Download Solutions

1
Problem 9.2
Prove that
~
dA
kLk
=
dt
2m
1.1
Solution
One of the most overlooked features of the cross product is that it can be used
to figure out the area of a parallelogram defined by two vectors. This can be
seen in Figure 1.
Figure 1: The magnitude of the cross product is the area of the parallelogram
that they make. Taken from Wikipedia.
In Figure 2 (a), we see how ~r and d~r form a triangle. In Figure 2(b) we see
how this triangle will have half the area of a parallelogram defined by the cross
product of ~r and d~r.
This leads to the relation
1
k~r × d~rk
2
Now, let’s take the time derivative of both sides
dA =
dA
1 d
=
k~r × d~rk
dt
2 dt
We must make sure to use the chain rule
dA
1 d~r
d~r
= k × d~r + ~r × k
dt
2 dt
dt
1
Figure 2: (a) The triangle swept out by ~r and d~r. (b) This triangle is half of
the parallelogram formed by taking the cross product of ~r and d~r.
If the mass of a system is not changing, then we have
p~ = m~v = m
d~r
dt
Plugging this in gives
dA
1
=
k~r × p~k
dt
2m
But the terms in the absolute value brackets are just the angular momentum
since it is defined as
~ = ~r × p~
L
Plugging this in gives
~
dA
kLk
=
dt
2m
2
Problem 9.4
List all possible sets of quantum numbers for an electron in
(a) the 3d subshell
(b) the 3p subshell
2
2.1
2.1.1
Solution
Part (a)
We have n = 3 and l = 2. This means the possible values for ml are:
ml = 0, ±1, ±2
Since it’s an electron, the allowed values for ms are
1
2
This means that there are (2l+1)x2=10 possible combinations of quantum
numbers:
ms = ±
n = 3, l = 2, ml = 0, ms =
1
2
n = 3, l = 2, ml = 0, ms = −
1
2
n = 3, l = 2, ml = 1, ms =
1
2
n = 3, l = 2, ml = 1, ms = −
1
2
n = 3, l = 2, ml = −1, ms =
1
2
n = 3, l = 2, ml = −1, ms = −
n = 3, l = 2, ml = 2, ms =
1
2
n = 3, l = 2, ml = 2, ms = −
n = 3, l = 2, ml = −2, ms =
1
2
n = 3, l = 2, ml = −2, ms = −
2.1.2
1
2
1
2
1
2
Part (b)
We have n = 3 and l = 1. This means the possible values for ml are:
ml = 0, ±1
Since it’s an electron, the allowed values for ms are
1
2
This means that there are (2l+1)x2=6 possible combinations of quantum
numbers:
ms = ±
n = 3, l = 1, ml = 0, ms =
1
2
n = 3, l = 1, ml = 0, ms = −
3
1
2
n = 3, l = 1, ml = 1, ms =
1
2
n = 3, l = 1, ml = 1, ms = −
n = 3, l = 1, ml = −1, ms =
1
2
n = 3, l = 1, ml = −1, ms = −
3
1
2
1
2
Problem 9.8
A cylinder with mass M and radius R is rotating about it’s longitudinal axis at
angular velocity ω. There is a charge Q spread evenly across the curved surface
of the cylinder.
~
(a) What is the angular momentum, L?
(b) What is the magnetic moment µ
~?
(c) What is the g factor?
3.1
Solution
3.2
Part (a)
The moment of inertia for a cylinder (or a solid disk) rotating about it’s central
axis is
1
M R2
2
This means the angular momentum will be
I=
~ = Iω = 1 M R2 ω
L
2
Where the angular momentum is perpendicular to the plane of rotation and
aligns with the angular velocity.
3.3
Part (b)
Remember that the definition of µ
~ is
µ
~ = IA
With the direction determined by the righthand rule. In our case, the current
will be
I=
Q
T
T =
2π
ω
Since
4
We have
I=
Qω
2π
And the area will just be
A = πR2
Putting these together, we get
1
QR2 ω
2
Or, expressing this in terms of the angular momentum
µ
~=
Q~
L
M
µ
~=
3.3.1
Part (c)
The definition for magnetic moment is
Q ~
L
2M
This means that in our case, g = 2. This confirms that the spinning cylinder
with charge only on its surface is the ”bizarre picture that cannot be taken
seriously” described on page 307 of the book.
µ
~ =g
4
Problem 9.9
3
The Ω− particle has s = . What is the magnitude of its spin? What are
2
the allowed angles for it to make with the z axis? Does it obey the exclusion
principle?
