Jeevanjee N. An Introduction to Tensors and Group Theory for Physicists

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30 2 Vector Spaces

completeness and noting that it is responsible for many of the nice features of Hilbert

spaces, in particular the generalized notion of a basis which we now describe.

Given a Hilbert space H and an orthonormal (and possibly inﬁnite) set {e

}⊂H,

the set {e

} is said to be an orthonormal basis for H if

|f)=0 ∀i ⇒ f =0. (2.28)

You can check (see Exercise 2.14 below) that in the ﬁnite-dimensional case this

deﬁnition is equivalent to our previous deﬁnition of an orthonormal basis. In the

inﬁnite-dimensional case, however, this deﬁnition differs substantially from the old

one in that we no longer require Span{e

}=H (recall that spans only include ﬁnite

linear combinations). Does this mean, though, that we now allow arbitrary inﬁnite

combinations of the basis vectors? If not, which ones are allowed? For L

([−a,a]),

for which {e

nπx

}

n∈Z

is an orthonormal basis, we mentioned in Example 2.13 that

any f ∈L

([−a,a]) can be written as

f =

∞



n=−∞

nπx

(2.29)

where



−a

|f |

dx =

∞



n=−∞

< ∞. (2.30)

(The ﬁrst equality in (2.30) should be familiar from quantum mechanics and follows

from Exercise 2.15 below.) The converse to this is also true, and this is where the

completeness of L

([−a,a]) is essential: if a set of numbers c

satisfy (2.30), then

the series

g(x) ≡

∞



n=−∞

nπx

(2.31)

converges, yielding a square-integrable function g.SoL

([−a,a]) is the set of all

expressions of the form (2.29), subject to the condition (2.30).Nowweknowhowto

think about inﬁnite-dimensional Hilbert spaces and their bases: a basis for a Hilbert

space is an inﬁnite set whose inﬁnite linear combinations, together with some suit-

able convergence condition, form the entire vector space.

Exercise 2.14 Show that the deﬁnition (2.28) of a Hilbert space basis is equivalent to our

original deﬁnition of a basis for a ﬁnite-dimensional inner product space V .

Exercise 2.15 Show that for f =



∞

n=−∞

nπx

,g=



∞

m=−∞

mπx

∈L

([−a,a]),

(f |g) =

∞



n=−∞

¯c

(2.32)

so that (·|·) on L

([−a,a]) can be viewed as the inﬁnite-dimensional version of (2.21),

the standard Hermitian scalar product on C

2.7 Non-degenerate Hermitian Forms and Dual Spaces 31

2.7 Non-degenerate Hermitian Forms and Dual Spaces

We are now ready to explore the connection between dual vectors and non-

degenerate Hermitian forms. Given a non-degenerate Hermitian form (·|·) on a

ﬁnite-dimensional vector space V , we can associate to any v ∈ V a dual vector

˜v ∈V

∗

deﬁned by

˜v(w) ≡(v|w). (2.33)

This deﬁnes a very important map,

L :V →V

∗

v → ˜v

which will crop up repeatedly in this chapter and the next. We will sometimes write

˜v as L(v) or (v|·), and refer to it as the metric dual of v.

Now, L is conjugate-linear since for v = cx +z, v,z,x ∈V ,

˜v(w) =(v|w) =(cx +z|w) =¯c(x|w) +(z|w) =¯c ˜x(w) +˜z(w) (2.34)

˜v =L(cx +z) =¯c ˜x +˜z =¯cL(x) +L(z). (2.35)

In Exercise 2.16 below you will show that the non-degeneracy of (·|·) implies that

L is one-to-one and onto, so L is an invertible map from V to V

∗

. This allows us to

identify V with V

∗

, a fact we will discuss further below.

Before we get to some examples, it is worth pointing out one potential point of

confusion: if we have a basis {e

}

i=1,...,n

and corresponding dual basis {e

}

i=1,...,n

it is not necessarily true that a dual vector e

in the dual basis is the same

as the metric dual L(e

); in particular, the dual basis vector e

is deﬁned rela-

tive to a whole basis {e

}

i=1,...,n

, whereas the metric dual L(e

) only depends on

what e

is, and does not care if we change the other basis vectors. Furthermore,

L(e

) depends on your non-degenerate Hermitian form (that is why we call it a met-

ric dual), whereas e

does not (remember that we introduced dual basis vectors in

Sect. 2.5, before we even knew what non-degenerate Hermitian forms were!). You

may wonder if there are special circumstances when we do have e

=L(e

);thisis

Exercise 2.17.

