Download Solution #3 - Temple University Department of Physics

Department of Physics
Temple University
Quantum Mechanics II, 5702
Instructor: Z.-E. Meziani
Homework Solution Set # 3
Thursday, September 22, 2016
Textbook: Claude Cohen Tannoudji, Bernard Diu and Franck Lalo, Second Volume Complement GX
Exercise 1 , page 1086
Consider a deuterium atom (composed of a nucleus of spin I = 1 and an electron). The electronic angular
~ + S,
~ where L
~ is the orbital angular momentum of the electron and S
~ is its spin. The
momentum is J~ = L
~
~
~
total angular momentum of the atom is F = J + I, where I is the nuclear spin. The eigenvalues of J 2 and
F 2 are J(J + 1)~2 and F (F + 1)~2 respectively.
a. What are the possible values of the quantum number J and F for a deuterium atom in the 1s ground
state?
In the 1s ground state of the deuterium atom the values of n and l are n = 1, l = 0 and s = 1/2.
Therefore the possible values of J are such that |L − 1/2| 6 L + 1/2 ⇒ 1/2 6 J 6 1/2 leaving the
only value possible for J to be J = 1/2. With this value of J we can find the possible values of F
assuming that I = 1, |J − I| 6 F 6 J + I ⇒ |1/2 − 1| 6 F 6 1/2 + 1. Thus F = 1/2 or F = 3/2.
b. Same question for the deuterium in the 2p excited state.
In the case of the 2p state of deuterium we have
n = 2,
l=1
(1)
This implies J values such that
|L − 1/2| 6 J 6 L + 1/2 ⇒ |1 − 1/2| 6 J ≤ 1 + 1/2
(2)
resulting in possible values of J = 3/2 and J = 1/2.
Now for the case of J = 1/2 we have
⇒ |I − 1/2| 6 F 6 I + 1/2 ⇒ |1 − 1/2| 6 F 6 1 + 1/2
(3)
leading to F = 1/2 and F = 3/2
For the case where J = 3/2 we find
⇒ |I − 3/2| 6 F 6 I + 3/2 ⇒ |1 − 3/2| 6 F 6 1 + 3/2
(4)
leading to F = 1/2, F = 3/2 and F = 5/2
Exercise 2, page 1086
The hydrogen atom nucleus is a proton of spin I = 1/2.
a) In the notation of the preceding exercise, what are the possible values of the quantum numbers J and
F for a hydrogen atom in the 2p level?
In the case of the 2p state for hydrogen we have n = 2 and l = 1 and the electron spin s = 1/2. This
implies that
|L − s| 6 J 6 L + s ⇒ |L − 1/2| 6 J 6 L + 1/2 ⇒ |1 − 1/2 6 J 6 1 + 1/2
1
(5)
Thus the two possible values of J are J = 3/2 and J = 1/2.
Now for the case of J = 1/2 we have
⇒ |I − 1/2| 6 F 6 I + 1/2 ⇒ |1/2 − 1/2| 6 F 6 1/2 + 1/2
(6)
leading to F = 0 and F = 1
For the case where J = 3/2 we find
⇒ |I − 3/2| 6 F 6 I + 3/2 ⇒ |1/2 − 3/2| 6 F 6 1/2 + 3/2
(7)
leading to F = 1 and F = 2
~ and S
~ to form J~ (MJ ~ is the eigenvalue of
b) Let {|n, l, s, J, MJ i } be the basis obtained by adding L
Jz ); and let {|n, l, s, J, I, F, MF i } be the basis obtained by adding J~ and I~ to form F~ (MF ~ is the
eigenvalue of Fz )
The magnetic operator of the electron is:
~ = µB (L
~ + 2S)/~
~
M
In each of the subspaces E(n = 2, l = 1, s = 1/2, J, I = 1/2, F ) arising from the 2p level and subtended
by the 2F + 1 vectors
|n = 2, l = 1, s =
1
1
, J, I = , F, MF i
2
2
(8)
corresponding to the fixed values of J and F , the projection theorem enables us to write
~ = gJF µB F~ /~
M
(9)
Calculate the various possible Landé factors gJF corresponding to the 2p level.
