Symmetries PS4 - DAMTPdamtp.cam.ac.uk/user/eal40/teach/symmetry/Sheet4_solutions.pdf · 88-1, 0

Symmetries PS413 May 2013 10:38

Exercise Solutions Page 1





Question 8 Exercise sheet 4

r = 881, 0, 0 <, 81 , 0, -1<, 8-1, 1, 0<<881, 0, 0<, 81, 0, -1<, 8-1, 1, 0<<

881, 0, 0<, 81, 0, -1<, 8-1, 1, 0<< �� MatrixForm

1 0 0

1 0 -1

-1 1 0

m = 881, 0, 0<, 80, 1, 0<, 80, 0, -1<<881, 0, 0<, 80, 1, 0<, 80, 0, -1<<

881, 0, 0<, 80, 1, 0<, 80, 0, -1<< �� MatrixForm

1 0 0

0 1 0

0 0 -1

Not unitary since r.Transpose(r) is not the identity:

r.Transpose@rD881, 1, -1<, 81, 2, -1<, 8-1, -1, 2<<

881, 1, -1<, 81, 2, -1<, 8-1, -1, 2<< �� MatrixForm

1 1 -1

1 2 -1

-1 -1 2

Can find the other group elements by matrix multiplication of generators

m2 = r.m

881, 0, 0<, 81, 0, 1<, 8-1, 1, 0<<

881, 0, 0<, 81, 0, 1<, 8-1, 1, 0<< �� MatrixForm

1 0 0

1 0 1

-1 1 0

m3 = r.r.m

881, 0, 0<, 82, -1, 0<, 80, 0, 1<<

881, 0, 0<, 82, -1, 0<, 80, 0, 1<< �� MatrixForm

1 0 0

2 -1 0

0 0 1

m4 = r.r.r.m

881, 0, 0<, 81, 0, -1<, 81, -1, 0<<

881, 0, 0<, 81, 0, -1<, 81, -1, 0<< �� MatrixForm

1 0 0

1 0 -1

1 -1 0

r2 = r.r

881, 0, 0<, 82, -1, 0<, 80, 0, -1<<

881, 0, 0<, 82, -1, 0<, 80, 0, -1<< �� MatrixForm

1 0 0

2 -1 0

0 0 -1

r3 = r.r.r

881, 0, 0<, 81, 0, 1<, 81, -1, 0<<

881, 0, 0<, 81, 0, 1<, 81, -1, 0<< �� MatrixForm

1 0 0

1 0 1

1 -1 0

e = IdentityMatrix@3D881, 0, 0<, 80, 1, 0<, 80, 0, 1<<

881, 0, 0<, 80, 1, 0<, 80, 0, 1<< �� MatrixForm

1 0 0

0 1 0

0 0 1

Find H as sum of terms, recognising that m2 m2

T can be simplified into r r

Tetc

H = 2 * He.e + r.Transpose@rD + r2.Transpose@r2D + r3.Transpose@r3DL888, 8, 0<, 88, 20, 0<, 80, 0, 12<<

888, 8, 0<, 88, 20, 0<, 80, 0, 12<< �� MatrixForm

8 8 0

8 20 0

0 0 12

Find eigenvalues and normalised eigenvectors to construct U

Eigensystem@888, 8, 0<, 88, 20, 0<, 80, 0, 12<<D8824, 12, 4<, 881, 2, 0<, 80, 0, 1<, 8-2, 1, 0<<<

u = Transpose@881 � Sqrt@5D, 2 � Sqrt@5D, 0<, 80, 0, 1<, 8-2 � Sqrt@5D, 1 � Sqrt@5D, 0<<D