With that caveat let us proceed to some examples, where you will see that you

are already familiar with L from a couple of different contexts.

Exercise 2.16 Use the non-degeneracy of (·|·) to show that L is one-to-one, i.e. that

L(v) =L(w) ⇒ v = w. Combine this with the argument used in Exercise 2.7 to show

that L is onto as well.

Exercise 2.17 Given a basis {e

}

i=1,...,n

, under what circumstances do we have e

=˜e

for

all i?

The · in the notation (v|·) signiﬁes the slot into which a vector w is to be inserted, yielding the

number (v|w).

32 2 Vector Spaces

Example 2.21 Bras and kets in quantum mechanics

Let H be a quantum-mechanical Hilbert space with inner product (·|·).InDirac

notation, a vector ψ ∈H is written as a ket |ψ and the inner product (ψ, φ) is writ-

ten ψ|φ. What about bras, written as ψ|? What, exactly, are they? Most quantum

mechanics texts gloss over their deﬁnition, just telling us that they are in 1–1 cor-

respondence with kets and can be combined with kets as ψ|φ to get a scalar. We

are also told that the correspondence between bras and kets is conjugate-linear, i.e.

that the bra corresponding to c|ψ is ¯cψ|. From what we have seen in this section,

it is now clear that bras really are dual vectors, labeled in the same way as regular

vectors, because the map L allows us to identify the two. In short, ψ| is really just

L(ψ), or equivalently (ψ|·).

Example 2.22 Raising and lowering indices in relativity

Consider R

with the Minkowski metric, let B ={e

}

μ=1,...,4

and B



={e

}

μ=1,...,4

be the standard basis and dual basis for R

(we use a Greek index to conform with

standard physics notation

), and let v =



μ=1

∈ R

. What are the compo-

nents of the dual vector ˜v in terms of the v

? Well, as we saw in Sect. 2.5,the

components of a dual vector are just given by evaluation on the basis vectors, so

˜v

=˜v(e

) =(v|e

) =



) =



νμ

. (2.36)

In matrices, this reads

[˜v]



=[η]

[v]

(2.37)

so matrix multiplication of a vector by the metric matrix gives the corresponding

dual vector in the dual basis. Thus, the map L is implemented in coordinates by [η].

Now, we mentioned above that L is invertible; what does L

−1

look like in coordi-

nates? Well, by the above, L

−1

should be given by matrix multiplication by [η]

−1

the matrix inverse to [η]. Denoting the components of this matrix by η

μν

(so that

τμ

ντ

=δ

) and writing

f ≡L

−1

(f ) where f is a dual vector, we have

[

f ]=[η]

−1

[f ] (2.38)

or in components



νμ

. (2.39)

The expressions in (2.36) and (2.39) are probably familiar to you. In physics

one usually works with components of vectors, and in relativity the numbers v

As long as we are talking about ‘standard’ physics notation, you should also be aware that in

many texts the indices run from 0 to 3 instead of 1 to 4, and in that case the zeroth coordinate

corresponds to time.

2.8 Problems 33

are called the contravariant components of v and the numbers v

≡



νμ

(2.36) are referred to as the covariant components of v. We see now that the con-

travariant components of a vector are just its usual components, while its covariant

components are actually the components of the associated dual vector ˜v.Fora

dual vector f , the situation is reversed—the covariant components f

are its actual

components, and the contravariant components are the components of

f . Since L

allows us to turn vectors into dual vectors and vice versa, we usually do not bother

trying to ﬁgure out whether something is ‘really’ a vector or a dual vector; it can be

either, depending on which components we use.

The above discussion shows that the familiar process of “raising” and “lowering”

indices is just the application of the map L (and its inverse) in components. For an

interpretation of [η]

−1

as the matrix of a metric on R

4∗

, see Problem 2.7.

Exercise 2.18 Consider R

with the Euclidean metric. Show that the covariant and con-

travariant components of a vector in an orthonormal basis are identical. This explains why

we never bother with this terminology, nor the concept of dual spaces, in basic physics

where R

is the relevant vector space. Is the same true for R

with the Minkowski metric?

