763313A QUANTUM MECHANICS II
Solutions 5
Spring 2017
1.
a) Show that
J 2 = J± J∓ + Jz2 ∓ ~Jz .
b) The raising and lowering operators change the value of m by one unit:
J± |jmi = Ajm
± |j(m ± 1)i ,
jm
where Ajm
± is some constant. What is A± if the eigenstates are normalized? (Hint:
First show that J∓ = J±† , and then apply a).)
Solution:
a) Angular momentum operator J = (Jx , Jy , Jz ) is defined as an operator which
satisfies the following commutation relations,
[Jx , Jy ] = i~Jz ,
[Jy , Jz ] = i~Jx ,
[Jz , Jx ] = i~Jy .
The raising (J+ ) and lowering (J− ) operators are defined as
J± = Jx ± iJy .
Now
J 2 = J · J = Jx2 + Jy2 + Jz2
= Jx2 + Jy2 + Jz2 + iJx Jy − iJx Jy + iJy Jx − iJy Jx
= Jx (Jx ∓ iJy ) ± iJy (Jx ∓ iJy ) ± i(Jx Jy − Jy Jx ) + Jz2
= Jx J∓ ± iJy J∓ ± i[Jx , Jy ] + Jz2
= (Jx ± iJy )J∓ + Jz2 ∓ ~Jz
= J± J∓ + Jz2 ∓ ~Jz .
b) Since Jx and Jy are hermitian operators
J±† = (Jx ± iJy )† = Jx† ∓ iJy† = Jx ∓ iJy = J∓ .
jm
Now J+ |jmi = Ajm
+ |j(m + 1)i and J− |jmi = A− |j(m − 1)i. Next we multiply
both equation from the left with hj 0 m0 |. We obtain
hj 0 m0 |J+ |jmi = Ajm
+ δj,j 0 δm0 ,m+1
hj 0 m0 |J− |jmi = Ajm
− δj,j 0 δm0 ,m−1 ,
1
since states |jmi are orthonormal. Since J+† = J−
0 0
0 0 †
Ajm
− δj,j 0 δm0 ,m−1 = hj m |J− |jmi = hj m |J+ |jmi
∗
= hjm|J+ |j 0 m0 i∗ = Aj+m−1 δj,j 0 δm−1,m0 .
∗
Thus Aj+m−1 = Ajm
− . Now
jm j m+1
J− J+ |jmi = Ajm
|jmi
+ J− |j(m + 1)i = A+ A−
∗
jm
jm 2
= Ajm
+ A+ |jmi = |A+ | |jmi .
Using the result from a) we obtain
J− J+ |jmi = (J 2 − Jz2 − ~Jz )|jmi = ~2 (j(j + 1) − m2 − m)|jmi ,
since J 2 |jmi = ~2 j(j + 1)|jmi and Jz |jmi = m~|jmi . Thus
2
2
2
|Ajm
+ | = ~ (j(j + 1) − m(m − 1)) = ~ (j − m)(j + m + 1) .
We choose Ajm
+ to be real and positive so
p
Ajm
(j − m)(j + m + 1)
+ = ~
p
jm
A− = ~ (j + m)(j − m + 1) .
2. An electron is in the spin state
|χi = A(3i|↑i + 4|↓i) .
a) Determine the normalization constant A.
b) Find the expectation values of Sx , Sy , and Sz .
c) Find the uncertainties σSx , σSy , and σSz . (The sigma’s are standard deviations,
not Pauli spin matrices.)
d) Confirm that your results are consistent with all three uncertainty principles associated with the operators Sx , Sy , and Sz .
Solution:
a) We determine A by demanding hχ|χi = 1. This gives
1 = hχ|χi = |A|2 (9h↑|↑i − 12ih↑|↓i + 12ih↓|↑i + 16h↓|↓i)
= |A|2 (9 + 16) = 25|A|2 .
We choose A positive and real. A = 51 .
