SQM problem set 1:

Groups, representations and their applications

1 Theory
1.1 Textbook refs
251b lecture notes chapter 8 gives a wide overview.
Ballentine chapter 3: Goes in considerable detail deriving the usual identications of cer-
tain operators with certain observables in quantum mechanics, from representation of Galilei
group. Studying this will not be needed for this course, but may be useful for students
wondering about foundational issues.
[Note: He also considers projective representations, which are representations U of a group
G, up to a phase factor: Instead of requiring U(gh) = U(g) U(h), one a priori allows U(gh) =
e
i(g,h)
U(g) U(h); at the level of Lie algebras this means allowing central extensions, i.e.
adding terms proportional to the identity to the commutation relations. In the end, this
turns out to aect only the commutator of innitesimal Galilei boosts G and translations P,
which is proportional to the mass times the identity operator I in (nonrelativistic) quantum
mechanics, whereas classically boosts and translations commute. Indeed for example in cases
where p
i
= mv
i
, a momentum (i.e. translation) eigenstate [p transforms to [p +m under a
boost with velocity , implying [G
i
, P
j
] = im
ij
I. Generalizations of this kind turn out to be
mostly irrelevant in applications.]
Ballentine chapter 7 and 13 gives a more systematic treatment of SO(3) and discrete spacetime
transformations like parity and time reversal. However the focus of this problem set will be
on more general concepts and ideas related to group theory and its applications.
In what follows I will give some additional comments which may be of use in solving the
problems.
1.2 The regular representation and nding all irreducible representations
The regular representation U
r
of a nite group G acts on the vector space V
r
constructed by letting
each group element g correspond to a orthonormal basis vector [g, that is the space of vectors
v =

g
v
g
[g with v
g
C. The action is given by
U
r
(g)[h = [gh .
So for example the regular representation of Z
3
= 1, ,
2
, the cyclic group of 3 elements generated
by = e
2i/3
, is given by U
r
()[1 = [, U
r
()[ = [
2
, U
r
()[
2
= [1, cyclically permuting
the three basis vectors.
The regular representation is interesting because it can be used to nd all irreps of a group, due
to the following theorem:
1
The decomposition of the regular representation into irreducible representations con-
tains every irreducible representation, with multiplicity given by the dimension of the
irrep.
As a check, recall that for the cyclic group of three elements Z
3
we have seen in the
notes that U
r
= U
1
U

2, where U

is the 1d irrep generated by multiplication with (i.e.

U

(
k
) =
k
), here living on the 1d subspace V

of V
r
spanned by [1 +
1
[ +
2
[.
As an example of a nonabelian group consider again the dihedral group D
4
, the symmetry
group of a square in the plane, generated by the /2 rotation R : (x, y) (y, x) and the
reection C : (x, y) (x, y), satisfying R
4
= 1, C
2
= 1 and CR = R
1
C. Then D
4
=
I, R, R
2
, R
3
, C, CR, CR
2
, CR
3
, and the regular representation is given by U
r
(R)[R
k
= [R
k+1
,
U
r
(R)[CR
k
= [CR
k1
, U
r
(C)[R
k
= [CR
k
, U
r
(C)[CR
k
= [R
k
. To minimize cluttering we will
abbreviate U
r
(g) as g in what follows.
Let us now decompose U
r
in irreps. We begin by diagonalizing R (and thus the full Z
4
subgroup
generated by R). Since R
4
= 1 the eigenvalues must be fourth roots of unity , i.e. 1, i, 1, i.
Dening
[
3

k=0

k
[R
k
,
the set [, C[
1
forms a basis of eigenvectors with eigenvalue . Indeed it is easy to check that
R[ = [ and RC[
1
= CR
1
[
1
= C[
1
.
When
1
= , i.e. = 1, the vectors [ C[
1
are in addition also eigenvectors of C,
with eigenvalue 1. Hence these produces 4 1d irreps R

, with basis vectors v

++
= [1 + C[1,
v
+
= [1 C[1, v
+
= [ 1 + C[ 1, v

= [ 1 C[ 1. The indices of v indicate the

eigenvalues of R and C, respectively, for example Rv
+
= v
+
, Cv
+
= v
+
.
When ,=
1
, i.e. = i, we get for each a 2d irrep U

, with basis vectors [, C[, and

with respect to this basis we have R =
_
0
0
1
_
and C =
_
0 1
1 0
_
. The representations U

for
= i and = i are clearly equivalent (ipping the order of the basis elements turns one in the
other).
Thus we see U
r
decomposes as
U
r
= U
++
U
+
U
+
U

