Cap 6 - 6.1, 6.6

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Department of Physics Temple University
Introduction to Quantum Mechanics, Physics 306 Instructor: Z.-E. Meziani
Solution set for homework # 6
Tuesday December 09, 2003
Exercise #1, Complement FVI, page 765
We consider a system of angular momentum j = 1 whose state space is spanned by the three eigenvectors
of J2 and Jz, {|+ 1 >, |0 >, | − 1 >} with respective eigenvalues +h¯, 0 and −h¯. The state of the system is:
|ψ >= α|+ 1 > +β|0 > +γ| − 1 > (1)
where α, β and γ are three complex parameters
a) We calculate here the mean value of the components of < J > using equation C60 page 659 in the
textbook.
< ψ|Jx|ψ > =
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
]
1
2
(J+ + J−)
[
α|+ 1 > +β|0 > +γ| − 1 >
]
=
1
2
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
][
h¯
√
2β|+ 1 > +h¯
√
2γ|0 > +αh¯
√
2|0 > +βh¯
√
2| − 1 >
]
=
1
2
[√
2h¯α∗β +
√
2h¯β∗γ +
√
2h¯β∗α +
√
2h¯γ∗β
]
=
1√
2
h¯
[
(αβ∗ + βα∗) + (βγ∗ + γβ∗)
]
(2)
< ψ|Jy|ψ > =
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
]
1
2i
(J+ − J−)
[
α|+ 1 > +β|0 > +γ| − 1 >
]
=
1
2
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
][
−ih¯
√
2β|+ 1 > −ih¯
√
2γ|0 > +iαh¯
√
2|0 > +iβh¯
√
2| − 1 >
]
=
1
2
[
−i
√
2h¯α∗β − i
√
2h¯β∗γ + i
√
2h¯β∗α + i
√
2h¯γ∗β
]
=
1√
2
h¯
[
i(αβ∗ − βα∗) + i(βγ∗ + γβ∗)
]
(3)
< ψ|Jz|ψ > =
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
]
Jz
[
α|+ 1 > +β|0 > +γ| − 1 >
]
=
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
][
h¯α|+ 1 > −h¯γ| − 1 >
]
= h¯
[
|α|2 − |γ|∗2
]
(4)
b) We turn to the mean values of < J2x >, < J
2
y > and < J
2
z >
< ψ|J2y |ψ > =
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
]
1
4
(J+ + J−)2
[
α|+ 1 > +β|0 > +γ| − 1 >
]
=
1
4
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
] (
J2+ + J
2
− + J+J− + J−J+
) [
α|+ 1 > +β|0 > +γ| − 1 >
]
=
1
4
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
]
×
[
2h¯2(α + γ)|+ 1 > +2h¯2α| − 1 > +4βh¯2|0 > +2(α + γ)h¯2| − 1 >
]
=
h¯2
2
[
(α∗α + α∗γ + 2β∗β + γ∗α + γ∗γ
]
=
h¯2
2
[
|α|2 + 2|β|2 + |γ|2 + α∗γ + γ∗α
]
(5)
< ψ|J2y |ψ > =
h¯2
2
[
|α|2 + 2|β|2 + |γ|2 − α∗γ − γ∗α
]
(6)
< ψ|J2z |ψ > =
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
]
J2z
[
α|+ 1 > +β|0 > +γ| − 1 >
]
=
[
α∗ < +1|+ β∗ < 0|+ γ∗ < −1|
][
h¯2α|+ 1 > +h¯2γ| − 1 >
]
= h¯2
[
|α|2 + |γ|∗2
]
(7)
Exercise #61, Complement FVI, page 768
Consider a system of angular momentum l = 1. A basis of its state space is formed by the three
eigenvectors of Lz: |+ 1 >, |0 >, | − 1 >, whose eigenvalues are , respectively +h¯, 0, −h¯ and which satisfy:
L± = h¯
√
2|m± 1 >
L+|+ 1 > = L−| − 1 >= 0 (8)
This system posses a quadrupole moment and is placed in an electric field gradient such that its Hamiltonian
can be written:
H =
ω0
h¯
(
L2u − L2v
)
(9)
where L2u and L
2
v are the components of �L along the two directions Ou and Ovof the XOZ plane which
form an angle of 45◦ with Ox and Oz.
x
y
z
u
v
45º
45º
Figure 1: Choice fo cartesian system to describe the components of Lu and Lv.
a We can first express the components Lu and Lv in terms of the components Lx and Ly. Looking at
figure 1 we see clearly that
Lu =
1√
2
(Lx + Lz)
Lv =
1√
2
(−Lx + Lz) (10)
Since we know the action of L± on the three eigenvectors of Lz we shall express Lx in terms of L±,
Lx =
1
2
(L+ + L−) (11)
Now we rewrite the Hamiltonian in terms of Lz and L± in order to evaluate the matix elements of
H in the basis of eigenvectors of Lz.
H =
ω0
2h¯
[
(Lx + Lz)
2 − (Lz − Lx)2
]
=
ω0
h¯
[LxLz + LzLx]
=
ω0
2h¯
[(L+ + L−)Lz + Lz (L+ + L−)] (12)
Lets evaluate the action of H on the eigenvectors of Lz.
H|+ 1 > = ω0
2h¯
[L+Lz + L−Lz + LzL+ + LzL−] |+ 1 >
H|+ 1 > = ω0
2h¯
[h¯L+ + h¯L− + 0 + LzL−] |+ 1 >
H|+ 1 > = ω0
2h¯
[h¯L+ + h¯L− + 0+] |+ 1 > +h¯
√
2Lz|0 >
H|+ 1 > = ω0h¯√
2
|0 >
(13)
H|0 > = ω0
2h¯
[L+Lz + L−Lz + LzL+ + LzL−] |0 >
H|0 > = ω0
2h¯
[
0 + 0 + 0 + h¯
√
2Lz|+ 1 > h¯
√
2Lz| − 1 >
]
H|0 > = ω0
2h¯
[
h¯2
√
2|+ 1 > −h¯2
√
2| − 1 >
]
H|0 > = h¯ω0√
2
[|+ 1 > −| − 1 >] (14)
H| − 1 > = ω0
2h¯
[L+Lz + L−Lz + LzL+ + LzL−] | − 1 >
H| − 1 > = ω0
2h¯
[−h¯L+ − h¯L−] | − 1 > + ω0√
2
Lz|0 >
H| − 1 > = ω0
2h¯
[
−h¯2
√
2
]
|0 >
H| − 1 > = − h¯ω0√
2
|0 >
(15)
It is now easy to evaluate all the matix eleements of the Hamiltonian in the basis of eigenvectors of
Lz.
H =
h¯ω0√
2