b) Since Sz |↑i = ~2 and Sz |↓i = − ~2 we can easily obtain
1
hSz i = hχ|Sz |χi =
(9h↑ |Sz | ↑i − 12ih↑ |Sz | ↓i + 12ih↓ |Sz | ↑i + 16h↓ |Sz | ↓i)
25 1
~
~
7
=
9 − 16
= − ~.
25
2
2
50
2
The expectation values of Sx and Sy are obtained by using
1
Sx = (S+ + S− )
2
i
Sy = − (S+ − S− )
2
(see lecture notes p. 21). Using these one gets
1
(hS+ i + hS− i)
2
1
= (9h↑ |S+ | ↑i − 12ih↑ |S+ | ↓i + 12ih↓ |S+ | ↑i + 16h↓ |S+ | ↓i
50
+ 9h↑ |S− | ↑i − 12ih↑ |S− | ↓i + 12ih↓ |S− | ↑i + 16h↓ |S− | ↓i)
1
=
(−12i~ + 12i~) = 0
50
hSx i =
and
i
hSy i = − (hS+ i − hS− i)
2
12
=− ~
25
since S− |↑i = ~|↓i, S+ |↓i = ~|↑i and S+ |↑i = S− |↓i = 0.
c) The standard deviations are defined as
q
σSi = hSi2 i − hSi i2 ,
where i = x, y, z. Let’s calculate hSi2 i.
1
1
hS+2 i + hS+ S− i + hS− S+ i + hS−2 i = (hS+ S− i + hS− S+ i)
4
4
1
=
(9h↑ |S+ S− | ↑i − 12ih↑ |S+ S− | ↓i + 12ih↓ |S+ S− | ↑i + 16h↓ |S+ S− | ↓i
100
+ 9h↑ |S− S+ | ↑i − 12ih↑ |S− S+ | ↓i + 12ih↓ |S− S+ | ↑i + 16h↓ |S− S+ | ↓i)
1
~2
=
(9~2 + 16~2 ) =
,
100
4
hSx2 i =
hSy2 i = −
~2
1
hS+2 i − hS+ S− i − hS− S+ i + hS−2 i =
,
4
4
and
1
(9h↑ |Sz2 | ↑i − 12ih↑ |Sz2 | ↓i + 12ih↓ |Sz2 | ↑i + 16h↓ |Sz2 | ↓i
25
2
2 !
~
1
~
~2
=
9
+ 16 −
=
.
25
2
2
4
hSz2 i =
3
Now the standard deviations are
r
σSx =
p
hSx2 i − hSx i2 =
~
~2
− 02 =
4
2
s
σSy
σSz
2
12
~2
7
=
− − ~ = ~
4
25
50
s
2
~2
7
12
=
− − ~ = ~.
4
50
25
d) Uncertainty relations for Sx , Sy and Sz are
σSi σSj ≥
~
|hSk i|εijk ,
2
where εijk is the Levi-Civita symbol. Now
~ 7
7 2 ~
~ 7
7 2
~=
~ ≥ |hSz i| =
~=
~
2 50
100
2
2 50
100
7 12
42 2 ~
~
= ~ ~=
~ ≥ |hSx i| = 0 = 0
50 25
625
2
2
12 ~
6 2 ~
~ 12
6
= ~ = ~ ≥ |hSy i| =
~ = ~2
25 2
25
2
2 25
25
σSx σSy =
σSy σSz
σSz σSx
3. Construct the spin matrices for a particle of spin 1.
Solution:
We shall use the basis {|↑i, |↓i}, i.e. the eigenstates of Sz , for the matrix representations.
Whenever one constructs a matrix representation in a basis composed of the states |smi,
it is customary to choose the ordering of the basis according to the quantum number m.
That is, the state with m = s comes first, the state with m = s − 1 second and so on.
Following this convention, the ordered basis in this problem is
{|1, 1i, |1, 0i, |1, −1i} .
(1)
Let us denote the matrix representation of an operator by the same symbol as the operator
itself. Then, the matrix representation of the operator Sz in the ordered basis (1) is


