Quantum Mechanics 2
Solution to problem set 5
1. First remember that the hydrogen energy eigenstates |nlmi must satisfy n > l and
|m| l, so there arent too many states to acoount for. Remember the form of the
Wigner-Eckart theorem
hn0 l0 m0 | Tq(k) |nlmi = (lk; mq| l0 m0 ) hn0 l0 k T (k) knli
~ = X x + Y y + Z z, but we know that we can construct a spherical
In addition X
irreducible tensor of rank 1 out of X, Y, Z
(1)
(1)
T1 = 12 (X iY ) ; T0
=Z .
If we look at the C.G. coefficient in the Wigner-Eckart theorem, we see that, since
in our case k = 1, then all matrix elements with l = l0 = 0 vanish identically.
Let us examine the case
(1)
h210| T0 |210i = h210| Z |210i
We can write this expectation value explicitly as
=
d3 x |210 (~x)|2 z.
Remember that d3 x = dxdydz and that |(~x)|2 must be (parity) even for z
z (regardless of nlm). Therefore the expectation value must vanish. If we pay
attention to the form of the Wigner-Eckart theorem, since (11; 00| 10) 6= 0, then in
order for the expectation value to vanish, it must be that
hn0 1k T (1) kn1i = 0
so we have
hn0 1m0 | Tq(1) |n1mi = 0.
Next, examine
(1)
h210| T0 |100i = h210| r cos |100i
The angular part of this expression is (just put the right coefficient of Y00 , and of
cos in terms of Y10 )
Z
1
1
dY10 () Y10 () =
3
3
1
so using the given integral, we get
(1)
h210| T0
2 6
a0
|100i = 8
3
and from the Wigner-Eckart theorem we know this means that
2 6
8 3 a0
h21k T (1) k10i =
.
(01; 00| 10)
The last Clebsh-Gordan coefficient is easy, since there is only one way to get a
|j = 1, m = 0i of |1, m0 i and |0, 0i, which means that (01; 00| 10) = 1. We have then
(1)
h210| T0
|100i =
(1)
h211| T1
|100i = h21
(1)
1| T1
2 6
a0
|100i = 8
3
(all the rest of the type h21m| T (1) |100i vanish, because of the Clebsh-Gordan coefficients, check it!). Next, and lastly, examine
(1)
h210| T0 |200i
The relevant
Clebsh-Gordan coefficient is (01; 00| 10) = 1, so we use the given
R 3
integral 0 r drR21 (r)R20 (r) to find
1
(1)
h210| T0 |200i = 27a0
3
= 3a0
and therefore
3a0
(01; 00| 10)
= 3a0 .
h21k T (1) k20i =
With our knowledge of the Clebsh-Gordan coefficient deduce then
(1)
(1)
h211| T1 |200i = h21 11| T1 |200i = 3a0 .
We have found the matrix element in terms of Tq(k) . Switching to the the original
~ is simple through
X
1 (1)
(1)
X = T1 T1
2
i (1)
(1)
Y = T1 + T1
2
(1)
Z = T0
2. First we write our operator in terms of its 0, 1 components
+iVy
Vx
2
.
V
=
z
Vq(1)
Vx iVy
Now, we use the given form of the rotaion matrix to get
+iVy
Vx
2
sin()
1
cos()
cos()
+
1
2
X (1)
1
(1)
V
dqq0 ()Vq0 =
2 sin() 2 cos()
2 sin()
2
Vx iVy
q0
1 cos()
2 sin() cos() + 1
2
V ( cos()) iVy Vz sin()
1 x
=
2 (Vz cos() Vx sin())
2
Vx cos() iVy + Vz sin()
If now we use
1 (1)
(1)
Vx = V1 V1
2
i (1)
(1)
Vy = V1 + V1
2
(1)
Vz = V0
we get
Vx = Vx cos() + Vz sin()
Vz = Vz cos() Vx sin()
Vy = Vy
which is just what is expected for a cartesian operator under a rotation about the
y axis.
~ , V~ we can construct spherical tensors with components
3. From the vector operators U
W1 =
(Wx iWy ) ; W0 = Wz
(W being either U or V ). We can construct a spherical tensor of rank k according
to
X
(q)
2)
Tk =
(k1 k2 ; q1 q2 | kq) Uq(k1 1 ) Vq(k
2
q1 q2
and in our case k1 = k2 = 1. Remember that in the previous homework, we have
found the Clebsh-Gordan coefficients corresponding to the addition of two spin 1,
3
so
(2)
= (11; 11| 22) U1 V1
1
(Ux + iUy ) (Vx + iVy )
=
2
1
=
(Ux Vx Uy Vy + iUx Vy + iUy Vx )
2
(2)
= (11; 01| 21) U0 V1 + (11; 10| 21) U1 V0
1
= (U0 V1 + U1 V0 )
2
1
(Uz Vx + Ux Vz + iUz Vy + iUy Vz )
=
2
T2
Now, for q = 1
T1
Similarly one gets
(2)
T0
(2)
T1
(2)
T2
1
= (2Uz Vz Ux Vx Uy Vy )
6
1
=
(Uz Vx + Ux Vz iUz Vy iUy Vz )
2
1
=
(Ux Vx Uy Vy iUx Vy iUy Vx )
2
4. (a) Let be any permutation, and denote |ui = (|1 i |2 i . . . |n i) and |vi =
(|1 i |2 i . . . |n i) then
E
hv |ui = (h1 | h2 | . . . hn |) (1) (2) . . . (n)
=
D
E
1 (1) 1 (2) . . . 1 (n) (|1 i |2 i . . . |n i)
v 1 ui .
