PHY6095/PHZ6166: homework assignment # 4:
Solutions
due Monday, April 22
2
1. Problem 1
Fill in the blanks (complete proof is required!)
pseudoscalar]:
[v ≡ polar vector, p ≡ pseudovector, s ≡ scalar, a ≡
a) v × p = v
b) p × p0 = p
c) v · v0 = s
d) p · p0 = s
e) s(v · p) = a
f) ap = v
g) v · (p × p0 ) = a
Under reflection in a plane with normal n̂, polar- and pseudovectors
v = v · n̂ + v − (v · n̂)n̂
p = p · n̂ + p − (p · n̂)n̂
transform as
v → v1 = v − 2(v · n̂)n̂
p → p1 = 2(p · n̂)n̂ − p
The cross-product transforms into
v1 × p1 = (v − 2(v · n̂)n̂) × (2(p · n̂)n̂ − p)
= −v × p + 2(v · n̂)(n̂ × p) + 2(p · n̂)(v × n̂)
Check the component of v1 × p1 parallel to n̂:
(v1 × p1 ) · n̂ = −(v × p) · n̂.
OK, looks like a polar vector. To be sure, check the component ⊥ to n̂:
n̂ × (v1 × p1 ) = −v(n̂ · p) + p(n̂ · v) + 2(v · n̂)(n̂(n̂ · p) − p)
+ 2(p · n̂)(v − n̂(v · n̂) = n̂ × (v × p)
Therefore, v × p is a polar vector.
The rest of the examples is done in the same way.
2. Problem 2
Consider the Hamiltonian for decay of K0 and K̄0 into pions
HKπ = g1 π † π † K0 + g2 π † π † K̄0 + g3 π † π † π † K0 + g4 π † π † π † K̄0 + H.c.
• Show that CPT requires that g1 = −g2∗ , g3 = g4∗ .
Recall that
C(π) = π; P (π) = −π; T (π) = π
C(K0 ) = K̄0 ; P (K0 ) = −K0 ; P (K̄0 ) = −K̄0 ; T (K0 ) = K0 ; T (K̄0 ) = K̄0 .
(1)
We work in the rest frame of K0 s, when the momenta of all π’s lie in the same plane. In this case, P and
T do not reverse the signs of the momenta. Also, recall that T (aA + bB) = a∗ T (A) + b∗ T (B). Applying
CPT to H, we find
CP T (HKπ ) = −g1∗ π † π † K̄0 − g2∗ π † π † K0 + g3∗ π † π † π † K̄0 + g4∗ π † π † π † K0 + H.c.
Therefore, CPT requires that
g1 = −g2∗ ; g3 = g4∗ .
(2)
3
• Derive the transformation properties of the short-living (S) and long-living (L) superpositions of K0 and
K̄0
1
KS† = p
g1 K0† + g2 K̄0†
|g1 |2 + |g2 |2
1
KL† = p
g2 K0† − g1 K̄0† .
|g1 |2 + |g2 |2
Using (2),
KS† = p
KL† = p
1
|g1 |2 + |g2 |2
1
|g1 |2 + |g2 |2
g1 K0† − g1∗ K̄0†
−g1∗ K0† − g1 K̄0† .
1
CP T (KS† ) = p
CP T g1 K0† − g1∗ K̄0†
|g1 |2 + |g2 |2
1 ∗ †
=√
−g1 K̄0 + g1∗ K0† = KS†
...
1
CP T (KL† ) = √
g1 K̄0† + g1∗ K0† = −KL† .
...
3. Problem 3
In certain semiconductor heterostructures, a two-dimensional gas contains holes rather electrons. Strong spinorbit interaction locks the angular momenta and spins of holes into the total momentum J = 3/2 with projections
Jz = ±1/2, ±3/2. The lowest energy state happens to be that with Jz = ±3/2. These states can be thought
of as having effective spin 3/2 rather than 1/2. The wavefunction of spin 3/2 state is symmetric only under
three-fold rotations, i.e., the Pauli matrices transform under rotations as σ± → σ± e∓3iθ . Suppose that a layer
of holes is sandwiched between two different materials such that the top-down symmetry is broken, but the
system is still invariant under a) rotations about the normal by an arbitrary angle; b) reflections in a plane
that contains the normal; and c) time-reversal. Construct the Rashba Hamiltonian for this system and find the
eigenstates and eigenergies.