Using the projection theorem Complement DX we can write
~ =
L
~ · Ji
~
hL
J~
J(J + 1)~2
~=
S
~ · Ji
~
hS
J~
J(J + 1)~2
Using the results of Complement DX of equations 41-a and 41-b after adding the two terms we can
write:
2
~ · Ji
~ E ,l,s,J = ~ [J(J + 1) + L(L + 1) − S(S + 1)]
hL
0
2
2
~
~ · Ji
~ E ,l,s,J =
hS
[J(J + 1) − L(L + 1) + S(S + 1)]
0
2
Therefore we obtain that
2
µB
~
~2
~ =
M
[J(J
+
1)
+
L(L
+
1)
−
S(S
+
1)]
+
2
×
[J(J
+
1)
−
L(L
+
1)
+
S(S
+
1)]
J~
J(J + 1)~3 2
2
µB
3
1
1
=
J(J + 1) − L(L + 1) + S(S + 1) J~
J(J + 1)~ 2
2
2
~ and J~ we use the projection theorem again to find the relation
Now armed with the relation between M
~
~
between M and F and deduce the Landé factor.
2
~ therefore
First lets remember that F~ = I~ + J,
J~ =
hJ~ · F~ i ~
F
F (F + 1)~2
~2
hJ~ · F~ iE0 ,l,s,J,I,F =
[F (F + 1) + J(J + 1) − I(I + 1)]
2
Thus
[F (F + 1) + J(J + 1) − I(I + 1)] ~
F
J~ =
2F (F + 1)
~ to obtain:
We now replace the expression of J~ in that of M
~
M
=
~
M
=
[F (F + 1) + J(J + 1) − I(I + 1)] ~
µB
[3J(J + 1) − 2L(L + 1) + S(S + 1)] ×
F
2J(J + 1)~
2F (F + 1)
µB
{3J(J + 1) − 2L(L + 1) + S(S + 1)} {F (F + 1) + J(J + 1) − I(I + 1))}F~
4J(J + 1)F (F + 1)~
Since s = 1/2 and I = 1/2, the above expression becomes
1
~ = µB
{3J(J + 1) − 3)} {F (F + 1) + J(J + 1) − 3/4)}F~
M
~ 4J(J + 1)F (F + 1)
(10)
We can write the Landé factor as :
gJF =
1
{3J(J + 1) − 3)} {F (F + 1) + J(J + 1) − 3/4)}
4J(J + 1)F (F + 1)
(11)
We can summarize the result for all values of J and F
l
l
l
l
1
,J
2
1
= 1, s = , J
2
1
= 1, s = , J
2
1
= 1, s = , J
2
=
1, s =
1
,
2
1
= ,
2
3
= ,
2
3
= ,
2
=
and F = 0, g 21 0 = not defined
and F = 1, g 21 1 = −
4
3
4
=
5
1
6
and F = 1, g 23 1 =
and F = 2, g 23 2
(12)
Exercise 3, page 1086-1087
Consider a system composed of two spins 1/2 particles whose orbital variables are ignored. The Hamiltonian of the system is:
b = ω1 Sb1z + ω2 Sb2z
H
(13)
~1 and S
~2 of the two particles onto Oz, and ω1 and ω2 are
where S1z and S2z are the projections of the spin S
real constants.
a. The initial state of the system , at time t = 0 is:
1
|ψ(0) >= √ [| + − > +| − + >]
2
(14)
where | + − >= | 12 , 12 , + 12 , − 12 > and | − + >= | 21 , 12 , − 12 , + 12 > are two of the common eigenvectors of
~1 + S
~2 )2 is measured, what results can be found and with what
Ŝ12 , Ŝ22 Ŝ1z and Ŝ2z . At time t, Ŝ 2 = (S
probabilities?
3
First, when a measurement is performed on Ŝ 2 the only possible results of the measurement are
eigenvalues of Ŝ 2 . In this case, the possible eigenvalues are S(S + 1)~2 with S = 1 and S = 0.
Therefore the possible results are 2~2 or 0~2 .