::1

5

, 0, -2

5

>, :2

5

, 0,

1

5

>, 80, 1, 0<>

2 Question8_Symmetries.nb

::1

5

, 0, -2

5

>, :2

5

, 0,

1

5

>, 80, 1, 0<> �� MatrixForm

1

5

0 -2

5

2

5

01

5

0 1 0

lambda = DiagonalMatrix@824, 12, 4<D8824, 0, 0<, 80, 12, 0<, 80, 0, 4<<

lambdahalf = 88Sqrt@24D, 0, 0<, 80, Sqrt@12D, 0<, 80, 0, 2<<

::2 6 , 0, 0>, :0, 2 3 , 0>, 80, 0, 2<>

::2 6 , 0, 0>, :0, 2 3 , 0>, 80, 0, 2<> �� MatrixForm

2 6 0 0

0 2 3 0

0 0 2

lambdainv = 881 � Sqrt@24D, 0, 0<, 80, 1 � Sqrt@12D, 0<, 80, 0, 1 � 2<<

::1

2 6

, 0, 0>, :0,1

2 3

, 0>, :0, 0,

1

2

>>

::1

2 6

, 0, 0>, :0,1

2 3

, 0>, :0, 0,

1

2

>> �� MatrixForm

1

2 6

0 0

01

2 3

0

0 01

2

uT = Transpose@uD

::1

5

,

2

5

, 0>, 80, 0, 1<, :-2

5

,

1

5

, 0>>

::1

5

,

2

5

, 0>, 80, 0, 1<, :-2

5

,

1

5

, 0>> �� MatrixForm

1

5

2

5

0

0 0 1

-2

5

1

5

0

Check H diagonalises properly and U is unitary

Htwid = uT.H.u �� Simplify

8824, 0, 0<, 80, 12, 0<, 80, 0, 4<<

Question8_Symmetries.nb 3

8824, 0, 0<, 80, 12, 0<, 80, 0, 4<< �� MatrixForm

24 0 0

0 12 0

0 0 4

uT.u

881, 0, 0<, 80, 1, 0<, 80, 0, 1<<

881, 0, 0<, 80, 1, 0<, 80, 0, 1<< �� MatrixForm

1 0 0

0 1 0

0 0 1

Construct elements of new unitary rep

rtwid = lambdainv.uT.r.u.lambdahalf �� Simplify

::3

5

, -2

5

, -6

5

>, :2

5

, 0,

3

5

>, :-6

5

, -3

5

,

2

5

>>

::3

5

, -2

5

, -6

5

>, :2

5

, 0,

3

5

>, :-6

5

, -3

5

,

2

5


3

5-

2

5-

6

5

2

50

3

5

-6

5-

3

5

2

5

Check unitary

rtwid.Transpose@rtwidD �� Simplify

881, 0, 0<, 80, 1, 0<, 80, 0, 1<<

881, 0, 0<, 80, 1, 0<, 80, 0, 1<< �� MatrixForm

1 0 0

0 1 0

0 0 1

mtwid = lambdainv.uT.m.u.lambdahalf �� Simplify

881, 0, 0<, 80, -1, 0<, 80, 0, 1<<

881, 0, 0<, 80, -1, 0<, 80, 0, 1<< �� MatrixForm

1 0 0

0 -1 0

0 0 1


mtwid.Transpose@mtwidD �� Simplify

881, 0, 0<, 80, 1, 0<, 80, 0, 1<<

881, 0, 0<, 80, 1, 0<, 80, 0, 1<< �� MatrixForm

1 0 0

0 1 0

0 0 1

Generate other elements either by multiplying rtwid and mtwid or applying transforms

r2twid = rtwid.rtwid �� Simplify

::1

5

, 0, -2 6

5

>, 80, -1, 0<, :-2 6

5

, 0, -1

5

>>

::1

5

, 0, -2 6

5

>, 80, -1, 0<, :-2 6

5

, 0, -1

5


1

50 -

2 6

5

0 -1 0

-2 6

50 -

1

5

r3twid = lambdainv.uT.r3.u.lambdahalf �� Simplify

::3

5

,

2

5

, -6

5

>, :-2

5

, 0, -3

5

>, :-6

5

,

3

5

,

2

5

>>

::3

5

,

2

5

, -6

5

>, :-2

5

, 0, -3

5

>, :-6

5

,

3

5

,

2

5


3

5

2

5-

6

5

-2

50 -

3

5

-6

5

3

5

2

5

mtwid = lambdainv.uT.m.u.lambdahalf �� Simplify

881, 0, 0<, 80, -1, 0<, 80, 0, 1<<

881, 0, 0<, 80, -1, 0<, 80, 0, 1<< �� MatrixForm

1 0 0

0 -1 0

0 0 1


m2twid = lambdainv.uT.m2.u.lambdahalf �� Simplify

::3

5

,

2

5

, -6

5

>, :2

5

, 0,

3

5

>, :-6

5

,

3

5

,

2

5

>>

::3

5

,

2

5

, -6

5

>, :2

5

, 0,

3

5

>, :-6

5

,

3

5

,

2

5


3

5

2

5-

6

5

2

50

3

5

-6

5

3

5

2

5


::1

5

, 0, -2 6

5

>, 80, 1, 0<, :-2 6

5

, 0, -1

5

>>

::1

5

, 0, -2 6

5

>, 80, 1, 0<, :-2 6

5

, 0, -1

5


1

50 -

2 6

5

0 1 0

-2 6

50 -

1

5


::3

5

, -2

5

, -6

5

>, :-2

5

, 0, -3

5

>, :-6

5

, -3

5

,

2

5

>>

::3

5

, -2

5

, -6

5

>, :-2

5

, 0, -3

5

>, :-6

5

, -3

5

,

2

5


3

5-

2

5-

6

5

-2

50 -

3

5

-6

5-

3

5

2

5

Find the sum of the squares of the characters of the representation - this is 16 which tells us it is made

up of two irreps as the group is of order 8 - therefore as it is a 3x3 rep it must be made up of one 1D

irrep (F1) and one 2D irrep (F2). See lecture notes pg 72 for 1D irreps of D4.


Tr@m4twidD^2 + Tr@m3twidD^2 + Tr@m2twidD^2 + Tr@mtwidD^2 +

Tr@eD^2 + Tr@rtwidD^2 + Tr@r2twidD^2 + Tr@r3twidD^2 �� Simplify

16

The 1D irrep is the trivial irrep as we can see by considering diagonalisation of r. (Also see lecture notes

pg 72 for 1D irreps of D4.)

