Download Separation of the hydrogen atom Schrödinger equation

Schrödinger equation for the hydrogen atom
Date: January 17, 2014
We first treat the classical case of two particles interacting through a gravitational or electrostatic
force. We write the Cartesian coordinates of the two particles as x1 = x1 , y1 , z1 and x2 = x2 , y2 , z2 ,
respectively, and their masses as m1 and m2 . Two vectors (and hence six numbers) are required to
specify the co-ordinates of the two bodies at any instant.
Classically, the motion of each particle is determined by solving the corresponding equation of motion,
i.e.
1
F(r)
m1
1
F(r)
x¨2 = −
m2
x¨1 = −
(1)
(2)
where r = x2 −x1 is the relative position vector. These appear to be two coupled second order differential
equations because the relative position depends upon x1 and x2 ). Coupled differential equations require
some effort to solve. The problem can be simplified if we appreciate that no external forces act on the
centre of mass of the system and hence moves with a constant velocity. The six numbers which specify
the co-ordinates of the two particles can be replaced with three numbers specifying the co-ordinates of
the centre of mass, R, and three more numbers specifying the relative positions of the particles, r. All
of the interesting dynamics resides in the behaviour of the vector r.
Multiplying equation 1 with m1 and adding it to equation 2 multiplied with m2 gives
m1 x¨1 + m2 x¨2 = 0,
(3)
which is a re-expression of momentum conservation. We look for a point R which does not accelerate,
i.e. the centre of mass satisfies the equation
(m1 + m2 )R̈ = 0
(4)
These two equations have the solution
R=
m1 x1 + m2 x2
m1 + m22
From equation 1 and 2 we also get
r̈ =
1
1
+
m1
m2
F(r)
(5)
or
F(r) = µr̈
where
µ=
(6)
m1 m2
m1 + m2
is called the reduced mass. This way of writing the equation of motion makes explicit the fact that the
motion of this two body system is mathematically equivalent to the motion of a single particle of mass
µ under the action of the force F. That is, the problem reduces to one whose mathematical analysis is
exactly the same as the analysis of the motion of a single particle. Do not forget that there are really
two particles and that we are solving for their relative position.
The hydrogen atom is also a two particle problem. The two particles interact through the Coulomb
attraction of the electrical charges +e of the nucleus and −e of the electron. The potential energy is
e2
in which r is the distance between the electron (m2 ) and the nucleus (m1 .
− 4π
0r
This general procedure outlined above can be directly extended to the hydrogen atom. We have
the center of mass (m1 + m2 ) undergoing force free translational motion and the reduced mass (µ)
experiencing a force
dV
e2
=−
F(r) = −
,
(7)
4π0 r2
dr
which implies that
Z
V =−
−
e2
e2
dr
=
−
4π0 r2
4π0 r
(8)
The total energy is then
E = TM + Tµ + V,
where the three terms are the kinetic energy of the center of mass, the kinetic of the reduced mass,
and the potential energy respectively. Each of these three energies are replaced by the corresponding
operators to get the Hamiltonian of the hydrogen atom.
Ĥ = −
h̄2 2
h̄2 2
e2
∇M −
∇µ −
2M
2µ
4π0 r
All that is left is to use the appropriate form of ∇. Here it is best to use the Cartesian coordinates for
the COM and spherical coordinates for the reduced mass.
We also outline an alternative approach starting from the Schrödinger equation. The wave equation
for the hydrogen atom has the form
2
2
1
∂ ψT
∂ 2 ψT
∂ 2 ψT
1
∂ ψT
∂ 2 ψT
∂ 2 ψT
2
+
+
+
+
+
+ 2 (ET − V ) ψT = 0
m1
∂x21
∂y12
∂z12
m2
∂x22
∂y22
∂z22
h̄
in which
V =−
e2
p
4π0 (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2
This equation can be immediately separated into two, one which represents the translational motion
of the center of mass and the other the relative motion of the electron with respect to the nucleus. This
separation can be accomplished in a general case when the potential energy is a function of the relative
positions of the two particles. We will restrict ourselves to the special case of the hydrogen atom.
To effect the separation, we introduce the new variables x, y,, and z, which are the Cartesian coordinates of the center of mass of the system, and r, θ, and φ, the polar coordinates of the second particle
relative to the first. These coordinates are related to the Cartesian coordinates of the two particles by
the equations
m1 x1 + m2 x2
x=
m1 + m2
m1 y1 + m2 y2
y=
m1 + m2
m1 z1 + m2 z2
z=
m1 + m2
r sin θ cos φ = x2 − x1
r sin θ sin φ = y2 − y1
r cos θ = z2 − z1
Introducing these new independent variables in the “usual” way, we get
2
∂ 2 ψT
∂ 2 ψT
h̄2
∂ ψT
+
+
ĤψT = −
2(m1 + m2 ) ∂x2
∂y 2
∂z 2
2 h̄2 1 ∂
1
∂
∂ψT
1
∂ ψT
2 ∂ψT
−
r
+
sin
θ
+
2µ r2 ∂r
∂r
r2 sin θ ∂θ
∂θ
∂φ2
r2 sin2 θ
e2
−
ψT
4π0 r
In this equation the symbol µ has been introduced to represent the quantity
m1 m2
1
1
1
µ=
or =
+
;
m1 + m + 2
µ
m1
m2
which is the reduced mass of the system. Notice that the quantity in the first set of parentheses is
the Laplacian of ψT in the Cartesian coordinates and the quantity in the braces is Laplacian in polar
coordinates.