T2: Quantenmechanik
Sheet 3, Course 17019
Professor: H. Ruhl, Exercises: N. Elkina, S. Rykovanov
Discussion of solutions: Oct. 25-29, 2010
Problem 1: Projection (Sakurai 1.9)
� · n̂; +� such that
Construct |S
�
� � ��
�
� · n̂��S
� · n̂; + = h̄ ��S
� · n̂; +
S
2
(1)
where n̂ is the unit vector with polar angle β and azimuthal angle α with respect to the z axis.
Solution 1:
Cartesian representation for n̂ is given by
(sin β cos α, sin β sin α, cos β)
(2)
� · n̂ as
so we can constract the matrix for the operator S
�
�
�
�
h̄
h̄
0 1
0
�
S · n̂ = (σx nx + σy ny + σz nz ) =
sin β cos α
+ sin β sin α
1
0
i
2
2
−i
0
�
h̄
cos β − λ
2
�
+ cos β
�
h̄
cos β
=
iα
e
sin β
2
�
�
� · n̂ − λI = 0, so
The eigenvalues of this matrix can be found by solving det S
0=
�
1
0
0
−1
��
e−iα sin β
− cos β
��
� � �2
� �2
h̄
h̄
h̄
h̄
− cos β − λ −
sin2 β = −
+ λ2 → λ = ±
2
2
2
2
= (3)
�
(4)
If we are intersted in the normalized eigenket corresponding to the positive eigenvalue, we can write
�
�
��
�S · n̂; + = A|+� + B|−�
Thus we have
h̄
2
�
cos β
eiα sin β
eiα sin β
− cos β
��
A
B
�
h̄
=
2
�
A
B
�
(5)
Using first equation in this system, one can get:
(cos β − 1)A + (e−iα sin β)B = 0
(6)
To fix the actual values of A and B we can use the normalization condition
|A|2 + |B|2 = 1
(7)
so,
|B| = |A|
1 − cos β iα
e
sin β
|A| = cos
β
2
(8)
sin β = 2 sin
β
β
cos
2
2
β
2
(9)
1 − cos β
β
A = eiα sin
sin β
2
(10)
1 − cos β = 2 sin2
We choose the phase of A to be real and positive, i.e.
A = cos
β
2
B = eiα
The final result is
� · n̂; +� = cos
|S
β
β
|+� + eiα sin |−�
2
2
(11)
Problem 2: Hamiltonian, (Sakurai 1.10)
Hamiltonian operator for a two-state system is given by
H = a(|1��1| − |2��2| + |1��2| + |2��1|)
(12)
here a is the number with dimension of energy. Find the energy eigenvalues and the corresponding eigenkets
(as linear combination of |1� and |2�)
Solution 2:
The matrix of H is
�
�
a a
H=
(13)
a −a
the sum of the eigenvalues λ1 , λ2 of H equals its trace, and their product equals its determinant, i.e.
√
√
λ1 + λ2 = 0, λ1 λ2 = −2a2 → λ1 = a 2 λ2 = −a 2
√
For λ1 = a 2 we can write the eigenket as c1 |1� + c2 |2�, then
√
�
��
�
1− 2
1√
c1
a
=0
c2
1
−1 − 2
(14)
(15)
so, we have
�
√
(1 − 2)c1 + c2 = 0
|c1 |2 + |c2 |2 = 1
(16)
Choosing c1 to be real and positive, we find for the normalized eigenket
�
c1
c2
�
=�
1
√
4−2 2
�
√ 1
2−1
�
(17)
√
for λ2 = − 2 we write the eigenket as d1 |1� + d2 |2�, then
�
��
�
√
d1
1+ 2
1
a
(18)
=0
d2
1
−1 + sqrt2
√
we we have (1 + 2)d1 + d2 = 0, plus normalization condition: |d1 |2 + |d2 |2 = 1. Choosing d1 to be real and
positive, we find for the normalized eigenket
�
�
�
�
1
1
d1
√
(19)
=�
√
d2
− 2−1
4+2 2
Problem 3: Dispersion of operator, (Sakurai 1.12)
