Download Separation of internal and center-of

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CONTENTS
I. Separation of motion: Two particles
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A. Classical Treatment
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B. Quantum treatment
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I.
SEPARATION OF MOTION: TWO PARTICLES
Consider two particles, with coordinates defined in Fig. 1 We will assume the particles
y
1
r1
r
R
2
r2
x
FIG. 1. Coordinate system for two particles of mass m1 and m2 . For simplicity, we have suppressed
the z axis.
interact with a potential V (r) which depends only on the magnitude of the vector which
connects them. This is called a “central” force.
A.
Classical Treatment
The (scalar) kinetic energy of the two particles is
1
1
1
T (~r1 , ~r2 ) = m1~v12 + m2~v22 = m1
2
2
2
2
1
2
1
= m1~r˙1 + m2~r˙2
2
2
∂~r1
∂t
2
1
+ m2
2
∂~r2
∂t
2
(1)
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Now, let us define two new coordinates: the position of the center-of-mass
~ = m1 ~r1 + m2 ~r2
R
M
M
~ is the mass-weighted average of the positions of the two
where M = m1 + m2 (note that R
particles) and the separation between the two particles
~r = ~r2 − ~r1
~ and ~r, the positions of the two particles are
You can show that in terms of R
~ − m2 ~r
~r1 = R
M
and
~ + m1 ~r
~r2 = R
M
Thus, the vector position of the first particle is equal to the vector position of the centerof-mass minus a fraction of the separation vector ~r (and similarly for the vector position
of the second particle. Consequently, from Fig. 1 you see that the position of the center of
~ must like between the particles along the separation
mass (the arrow point of the vector R)
distance.
Differentiation with respect to time gives
~˙ − m2 ~r˙
~r˙1 = R
M
and, similarly for ~r˙2 . Thus, the kinetic energy is
h m i2 2
1 m2 ~
1 ~˙ 2 1
2
R · ~r
~r˙ − 2 m1
+ m1
T = m1 R
2
2
M
2
M
h m i2 2
1 ~˙ 2 1
1 m1 ~
1
+ m2 R
~r˙ + 2 m2
R · ~r
+ m2
2
2
M
2
M
3
The cross terms vanish, which leaves
2
2
2
2
1
~˙ + 1 m1 m2 + m2 m1 ~r˙ 2 = 1 M R
~˙ + 1 m1 m2 ~r˙ 2
T = (m1 + m2 )R
2
2
M2
2
2 M
2
1 ˙2
1 ~˙
= M R + µ~r
2
2
(2)
where µ ≡ m1 m2 /(m1 + m2 ) is the “reduced” mass.
Thus, since the potential depends only on |~r|, the Hamiltonian separates into a term
describing the kinetic energy of the center-of-mass, moving through space with mass M,
and the kinetic energy and potential energy of the relative motion of particle 2 with respect
to particle 1, with the kinetic energy proportional to the reduced mass.
B.
Quantum treatment
In quantum mechanics the kinetic energy operator is proportional to the Laplacian so
that (in atomic units)
T̂ (~r1 , ~r2 ) = −
1
1
∇2r1 −
∇2r2
2m1
2m2
By the chain rule
∂
∂ ∂X
∂ ∂x
=
+
∂x1
∂X ∂x1 ∂x ∂x1
=
∂
m1 ∂
−
M ∂X
∂x
(3)
By squaring this, we find
m21 ∂ 2
m1 ∂ 2
∂2
∂2
= 2
−2
+
∂x21
M ∂X 2
M ∂X∂x ∂x2
and, similarly
∂2
m2 ∂ 2
∂2
m22 ∂ 2
+
2
+
=
∂x22
M 2 ∂X 2
M ∂X∂x ∂x2
Entirely equivalent relations apply to the y and z components of the Laplacian operator.
~ and ~r, the kinetic energy operator becomes (here we show
Consequently, in terms of R
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explicitly only differentiation with respect to x)
1 − 1 ∂2
1
m1 + m2 ∂ 2
−
−
T̂ (~r1 , ~r2 ) = −
2
2
2M
∂X
M ∂X∂x 2
=−
1
1
+
m1 m2
∂2
+···
∂x2
1 ∂2
1 ∂2
−
+···
2M ∂X 2 2µ ∂x2
(4)
where the dots indicate the corresponding terms in y and z. Thus, in terpms of the coordi~ and ~r, the kinetic energy operator is
nates R
T̂ (~r1 , ~r2 ) = −
1 2
1 2
∇R −
∇
2M
2µ r
The total Hamiltonian is then
Ĥ(~r1 , ~r2 ) = −
1 2
1 2
∇R −
∇ + V (r)
2M
2µ r
~ and on
Because the Hamiltonian is separable into terms which depend separately on R
~ and a function of ~r,
~r, the wave function can be written as a product of a function of R
namely
~ ~r) = Φ(R)ψ(~
~
Ψ(R,
r)
The wave function for motion of the center of mass satisfies the equation
−
1 2 ~
~
∇ Φ(R) = Ecm Φ(R)
2M R
whereas the wave function for internal motion satisfies the equation
1 2
− ∇r + V (r) ψ(~r) = Eint ψ(~r)
2µ
The total energy is the sum of the internal energy Eint plus the energy associated with the
motion of the center of mass Ecm .