U
i
U
i
,
with U
i
= U
i
, that is to say into four 1d irreps appearing with multiplicity 1 and one 2d irrep
appearing with multiplicity 2, conrming the general theorem for this case.
The dihedral group D
n
, i.e. the symmetries of a regular n-gon in the plane, is generated by a
rotation R for which R
n
= 1 and a reection C, for which CR = R
1
C. The group has 2n elements,
so the regular representation is 2n-dimensional. For every n-th root of unity we construct very
similarly vectors [ =

k

k
[R
k
, and C[, on which R and C act with the same 2 2 matrices
as given above. These 2-dimensional representations U

are again irreducible if ,=

1
, and U

is
equivalent to U

1. If =
1
, the representation U

is still further reducible in 1d representations

2
U

, exactly as before. Thus for n even we nd, with e

2i/n
:
U
r
= U
++
U
+
U
+
U

2 U

n1 ,
and for n odd
U
r
= U
++
U
+
U

2 U

n1 .
In this way we have found all irreducible representations of all the dihedral groups D
n
.
2 Problems
The group theoretic part of this nice problem was discussed in class. Getting to the bottom line
turned out a little harder than anticipated though, courtesy of a tougher than expected overlap
integral. I will give the solution below modulo a satisfactory resolution of this issue. If you have
ideas please let me or Gabriele know. (Gabriele Conti is the mystery TA operating behind the
scenes you can contact him at gbr.conti@gmail.com).
1. Applying group theory to nd energy level splittings:
Four protons are xed to the corners of a square of size L. When L , the ground state
for a single electron in the presence of these protons is fourfold degenerate. When L becomes
nite these degeneracies get lifted.
(a) What is the symmetry group at L = ? What is it at nite L?
(b) Describe in detail the splitting of the ground state energy level to leading order at large
L. Use representation theory to simplify the analysis.
(c) Generalize to n protons on a regular n-gon.
The following integrals may be useful (a [r
1
r
2
[):

d
3
r e
2|rr
1
|
= ,

d
3
r e
2|rr
1
| 1
|rr
2
|
=

a
_
1 (a + 1) e
2a
_

d
3
r e
|rr
1
||rr
2
|
=
_
1
3
a
2
+a + 1
_
e
a
,

d
3
r e
|rr
1
||rr
2
| 1
|rr
1
|
= (a + 1) e
a
.
The following integral appears to be needed to solve the problem completely analytically to
leading order in e
L
, but we have not been able so far to do it:

d
3
r e
|rr
1
||rr
2
| 1
|rr
3
|
= ?
3
3 Solutions
3.1 Problem 1a
The Hamiltonian is
H =
p
2
2m

3

k=0
g
[r r
k
[
where g =
e
2
4
0
and r
k
are the positions of the protons, which we can take to be at
r
0
= L(
1
2
,
1
2
, 0), r
1
= L(
1
2
,
1
2
, 0), r
2
= L(
1
2
,
1
2
, 0), r
3
= L(
1
2
,
1
2
, 0).
(a): When L , the neighborhoods of the four protons become eectively disconnected from each
other and we get an O(3) symmetry around each proton. In addition there is an S
4
permutation
symmetry permuting the four parts of space. More formally, we can write (r) =

k

k
(r r
k
),
where
k
vanishes outside of the quadrant in which r
k
is located. The Hamiltonian acts as H =

k
_
p
2
2m
+

g
|rr
k
|
_

k
(r r
k
). When L , we can drop the term with k

,= k in the
potential (since it vanishes wherever
k
is nonvanishing), so the action of H becomes H =

k
_
p
2
2m
+
g
|rr
k
|
_

k
(r r
k
), making the decoupling of the four quadrants manifest. The symmetry
R O(3)
k
then acts as U(R) (r) =

k

k
(R
1
k
(r r
k
)), and the permutation symmetry by
shuing the order of k = 0, 1, 2, 3, which is indeed irrelevant in the limit L .
In fact, there is even more symmetry, for which there is no classical counterpart: we can
independently rotate the phases of the individual components
k
, i.e.
k
e
i
k

k
, giving us a
U(1)
4
. Thus, all in all, we get a SO(3)
4
U(1)
4
S
4
symmetry group.
At nite L, the symmetry group is drastically reduced, to D
4
. The rotation R and reection C
are represented by the following O(3) matrices:
R =
_
_
_
0 1 0
1 0 0
0 0 1
_
_
_ , C =
_
_
_
1 0 0
0 1 0
0 0 1
_
_
_ ,
chosen such that Rr
k
= r
k+1
. We have R
4
= 1, C
2
= 1 and RC = CR
1
. The transformations
g D
4
are represented on wave functions as
U(g)(r) = (g
1
r) .
3.2 Problem 1b
In the limit LE
(0)
the ground state is fourfold degenerate, with ground state wave functions

