the maximum solid angle that can be subtended
at any point is 4π steradians.
Concisely, the solid angle Ω subtended by a surface S is defined
as the surface area Ω of a unit sphere covered by the surface’s
projection onto the sphere.[L4Weisstein, E. W. "Solid Angle," From MathWorld - A
Wolfram Web Resource.] That’s a very complicated way of saying
the following: You take a surface, like the bright red circle in the figure on
the right. Then you project the edge of the circle (but in general it could be
any closed curve) to the center of a sphere of any radius R. The
projection intersects the sphere and forms a surface area A. Then you
calculate the surface area of your projection and divide the result by the
radius squared. That's it.
The infinitesimal element of a solid angle Ω is given by dΩ =
dA/R^{2}, where dA is the surface area element in spherical coordinates
defined in [N2Surface area element in
spherical coordinates.], so that
Volume of a spherical shell of infinitesimal thickness.
Consider two spheres, both centered in the origin: the inner with radius
r and the outer with radius r + dr. To compute the
volume of the spherical shell between their two surfaces, proceed as follows:
where, in the last equality, the terms dr^{n} with
n > 1 have been neglected.
In a different way, using differentiation of volume V =
4/3π r^{3}
w.r.t. radius, you get area (rate of change of volume is area)
where P_{ℓ}^{m}(cosθ) is the associate Legendre polynomial of
degree ℓ and order m ( ℓ ≥ 0 and
−ℓ≤ m ≤ ℓ). Spherical harmonics
are single-valued, smooth (infinitely differentiable), complex functions of two
variables, θ and φ, indexed by two integers,
ℓ and m, and form a complete set of orthonormal functions:
This normalization is used in quantum mechanics because it ensures that
probability is normalized, i.e. π∫0
|Y_{ℓ}^{m}(Ω)|^{2}
dΩ = 1, with
dΩ = sinθ
dθ and normalization factor:
N_{ℓm} =
√
2ℓ + 1/
4 π
⋅
(ℓ − |m|)!/
(ℓ + |m|)!
(N7.3)
The real forms of the spherical harmonics, also known as tesseral
harmonics, are obtained by taking the linear combinations [Y_{ℓ}^{m}(θ,φ) ± Y_{ℓ}^{−m}(θ,φ)]/√2.
They are defined as follows
Y_{ℓm}(θ,φ) =
M_{ℓm}U_{ℓm}(θ,φ)
(N7.4)
where
U_{ℓm}(θ,φ)
= P_{ℓ}^{|m|}(cosθ)
{cos(|m|φ),
m ≥ 0
/
sin(|m|φ),
m < 0
(N7.5)
are the un-normalized tesseral harmonics, and M_{ℓm} is
the normalization factor
“Again an idea of
Einstein's gave me the lead. He had tried to make the duality of
particles - light quanta or photons - and waves comprehensible by
interpreting the square of the optical wave amplitudes as probability
density for the occurrence of photons. This concept could at once be
carried over to the ψ-function: |ψ|^{2}
ought to represent the probability density for electrons (or other
particles).”