Download 1. Schrödinger`s Equation for the Hydrogen Atom

1. Schrödinger’s Equation for the Hydrogen Atom
We solved the Schrödinger equation for the case of a three-dimensional infinite
square well and found that for each dimension we introduce a quantum number.
In the case of the infinite square well the quantum numbers were independent
of one another - i.e., the value of one did not influence the value of another.
In addition, we found that the energy states are often degenerate when there is
a degree of symmetry in the system. We now want to look at the case of the
hydrogen atom. Here the potential energy of interaction between the electron
and the proton arising from a radial force depends only upon the distance r from
the proton to the electron, and not on the particular location of the electron (i.e.,
the particular value of θ or φ).
z
θ
r
y
φ
x
Polar Coordinate System used for the Hydrogen Atom
The Schrodinger equation
h̄2
−
2m
∂2
∂2
∂2
Ze2
+
+
ψ(r) + −k
ψ(r) = Eψ(r)
∂x2 ∂y 2 ∂z 2
r
can be written in terms of r, θ, and φ, by using the relations
x = r sin θ cos φ
y = r sin θ sin φ
z = r cos θ
The result, after some algebra, is:
h̄2
−
2m
1 ∂
∂
r2
2
r ∂r
∂r
1
h̄2
−
2
2mr sin θ
∂
∂
sin θ
∂θ
∂θ
1 ∂2
+
sin2 θ ∂φ2
ψ(r)+U (r)ψ(r) = Eψ(r)
Now this equation can be written in the form:



L̂2 
p̂2
− r − θ,φ2 ψ(r) + U(r)ψ(r) = Eψ(r)
 2m
2mr 
where
p̂2r
2
= h̄
1 ∂
∂
r2
2
r ∂r
∂r
is the square of the radial part of the momentum, and
L̂2θ,φ
2
= h̄
1
sin θ
∂
∂
sin θ
∂θ
∂θ
1 ∂2
+
sin2 θ ∂φ2
is square of the angular momentum Lθ,φ which is only a function of the angles θ
and φ.
To solve this equation, we assume that ψ(r) is separable so that it can be
written as the product of a function of the radius only with a function which is
only dependent upon θ and φ, i.e., we assume
ψ(r) = R(r)Y (θ, φ).
As shown below, Schrodinger’s equation in r, θ, and φ can then be separated
and written in terms of two equations, one dependent only upon the radius and
containing the electrostatic potential energy term, and the other dependent only
upon the angles θ and φ but which is independent of the force. We write
L̂2θ,φ
p̂2r
R(r)Y (θ, φ) −
R(r)Y (θ, φ) + U (r)R(r)Y (θ, φ) = ER(r)Y (θ, φ)
−
2m
2mr2
which simplifies to give
p̂2
R(r) 2
Y (θ, φ) − r R(r) −
L̂ Y (θ, φ) + U(r)R(r)Y (θ, φ) = ER(r)Y (θ, φ)
2m
2mr2 θ,φ
2
and, dividing by ψ(r) = R(r)Y (θ, φ), we obtain
1
1
p̂2
1
L̂2θ,φ Y (θ, φ) + U (r) = E.
− r R(r) −
2
R(r)
2m
2mr Y (θ, φ)
Multiplying this last equation by 2mr2 we obtain
r2 2
1
−p̂r R(r) −
L̂2 Y (θ, φ) = 2mr2 [E − U (r)]
R(r)
Y (θ, φ) θ,φ
which we can separate into terms which are functions only of the radius, and
terms dependent only on the angles:
r2 2
1
p̂r R(r) + 2mr2 [E − U (r)] = −
L̂2 Y (θ, φ)
R(r)
Y (θ, φ) θ,φ
Both sides of this last equation must be equal to some constant for the equation
to be valid for all r, θ, and φ. We will express this constant as h̄2 λ. This gives us
two equations,
r2 p̂2r R(r) + 2mr2 [E − U(r)] − h̄2 λ R(r) = 0
and
L̂2θ,φ Y (θ, φ) = −h̄2 λY (θ, φ)
This last equation is an eigenvalue equation for the angular momentum operator. Since there is no external torque in this system, the angular momentum
of the system must remain constant. This last equation simply states that the
square of the angular momentum operator acting on the eigenfunction of the angles is some constant times that function. The value of the constant will change
as the function Y (θ, φ) changes, but for a given function the angular momentum
is fixed! The value of the quantum number λ, then, is related to the total angular
momentum of the system. When the complete solution of the differential equation
is determined, we find that the values of λ are given by
λ = ( + 1)
where is an integer. This means that the total angular momentum of the system
is given by
|L| = h̄ ( + 1)
3
which is a little different from the form of the solution assumed by Bohr. Bohr
assumed that the angular momentum was given by
L = nh̄
where n was an integer. You will notice, however, that for large values of the
constant , these two expression give approximately the same result.