k
(r) =
(0)
(r r
k
) =
1
_
a
3
0

e
|rr
k
|/a
0
.
4
In what follows we will choose length units such that the Bohr radius a
0
1. The symmetries g
act on these as
U(g)
k
(r) =
k
(g
1
r) =
(0)
((g
1
r) r
k
) =
(0)
(g
1
(r gr
k
)) =
(0)
(r gr
k
) =
g(k)
(r).
Thus we see that D
4
simply permutes the wave function indices like the corners of the square;
for example U(R)
0
=
1
and U(C)
1
=
3
. Note that in the fourth step we used the spherical
symmetry of the hydrogen ground state; for excited states this would be dierent, and we would
end up with a dierent representation of D
4
.
When L is nite but still large, the lowest energy eigenstates will to leading order be given by
linear combinations of the
k
. In principle one could compute these by rst computing the matrix
H
kl
=
k
[H[
l
and then diagonalizing it. However, using the symmetry, we can simplify this task
considerably.
The Z
4
subgroup generated by R is diagonalized by the linear combinations (writing U(R) just
as R from now on):
[ =
3

k=0

k
[
k
R[ = [ ,
where
4
= 1, i.e. 1, i, 1, i (since R
4
= 1). Reection ips between eigenvalues and

1
, as follows directly from the group structure: RC[ = CR
1
[ = C
1
[ =
1
C[. More
precisely by inspection we see that
C[ = [
1
.
For = 1 this maps the state to itself, for = 1 to minus itself, and the states with = i
get paired up. Thus we see that our four dimensional representation splits in the sum of two one
dimensional irreps and a two dimensional irrep:
V
(4d)
= V
(1d)
++
V
(1d)

V
(2d)
i
,
where V
++
is spanned by [1, V

by [1 and V
i
by [i and [i. This decomposition automatically
diagonalizes H and the scalar product:
[H[ =

[H[ , [ =

[ .
All that remains to be done is compute the energies
E

=
[H[
[
.
Notice that because V
i
forms an irreducible representation, the energies E
i
are necessarily the
same. So the energy level splits (at most) in 3 perturbed energy levels, one of which is still doubly
degenerate.
We compute
[ =

lk

lk

l
[
k
=

lk

lk

0
[
kl
= N

0
[
k

5
where N = 4 in the case at hand. (We will keep the expressions general in what follows, so they
will be immediately applicable to the generalization to arbitrary N). Similarly
[H[ =

lk

lk

l
[H[
k
=

lk

lk

0
[H[
kl
= N

0
[H[
k
.
Using the denition of the
k
as hydrogen ground state wave functions, we can furthermore write
H[
0
= E
(0)
+

k=0
V
k
[
0
,
where E
(0)
is the hydrogen ground state energy and V
k
(r) =
g
|rr
k
|
. In fact E
(0)
=
g
2a
0
, so in the
a
0
1 units we are using we can write
E
(0)
=
g
2
.
Thus if k = 0,

0
[H[
k
=
g
2
+

l=0

0
[V
l
[
0
.
and if k ,= 0:

0
[H[
k
=
g
2

0
[
k
+
0
[V
k
[
k
+

l=0,k

0
[V
l
[
k
,
where we could also substitute
0
[V
k
[
k
=
0
[V
0
[
k
. Using these expressions in the expression
found earlier for H[, we thus get
[H[ = N
_

g
2
+

l=0

0
[V
l
[
0
+

k=0

k
_

g
2

0
[
k
+
0
[V
0
[
k
+

l=0,k

0
[V
l
[
k

_
_
.
In this expression, we can use the integrals provided to compute
s
k

0
[
k
= (
1
3
a
2
k
+a
k
+ 1)e
a
k
with a
k
= [x
0
x
k
[ ,
w
0

0
[

l=0
V
l
[
0
= g

l=0
1
a
l
(1 (a
l
+ 1) e
2a
l
)
v
k

0
[V
0
[
k
= g(a
k
+ 1) e
a
k
.
Unfortunately for k ,= 0 we cant seem to gure out how to analytically compute
w
k