Example 2.23 L

([−a,a]) and its dual

In our above discussion of the map L we stipulated that V should be ﬁnite-

dimensional. Why? If you examine the discussion closely, you will see that the only

place where we use the ﬁnite-dimensionality of V is in showing that L is onto. Does

this mean that L is not necessarily onto in inﬁnite dimensions? Consider the Dirac

Delta functional δ ∈L

([−a,a])

∗

. Does

δ(g) =g(0)



δ(x)|g



(2.40)

for some function δ(x)? If we write g as g(x) =



∞

n=−∞

nπx

, then simply eval-

uating g at x =0gives

g(0) =

∞



n=−∞



δ(x)|g



(2.41)

which, when compared with (2.32), tells us that the function δ(x) must have Fourier

coefﬁcients c

=1 for all n. Such c

, however, do not satisfy (2.30) and hence δ(x)

cannot be a square-integrable function. So the dual vector δ is not in the image of

the map L, hence L is not onto in the case of L

([−a,a]).

2.8 Problems

2.1 Prove that L

([−a,a]) is closed under addition. You will need the trian-

gle inequality, as well as the following inequality, valid for all λ ∈ R:0≤



−a

(|f |+λ|g|)

dx.

34 2 Vector Spaces

2.2 In this problem we show that {r

} is a basis for H

), which implies that

} is a basis for

. We will gloss over a few subtleties here; for a totally rigorous

discussion see Sternberg [16] or our discussion in Chap. 4.

(a) Let f ∈H

), and write f as f =r

Y(θ,φ). Then we know that



Y =−l(l +1)Y. (2.42)

If you have never done so, use the expression (2.4)for

and the expressions

(2.11) for the angular momentum operators to show that



≡L

so that (2.42) says that Y is an eigenfunction of L

, as expected. You will need

to convert between cartesian and spherical coordinates. The theory of angular

momentum

then tells us that H

) has dimension 2l +1.

(b) Exhibit a basis for H

) by considering the function f

≡(x +iy)

and show-

ing that





=lf





≡(L

+iL

)





=0.

The theory of angular momentum then tells us that (L

−

)

≡ f

satisﬁes

) =(l −k)f

and that {f

}

0≤k≤2l

is a basis for H

l−k

we see that Y

satisﬁes L

=−l(l +1)Y

and L

as expected. Now use this deﬁnition of Y

to compute all the spherical

harmonics for l = 1, 2 and show that this agrees, up to normalization, with

the spherical harmonics as tabulated in any quantum mechanics textbook. If

you read Example 2.6 and did Exercise 2.5 then all you have to do is compute

, 0 ≤ k ≤ 2 and f

, 0 ≤ k ≤ 4 and show that these functions agree with the

ones given there.

2.3 In discussions of quantum mechanics you may have heard the phrase “angular

momentum generates rotations”. What this means is that if one takes a component

of the angular momentum such as L

and exponentiates it, i.e. if one considers the

operator

exp(−iφL

) ≡

∞



n=0

(−iφL

)

=I −iφL

(−iφL

)

(−iφL

)

+···

(the usual power series expansion for e

) then one gets the operator which represents

a rotation about the z axis by an angle φ. Conﬁrm this in one instance by explicitly

summing this power series for the operator [L

]

{x,y,z}

of Example 2.15 to get

See Sakurai [14] or Gasiorowicz [5] or our discussion in Chap. 4.

2.8 Problems 35

exp



−iφ[L

]

{x,y,z}



⎛

⎝

cosφ −sin φ 0

sin φ cos φ 0

001

⎞

⎠

the usual matrix for a rotation about the z-axis.

2.4 Let V be ﬁnite-dimensional, and consider the ‘double dual’ space (V

∗

)

∗

.De-

ﬁne a map

J :V →



∗



∗

v →J

where J

acts on f ∈V

∗

(f ) ≡f(v).

Show that J is linear, one-to-one, and onto. This means that we can identify (V

∗

)

∗

with V itself, and so the process of taking duals repeats itself after two iterations.

Note that no non-degenerate Hermitian form was involved in the deﬁnition of J !

2.5 Consider a linear operator A on a vector space V . We can deﬁne a linear oper-

ator on V

∗

called the transpose of A and denoted by A

as follows:



(f )



(v) ≡f(Av) where v ∈V, f ∈V

∗

If B is a basis for V and B

∗

the corresponding dual basis, show that





∗

=[A]

Thus the transpose of a matrix really has meaning; it is the matrix representation of

the transpose of the linear operator represented by the original matrix!

2.6 This problem will explore the notion of the Hermitian adjoint of a linear oper-

ator.