 0 1 01 0 −1
0 −1 0

 (16)
Solving for the eigenvalues of the Hamiltonian we find E1,2,3 = −h¯ω0, 0, h¯ω0 the corresponding
eigenvectors are found by solving for the eigenvectors for each eigenvalue. We find
|E1 > = 12
[
|+ 1 > +
√
2|0 > −| − 1 >
]
|E2 > = 1√
2
[|+ 1 > +| − 1 >]
|E3 > = 12
[
|+ 1 > −
√
2|0 > −| − 1 >
]
(17)
b At time t, the system is the state
|ψ(0) >= 1√
2
[|+ 1 > −| − 1 >] (18)
We first express the state |ψ(0) > in the basis of eigenvectors of the Hamiltonian in order to write
the time evolution of this state in a straightforward way. Remember that these eigenstates of the
Hamiltonian are know as stationary states.
|ψ(0) > = 1√
2
[|E1 > +0|E2 > +|E3 >]
|ψ(0) > = 1√
2
[|E1 > +|E3 >]
(19)
At time t the state |ψ(t) > is given thus by:
|ψ(t) >= 1√
2
[
e−iω0t|E1 > +eiω0t|E3 >
]
(20)
If Lz is measured at time t the results that can be found are only the eigenvalues of Lz, namely
+h¯, 0, − h¯. Now we evaluate the probabilities to find each of these eigenvalues:
P(+h¯) =
∣∣∣∣ 1√2 < +1|Lz
[
e−iω0t|E1 > +eiω0t|E3 >
]∣∣∣∣
2
=
∣∣∣∣ 1√2 < +1|Lz
[
e−iω0t
1
2
(
|+ 1 > +
√
2|0 > −| − 1 >
)
+ eiω0t
1
2
(
|+ 1 > −
√
2|0 > −| − 1 >
)]∣∣∣∣
2
=
∣∣∣∣∣
1√
2
(
e−iω0t + eiω0t
2
)∣∣∣∣∣
2
=
1
2
cos2 (ω0t) (21)
similarely
P(0h¯) =
∣∣∣∣ 1√2 < 0|Lz
[
e−iω0t|E1 > +eiω0t|E3 >
]∣∣∣∣
2
=
∣∣∣∣ 1√2 < 0|Lz
[
e−iω0t
1
2
(
|+ 1 > +
√
2|0 > −| − 1 >
)
+ eiω0t
1
2
(
|+ 1 > −
√
2|0 > −| − 1 >
)]∣∣∣∣
2
=
∣∣∣∣∣
(
e−iω0t − eiω0t
2
)∣∣∣∣∣
2
= sin2 (ω0t) (22)
P(−h¯) =
∣∣∣∣ 1√2 < −1|Lz
[
e−iω0t|E1 > +eiω0t|E3 >
]∣∣∣∣
2
=
∣∣∣∣ 1√2 < −1|Lz
[
e−iω0t
1
2
(
|+ 1 > +
√
2|0 > −| − 1 >
)
+ eiω0t
1
2
(
|+ 1 > −
√
2|0 > −| − 1 >
)]∣∣∣∣
2
=
∣∣∣∣∣
1√
2
(
−e−iω0t − eiω0t
2
)∣∣∣∣∣
2
=
1
2
cos2 (ω0t) (23)
c Now we calculate the mean values of each component of the angular momentum. We first re-write
the expression of |ψ(t) > in terms of the eigenvectors of Lz.
|ψ(t) > = 1√
2
[
1
2
e−iω0t
(
|+ 1 > +
√
2|0 > −| − 1 >
)
+