� · n̂ with eigenvalue h̄/2, where n̂ is a unit vector lying
A spin 1/2 system is known to be in an eigenstate of S
in the xz-plane that makes an angle γ with the positive z-axis
1. Suppose Sx is measured. What is the probability of getting +h̄/2?
2. Evaluate the dispersion in Sx that is,
(20)
�(Sx − �Sx �)2 �
Check special cases γ = 0, π/2, π
Solution 3:
Since
1
1
|Sx ; +� = √ |+� + √ |−�
2
2
So the probability to get +h̄/2 when Sz is measured is given by
�
�
��
�
�2 � � 1
� �2
1
γ
γ
�
�
�
� · n̂; +�� = � √ �+| + √ �−| cos |+� + sin |−� �� =
��Sx ; +|S
�
�
2
2
2
2
�
�2
�
1
1
γ ��
1
γ
1
γ
γ
γ
1 1
1
�
+ √ sin � = cos2 + sin2 + cos sin = + sin γ = (1 + sin γ)
�√
γ
� 2 cos 2
2�
2
2 2
2
2
2
2 2
2
2
• γ = π/2, |Sx ; +�:
(22)
1
2
(23)
|�Sx ; +|Sx ; +�|2 = 1
(24)
1
2
(25)
|�Sx ; +|Sz ; +�|2 =
• γ = π/2, |Sx ; +�
(21)
• γ = π, |Sz ; −�
|�Sx ; +|Sz ; −|2 =
(2.) Using
�(Sx − �Sx �)2 � = �Sx2 � − (�Sx �)2
using
Sx =
Sx2 =
h̄
(|+��−| + |−��+|)
2
h̄2
(|+��−| + |−��+|)(|+��−| + |−��+|)
4
h̄2
h̄2
Sx2 =
(|+��+| + |−��−|) =
4
4
�
� h̄
�
�
γ
γ
γ
γ
�Sx � = cos �+| + sin �−| (|+��−| + |−��+|) cos |+� + sin |−� =
2
2
2
2
2
h̄
γ
γ
h̄
γ
γ
h̄
cos sin + sin cos = sin γ
2
2
2
2
2
2
2
h̄
2
(�Sx �) = sin2 γ
4
�
� h̄2 �
� h̄2 �
� h̄2
γ
γ
γ
γ
γ
γ
�Sx2 � = cos �+| + sin �−|
cos |+� + sin |−� =
cos2 + sin2
=
2
2
4
2
2
4
2
2
4
(26)
(27)
(28)
(29)
(30)
(31)
(32)
Substituting in (20) we will have
�(Sx − �Sx �)2 � =
h̄2
h̄2
(1 − sin2 γ) =
cos2 γ
4
4
(33)
•
h̄2
4
(34)
�(∆Sx )2 �γ=π/2;|Sx ;+� = 0
(35)
h̄2
4
(36)
�(∆Sx )2 �γ=0;|Sz ;+� =
•
•
�(∆Sx )2 �γ=0;|Sz ;−� =
Problem 4: Uncertainty principle formomentum and coordinate
Using uncertaintly principle for momentum and coordinate, find minimal mean valuer for mean of Hamiltonian
of harmonic oscillator
p̂2
mω 2 x̂2
+
2m
2
(37)
�
ˆ 2 � �(∆p)
ˆ 2 � ≥ h̄
�(∆x)
2
(38)
Ĥ(p, x) =
Solution 4:
The unceraintly relattions is given by
�
ˆ = x − �x� and ∆x
ˆ = x − �x�. From physical point of view, the mean value for momentum and
where ∆x
coordinate are equal zero. In this case, one can assume
�
�
ˆ 2 � = �x̂2 � = a
�(∆x)
(39)
�
�
ˆ 2 � = �p̂2x � = b
�(∆p)
(40)
then, we obtain
�Ĥ� =
b2
mω 2 2
+
a
2m
2
(41)
Then,
1 h̄2
mω 2 2
h̄ω
�Ĥ� ≥
+
a =
2m 4a2
2
4
�
h̄
2mωa2
+
2mωa2
h̄
�
(42)
this function has extremum (min) at a2 = h̄/2mω. Thus
min�Ĥ� =
h̄ω
2
(43)
Problem 5: Density matrix
Let
ψ� =
�
cos θ
eiφ sin θ
�
θ, φ ∈ R
(44)
Does
ρ = |ψ��ψ|
define a density matrix?
Solution 5:
We find the 2 × 2 matrix for ρ
ρ=
�
cos2 θ
iφ
e cos θ sin θ
(45)
e−iφ cos θ sin θ
sin2 θ
�
(46)
Thus
ρ = ρ†
T rρ = cos2 θ + sinθ = 1
(47)
(48)
and
�x|ρ|x� = |x1 |2 cos2 θ + x1 x2 e−iφ cos θ sin θ + x1 x2 eiφ cos θ sin θ + |x2 |2 sin2 θ
(49)
≥ |x1 | cos θ + 2Re(x1 x2 e ) cos θ sin θ + |x2 | sin θ
2
2
iφ
2
2
≥ (|x1 | cos θ + |x2 | sin θ)2 ≥ 0
for all x = (x1 , x2 )T ∈ C 2 . Thus ρ defines a density matrix.
Problem 6: Mixed states
A mixed state is a statistical misture of pure states, i.e., the state is described by pairs of probabilities and pure
states. Given a mixture
{(p1 , |ψ1 �), ..., (Pn , |ψn �)}
we define its density matrix to be the positive hermitian matrix
ρ=
n
�
j=1
pj |ψj ��ψj |
(50)
where the pure states |ψ� are normalized (i.e., �ψj |ψj � = 1), and pj ≥ 0 for j = 1, 2, ..., n with
n
�
pj = 1
(51)
j=1
1. Find the probability that measurement in the orthonormal basis
{|k1 �, ..., |kn �}
will yield |kj �
2. Find the density matrix ρU when the mixture is transformed according to the unitary matrix U
Solution 6:
(1). From the probability distribution of states in the mixture we have for the probability P (kj ) of measuring
the state |kj �:
P (kj ) =
n
�
l=1
pl |�kj |ψj �|2 =
n
�
l=1
pl �kj |ψl ��ψl |kj � = �kj |ρ|kj �
(52)
(2). After applying the transform U to the states in the mixture we have the new mixture
{(p1 , U |ψ�), ..., (pn , U |ψn �)}
with density matrix
ρU =
n
�
j=1

pj U |ψj ��ψj |U ∗ = U 
n
�
j=1

pj |ψj ��ψj | U ∗ = U ρU ∗
(53)