In quantum mechanics free particles are described by wave functions which satisfy a wave equation. So, there is a waveparticle duality with a relationship between particle properties (energy, E, and momentum, p) and wave properties (frequency, ν, and wavelength, λ) given by Plank's law (E = h ν) and de Broglie equation (λ = h p ), both of which involve the Planck's constant, h.
The wave function which determines the motion of a single electron is called orbital, χ(r), and depends on the position vector r. If the oneelectron wave function is for an atomic system, it is called an atomic orbital (AO).
The oneelectron wave function, χ(r), is interpreted as a “probability amplitude” for the electron. The square modulus of the wave function is interpreted as the “probability density” to find an electron:
ρ(r) = χ(r)^{2}  (1) 
Quantum mechanics only tells us the probability for some quantum events to occur. The probability to find an electron over all the threedimensional space is equal to 1, and it is given by the integral of the probability density over ℝ^{3}
∫ ℝ^{3} ρ(r) d^{3}r = +∞ ∫ 0 χ(r)^{2} d^{3}r = 1  (2) 
In other words, the previous equation provides the number of electrons and the electronic charge in ℝ^{3}. The product ρ(r) d^{3}r gives the probability of finding the electron in an infinitesimal volume element d^{3}r (or dV). If the components of r are the cartesian coordinates x, y, and z, then the volume element is dV = dx dy dz.
Though most atoms, when they are free, are not spherical, still it may be useful to think of an atom as some sort of fuzzy ball of electrons around a small, central, heavy nucleus. And it easy to extend such a mental picture even to bonded atoms in molecules and crystals. For this reason, it is more convenient to express the electron density in spherical coordinates, r, θ and φ, which are related to the cartesian coordinates x, y and z by the transformation

The expression of the density function under the transformation of the
random vector variable r is discussed in
Section 4.7 of
the chapter on "Charge Distributions"“Transformation of the Random
Vector Variable”
PAMoC Manual: Charge Distributions, Section
4.7, Equations (4.7.4) and (4.7.10).. The conclusion is that,
passing from the reference system of Cartesian axes x, y and
z to the reference system of curvilinear coordinates r,
θ and φ, the value of the probability density must be
multiplied by the determinant of the Jacobian matrix of the transformation of
coordinates, which in the case of spherical coordinates is r^{2} sin θ:
ρ(r,θ,φ) =
ρ(x,y,z) det

∂ r
∂
r
∂ r
∂
θ
∂ r
∂
φ

= ρ(x,y,z) det
= ρ(x,y,z) r^{2} sinθ  (4) 
In general an orbital has the structure
χ_{n,ℓ,m}(r) = R_{n}(r) Y_{ℓ,m}(θ,φ)  (5) 
where R_{n}(r) are radial functions and
Y_{ℓ,m}(θ,φ) are spherical
harmonics [L1(a) Wikipedia contributors,
"Spherical harmonics,"
Wikipedia, The Free Encyclopedia.
(b)
Citizendium contributors, "Spherical harmonics."
Citizendium,
The Citizens' Compendium.
(c) Weisstein, E. W. "Spherical
Harmonic,"
From MathWorld  A Wolfram Web Resource.,
L2Wikipedia contributors, "Table of spherical
harmonics,"
Wikipedia, The Free Encyclopedia., N1Real Spherical Harmonics.] of degree ℓ
and order m. Spherical harmonics can be represented using spherical
coordinates (which are convenient when computing integrals) or polynomials
(as is commonly done when evaluating them).
The integer n = 1, 2, … is the principal quantum number, the
integer ℓ = 0, 1, … n − 1 is the azimuthal
quantum number, and the integer m = −ℓ, −ℓ
+ 1, …, ℓ − 1, ℓ is the magnetic quantum number.