We now want to look in more detail at the the angular equation, which, when
fully expanded, can be written:
L̂2θ,φ Y
(θ, φ) = h̄
2
1
sin θ
∂
∂
sin θ
∂θ
∂θ
1 ∂2
+
Y (θ, φ) = −h̄2 λY (θ, φ)
sin2 θ ∂φ2
Just as was done above, this equation can be separated into two terms, each of
which depends upon a different angle. Writing the equation
∂
∂
sin θ
sin θ
∂θ
∂θ
∂2
+ 2 Y (θ, φ) = −λ Y (θ, φ) sin2 θ
∂φ
we can move all the partials with respect to θ to one side of the equation and all
the partials with respect to φ to the other side
∂
∂
sin θ
sin θ
∂θ
∂θ
∂2
+ λ sin θ Y (θ, φ) = − 2 Y (θ, φ)
∂φ
2
We now assume that the function Y (θ, φ) is separable so that Y (θ, φ) = Θ(θ)Φ(φ),
giving
∂
∂
sin θ
sin θ
∂θ
∂θ
or
2
+ λ sin θ Θ(θ)Φ(φ) = −
1
∂
∂
sin θ
sin θ
Θ(θ)
∂θ
∂θ
2
+ λ sin θ Θ(θ) = −
∂2
Θ(θ)Φ(φ)
∂φ2
1 ∂2
Φ(φ)
Φ(φ) ∂φ2
Again we see that each side of this equation must be equal to some constant which
we will designate as m2 giving the two equations
1
∂
∂
sin θ
sin θ
Θ(θ)
∂θ
∂θ
4
2
+ λ sin θ Θ(θ) = m2
and
1 ∂2
Φ(φ) = m2
Φ(φ) ∂φ2
The second of these two equations is trivial to solve. We obtain a set of
solutions of the form:
Φm (φ) = Ae±imφ
−
where m must be an integer for the function to be physical (i.e., the value of the
function must repeat for every complete cycle of the angle φ). The ± indicates
that the electron movement is equally likely in either the + direction of the −
direction (clockwise or conter-clockwise). But since it cannot move in both at
the same time, then we choose the + sign and allow m to take on values both
positive and negative. The fact that m is restricted on a physical basis, means
that the solutions of the Θ(θ) equation are influenced by this quantum number
and therefore that the values of λ which give reasonable results depend upon the
values of m.
One can show that the z component of the angular momentum is given by
Lz = −ih̄
so that
L2z Y (θ, φ) = −h̄2
∂
∂φ
∂2
Y (θ, φ)
∂φ2
which leads to the eigenvalue equation
L2z Φ(φ) = −h̄2 m2 Φ(φ)
We see, then, that the z component of the angular momentum is quantized and
can take on only certain values. This has some very interesting consequences
which we shall discuss later.
The Θ(θ) equation is not quite so simple. The solution to this equation is the
associated Legendre Polynomials. The normalized solutions for Y (θ, φ) are the
so-called spherical harmonics. The first few of these are given below:
1
4π
3
=
cos θ
4π
Y0,0 =
Y1,0
5
3
sin θ e±iφ
8π
5 =
3 cos2 θ − 1
16π
15
= ∓
sin θ cos θ e±iφ
8π
15
=
sin2 θ e±i2φ
32π
Y1,±1 =
Y2,0
Y2,±1
Y2,±2
and a representative set are plotted on the next page. The general formula for
the spherical harmonics is
(2 + 1) ( − |m|)! imφ
e Pl,m (cos θ)
Yl,m (θ, φ) = 4π ( + |m|)!
where = (−1)m for m ≥ 0 and = 1 for m ≤ 0, and where Pl,m (cos θ) are the
Legendre Polynomials. The values which the quantum number m can take on
must be integers to satisfy the condition that the wave function be periodic in
the angle φ. For the angular wave functions Yl,m (θ, φ) to be normalizable, we find
that the value of the quantum numbers m and must be related by the equation
− ≤ m ≤ +
Remember that the quantum number is related to the magnitude of the
angular momentum vector by the equation
|L| = h̄ ( + 1)
and m is related to the z-component of the angular momentum vector by the
equation
L2z Φ(φ) = −h̄2 m2 Φ(φ) → Lz = h̄m
where m can take on all integer values from − to +. In the case where = 0 (i.e.,
the angular momentum equal to zero) m is√also zero as you would expect. In the
case where = 1, however, we have |L| = h̄ 2 and Lz = h̄m where m = −1, 0, +1.
This means that the z-component of the angular momentum can never be as great
as the magnitude of the angular momentum! The angular momentum vector can
never be completely lined up along the z-axis! Now, the angle that the total
6
angular momentum vector makes with the z-axis can be determined using the
definitions and the diagram shown below.
z
Lz
θ
L
y
x
As can be seen,
cos θ =
or
Lz
L
m
cos θ = ( + 1)
When = 1, and m = 0 the angle is 90◦ . But when m = ±1 we obtain
cos θ = √12 ⇒ θ = 45◦ is the minimum value of θ. The angular momentum
vector precesses about the z-axis, maintaining the constant value of θ for a particular value of the quantum number m. You can see that as L becomes larger
(i.e., as becomes larger), the minimum value of the angle θ approaches zero.
Thus, for large quantum numbers, the angular momentum can be aligned with
the z-axis, as is the case in classical physics.
Continuing our discussion of the solution to the Hydrogen atom, one finds that
the solution to the radial equation, R(r), depends upon the quantum number λ
(and therefore ), and also upon another quantum number that arises from the
demand that the wave function be finite. This latter quantum number we will
designate as n. The first few solutions to the radial wave equation are:
7
2
R1,0 = e−r/ao
a3o
1
r
R2,0 =
1−
e−r/2ao
3
2ao
2ao
1
r −r/2ao
R2,1 = e
24a3o ao
Plots of the first few solutions are shown on the next page, and on the following
page are the plots of the probability of finding an electron at a particular value
of the radius (independent of the angles θ and φ). To obtain this latter function
one must integrate the radial wave function over all angles.
The general solution to the three-dimensional hydrogen atom is of the form
ψ(r) = Rn,l (r)Y,m (θ, φ)
where the values of the quantum numbers which are associated with realizable
wave functions are given by:
n = 0, 1, 2, 3, . . .
= 0, 1, 2, . . . , n − 1
− ≤ m ≤ +
Some representative diagrams are shown on the following pages which attempt to
illustrate the three-dimensional nature of the solutions.
8