Rotational invariants are p3+ σ− and p3− σ− . Under reflection in a vertical plane that contains the x axis, px →
px , py → −py ⇒ p± → p∓ while σx → −σx ; σy → −σy ⇒ σ± → −σ∓ . Therefore, the two rotational
invariants must enter with a relative minus sign. Also, HSO must be Hermitian, which requires a factor of i up
front:
α 3
HSO = i
p+ σ− − p3− σ+
2
. The full Hamiltonian
p2 ˆ
H=
I + HSO =
2m
p2 /2m
−α(py + ipx )3
3
−α(py − px )
p2 /2m
(3)
Eigenenergies
E=
p2
± αp3
2m
Eigenvectors
|p, si =
1
−ise3iφ
; s = ±1
4. Problem 4
The symmetry group of a square, D4 , consists of eight elements: rotations by 0◦ , 90◦ , 180◦ , and 270◦ about
an axis perpendicular to the plane of the square and passing through its center (chosen as the z−axis) plus
rotations by 180◦ about the lines that bisect opposite sides of the square (chosen as the x and y axes) plus
rotations by 180◦ about the square diagonals.
4
• List the effect of all operations of D4 on the coordinates x, y, z.
–
–
–
–
–
–
–
Identity: x, y, z → x, y, z
C4 about z (π/2 rotation): x, y, z → y, −x, z
C43 about z (3π/4 rotation): x, y, z → −y, x, z
C2 = C42 about z: x, y, z → −x, −y, z
C2 about x (U2x ): x, y, z → x, −y, −z
C2 about y (U2y ): x, y, z → −x, y, z
C2 about y = x (U2y=x ): x, y, z → y, x, −z
– C2 about y = −x (U2y=−x ): x, y, z → −y, −x, −z
• Construct a table, similar to that we constructed for group D2 in the class, for the effects of all these
operations on the following functions, which can represent 3d-orbitals:
u1 = z 2 (r), u2 = (x2 − y 2 )f (r), u3 = xyf (r),
u4 = xzf (r), u5 = yzf (r),
where r is the distance from the origin.
Functions u1−4 and z form 1D representations, because they transform into themselves up to a possible
sign change. Subjecting these functions to all operations listed above, we obtain characters (= ±1) for
the representations. Functions u4 and u5 transform as components of a 2D vector. For example, on C4 ,
u4 = xz → yz = u5 and u5 = yz → −xz = −u4 , which means that the transformation is described by the
matrix
0 1
(4)
−1 0
with Tr= 0. Likewise, C43 corresponds to a matrix
0 −1
1 0
(5)
also with Tr= 0 ,while C2 corresponds to
−1 0
0 −1
(6)
with Tr= −2. The table looks like this:
D4
u1 = z 2 f (r)
z
u2 = (x2 − y 2 )f (r)
u3 = xyf (r)
{u4 , u5 } = {xzf (r), yzf (r)}
A1
A2
B1
B2
E
I
1
1
1
1
2
C2
1
1
1
1
−2
2C4
1
1
−1
−1
0
U2
1
−1
1
−1
0
U20
1
−1
−1
1
0
• Show that the functions u1 , u2 , and u3 satisfy
Pα un = cαn un,
where Pα is each of the eight symmetry transformations, and thus that u1 , u2 , and u3 generate onedimensional representations of D4 . Show also that u4 and u5 generate two-dimensional representations,
i.e., that they transform as vectors.
5. Problem 5
Find which transitions between the states u1 , u2 , and u3 can be induced, to first order in the field, by an electric
field directed along the z axis.
Electric field along z transforms as A2 of D4 . Choose u1 (A1 ) as the initial state. Then A2 ⊗ A1 = A2 , which is
not one the u1−3 states. Choose u2 (B1 ) as the initial state. The characters of A2 ⊗ B1 are shown in the table
below
5
D4
2
u1 = z f (r)
z
2
u2 = (x − y 2 )f (r)
u3 = xyf (r)
{u4 , u5 } = {xzf (r), yzf (r)}
I
A1
1
A2
1
B1
1
B2
1
E
2
A2 ⊗ B 1 1
A2 ⊗ B 2 1
C2
1
1
1
1
−2
1
1
2C4
1
1
−1
−1
0
−1
1
U2
1
−1
1
−1
0
−1
−1
U20
1
−1
−1
1
0
1
−1
Without doing any further calculations, we see that A2 ⊗ B1 = B2 . Likewise, A2 ⊗ B2 = B1 . Thus, transitions
u2 ↔ u3 are allowed.