0
[

l=0,k
V
l
[
k
= ?
If we knew w
k
, this exercise would be a done deal, as
E

=
[H[
[
=

g
2
+w
0
+

N/2
k=1
(
k
+
k
)(
g
2
s
k
+v
k
+w
k
)
1 +

N/2
k=1
(
k
+
k
) s
k
,
where we also used s
k
= s
Nk
m v
k
= v
Nk
, w
k
= w
Nk
(in the case at hand s
1
= s
3
etc). Before
6
resorting to numerical evaluation, let us rewrite this in a a way that makes more explicit what
will be small in the large L limit. Note that for k ,= 0, s
k
, w
k
and v
k
are all exponentially small,
proportional to e
a
k
. So if we write
v
k
= v
k
s
k
, w
k
= w
k
s
k
,
then the hatted quantities are not exponentially suppressed and using in addition
= e
2i/N
, = 0, 1, . . . , N 1 ,
we get
E

g
2
+w
0
+ 2

N/2
k=1
cos(
2k
N
)(
g
2
+ v
k
+ w
k
) s
k
1 + 2

N/2
k=1
cos(
2k
N
) s
k
.
Notice that a
1
< a
2
< , so if we only keep the least exponentially suppressed terms, i.e. k = 1,
this becomes
E

g
2
+w
0
+ 2 cos(
2
N
)(
g
2
+ v
1
+ w
1
) s
1
1 + 2 cos(
2
N
) s
1

g
2
+w
0
+ 2 cos(
2
N
)(w
0
+ v
1
+ w
1
) s
1
where
s
1
= (
1
3
L
2
+L + 1)e
L
, w
0
=
g
L

l=0
[1 e
2i/N
[
[1 e
2il/N
[
, v
1
= g
L + 1
1
3
L
2
+L + 1
,
where for N = 4 the sum evaluates to 2 +
1

2
. The remaining unknown is
w
1
=

0
[

l=0,1
V
l
[
1

0
[
1

.
The product of
0
and
1
looks like a narrow almost-constant ridge between r
0
and r
1
, exponentially
dropping beyond this. Thus it would appear to be a good approximation to replace the above
expression with the average value of

l=0,1
V
l
on the ridge, which for N = 4 is
w
1
2g log(1 +

2)
1
L
1.763
g
L
.
Numerically we nd something quite close to this: for large L, w
1
1.77352
g
L
, with a 5%
deviation from this at L = 2, 0.1% at L = 5 and 0.00001% at L = 10. We will use the numerical
result in what follows.
Putting everything together this becomes for N = 4:
E

=
g
2

g
L
(2 +
1

2
)
2g
L
cos(
2
N
)
_
(2 +
1

2
) +
L(L + 1)
1
3
L
2
+L + 1
+ 1.77352
_
(
1
3
L
2
+L + 1) e
L
.
and to leading order in 1/L:
E

=
g
2

g
L
(2 +
1

2
)
2gL
3
cos(
2
N
)
_
(2 +
1

2
) + 3 + 1.77352
_
e
L
,
7
that is
E

g
2
2.707
g
L
1.3776 gL cos(
2
N
) e
L
.
Thus, we nd indeed 3 energy levels E
g
2
2.7
g
L
+ E, with
E
++
1.38 gLe
L
, E
i
0 , E

+1.38 gLe
L
,
where we used the irreducible representation labels. Reinstating the Bohr radius and recalling that
g/a
0
= 2E
(0)
with E
(0)
= 13.6 eV, we have
E
++
2.76 13.6 eV
L
a
0
e
L/a
0
, etc.
Notice that self-consistency of our approximations requires that E is much smaller than the
spacing between the ground state and the rst excited state, requiring L to be larger than a few
Bohr radii.
3.3 Problem 1c
The way we wrote down the solution for N = 4 makes generalization to N > 4 straightforward in
principle. Notice the appearance of bands in the large N limit.
For N = 2 the above describes the hydrogen molecular ion. In this case w
0
=
g
L
(1(L+1)e
2L
)
(now retaining also exponentially suppressed terms) and w
1
= 0, and we get for the symmetric and
antisymmetric states:
E

=
1 +
2
L
(1 (L + 1)e
2L
) (
L
2
3
+ 3L + 3) e
L
1 (
L
2
3
+L + 1) e
L
E
(0)
.
Combined with the proton-proton repulsion energy E
pp
= 2E
(0)
/L, this gives a total binding energy
E
b
= E

+E
pp
E
(0)
given by
E
b
=
2
L
(
2L
2
3
1)e
L
(L + 1)e
2L
1 (
L
2
3
+L + 1) e
L
For the symmetric state (plus sign) this has a minimum at L 2.5, where E
b
0.13 E
(0)
, suggesting
the formation of a bound state with a binding energy of 1.76 eV. The measured value is 2.8 eV.
The roughness of the approximation was to be expected since at this value of L the energy level
splitting E
+
E

0.55 E
(0)
and therefore mixing with the unperturbed excited hydrogen states
cannot be neglected.
8