By virtue of eq. (4) combined with eq. (5), the electron density function
in the system of curvilinear coordinates r, θ and
φ is given by
ρ_{n,ℓ,m}(r,θ,φ) =
 χ_{n,ℓ,m}(r,θ,φ)
^{2} r^{2} sinθ
= [ r^{2} R_{n}(r)^{2} ] [ sinθ Y_{ℓ,m}(θ,φ)^{2} ] = ρ_{n}(r) ρ_{ℓ,m}(θ,φ)  (6) 
This is the joint probability
density“Joint and Marginal Probability Density Function”
PAMoC User's Manual which is related to the probability
ρ_{n,ℓ,m}(r,θ,φ) dr dθ dφ of finding the electron in an
infinitesimal volume element, which in spherical coordinatess, is given by
dV = r^{2} sinθ
dr dθ dφ.[N2Volume element in spherical coordinates.]
The joint probability density is factorized into the product of two
marginal densities,
ρ_{n}(r) and
ρ_{ℓ,m}(θ,φ) .
The marginal density
function“Joint and Marginal Probability Density Function”
PAMoC User's Manual
ρ_{ℓ,m}(θ,φ) is defined by
ρ_{ℓ,m}(θ,φ) =
Y_{ℓ,m}(θ,φ)^{2}
sinθ
 (7) 
and is related to the probability ρ_{ℓ,m}(θ,φ) dθ dφ of finding the electron on a surface element of the unit sphere,[N3Surface area element in spherical coordinates.] i.e. the element of solid angle dΩ = sinθ dθ dφ.[N4Solid angle element in spherical coordinates.]
The radial electron
density“Joint and Marginal Probability Density
Function”
PAMoC User's Manual
ρ_{n}(r), also referred to as the radial
distribution, is defined by
ρ_{n}(r) =
r^{2} R_{n}(r)^{2}
 (8) 
and is related to the probability ρ_{n}(r) dr that a measurement of the electron's position yields a value anywhere in a spherical shell of infinitesimal thickness at a distance r from the nucleus as shown in the following figure for the radial distribution with n = 1.[N5Volume of a spherical shell of infinitesimal thickness.]
What the radial probability distribution shows is that the electron cannot
be sucked into the nucleus because ρ_{n}(0) = 0.
Hence, as we shrink the radial shell into the nucleus, the probability of
finding the electron in that shell goes to 0.[L3Mark E. Tuckerman
“The Schrödinger equation
for the hydrogen atom and hydrogenlike cations”
20111026.]
The density functions in eqs (79) are normalized to 1, meaning that the probability to find the electron in ℝ^{3} as well as the number of electrons and the electronic charge in ℝ^{3} are equal to 1.
+∞ ∫ r=0 π ∫ θ=0 2π ∫ φ=0 ρ_{n,ℓ,m}(r,θ,φ) dr dθ dφ = +∞ ∫ r=0 π ∫ θ=0 2π ∫ φ=0  χ_{n,ℓ,m}(r,θ,φ) ^{2} r^{2} sinθ dr dθ dφ = 1  (9) 
π ∫ θ=0 2π ∫ φ=0 ρ_{ℓ,m}(θ,φ) dθ dφ = π ∫ θ=0 2π ∫ φ=0 Y_{ℓ,m}(θ,φ)^{2} sinθ dθ dφ = 1  (10) 
+∞ ∫ r=0 ρ_{n}(r) dr = +∞ ∫ r=0 r^{2} R_{n}(r)^{2} dr = 1  (11) 
The normalization property of the radial density function, eq. (11), is strictly related to the radial cumulative distribution function D_{n}(r), which is defined as the probability to find the electron in the interval [0,r] and has the following expression
D_{n}(r) = r ∫ 0 ρ_{n}(r') dr' = r ∫ 0 r'^{2} R_{n}(r')^{2} dr'  (12) 
(see section 4.2 of the chapter on “Charge Distributions” for more details). It satisfies the relationships D(0) = 0, lim_{r→+∞} D(r) = 1, and dD(r) dr = ρ(r). In addition, the probability to find the electron in any interval [a,b] (∀ 0 ≤ a < b ∈ ℝ) is given by the difference D(b) − D(a). The joint cumulative distribution functions D_{n,ℓ,m}(r,θ,φ) and D_{ℓ,m}(θ,φ) are defined in a similar way.
