Download Consider two point particles of mass m1 and m2 with position

Consider two point particles of mass m1 and m2 with position vectors ~r1 and ~r2 and
momenta p~1 and p~2 respectively interacting via a central potential. The energy is given by
H =
p~1 · p~1
p~2 · p~2
+
+ V (|~r1 − ~r2 |) .
2m1
2m2
We will consider this problem classically and then quantize.
~ ≡ [ m1~r1 + m2~r2 ]/M where M = m1 + m2
We define the center-of-mass coordinate R
is the total mass and the relative coordinate is given by ~r = ~r1 − ~r2 . The so-called reduced
mass is defined by µ = m1 m2 /M . We will show the key identity
p~1 · p~1
p~2 · p~2
P~ · P~
p~ · p~
+
=
+
.
2m1
2m2
2M
2µ
The center of mass velocity and the relative velocity are obtained by differentiating the
corresponding coordinates:
~˙ ≡ V
~ = m1~v1 + m2~v2
R
m1 + m2
and ~r˙ ≡ ~v = ~v1 − ~v2 .
The velocities can be solve to obtain
~ +
~v1 = V
m2
~ + µ ~v and ~v2 = V
~ − µ ~v .
~v = V
m1 + m2
m1
m2
Therefore, the kinetic energy is given by
KE ≡
1
1
1
m1~v1 · ~v1 + m2~v2 · ~v2 = m1
2
2
2
µ
~ + µ ~v
V
m1
µ
¶2
+
1
~ − µ ~v
m2 V
2
m2
¶2
.
Simplifying we note that the cross terms vanish1
1
~ ·V
~ + 1
KE = (m1 + m2 )V
2
2
Ã
µ2
µ2
+
m1
m2
!
~v · ~v =
M~ ~
µ
V · V + ~v · ~v .
2
2
where the last equality follows since
µ =
m1 m2
1
1
1
⇒
=
+
.
m1 + m2
µ
m1
m2
~ and is given by P~ = m1~v1 + m2~v2 and the
The center of mass momentum is (m1 + m2 )V
relative momentum is defined to be p~ = µ (~v1 − ~v2 ) . The expression for the kinetic energy
in terms of the momenta is
1 ~ ~
1
KE =
P ·P +
p~ · p~ .
2M
2µ
1
m1 µ ~
m2 µ ~
2
V · ~v −
2
V · ~v = 0 .
2 m1
2 m2
1
For doing quantum mechanics it is useful to write the center-of-mass and relative momenta
in terms of the individual momenta:
µ
µ
P~ = p~1 + p~2 and p~ =
p~1 −
p~2 .
m1
m2
How does one go over to quantum mechanics? We know that the components of p~1 , p~2 ,
~r1 , and ~r2 are now operators and obey the standard commutation relations. The momenta
and coordinates of different particles commute: for example, [x1 , p2x ] = 0. Now we have
to convince ourselves that the the components of ~r and p~ obey the correct commutation
relations. Calculate [x, px ]:
·
¸
x1 − x2 ,
µ
µ
µ
µ
p1x −
p2x =
[x1 , p1x ] +
[x2 , p2x ] = ih̄ µ
m1
m2
m1
m2
µ
1
1
+
m1 m2
¶
= ih̄
where we have neglected all commutators that involve different particles since they vanish.
Note that center of mass momentum and the relative coordinates commute: we compute
[x1 − x2 , Px ]:
[x1 − x2 , p1x + p2x ] = [x1 , p1x ] − [x2 , p2x ] = ih̄ − ih̄ = 0 .
All the other relations can be similarly verified.
Thus we can focus only on the relative coordinates and the Hamiltonian is
H =
1
p~ · p~ + V (r) .
2µ
Now we can write the kinetic energy as a radial and angular part just as we did before. The
m2
only difference is that we use the reduced mass µ = mm11+m
.
2
Thus for the hydrogen atom the energy levels are given by (Equation 4.70 in the text)
En = −
2
m m
µq 4 1
2h̄2 n2
e
where q 2 = 4π²
and µ = mee+mpp using obvious notation for the electron and proton masses.
0
If we had a muonic hydrogen atom with a µ− in a bound state with a proton we will use the
corresponding masses.
2