Chapter 1 Lorentz Group and Lorentz Invariance where one second is deﬁned (1967) to be 9192631770...

Chapter 1

Lorentz Group and LorentzInvariance

In studying Lorentz-invariant wave equations, it is essential that we put our under-standing of the Lorentz group on firm ground. We first define the Lorentz transfor-mation as any transformation that keeps the 4-vector inner product invariant, andproceed to classify such transformations according to the determinant of the transfor-mation matrix and the sign of the time component. We then introduce the generatorsof the Lorentz group by which any Lorentz transformation continuously connected tothe identity can be written in an exponential form. The generators of the Lorentzgroup will later play a critical role in finding the transformation property of the Diracspinors.

1.1 Lorentz Boost

Throughout this book, we will use a unit system in which the speed of light c is unity.This may be accomplished for example by taking the unit of time to be one secondand that of length to be 2.99792458 × 1010 cm (this number is exact1), or taking theunit of length to be 1 cm and that of time to be (2.99792458 × 1010)−1 second. Howit is accomplished is irrelevant at this point.

Suppose an inertial frame K (space-time coordinates labeled by t, x, y, z) is mov-ing with velocity β in another inertial frame K ′ (space-time coordinates labeled byt′, x′, y′, z′) as shown in Figure 1.1. The 3-component velocity of the origin of K

1One cm is defined (1983) such that the speed of light in vacuum is 2.99792458 × 1010 cm persecond, where one second is defined (1967) to be 9192631770 times the oscillation period of thehyper-fine splitting of the Cs133 ground state.

5

6 CHAPTER 1. LORENTZ GROUP AND LORENTZ INVARIANCE

K' K

y'

x'

y

xβ− K'βK

(E',P') (E,P)

K' Kframe frame

Figure 1.1: The origin of frame K is moving with velocity �β = (β, 0, 0) in frame K ′,

and the origin of frame K ′ is moving with velocity −�β in frame K. The axes x andx′ are parallel in both frames, and similarly for y and z axes. A particle has energymomentum (E, �P ) in frame K and (E ′, �P ′) in frame K ′.

measured in the frame K ′, �β′K , is taken to be in the +x′ direction; namely,

�β′K (velocity of K in K ′) = (β, 0, 0)

def≡ �β. (1.1)

Assume that, in the frame K ′, the axes x, y, z are parallel to the axes x′, y′, z′. Then,the velocity of the origin of K ′ in K, �βK′ , is

�βK′ = −�β′K = (−β, 0, 0) (velocity of K ′ in K). (1.2)

Note that �β′K (�βK′) is measured with respect to the axes of K ′ (K).

If a particle (or any system) has energy and momentum (E, �P ) in the frame K,

then the energy and momentum (E ′, �P ′) of the same particle viewed in the frame K ′

are given by

E ′ =E + βPx√

1 − β2, P ′

y = Py ,

P ′x =

βE + Px√1 − β2

, P ′z = Pz, .

(1.3)

This can be written in a matrix form as(E ′

P ′x

)=

(γ η

η γ

) (E

Px

),

(P ′

y

P ′z

)=

(Py

Pz

)(1.4)

with

γ ≡ 1√1 − β2

, η ≡ βγ =β√

1 − β2. (1.5)

Note that γ and η are related by

γ2 − η2 = 1 . (1.6)

1.1. LORENTZ BOOST 7

K' K

y'

x'

y

x

K'βK

(E',P') (E,P)

K' frame K frame

β−

Figure 1.2: Starting from the configuration of Figure 1.1, the same rotation is appliedto the axes in each frame. The resulting transformation represents a general Lorentzboost.

Now start from Figure 1.1 and apply the same rotation to the axes of K andK ′ within each frame without changing the motions of the origins of the frames andwithout touching the paqrticle (Figure 1.2). Suppose the rotation is represented by

a 3 × 3 matrix R. Then, the velocity of K ′ in K, �βK′ , and and the velocity of K inK ′, �β′

K , are rotated by the same matrix R,

�β′K → R�β′

K , �βK′ → R�βK′ , (1.7)

and thus we still have�β′

K = −�βK′def≡ �β , (1.8)

where we have also redefined the vector �β which is well-defined in both K and K ′

frames in terms of �βK′ and �β′K , respectively. The transformation in this case can

be obtained by noting that, in (1.4), the component of momentum transverse to �β

does not change and that Px, P ′x are the components of �P , �P ′ along �β in each frame.