(a) Let A be a linear operator on a ﬁnite-dimensional, real or complex vector space

V with inner product (·|·). Using the transpose A

from the previous problem,

as well as the map L : V → V

∗

deﬁned in Sect. 2.7, we can construct a new

linear operator A

†

; this is known as the Hermitian adjoint of A, and is deﬁned

†

≡L

−1

◦A

◦L :V →V. (2.43)

Show that A

†

satisﬁes



†

v|w



=(v|Aw), (2.44)

which is the equation that is usually taken as the deﬁnition of the adjoint opera-

tor. The advantage of our deﬁnition (2.43) is that it gives an interpretation to A

†

;

it is the operator you get by transplanting A

, which originally acts on V

∗

, over

to V via L.

36 2 Vector Spaces

(b) Show that in an orthonormal basis {e

}

i=1,...,n



†



=[A]

†

where the dagger outside the brackets denotes the usual conjugate transpose of

amatrix(ifV is a real vector space, then the dagger outside the brackets will

reduce to just the transpose). You may want to prove and use the fact A

(Ae

†

, A is then said to be self-adjoint or Hermitian. Since A

†

is deﬁned with respect to an inner product, self-adjointness indicates a certain

compatibility between A and the inner product. Show that even when V is a

complex vector space, any eigenvalue of A must be real if A is self-adjoint.

(d) In part (b) you showed that in an orthonormal basis the matrix of a Hermitian

operator is a Hermitian matrix. Is this necessarily true in a non-orthonormal

basis?

2.7 Let g be a non-degenerate bilinear form on a vector space V (we have in

mind the Euclidean metric on R

or the Minkowski metric on R

). Pick an arbitrary

(not necessarily orthonormal) basis, let [g]

−1

be the matrix inverse of [g] in this

basis, and write g

μν

for the components of [g]

−1

.Alsoletf,h ∈V

∗

. Deﬁne a non-

degenerate bilinear form ˜g on V

∗

˜g(f,h) ≡g(

h) (2.45)

where

f =L

−1

(f ) as in Example 2.22. Show that

˜g

μν

≡˜g





μν

(2.46)

so that [g]

−1

is truly a matrix representation of a non-degenerate bilinear form

on V

∗

2.8 In this problem we will acquaint ourselves with P(R), the set of polynomials in

one variable x with real coefﬁcients. We will also meet several bases for this space

which you should ﬁnd familiar.

(a) P(R) is the set of all functions of the form

f(x)=c

x +c

+···+c

(2.47)

where n is arbitrary. Verify that P(R) is a (real) vector space. Then show that

P(R) is inﬁnite-dimensional by showing that, for any ﬁnite set S ⊂P(R), there

is a polynomial that is not in SpanS. Exhibit a simple inﬁnite basis for P(R).

(b) Compute the matrix corresponding to the operator

∈L(P (R)) with respect

to the basis you found in part (a).

of the form

(f |g) ≡



f(x)g(x)W(x)dx (2.48)

2.8 Problems 37

where W(x) is a nonnegative weight function. One can then take the basis

B ={1,x,x

,...} and apply the Gram–Schmidt process to get an orthogo-

nal basis. With the proper choice of range of integration [a,b] and weight func-

tion W(x), we can obtain (up to normalization) the various orthogonal poly-

nomials one meets in studying the various differential equations that arise in

electrostatics and quantum mechanics.

(i) Let [a,b]=[−1, 1]and W(x)=1. Consider the set S ={1,x,x

}⊂B.

Apply the Gram–Schmidt process to this set to get (up to normalization)

the ﬁrst four Legendre Polynomials

(x) =1

(x) =x

(x) =



−1



(x) =



−3x



The Legendre Polynomials show up in the solutions to the differential

equation (2.42), where we make the identiﬁcation x = cosθ . Since −1 ≤

cosθ ≤1, this explains the range of integration in the inner product.

(ii) Now let [a,b]=(−∞, ∞) and W(x)=e

−x

. Use Gram–Schmidt on S to

get the ﬁrst four Hermite Polynomials

(x) =1

(x) =2x

(x) =4x

−2

(x) =8x

−12x.

These polynomials arise in the solution to the Schrödinger equation for

a one-dimensional harmonic oscillator. Note that the range of integration

corresponds to the range of the position variable, as expected.

(iii) Finally, let [a,b]=(0,∞) and W(x) = e

−x

. Again, use Gram–Schmidt

on S to get the ﬁrst four Laguerre Polynomials

(x) =1

(x) =−x +1

(x) =



−4x +2



(x) =



−x

+9x

−18x +6



These polynomials arise as solutions to the radial part of the Schrödinger

equation for the hydrogen atom. In this case x is interpreted as a radial

variable, hence the range of integration (0, ∞).

