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Homework 6: # 4.1, 4.2, 4.10, 4.14, 4.15

Michael Good

Oct 4, 2004

4.1Prove that matrix multiplication is associative. Show that the product of twoorthogonal matrices is also orthogonal.

Answer:

Matrix associativity means

A(BC) = (AB)C

The elements for any row i and column j, are

A(BC) =k

Aik(m

BkmCmj)

(AB)C =m

(k

AikBkm)Cmj

Both the elements are the same. They only differ in the order of addition.As long as the products are defined, and there are finite dimensions, matrixmultiplication is associative.

Orthogonality may be defined by

AA = IThe Pauli spin matrices, x, and z are both orthogonal.

xx =

0 11 0

0 11 0

=

1 00 1

= I

zz = 1 00 1 1 00 1 = 1 00 1 = IThe product of these two:

xz =

0 11 0

1 00 1

=

0 11 0

q

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is also orthogonal:

qq = 0 11 0

0 11 0

= 1 00 1

= IMore generally, if

AA = 1 BB = 1then both A, and B are orthogonal. We can look at

ABAB =k

(AB)ik(AB)kj =k

ABkiABkj =k,s,r

aksbsiakrbrj

The elements are

k,s,r

aksbsiakrbrj = bsiaksakrbrj = bsi(AA)srbrj

This is

ABAB = bsisrbrj = BBij = ijTherefore the whole matrix is I and the product

ABAB = Iis orthogonal.

4.2

Prove the following properties of the transposed and adjoint matrices:AB = B A(AB) = BA

Answer:

For transposed matrices

AB = ABij = ABji = ajsbsi = bsiajs = Bis Asj = ( B A)ij = B AAs for the complex conjugate,

(AB) = (AB)From our above answer for transposed matrices we can say

AB = B A2

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And so we have

(AB) = (AB) = ( B A) = B A = BA4.10If B is a square matrix and A is the exponential of B, defined by the infiniteseries expansion of the exponential,

A eB = 1 + B +1

2B2 + ... +

Bn

n!+ ...,

then prove the following properties:

eBeC = eB+C, providing B and C commute.

A1 = eB

eCBC1

= CAC1

A is orthogonal if B is antisymmetric

Answer:

Providing that B and C commute;

BC CB = 0 BC = CB

we can get an idea of what happens:

(1+B +B2

2

+O(B3))(1+C+C2

2

+O(C3)) = 1+ C+C2

2

+B +BC+B2

2

+O(3)

This is

1+ (B +C)+1

2(C2+2BC+B2)+ O(3) = 1+ (B +C)+

(B + C)2

2+O(3) = eB+C

Because BC = CB and where O(3) are higher order terms with products of3 or more matrices. Looking at the kth order terms, we can provide a rigorousproof.

Expanding the left hand side of

eBeC = eB+C

and looking at the kth order term, by using the expansion for exp we get,noting that i + j = k

k0

BiCj

i!j!=

k0

BkjCj

(k j)!j!

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and using the binomial expansion on the right hand side for the kth orderterm, (a proof of which is given in Riley, Hobsen, Bence):

(B + C)k

k!=

1

k!

k0

k!

(k j)!j!BkjCj =

k0

BkjCj

(k j)!j!

we get the same term. QED.To prove

A1 = eB

We remember that

A1A = 1

and throw eB on the right

A1AeB = 1eB

A1eBeB = eB

and from our above proof we know eBeC = eB+C so

A1eBB = eB

Presto,

A1 = eB

To prove

eCBC1

= CAC1

its best to expand the exp

0

1

n!(CB C1)n = 1+CB C1+

CB C1CB C1

2+....+

CB C1CB C1CB C1...

n!+...

Do you see how the middle C1C terms cancel out? And how they canceleach out n times? So we are left with just the C and C1 on the outside of theBs.

0

1n!

(CB C1)n =0

1n!

CBnC1 = CeBC1

Remember A = eB and we therefore have

eCBC1

= CAC1

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To prove A is orthogonal

A = A1if B is antisymmetric

B = BWe can look at the transpose of A

A = 0

Bn

n!=

0

Bnn!

=0

(B)n

n!= eB

But from our second proof, we know that eB = A1, so

A = A1and we can happily say A is orthogonal.