Spherical STO  Spherical GTO  
Orbital  χ_{n,ℓ,m}(ζ; r,θ,φ) = R_{n}(ζ; r) Y_{ℓ,m}(θ,φ)  χ_{n,ℓ,m}(α; r,θ,φ) = R_{n}(α; r) Y_{ℓ,m}(θ,φ)  (D1) 
Radial function  R_{n}(ζ; r) = N_{n}(ζ) r^{n−1} e^{−ζr}  R_{n}(α; r) = N_{n}(α) r^{n−1} e^{−αr2}  (D2) 
Normalizing constant 
N_{n}(ζ) = [ (2ζ)^{2n+1} (2n)! ]^{1⁄2}  N_{n}(α) = [ 2^{4n+3} α^{2n+1} [(2n − 1)!!]^{2} π ]^{1⁄4}  (D3) 
Cartesian STO  Cartesian GTO  
Orbital  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^{−ζr}  χ_{a,b,c}(α; x,y,z) = N_{a,b,c}(α) x^{a} y^{b} z^{c} e^{−αr2}  (D4) 
Normalizing constant 
N_{a,b,c,k}(ζ) = [ (2ζ)^{2k+1} (2k)! (2a + 2b + 2c + 1)!! 4π (2a − 1)!! (2b − 1)!! (2c − 1)!! ]^{1⁄2}  N_{a,b,c}(α) = [ 2^{2(a + b + c) + 3⁄2} α^{a + b + c + 3⁄2} π^{3⁄2} (2a − 1)!! (2b − 1)!! (2c − 1)!! ]^{1⁄2}  (D5) 
where n, ℓ, and m are the wellknown quantum
numbers, in addition k = n − (a + b +
c) and a, b, and c are integers.
The sum a + b + c = ℓ is usually referred
to as s, p, d, etc., even though they generally contain components of lower
angular momentum.
The functions Y_{ℓ,m}(θ,φ) are
orthonormalized real spherical harmonics. Their expressions can be found in
reference [L2Wikipedia contributors, "Table of
spherical harmonics,"
Wikipedia, The Free Encyclopedia.].
Derivation of the normalization constants can be found on the tutorial pages of this manual (Exercise 1, Exercise 2, Exercise 3, Exercise 4, Exercise 5).
The Schrödinger equation (in atomic units) for the
hydrogen atom is[2Szabo, A.; Ostlund, N. S.
“Modern Quantum Chemistry  Introduction to Advanced
Electronic Structure Theory”
Macmillan Publishing Co., Inc.,
New York, 1982, p. 33.]
(− 1 2 ∇^{2} − Z r ) ❘Φ⟩ = E ❘Φ⟩  (13) 
where Z is the atomic number and is equal to the nuclear charge. The same equation also applies to hydrogenlike (or hydrogenic) ions, which are constituted by any atomic nucleus with a single electron, so that they are isoelectronic with hydrogen. These ions carry the positive charge (Z − 1) e, where Z is the atomic number of the atom. Left multiplying both sides of eq. (13) by ⟨Φ❘, yields the formal solution
E_{1s} = ⟨ Φ❘− 1 2 ∇^{2} − Z r ❘Φ ⟩ ⟨Φ❘Φ⟩  (14) 
If we use the 1sSTO ❘Φ⟩ ≡ χ_{1,0,0}(r) = R_{1}(r) Y_{0,0}(θ,φ) as a trial function for the ground state energy of the hydrogen atom, eq. (14) can be rewritten as
E_{1s} = ⟨ R_{1}(r)❘− 1 2 ∇^{2} − Z r ❘R_{1}(r) ⟩ ⟨ Y_{0,0}(θ,φ) ❘ Y_{0,0}(θ,φ) ⟩ ⟨Φ❘Φ⟩  (15) 
Both the radial part, R_{1}(r), and the angular part, Y_{0,0}(θ,φ), of the 1sSTO χ_{1,0,0}(r) as well as the 1sSTO itself are normalized separately, i.e.