1
2
eiω0t
(
|+ 1 > −
√
2|0 > −| − 1 >
)]
=
1√
2
[
cos (ω0t)|+ 1 >
√
2
2
(
e−iω0t − eiω0t
)
|0 > − cos (ω0t)| − 1 >
]
=
1√
2
[
cos (ω0t)|+ 1 > −i
√
2 sin (ω0t)|0 > − cos (ω0t)| − 1 >
]
(24)
Now we proceed to calculate < Lx >, < Ly > and < Lz >.
< Lx > (t) = < ψ(t)|Lx|ψ(t) >= 12 < ψ(t)|(L+ + L−)|ψ(t) >
=
1
2
< ψ(t)|
[
h¯ cos (ω0t)|0 > −ih¯
√
2 sin (ω0t)| − 1 > −ih¯
√
2 sin (ω0t)|+ 1 > −h¯ cos (ω0t)|0 >
]
=
1
2
[
− ih¯
√
2√
2
cos (ω0t) sin (ω0t) +
ih¯
√
2√
2
cos (ω0t) sin (ω0t)
]
= 0 (25)
< Ly > (t) = < ψ(t)|Lx|ψ(t) >= 12i < ψ(t)|(L+ − L−)|ψ(t) >
=
1
2
< ψ(t)|
[
−h¯ cos (ω0t)|0 > +ih¯
√
2 sin (ω0t)| − 1 > −ih¯
√
2 sin (ω0t)|+ 1 > −h¯ cos (ω0t)|0 >
]
=
1
2i
[−ih¯ cos (ω0t) sin (ω0t)− 2ih¯ cos (ω0t) sin (ω0t)−−ih¯ cos (ω0t) sin (ω0t)]
= −2h¯ cos (ω0t) sin (ω0t)
= −h¯ sin (2ω0t) (26)
< Lz > (t) = < ψ(t)|Lz|ψ(t) >
=
1
2
< ψ(t)|
[
h¯√
2
cos (ω0t)|+ 1 > h¯√
2
sin (ω0t)| − 1 >
]
=
h¯
2
cos2 (ω0t)− h¯2 cos
2 (ω0t)
= 0 (27)
From the above resultswe can conclude that the motion of < �L > will lie along the y − axis only
since < Lx > = < Lz > = 0. It oscillate with a frequency of 2ω0 from −h¯ to +h¯ starting from
< �L > (t = 0) = 0
d i) At time t, a measurement of L2z is measured. The matrix of L
2
z is given by:
L2z = h¯
2
√
2

 1 0 00 0 0
0 0 1

 (28)
Therefore the eigenvalues are h¯2 doubly degenerate and 0 non degenerate. Evaluating the probabilities
to find each of these eigenvalues we obtain:
P(h¯2) = cos2(ω0t)
P(0) = sin2(ω0t) (29)
Therefore when ω0t = nπ/2 , t = nπ/2ω0 only one result is possible: h¯2 for n even and 0 for n odd.
d ii) If the measurement has yielded the result h¯2 the state of the system will collapse onto:
|ψ(t′0 = 0) >=
1√
2
[|+ 1 > −| − 1 >] (30)
From t′0 = 0 where the measurement of L2z is performed leading to a value of h¯
2 the system will
evolve following:
|ψ(t′) >= 1√
2
[
cos (ω0t′)|+ 1 > +i
√
2 sin (ω0t′)|0 > − cos (ω0t′)| − 1 >
]
(31)
t′ is time beyond t′0.