Solution for Homework 22
43) Total and Relative Momentum
a) We write
P~ = a~p1 + b~p2 , p~ = c~p1 + d~p2 .
We also have
~ = M1~r1 + M2~r2 , ~r = ~r1 − ~r2 .
R
M1 + M2
The various commutation relations then take the form
M1 a[r1i , p1j ] + M2 b[r2i , p2j ]
M1 a + M2 b
=
i~δij ⇒
M1 + M2
M1 + M2
M1 a + M2 b
= 1,
M1 + M2
[ri , pj ] = c[r1i , p1j ] − d[r2i , p2j ] = (c − d)i~δij ⇒ c − d = 1,
M1 c + M2 d
M1 c[r1i , p1j ] + M2 d[r2i , p2j ]
=
i~δij ⇒
[Ri , pj ] =
M1 + M2
M1 + M2
M1 c + M2 d = 0,
[ri , Pj ] = a[r1i , p1j ] − b[r2i , p2j ] = (a − b)i~δij ⇒ a − b = 0.
[Ri , Pj ] =
This implies a = b = 1 and c = M2 /(M1 + M2 ), d = −M1 /(M1 + M2 ), such
that
M2 p~1 − M1 p~2
.
P~ = p~1 + p~2 , p~ =
M1 + M2
b) The energy of the center of mass motion is
ER =
P~ 2
,
2(M1 + M2 )
and the energy of the relative motion is
Er =
p~ 2
+ V (|~r|),
2µ
where µ = M1 M2 /(M1 + M2 ) is the reduced mass. Both energies are simultaneously measurable, i.e. [ER , Er ] = 0, because [ri , Pj ] = 0. The total
energy is E = ER + Er , which is hence also simultaneously measurable
1
with ER and Er . The total and relative momentum are also simultaneously
measurable because [Pi , pj ] = 0. The total momentum is simultaneously
measurable with both ER and Er and hence also with E, while the relative
momentum is simultaneously measurable only with ER but not with Er ,
and thus also not with the total E.
44) Angular Momentum Eigenfunctions
a) First we consider the normalization conditions
Z 1
Z 2π
Z 1
Z 2π
1
2
d cos θ
dϕ
= 1,
d cos θ |Y00 (θ, ϕ)| =
dϕ
4π
−1
0
−1
0
Z 2π
Z 1
Z 2π
Z 1
3
2
dϕ
d cos θ |Y10 (θ, ϕ)| =
dϕ
d cos θ
cos2 θ = 1,
4π
0
−1
0
−1
Z 2π
Z 1
Z 2π
Z 1
3
dϕ
d cos θ |Y1±1 (θ, ϕ)|2 =
dϕ
d cos θ
(1 − cos2 θ) = 1.
8π
0
−1
0
−1
Next we investigate the orthogonality conditions. Two functions with different m values are orthogonal, because the ϕ-integration then gives
Z 2π
dϕ exp(i(m0 − m)ϕ) = 0.
0
Hence all that is left to show is
Z
Z 1
Z 2π
∗
∗
d cos θ Y00 (θ, ϕ) Y10 (θ, ϕ) =
dϕ
0
−1
0
2π
√
3
d cos θ
dϕ
cos θ = 0.
4π
−1
Z
1
b) First we act with Lz
1
Lz Y00 (θ, ϕ) = −i~∂ϕ √ = 0,
4π
r
3
Lz Y10 (θ, ϕ) = −i~∂ϕ
cos θ = 0,
4π
r
3
Lz Y1±1 (θ, ϕ) = ±i~∂ϕ
sin θ exp(±iϕ) =
8π
r
3
sin θ exp(±iϕ) = ±~Y1±1 (θ, ϕ).
− ~
8π
2
Next we evaluate
sin θ∂θ Y00 (θ, ϕ) = 0,
r
3
sin2 θ,
4π
r
3
sin θ∂θ Y1±1 (θ, ϕ) = ∓
sin θ cos θ exp(±iϕ),
8π
sin θ∂θ Y10 (θ, ϕ) = −
which implies
~ 2 Y00 (θ, ϕ) = 0,
L
r
r
3
3
1
~ 2 Y10 (θ, ϕ) = ~2
∂θ
sin2 θ = ~2
2 cos θ = 2~2 Y10 (θ, ϕ),
L
sin θ
4π
4π
r
~ 2 Y1±1 (θ, ϕ) = ±~2 3 [ 1 ∂θ (sin θ cos θ) − 1 ] exp(±iϕ)
L
8π sin θ
sin θ
r
3
= 2~2 [∓
sin θ exp(±iϕ)] = 2~2 Y1±1 (θ, ϕ).
8π
As expected l(l + 1) = 2 for l = 1.
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