⟨Y_{0,0}(θ,φ)❘Y_{0,0}(θ,φ)⟩ =
1
4 π
π
∫
0
sinθ dθ
2π
∫
0
dφ
= 1 4 π ⋅ [−cosθ ]^{π}_{0} ⋅ [φ ]^{2π}_{0} = 1 4 π ⋅ 2 ⋅ 2 π = 1  (16) 
where the spherical harmonic of degree ℓ = 0 and order
m = 0 is the constant
Y_{00} =
1
2
√
1
π
,
[L2Wikipedia contributors, "Table of spherical
harmonics,"
Wikipedia, The Free Encyclopedia.]
⟨R_{1}(r)❘R_{1}(r)⟩ = 4 ζ^{3} +∞ ∫ 0 e^{−2ζ r} r^{2} dr = 4 ζ^{3} ⋅ 1 4 ζ^{3} = 1  (17) 
where the radial part of the 1sSTO has the expression R_{1}(r) = 2 √ζ^{3} e^{−ζ r} (cf the table in the “details” section above), and
⟨Φ❘Φ⟩ ≡ ⟨χ_{1,0,0}(r)❘χ_{1,0,0}(r)⟩ = ⟨R_{1}(r)❘R_{1}(r)⟩ ⟨Y_{0,0}(θ,φ)❘Y_{0,0}(θ,φ)⟩ = 1  (18) 
where the 1sSTO has the espression χ_{1,0,0}(r) = √ ζ^{3} π e^{−ζ r}.
Substituting the Laplace operator ∇^{2} in eq. (15) with its expression in spherical coordinates, and remembering the normalization conditions (1618), yields
E_{1s} =
⟨
R_{1}(r)❘−
1
2
d^{2}
d
r^{2} −
1
r
d
d
r
−
Z
r
❘R_{1}(r)
⟩
= ⟨ R_{1}(r)❘− ζ^{2} 2 − 1 r (Z − ζ) ❘R_{1}(r) ⟩ = − ζ^{2} 2 ⟨R_{1}(r)❘R_{1}(r)⟩ + (ζ − Z) ⟨ R_{1}(r)❘ 1 r ❘R_{1}(r) ⟩ = − ζ^{2} 2 + 4 ζ^{3} (ζ − Z) +∞ ∫ 0 r e^{−2ζ r} dr = − ζ^{2} 2 + 4 ζ^{3} (ζ − Z) 1 (2 ζ)^{2} = − ζ^{2} 2 + ζ (ζ − Z)  (19) 
In deriving eqs (17) and (19) use of the integral
+∞
∫
0
r^{n} e^{−γ r}
dr =
n!
γ^{n+1}
has been made [3“A
Concise Handbook of Mathematics, Physics, and Engineering Sciences”
Polyanin, A. D., Chernoutsan, A. I., Eds.
Taylor & Francis Group: New York, 2011.] (cf the
Exercise 2 to see how the integral can be
worked out). The optimal value of the Slater exponent ζ is
determined by calculating the energy derivative with respect to ζ
and equating to zero, i.e.
d E_{1s} d ζ = ζ − Z = 0 ⇒ ζ = Z 
so that eq. (19) yields
E_{1s} = − Z^{2} 2  (20) 
The radial density function of the ground state of the hydrogen atom is given by eq.(8) with n = 1, i.e.