h1, 1|Sz |1, 1i
h1, 1|Sz |1, 0i
h1, 1|Sz |1, −1i
h1, 0|Sz |1, 0i
h1, 0|Sz |1, −1i  .
Sz =  h1, 0|Sz |1, 1i
h1, −1|Sz |1, 1i h1, −1|Sz |1, 0i h1, −1|Sz |1, −1i
Since the basis states are eigenstates of Sz , we easily obtain


1 0 0
S z = ~ 0 0 0  .
0 0 −1
4
As we expected, the matrix representation of Sz in this basis is diagonal with the eigenvalues as the diagonal elements.
The matrix representations of Sx and Sy can be calculated with the help of the creation
and annihilation operators. The point is that we do not know beforehand the effects of
the operators Sx and Sy on the states |smi. Instead, we know those of the operators S+
and S− :
p
S± |smi = ~ (s ∓ m)(s ± m + 1)|s, m ± 1i .
By definition, S+ = Sx + iSy and S− = Sx − iSy . Therefore, Sx =
Sy = 2i1 (S+ − S− ). Consequently,
1
(h1m0 |S+ |1mi + h1m0 |S− |1mi)
2
1 p
=
~ (1 − m)(1 + m + 1)h1m0 |1, m + 1i
2
p
0
+ ~ (1 + m)(1 − m + 1)h1m |1, m − 1i
1 p
=
~ (1 − m)(1 + m + 1)δm0 ,m+1
2
p
+ ~ (1 + m)(1 − m + 1)δm0 ,m−1 .
1
(S+
2
+ S− ) and
h1m0 |Sx |1mi =
(2)
Using Eq. (2), we obtain
√

0
2 √0
√
~
Sx =  2 √0
2 .
2
0
2 0

By repeating a corresponding calculation for the operator Sy , we obtain
√


0
2 √0
√
i~
Sy = − − 2
0
2 .
√
2
0
− 2 0
4. An electron is at rest in an oscillating magnetic field
B = B0 cos(ωt)k̂ ,
where B0 and ω are constants.
a) Construct the Hamiltonian matrix for this system.
b) The electron starts out (at t = 0) in the spin-up state with respect to the xaxis (|χi = |↑(x) i). Determine |χ(t)i at any subsequent time. Beware: This is
a time-dependent Hamiltonian, so you cannot get |χ(t)i in the usual way from
5
the stationary states. Fortunately, in this case you can solve the time-dependent
Schrödinger equation directly.
c) Find the probability of getting −~/2, if you measure Sx . Answer:
γB0
2
sin(ωt) .
sin
2ω
d) What is the minimum field (B0 ) required to force a complete flip in Sx ?
Solution:
a) The Hamiltonian of the system is
H = T + V = 0 + (−µ · B) = −µz B0 cos(ωt) ,
(3)
where µ is the magnetic moment of an electron and µz is the z-component of µ.
For an electron µ = γS, where γ is the gyromagnetic ratio. Thus
H = −γB0 cos(ωt)Sz = −
γB0 ~
cos(ωt)σz .
2
b) We express the |χ(0)i in the eigenstates of Sz
1
|χ(0)i = √ (|↑i + |↓i)
2
(see lecture notes p.22). To solve |χ(t)i we write it as a linear combination of |↑i
and |↓i,
|χ(t)i = C↑ (t)|↑i + C↓ (t)|↓i .
The time-evolution of a quantum system is determined by the Schrödinger equation
∂
|χ(t)i = H(t)|χ(t)i
∂t
γB0 ~
i~ Ċ↑ (t)|↑i + Ċ↓ (t)|↓i = −
cos(ωt) (C↑ (t)|↑i − C↓ (t)|↓i) .
2
i~
We get
γB0
cos(ωt)C↑ (t)
Ċ↑ (t) = i
2
γB0
Ċ↓ (t) = −i
cos(ωt)C↓ (t)
2
Thus
γB0
C↑ (t) = C↑ (0) exp i
sin(ωt)
2ω
γB0
C↓ (t) = C↓ (0) exp −i
sin(ωt) .
2ω
⇔
⇔
1
|χ(t)i = √ eiα sin(ωt) |↑i + e−iα sin(ωt) |↓i ,
2
0
where α = γB
and we have used initial condition |χ(0)i =
2ω
C↑ (0) = C↓ (0) = √12 .
6
√1 (|↑i
2
+ |↓i) to set
c) Probability of getting −~/2 when measuring Sx , i.e. being in state |↓(x) i is
P↓ (t) = |h↓(x) |χ(t)i|2
1
= | (h↑| − h↓|) eiα sin(ωt) |↑i + e−iα sin(ωt) |↓i |2
2
1 iα sin(ωt)
− e−iα sin(ωt) |2 = |i sin(α sin(ωt))|2
=| e
2 γB0
2
= sin
sin(ωt) .
2ω
d) For Sx to flip completely, that is having P↓ (t) = 1 for some t, requires
γB0
2
sin
sin(ωt) = 1
2ω
γB0
sin
sin(ωt) = ±1
2ω
1
γB0
sin(ωt) = n +
π
2ω
2
γB0
π
for minimum B0 , n = 0 and
=
sin(ωt) = 1
2ω
2
πω
B0 =
.
γ
7