Chapter 3

Tensors

Now that we are familiar with vector spaces we can ﬁnally approach our main sub-

ject, tensors. We will give the modern component-free deﬁnition, from which will

follow the usual transformation laws that used to be the deﬁnition.

From here on out we will employ the Einstein summation convention, which is

that whenever an index is repeated in an expression, once as a superscript and once

as a subscript, then summation over that index is implied. Thus an expression like

v =



i=1

becomes v =v

. We will comment on this convention in Sect. 3.2.

Also, in this chapter we will treat mostly ﬁnite-dimensional vector spaces and ignore

the complications and technicalities that arise in the inﬁnite-dimensional case. In the

few examples where we apply our results to inﬁnite-dimensional spaces, you should

rest assured that these applications are legitimate (if not explicitly justiﬁed), and that

rigorous justiﬁcation can be found in the literature.

3.1 Deﬁnition and Examples

A tensor of type (r, s) on a vector space V is a C-valued function T on

V ×···×V



 

r times

×V

∗

×···×V

∗

  

s times

(3.1)

which is linear in each argument, i.e.

T(v

+cw, v

,...,v

,...,f

)

=T(v

,...,v

,...,f

) +cT (w, v

,...,f

) (3.2)

and similarly for all the other arguments. This property is called multilinearity.Note

that dual vectors are (1, 0) tensors, and that vectors can be viewed as (0, 1) tensors:

v(f ) ≡f(v) where v ∈V, f ∈V

∗

. (3.3)

Similarly, linear operators can be viewed as (1, 1) tensors as

A(v, f ) ≡f(Av). (3.4)

N. Jeevanjee, An Introduction to Tensors and Group Theory for Physicists,

40 3Tensors

We take (0, 0) tensors to be scalars, as a matter of convention. You will show in

Exercise 3.1 below that the set of all tensors of type (r, s) on a vector space V ,

denoted T

(V ) or just T

, form a vector space. This should not come as much

of a surprise since we already know that vectors, dual vectors and linear operators

all form vector spaces. Also, just as linearity implies that dual vectors and linear

operators are determined by their values on basis vectors, multilinearity implies the

same thing for general tensors. To see this, let {e

}

i=1,...,n

be a basis for V and

}

i=1,...,n

the corresponding dual basis. Then, denoting the ith component of the

vector v

as v

and the jth component of the dual vector f

as f

,wehave(by

repeated application of multilinearity)

T(v

,...,v

,...,f

) =v

...v

...f



,...,e



≡v

...v

...f

,...,i

...j

where, as before, the numbers

...i

...j

≡T



,...,e



(3.5)

are called the components of T in the basis {e

}

i=1,...,n

. You should check that this

deﬁnition of the components of a tensor, when applied to vectors, dual vectors, and

linear operators, agrees with the deﬁnitions given earlier. Also note that (3.5)gives

us a concrete way to think about the components of tensors: they are the values of

the tensor on the basis vectors.

Exercise 3.1 By choosing suitable deﬁnitions of addition and scalar multiplication, show

that

(V ) is a vector space.

If we have a non-degenerate bilinear form on V , then we may change the type

of T by precomposing with the map L or L

−1

.IfT is of type (1, 1) with com-

ponents T

, for instance, then we may turn it into a tensor

T of type (2, 0) by

deﬁning

T(v,w)= T(v,L(w)). This corresponds to lowering the second index,

and we write the components of

T as T

, omitting the tilde since the fact that we

lowered the second index implies that we precomposed with L. This is in accord

with the conventions in relativity, where given a vector v ∈R

we write v

for the

components of ˜v when we should really write ˜v

. From this point on, if we have

a non-degenerate bilinear form on a vector space then we permit ourselves to raise

and lower indices at will and without comment. In such a situation we often do

not discuss the type of a tensor, speaking instead of its rank, equal to r +s, which

obviously does not change as we raise and lower indices.

Example 3.1 Linear operators in quantum mechanics

Thinking about linear operators as (1, 1) tensors may seem a bit strange, but in fact

this is what one does in quantum mechanics all the time! Given an operator H on

a quantum mechanical Hilbert space spanned by orthonormal vectors {e

} (which

in Dirac notation we would write as {|i}), we usually write H |i for H(e

), j |i

for ˜e

) =(e

), and j |H |i for (e

|He

).Thus,(3.4) would tell us that (using

orthonormal basis vectors instead of arbitrary vectors)