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The Hydrogen Atom
Michael Fowler 11/22/06
Factoring Out the Center of Mass Motion
The hydrogen atom consists of two particles, the proton and the electron, interacting via the
Coulomb potential
, where as usual
()
2
12
/
Vr r
e r
−=
GG
rrr
=
−
G
G
. Writing the masses of the two
particles as
Schrödinger’s equation for the atom is:
1
,
mm
2
22
2
1
2
,,
e
rr
E rr
r
ψψ
⎛⎞
−∇
−
∇
−
=
⎜⎟
⎝⎠
==
.
G
G
But
are
not
the most natural position variables for describing this system: since the
potential depends only on the relative position, a better choice is
,
,
rR
G
G
defined by:
11
2 2
,
mr mr
rrr R
+
=−
=
+
G
G
G
GGG
so
R
G
is the center of mass of the system. It is convenient at the same time to denote the total
mass by
,
M
=+
and the reduced mass by
.
m
=
+
Transforming in straightforward fashion to the variables ,
G
G
Schrödinger’s equation becomes
2
Rr
e
.
R
rE
R
r
Mm
r
−
∇
−
=
G
G
G
G
Writing the wave function
( ) ( ) ()
,
R
r
ψ
=Ψ
G
G
G
G
we can split the equation into two:
() ()
() () ()
2
2
2
2
2
2
RR
RE
R
M
Vr
r E r
m
Ψ=
Ψ
+
=
G
=
G
=
G
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and the total system energy is
Note that the motion of the center of mass is (of
course) just that of a free particle, having a trivial plane wave solution.
From now on, we shall
only be concerned with the
relative
motion of the particles.
Since the proton is far heavier than
the electron, we will almost always ignore the difference between the electron mass and the
reduced mass, but it should be noted that the difference is easily detectable spectroscopically: for
example, the lines shift if the proton is replaced by a deuteron (heavy hydrogen).
.
Rr
EE E
=+
We’re ready to write Schrödinger’s equation for the hydrogen atom, dropping the
r
suffixes in
the second equation above, and writing out
2
∇
G
explicitly in spherical coordinates:
2
22
2
2
2
2
11
1
sin
.
2s
i
n
s
i
n
e
rE
mr r
r
r
r
r
ψψ
ψ
θ
θθ
θϕ
⎛⎞
∂∂
∂
∂
∂
⎛
⎞
−+
+
−
⎜⎟
⎜
⎟
∂
∂
∂
⎝⎠
⎝
⎠
=
=
Factoring Out the Angular Dependence: the Radial Equation
Since the potential is spherically symmetric, the Hamiltonian
H
commutes with the angular
momentum operators
2
,
z
L
L
so we can construct a common set of eigenkets of the three
operators
2
,,
z
.
H
LL
The angular dependence of these eigenkets is therefore that of the
’s,
so the solutions must be of the form
m
l
Y
( ) ( ) ( )
, .
m
Elm
Elm
l
rR
r
Y
θφ
=
Now, notice that in the Schrödinger equation above, the angular part of
2
∇
G
is
exactly
the
differential operator
2
/2
2
L
mr
, so operating on
( ) ( ) ( )
,
m
Elm
Elm
l
r
Y
=
it will give
.
The spherical harmonic
can then be cancelled from the two sides of the
equation leaving:
()
2
1/2
ll
m
r
+
=
2
m
l
Y
( )
() ()
2
1
1
2
El
El
El
dd
e
r
R
r
E
m r dr
dr
r
r
+
−−
−
=
=
R
r
it now being apparent that
R
(
r
) cannot depend on
m
.
The radial derivatives simplify if one factors out 1/
r
from the function
R
, writing
( )
El
ur
r
=
and temporarily suppressing the
E
and
l
to reduce clutter.
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 Fall '07
 MichaelFowler
 Center Of Mass, Mass

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