4.14

Verify that the permutation symbol satisfies the following identity in termsof Kronecker delta symbols:

ijprmp = irjm imjr

Show thatijpijk = 2pk

Answer:To verify this first identity, all we have to do is look at the two sides of theequation, analyzing the possibilities, i.e. if the right hand side has

i = r j = m = i

we get +1. If

i = m j = r = i

we get 1. For any other set of i, j, r, and m we get 0.

For the left hand side, lets match conditions, if

i = r j = m = ithen ijp = rmp and whether or not ijp is 1 the product of the two gives

a +1. If

i = m j = r = i

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then rmp = jip = ijp and whether or not ijp is 1 the product is nowequal to 1.

These are the only nonzero values because for i,j,r,m, none can have thesame value as p. Since there are only three values, that any of the subscriptsmay take, the only non-zero values are the ones above. (not all four subscriptsmay be equal because then it would be = 0 as ifi = j or r = m).

To show thatijpijk = 2pk

we can use our previous identity, cast in a different form:

ijkimp = jmkp jpkm

This is equivalent because the product of two Levi-Civita symbols is foundfrom the deteriment of a matrix of deltas, that is

ijkrmp = irjmkp + imjpkr + ipjrkmimjrkpirjpkmipjmkr

For our different form, we set i = r. If we also set j = m, this is calledcontracting we get

ijkijp = jj kp jpkj

Using the summation convention, jj = 3,

ijkijp = 3kp kp

ijkijp = 2kp

4.15Show that the components of the angular velocity along the space set of axesare given in terms of the Euler angles by

x = cos + sin sin

y = sin sin cos

z = cos +

Answer:

Using the same analysis that Goldstein gives to find the angular velocityalong the body axes (x, y, z) we can find the angular velocity along the spaceaxes (x,y,z). To make a drawing easier, its helpful to label the axes of rotationfor , and .

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L.O.N.

z

z

We want

x = x + x + x

y = y + y + y

z = z + z + z

Lets start with z first to be different. If we look at the diagram carefully onpage 152, we can see that is along the line of nodes, that is revolves aroundthe line of nodes. Therefore because the line of nodes is perpendicular to the zspace axis there is no component of contributing to angular velocity aroundthe z space axis. z = 0. What about z? Well, revolves around z

. So thereis a component along z due to a changing . That component depends on howmuch angle there is between z and z, which is . Does this makes sense? Wefind the z part, which is the adjacent side to . Thus we have z = cos .Now lets look at z. We can see that just revolves around z in the first place!Right? So there is no need to make any transformation or make any changes.Lets take z = . Add them all up for our total z.

z = z + z + z = 0 + cos +

Now lets do the harder ones. Try x. What is x? Well, is along the lineof nodes, that is, changes and revolves around the line of nodes axis. To find

the x component of that, we just see that the angle between the line of nodesand the x axis is only , because they both lie in the same xy plane. Yes? Sox = cos . The adjacent side to with as the hypotenuse. Lets look at x.See how revolves around the z axis? Well, the z axis is perpendicular to thex axis there for there is no component of that contributes to the x space axis.x = 0. Now look at x. We can see that is along the z body axis, that is,it is in a whole different plane than x. We first have to find the component inthe same xy plane, then find the component of the x direction. So to get intothe xy plane we can take x,y = sin . Now its in the same plane. But whereis it facing in this plane? We can see that depends on the angle . If = 0we would have projected it right on top of the y axis! So we can make surethat if = 0 we have a zero component for x by multiplying by sin . So weget after two projections, x = sin sin . Add these all up for our total x,

angular velocity in the x space axis.

x = x + x + x = cos + sin sin + 0

Ill explain y for kicks, even though the process is exactly the same. Look

for y. is along the line of nodes. Its y component depends on the angle . So

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project it to the y axis. y = sin . Look for y. Its in a different plane again,so two projections are necessary to find its component. Project down to the xy

plane like we did before, x,y = sin and now we remember that if = 0 wewould have exactly placed it on top of the y axis. Thats good! So lets make it if = 0 we have the full sin , (ie multiply by cos because cos 0 = 1). But wealso have projected it in the opposite direction of the positive y direction, (throwin a negative). So we have y = sin cos . For y we note that revolvesaround the z axis, completely perpendicular to y. Therefore no component inthe y direction. y = 0. Add them all up

y = y + y + y = sin sin cos + 0

Here is all the s together

x = cos + sin sin

y = sin sin cos

z = cos +

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