4.1
Solution
We know that
~=
S
p
s(s + 1)~
So we get
r
3 3
( + 1)~
2 2
r
~ = 15 ~
S
2
The allowed angles can be found using
~=
S
5
Sz
~
kSk
−1
θ = cos
!
Rearranging gives
!
ms ~
−1
θ = cos
p
s(s + 1)~
Or
r
−1
θ = cos
ms
2
15
!
But now ms can take the values
ms = ±
1
3
or ±
2
2
This makes the allowed values:
3
Since s = is a half integer number, the Ω− is a fermion and will follow the
2
Pauli exclusion principle.
5
Problem 9.12
Give the spectroscopic notation for all of the following states.
(a) n = 7, l = 4 j = 9/2
(b) All possible states of an electron with n = 6 and l = 5.
5.1
5.1.1
Solution
Part (a)
The notation for this state is 7G9/2
5.1.2
Part (b)
There are only two possible states since the magnitude of l is fixed. They are
6H11/2
6
and
6H9/2
Problem 9.14
~ +S
~ derive an expression for L
~ ·S
~ in
(a) Starting with the expression J~ = L
terms of j, l, and s.
~ ·S
~ = kLkk
~ Sk
~ cos θ find θ for P1/2 , P3/2 , H9/2 , H11/2
(b) Using L
6
6.1
Solution
6.1.1
Part (a)
If we square both sides of
~ +S
~
J~ = L
We get
~ ·S
~
J 2 = L2 + S 2 + 2L
Rearranging gives
~ ·S
~ = 1 [J 2 − L2 − S 2 ]
L
2
Plugging in the values of J, L, and S will give
~ ·S
~ = 1 [j(j + 1)~2 − l(l + 1)~2 − s(s + 1)~2 ]
L
2
or
2
~ ·S
~ = ~ [j(j + 1) − l(l + 1) − s(s + 1)]
L
2
6.1.2
Part (b)
We have
~ ·S
~ = kLkk
~ Sk
~ cos θ
L
Or
~ ·S
~=
L
p
p
l(l + 1)~ s(s + 1)~ cos θ
We can use the result from before and see that
cos θ =
j(j + 1) − l(l + 1) − s(s + 1)
p
p
2 l(l + 1) s(s + 1)
Or
"
−1
θ = cos
j(j + 1) − l(l + 1) − s(s + 1)
p
p
2 l(l + 1) s(s + 1)
#
If we consider P1/2 we get
"
−1
θ = cos
#
r !
(1/2)(3/2) − 1(2) − (1/2)(3/2)
2
p
p
= cos −
≈ 144.7◦
3
2 1(2) (1/2)(3/2)
7
If we consider P3/2 we get
"
−1
θ = cos
#
r !
(3/2)(5/2) − 1(2) − (1/2)(3/2)
1
p
p
= cos
≈ 65.9◦
6
2 1(2) (1/2)(3/2)
If we consider H9/2 we get
"
θ = cos−1
#
r !
(9/2)(11/2) − 5(6) − (1/2)(3/2)
2
p
p
= cos −
≈ 129.2◦
5
2 5(6) (1/2)(3/2)
If we consider H11/2 we get
"
−1
θ = cos
7
#
(11/2)(13/2) − 5(6) − (1/2)(3/2)
p
p
cos
2 5(6) (1/2)(3/2)
1
3
r !
5
≈ 58.2◦
2
Problem 9.16
Show that the symmetric combination of two single particle wavefunctions
ψab (r~1 , r~2 ) = ψa (r~1 )ψb (r~2 ) + ψa (r~2 )ψb (r~1 )
displays the same exchange symmetry as bosons. Can two bosons occupy
the same quantum state?
7.1
Solution
ψab (r~2 , r~1 ) = ψa (r~2 )ψb (r~1 ) + ψa (r~1 )ψb (r~2 )
Changing the order of the terms gives
ψab (r~2 , r~1 ) = ψa (r~1 )ψb (r~2 ) + ψa (r~2 )ψb (r~1 )
But this is just the original wavefunction, so we have
ψab (r~2 , r~1 ) = ψab (r~1 , r~2 )
This is the exchange symmetry for bosons. The reason fermions cannot
occupy the same quantum state is that setting a = b sends their wavefunction
to zero. For bosons it just becomes
ψaa (r~1 , r~2 ) = ψa (r~1 )ψa (r~2 ) + ψa (r~2 )ψa (r~1 )
ψaa (r~1 , r~2 ) = 2ψa (r~1 )ψa (r~2 ) 6= 0
This means that bosons can occupy the same quantum state.
8