Hence, = 1 and therefore
1 X
1 X 1
1 X
=
=
= S.
S =
n!
n!
n!
Also
1 X
A =
sgn()
n!
1 X
sgn() 1
=
n!
1 X
=
sgn( 1 ) 1
n!
1 X
=
sgn()
n!
= A.
4
(b) We prove that S |ui is a symmetrical vector by showing that for an arbitrary
permutation ,
S |ui =
1 X
1 X 0
1 X
|ui =
|ui =
|ui = S |ui .
n!
n!
n!
Similarly
1 X
1 X
A |ui =
sgn() |ui =
sgn() |ui
n!
n!
1 X
= (sgn( ))1
sgn()sgn( ) |ui
n!
1 X
= (sgn( ))
sgn( ) |ui
n!
1 X
= (sgn( ))
sgn( 0 ) 0 |ui = (sgn( )) A |ui
n!
(c) For S2
S2 =
1 X
1 X n!
1 X
S = S = S.
S =
S =
n!
n!
n!
n!
For A2
A2 =
1 X
1 X
1 X
sgn() A =
sgn() A =
A = A.
n!
n!
n!
SA
And for AS,
1 X
1 X
AS =
sgn () S = S
sgn () = 0
n!
n!
and
1 X
1 X
A = A
sgn () = 0
SA =
n!
n!
(d) Remember that a permutation is equivalent to specifying the order of the
state functions, for example, if the state functions of the independent particles
are |i i then the permutation = {1, 3, 2} means that the state-function is
|1 i |3 i |2 i. It is clear that the permutation
= {1, 2, 3}
is the identity operator. The permutations also satisfy closure. There are six
different permutations of (3) , so there are 36 closure relations that need to be
verified, I wont go through them here, here is an example
(1, 3, 2)(2, 3, 1) = (2, 1, 3).
5
It may also be proven for any n, by induction, but it is quite involved. The
existance of a reciprocal for each element is easy to verify
(1, 2, 3)(1, 2, 3) = I; (1, 3, 2)(1, 3, 2) = I; (3, 1, 2)(2, 3, 1) = I;
(2, 1, 3)(2, 1, 3) = I; (2, 3, 1)(3, 1, 2) = I; (3, 2, 1)(3, 2, 1) = I;
Indeed, the permutations (3) form a group.
5. (a) An electron has spin 1/2; thus, the eigenstates and eigenvalues are
n+
2
L
sin
nx
L
1
0
2
L
sin
nx
L
0
1
where En = 2 h
2 n2 /2mL2 . The additional degree of freedom, namely, the spin,
allows us to put two electrons in the first energy level, since this energy level
corresponds now to two different eigenstates: spin up and spin down. Thus,
there are two possible configurations for the ground state; they are depicted in
figure (1)
(b) There are three basic functions for each diagram in Fig. 13-2. For the left
diagram we have 1+ , 1 , 2+ , and for the right diagram we have 1+ , 1 , 2 .
Using the slater determinant we get
lef t
1+ (x1 ) 1 (x1 ) 2+ (x1 )
1
= 1+ (x2 ) 1 (x2 ) 2+ (x2 )
6 + (x ) (x ) + (x )
3
3
3
1
1
2
right
+ (x ) (x ) (x )
1
1
1
2
1 1+ 1
= 1 (x2 ) 1 (x2 ) 2 (x2 )
6 + (x ) (x ) (x )
3
3
3
1
1
2
and
6. Consider two identical particles of spin 1 that have the same spatial function (r).
(a) This was actually calculated in the previous problem set. For S = 2
|2, 2i = |1, 1; 1, 1i
1
|2, 1i = (|1, 0; 1, 1i + |1, 1; 1, 0i)
2
1
|2, 0i = (|1, 1; 1, 1i + 2 |1, 0; 1, 0i + |1, 1; 1, 1i)
6
1
|2, 1i = (|1, 1; 1, 0i + |1, 0; 1, 1i)
2
|2, 2i = |1, 1; 1, 1i .
6
Figure 1: The two possible states of the system in its ground state.
For S = 1
1
|1, 1i = ( |1, 0; 1, 1i + |1, 1; 1, 0i)
2
1
|1, 0i = ( |1, 1; 1, 1i + |1, 1; 1, 1i)
2
1
|1, 1i = ( |1, 1; 1, 0i + |1, 0; 1, 1i) .
2
For S = 0
1
|0, 0i = (|1, 1; 1, 1i + |1, 1; 1, 1i |1, 0; 1, 0i)
3
(b) We know that the two particles have the same spatial wavefunction. In addition, from part (a) of the question, we know that under permutation of the
particles
|s = 2, mi |s = 2, mi
|s = 1, mi |s = 1, mi
|s = 0, 0i |s = 0, 0i
so in order to make the overall wavefunction symmetric, the only allowed spin
states are those with s = 0, 2