Namely, the tranformation can be written as

(E ′

P ′‖

)=

(γ η

η γ

) (E

P‖

), �P ′

⊥ = �P⊥ , (1.9)

where ‖ and ⊥ denote components parallel and perpecdicular to �β, repectively. Note

that �P ′⊥ and �P⊥ are 3-component quantities and the relation �P ′

⊥ = �P⊥ holds compo-nent by component because we have applied the same roation R in each frame.

The axes of K viewed in the frame K ′ are no longer perpendicular to each othersince they are contracted in the direction of �β′

K . Thus, the axes of K in generalare not parallel to the corresponding axes of K ′ at any time. However, since thesame rotation is applied in each frame, and since components transverse to �β arethe same in both frames, the corresponding axes of K and K ′ are exactly parallel


when projected onto a plane perpendicular to �β in either frames. The transformation(1.9) is thus correct for the specific relative orientation of two frames as defined here,and such transformation is called a Lorentz boost, which is a special case of Lorentztransformation defined later in this chapter for which the relative orientation of thetwo frames is arbitrary.

1.2 4-vectors and the metric tensor gµν

The quantity E2 − �P 2 is invariant under the Lorentz boost (1.9); namely, it has thesame numerical value in K and K ′:

E ′2 − �P ′2 = E ′2 − (P ′2‖ + �P ′2

⊥ )

= (γE + ηP‖)2 −

[(ηE + γP‖)

2 + �P 2⊥

]= (γ2 − η2)︸︷︷︸

1

E2 + (η2 − γ2)︸︷︷︸−1

P 2‖ − �P 2

⊥

= E2 − �P 2 ,

(1.10)

which is the invariant mass squared m2 of the system. This invariance applies to anynumber of particles or any object as long as E and �P refer to the same object.

The relative minus sign between E2 and �P 2 above can be treated elegantly asfollows. Define a 4-vector P µ (µ = 0, 1, 2, 3) by

P µ = (P 0, P 1, P 3, P 4)def≡ (E, Px, Py, Pz) = (E, �P ) (1.11)

called an energy-momentum 4-vector where the index µ is called the Lorentz index(or the space-time index). The µ = 0 component of a 4-vector is often called ‘timecomponent’, and the µ = 1, 2, 3 components ‘space components.’

Define the inner product (or ‘dot’ product) A ·B of two 4-vectors Aµ = (A0, �A)

and Bµ = (B0, �B) by

A ·B def≡ A0B0 − �A · �B = A0B0 − A1B1 − A2B2 − A3B3 . (1.12)

Then, P 2 ≡ P ·P is nothing but m2:

P 2 = P 02 − �P 2 = E2 − �P 2 = m2 (1.13)

which is invariant under Lorentz boost. This inner product P ·P is similar to the normsquared �x2 of an ordinary 3-dimensional vector �x, which is invariant under rotation,except for the minus signs for the space components in the definition of the innerproduct. In order to handle these minus signs conveniently, we define ‘subscripted’components of a 4-vector by

A0def≡ A0, Ai

def≡ −Ai (i = 1, 2, 3) . (1.14)

1.2. 4-VECTORS AND THE METRIC TENSOR Gµν 9

Then the inner product (1.12) can be written as

A ·B = A0B0 + A1B

1 + A2B2 + A3B

3 def≡ AµBµ = AµBµ , (1.15)

where we have used the convention that when a pair of the same index appears in thesame term, then summation over all possible values of the index (µ = 0, 1, 2, 3 in thiscase) is implied. In general, we will use Roman letters for space indices (take values1,2,3) and greek letters for space-time (Lorentz) indices (take values 0,1,2,3). Thus,

xiyi =3∑

i=1

xiyi (= �x · �y), (Aµ + Bµ)Cµ =3∑

µ=0

(Aµ + Bµ)Cµ , (1.16)

but no sum over µ or ν in

AµBν + CµD

ν (µ, ν not in the same term). (1.17)

When a pair of Lorentz indices is summed over, usually one index is a subscriptand the other is a superscript. Such indices are said to be ‘contracted’. Also, it isimportant that there is only one pair of a given index per term. We do not considerimplied summations such as AµBµCµ to be well-defined. [(Aµ +Bµ)Cµ is well-definedsince it is equal to AµCµ + BµCµ.]

Now, define the metric tensor gµν by

g00 = 1 , g11 = g22 = g33 = −1 , gµν = 0 (µ �= ν) (1.18)

which is symmetric:gµν = gνµ . (1.19)

The corresponding matrix G is defined as

{gµν}def≡

0 1 2 3 = ν

0

1

2

3

1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

||

µ

def≡ G (1.20)

When we form a matrix out of a quantity with two indices, by definition we take thefirst index to increase downward, and the second to increase to the right.