ρ_{1s}(r) = r^{2}
R(r)^{2} =
4 ζ^{3} r^{2}
e^{− 2 ζ r}
 (21) 
ψ_{n,ℓ,m}(α; r,θ,φ) = R_{n,ℓ}(α; r) Y_{ℓ,m}(θ,φ) 
R_{n,ℓ}(α; r) = N_{n,ℓ}(α) ( 2Z r n ) ^{ℓ } L^{2ℓ+1}_{n−ℓ−1} ( 2Z r n ) exp ( − Z r n ) 
N_{n,ℓ}(α) = √ (n − ℓ − 1)! 2n (n + ℓ)! ( 2Z n ) ^{3} 
χ_{n,ℓ,m}(ζ; r,θ,φ) = R_{n}(ζ; r) Y_{ℓ,m}(θ,φ) 
R_{n}(ζ; r) = N_{n}(ζ) r^{n−1} e^{−ζ r} 
R(s) = s^{ℓ} L(s) exp(−s/2) 
L^{j}_{p}(s) = p ∑ q=0 (−1)^{q} ( p + j p − q ) 1 q! s^{q} = p ∑ q=0 (−1)^{q} (p + j)! (p − q)! (j + q)! q! s^{q} 
[N_{n,ℓ}]^{2} +∞ ∫ 0 [R_{n,ℓ}(r)]^{2} r^{2} dr ≡ [N_{n,ℓ}]^{2} ( n 2Z ) ^{3} +∞ ∫ 0 [R_{n,ℓ}(s)]^{2} s^{2} ds = 1 
+∞ ∫ 0 [R_{n,ℓ}(s)]^{2} s^{2} ds = +∞ ∫ 0 s^{2ℓ} [L^{2ℓ+1}_{n−ℓ−1}(s) ]^{2} e^{−s} s^{2} ds = 2n (n + ℓ)! (n − ℓ − 1)! 
N_{nℓ} = √ (n − ℓ − 1)! 2n (n + ℓ)! ( 2Z n ) ^{3} 
M_{r}(s) = E(e^{s r}) = +∞ ∫ 0 e^{s r} ρ(r) dr = γ^{3} 2 +∞ ∫ 0 e^{−(γ−s)r} r^{2} dr = γ^{3} (γ − s)^{3} 
d^{n} d s^{n} M_{r}(s) = d^{n} d s^{n} [γ (γ − s)^{−1}]^{3} = n! 2 (γ − s)^{n} 
d^{n} d s^{n} M_{r}(s)❘_{s=0} = n! 2 γ^{n}. 
⟨r^{n}⟩ ≡ E(r^{n}) = +∞ ∫ 0 r^{n} ρ(r) dr = γ^{3} 2 +∞ ∫ 0 e^{−γ r} r^{n+2} dr = n! 2 γ^{n}. 