Eigensystem@rD88ä, -ä, 1<, 880, ä, 1<, 80, -ä, 1<, 81, 1, 0<<<

P = Transpose@880, I, 1<, 80, -I, 1<, 81, 1, 0<<D880, 0, 1<, 8ä, -ä, 1<, 81, 1, 0<<

Pminus1 = Inverse@PD

::ä

2

, -ä

2

,

1

2

>, :-ä

2

,

ä

2

,

1

2

>, 81, 0, 0<>

rdash = Pminus1.r.P

88ä, 0, 0<, 80, -ä, 0<, 80, 0, 1<<

88ä, 0, 0<, 80, -ä, 0<, 80, 0, 1<< �� MatrixForm

ä 0 0

0 -ä 0

0 0 1

mdash = Pminus1.m.P

880, -1, 0<, 8-1, 0, 0<, 80, 0, 1<<

880, -1, 0<, 8-1, 0, 0<, 80, 0, 1<< �� MatrixForm

0 -1 0

-1 0 0

0 0 1


Question 10 Exercise sheet 4

Sa = 881, 0<, 80 , I<<881, 0<, 80, ä<<

881, 0<, 80, ä<< �� MatrixForm

K 1 0

0 äO

Sb = 88-1, 0<, 80 , 1<<88-1, 0<, 80, 1<<

88-1, 0<, 80, 1<< �� MatrixForm

K -1 0

0 1O

Calculate other elements:

Sa2 = Sa.Sa

881, 0<, 80, -1<<

881, 0<, 80, -1<< �� MatrixForm

K 1 0

0 -1O

Sa3 = Sa.Sa.Sa

881, 0<, 80, -ä<<

881, 0<, 80, -ä<< �� MatrixForm

K 1 0

0 -äO

Sab = Sa.Sb

88-1, 0<, 80, ä<<

88-1, 0<, 80, ä<< �� MatrixForm

K -1 0

0 äO

Sa2b = Sa.Sa.Sb

88-1, 0<, 80, -1<<

88-1, 0<, 80, -1<< �� MatrixForm

K -1 0

0 -1O

Sa3b = Sa.Sa.Sa.Sb

88-1, 0<, 80, -ä<<

88-1, 0<, 80, -ä<< �� MatrixForm

K -1 0

0 -äO

e = Sb.Sb

881, 0<, 80, 1<<

881, 0<, 80, 1<< �� MatrixForm

K 1 0

0 1O

Find the sum of the squares of the characters of the representation - this is 16 which tells us it is

reducible as this is greater than the order of the group - also we can see that it is in block diagonal form.

Abs@Tr@eDD^2 + Abs@Tr@SaDD^2 + Abs@Tr@SbDD^2 + Abs@Tr@SabDD^2 + Abs@Tr@Sa2DD^2 +

Abs@Tr@Sa3DD^2 + Abs@Tr@Sa2bDD^2 + Abs@Tr@Sa3bDD^2 �� Simplify

16

Similarly for T

Ta = 88I, 0<, 81 , 1<<88ä, 0<, 81, 1<<

88ä, 0<, 81, 1<< �� MatrixForm

K ä 0

1 1O

Tb = 88-1, 0<, 8I + 1 , 1<<88-1, 0<, 81 + ä, 1<<

88-1, 0<, 81 + ä, 1<< �� MatrixForm

K -1 0

1 + ä 1O

Ta2 = Ta.Ta

88-1, 0<, 81 + ä, 1<<

88-1, 0<, 81 + ä, 1<< �� MatrixForm

K -1 0

1 + ä 1O

Ta3 = Ta.Ta.Ta

88-ä, 0<, 8ä, 1<<

88-ä, 0<, 8ä, 1<< �� MatrixForm

-ä 0

ä 1


Tab1 = Ta.Tb

88-ä, 0<, 8ä, 1<<

88-ä, 0<, 8ä, 1<< �� MatrixForm

-ä 0

ä 1

Ta2b = Ta.Ta.Tb

881, 0<, 80, 1<<

881, 0<, 80, 1<< �� MatrixForm

K 1 0

0 1O

Ta3b = Ta.Ta.Ta.Tb

88ä, 0<, 81, 1<<

88ä, 0<, 81, 1<< �� MatrixForm

K ä 0

1 1O

Abs@Tr@eDD^2 + Abs@Tr@TaDD^2 + Abs@Tr@TbDD^2 + Abs@Tr@Tab1DD^2 +

Abs@Tr@Ta2DD^2 + Abs@Tr@Ta3DD^2 + Abs@Tr@Ta2bDD^2 + Abs@Tr@Ta3bDD^216

Sum of the squares of the characters of the representation is 16 which is greater than the order of the

group so it is reducible. Comparing the characters of the elements of the group in each rep, they are not

the same for all elements (the ab elements gain a minus sign) and so the reps are inequivalent - if they

were related by a similarilty transform the trace of the matrix would be preserved, and so they would be

the same for all corresponding elements.


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