b =
Written in the basis of common vectors of Ŝ12 , Ŝ22 , Ŝ1z and Ŝ2z thus also a common basis of H
b
b
ω1 S1z + ω2 S2z
1
√ [| + −i + | − +i] = |10i
2
i
E−+
1 h i E+− t
√ e ~ | + −i + ei ~ t | − +i
2
i
1 h i(ω1 −ω2 )t
√ e
| + −i + e−i(ω1 −ω2 )t | − +i
2
|ψ(0)i =
|ψ(t)i =
|ψ(t)i =
where E+− = ~2 (ω1 − ω2 ) and E−+ = ~2 (−ω1 + ω2 ). The probability to find 2~2 when measuring S 2 at
time t is given by
P(2~2 ) = h1 + 1|ψ(t)i + h10|ψ(t)i + h1 − 1|ψ(t)i
In the next expression we switch from the basis |SMS i to the basis |mS1 , mS2 i to find
P(2~2 )
=
=
=
1
|h+ + |ψ(t)i + √ (h+ − | + h− + |)ψ(t)i + h− − ψ(t)i|2
2
2
1 h i(ω1 −ω2 )/2t i 1 h −i(ω1 −ω2 )/2t i
e
e
+
+ 0
0 +
2
2
ω1 − ω2
ω1 − ω2 2
2
t = cos
t
cos
2
2
while the probability to find 0~2 is given by
P(0~2 ) = |h00|ψ(t)i|2
switching basis like before we have
P(0~2 )
1
√ (h+ − | − h− + |)ψ(t)i
2
1 h
i 1h
i 2
= ei(ω1 −ω2 )/2t −
e−i(ω1 −ω2 )/2t 2 2
ω1 − ω2
ω1 − ω2 2
t = sin2
t
= i sin
2
2
=
b If the initial state of the system is arbitrary, what Bohr frequencies can appear in the time evolution
of < Sˆ2 >?
Using the Ehrenfest Theorem we can write in general that
i
dhSˆ2 i
1 h
= h Ĥ, Ŝ 2 i
dt
i~
Since Ĥ and Ŝ 2 commute then
h
i
dhSˆ2 i
Ĥ, Ŝ 2 = 0 ⇒
= constant
dt
Thus < Sˆ2 > does not evolve in time and thus no frequencies will be needed.
c Same question for < Ŝx >=< Ŝ1x + Ŝ2x >
If the initial state is arbitrary we can write it in the basis of eigenstates of Ĥ, S12 , S22 , S1z and S2z as
|ψ(0) >= α(0)| + +i + β(0)| + −i + γ(0)| − +i + δ(0)| − −i
4
since this state is an eigenstate of Ĥ the time evolution is written as
|ψ(t) >= α(0)eiE+ +t/~ | + +i + β(0)eiE+ −t/~ | + −i + γ(0)eiE− +t/~ | − +i + δ(0)eiE− −t/~ | − −i
where we have
E++ /~
E−+ /~
ω1 + ω2
2
ω1 − ω2
= −
2
=
ω1 − ω2
2
ω1 + ω2
E−− /~ = −
2
E+− /~ =
(15)
the corresponding bra is:
hψ(t)| = α(t)∗ h+ + | + β(t)∗ h+ − | + γ(t)∗ h− + | + δ(t)∗ h+ + |
for convenience we leave out in the notation of the coefficients the time dependence in what follows.
hŜx i =
hŜ1x + Ŝ2x i
=
(α∗ h+ + | + β ∗ h+ − | + γ ∗ h− + | + δ ∗ h− − |) S1x (α| + +i + β| + −i + γ| − +i + δ| − −i)
+
(α∗ h+ + | + β ∗ h+ − | + γ ∗ h− + | + δ ∗ h− − |) S2x (α| + +i + β| + −i + γ| − +i + δ| − −i)
1
(α∗ h+ + | + β ∗ h+ − | + γ ∗ h− + | + δ ∗ h− − |) (S1+ + S1− ) (α| + +i + β| + −i + γ| − +i + δ| − −i)
2
1
∗
∗
∗
∗
(α h+ + | + β h+ − | + γ h− + | + δ h− − |) (S2+ + S2− ) (α| + +i + β| + −i + γ| − +i + δ| − −i)
2
1
1
∗
∗
(α h+ + | + β h+ − |) S1+ (γ| − +i + δ| − −i) + (δ ∗ h− − | + γ ∗ h− + |) S1− (α| + +i + β| + −i)
2
2
1
1
∗
∗
∗
∗
(α h+ + | + γ h− + |) S2+ (β| + −i + δ| − −i) + (δ h− − | + β h+ − |) S2− (α| + +i + γ| − +i)
2
2
1 ∗
1 ∗
1 ∗
1 ∗
∗
∗
∗
∗
(α γ + β δ) + (δ β + γ α) + (α β + γ δ) + (δ γ + β α)
2
2
2
2
=
+
=
+
=
Both frequencies will appear in the time dependence of < Sˆx >.
5