❘Φ⟩ = ( 2α π )^{3⁄4} e^{−α r2}  (t) 
⟨Φ❘Φ⟩ =
(
2α
π
)^{3⁄2}
+∞
∫
0
e^{−2α r2}
r^{2} dr
π
∫
0
sinθ dθ
2π
∫
0
dφ
= ( 2α π )^{3⁄2} ⋅ π^{½} 4 (2 α)^{3/2} ⋅ 4 π = 1  (u) 
E =
⟨
Φ❘−
1
2
d^{2}
d
r^{2} −
1
r
d
d
r
−
Z
r
❘Φ
⟩
=
⟨
Φ❘− 2
α^{2} r^{2} + 3 α −
Z
r
❘Φ
⟩
= 4 π ( 2α π )^{3⁄2} ( − 2 α^{2} +∞ ∫ 0 e^{−2α r2} r^{4} dr + 3 α +∞ ∫ 0 e^{−2α r2} r^{2} dr − Z +∞ ∫ 0 r e^{−2α r2} dr ) = 4 π ( 2α π )^{3⁄2} [ 3 16 ( π 2 α )^{1⁄2} − Z 4 α ] = 1 2 ( 3 α − 8 Z √ α 2 π )  (v) 
3 − 4 Z √ 2 π α = 0 ⇒ α = 8 Z^{2} 9 π = 0.28294246 Z^{2}  (w) 
( 2α π )^{3⁄4} = 8 ( Z 3 π )^{3⁄2} = 0.27649271 Z^{3/2}  (x) 
ρ_{1s}(r) =
64
27
(
Z
π
)^{3}
r^{2} exp
(
−
16 Z^{2}
9 π
r^{2}
)
 (y) 
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
Name H Run Type SinglePoint Method ROHF Basis Set STO1G
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
Y_{ℓ}^{m}(θ,φ) = (−1)^{m} N_{ℓm} P_{ℓ}^{m}(cosθ) e^{imφ}  (N1.1) 
2π ∫ 0 dφ π ∫ 0 Y_{ℓ}^{m}(θ,φ) Y_{ℓ'}^{m'*}(θ,φ) sinθ dθ = δ_{ℓℓ'} δ_{mm'}  (N1.2) 
N_{ℓm} = √ 2ℓ + 1 4 π ⋅ (ℓ − m)! (ℓ + m)!  (N1.3) 
Y_{ℓm}(θ,φ) = M_{ℓm} U_{ℓm}(θ,φ)  (N1.4) 
U_{ℓm}(θ,φ) = P_{ℓ}^{m}(cosθ) { cos(mφ), m ≥ 0 sin(mφ), m < 0  (N1.5) 
M_{ℓm} = N_{ℓm} √ 2 − δ_{m,0} = √ 2ℓ + 1 4 π ⋅ (2 − δ_{m,0}) ⋅ (ℓ − m)! (ℓ + m)!  (N1.6) 
Working in spherical coordinates, the diagram on the left enable us to
calculate the infinitesimal volume element as

dA = r^{2} sinθ dθ dφ 
A =
∫
sphere dA =
R^{2}
π
∫
0
sinθ dθ
2π
∫
0
dφ
= R^{2} ⋅ [−cosθ ]^{π}_{0} ⋅ [φ ]^{2π}_{0} = 4 π R^{2} 
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),[8A. V. Arecchi, T. Messadi, and R. J. Koshel “Field Guide to Illumination” SPIE Press, Bellingham, WA, 2007, p. 2., L4"Solid Angle". Optipedia: SPIE Press books opened for your reference. Excerpt from "Field Guide to Illumination" by Angelo V. Arecchi, Tahar Messadi, R. John Koshel.] 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

8A. V. Arecchi, T. Messadi, and R. J. Koshel “Field Guide to Illumination” SPIE Press, Bellingham, WA, 2007, p. 2., L4"Solid Angle". Optipedia: SPIE Press books opened for your reference. Excerpt from "Field Guide to Illumination" by Angelo V. Arecchi,Tahar Messadi, R. John Koshel.] 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,
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.[L5(a) Weisstein, E. W. "Solid Angle," From MathWorld  A Wolfram Web Resource. (b) Wikipedia contributors, "Solid angle," Wikipedia, The Free Encyclopedia.] href="#L4">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. 
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,[
dΩ = sinθ dθ dφ 
Ω = ∫ Ω dΩ = π ∫ 0 sinθ dθ 2π ∫ 0 dφ = [−cosθ ]^{π}_{0} ⋅ [φ ]^{2π}_{0} = 4 π 
dV  = 4 3 π (r + dr)^{3} − 4 3 π r^{3}  
= 4 3 π (r^{3} + 3r^{2}dr + 3r dr^{2} + dr^{3} − r^{3}  (N5.1)  
= 4 3 π (3r^{2}dr + 3r dr^{2} + dr^{3})  
≈ 4 π r^{2}dr 
d V d r = 4 π r^{2} ⇒ d V = 4 π r^{2} d r  (N5.2) 