As defined in (1.14) for a 4-vector, switching an index between superscript andsubscript results in a sign change when the index is 1,2, or 3, while the sign isunchanged when the index is zero. We adopt the same rule for the indices of gµν . In


fact, from now on, we enforce the same rule for all space-time indices (unless otherwisestated, such as for the Kronecker delta below). Then we have

gµν = gµν , gµν = gµ

ν = δµν (1.21)

where δµν is the Kronecker’s delta (δµν = 1 if µ = ν, 0 otherwise) which we define tohave only subscripts. Then, gµν can be used together with contraction to ‘lower’ or‘raise’ indices:

Aν = gµνAµ, Aν = gµνAµ (1.22)

which are equivalent to the rule (1.14).The inner product of 4-vectors A and B (1.12) can also be written in matrix form

as

A ·B = AµgµνBν =

(A0 A1 A2 A3)

1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

B0

B1

B2

B3

= AT GB . (1.23)

When we use 4-vectors in matrix form, they are understood to be column vectorswith superscripts, while their transpose are row vectors:

Adef≡

A0

A1

A2

A3

=

A0

Ax

Ay

Az

, AT = (A0, A1, A2, A3) (in matrix form). (1.24)

1.3 Lorentz group

The Lorentz boost (1.4) can be written in matrix form as

P ′ = ΛP (1.25)

with

P ′ =

E ′

P ′x

P ′y

P ′z

, Λ =

γ η 0 0

η γ 0 0

0 0 1 0

0 0 0 1

, P =

E

Px

Py

Pz

. (1.26)

In terms of components, this can be written as

P ′µ = ΛµνP

ν , (1.27)

1.3. LORENTZ GROUP 11

where we have defined the components of the matrix Λ by taking the first index to besuperscript and the second to be subscript (still the first index increases downwardand the second index increases to the right):

Λdef≡ {Λµ

ν} =

0 1 2 3 = ν

0

1

2

3

γ η 0 0

η γ 0 0

0 0 1 0

0 0 0 1

||

µ

(1.28)

For example, Λ01 = η and thus Λ01 = −η, etc. Why do we define Λ in this way? The

superscripts and subscript in (1.27) were chosen such that the index ν is contractedand that the index µ on both sides of the equality has consistent position, namely,both are superscript.

We have seen that P 2 = E2 − �P 2 is invariant under the Lorentz boost given by(1.4) or (1.9). We will now find the necessary and sufficient condition for a 4 × 4matrix Λ to leave the inner product of any two 4-vectors invariant. Suppose Aµ andBµ transform by the same matrix Λ:

A′µ = ΛµαAα, B′ν = Λν

βBβ . (1.29)

Then the inner products A′ ·B′ and A ·B can be written using (1.22) as

A′ ·B′ = A′ν︸︷︷︸

gµν A′µ︸︷︷︸Λµ

αAα

B′ν︸︷︷︸Λν

βBβ

= (gµνΛµ

αΛνβ)AαBβ

A ·B = Aβ︸︷︷︸gαβAα

Bβ = gαβAαBβ .

(1.30)

In order for A′ ·B′ = A ·B to hold for any A and B, the coefficients of AαBβ shouldbe the same term by term (To see this, set Aν = 1 for ν = α and 0 for all else, andBν = 1 for ν = β and 0 for all else.):

gµνΛµ

αΛνβ = gαβ . (1.31)

On the other hand, if Λ satisfies this condition, the same derivation above can betraced backward to show that the inner product A ·B defined by (1.12) is invariant.Thus, (1.31) is the necessary and sufficient condition.


What does the condition (1.31) tell us about the nature of the matrix Λ? Using(1.22), we have gµνΛ

µα = Λνα, then the condition becomes

ΛναΛνβ = gαβ

raise α on both sides−→ Λν

αΛνβ = gα

β (= δαβ) . (1.32)

Compare this with the definition of the inverse transformation Λ−1:

Λ−1Λ = I or (Λ−1)ανΛ

νβ = δαβ , (1.33)

where I is the 4× 4 indentity matrix. The indexes of Λ−1 are superscript for the firstand subscript for the second as before, and the matrix product is formed as usual bysumming over the second index of the first matrix and the first index of the secondmatrix. We see that the inverse matrix of Λ is obtained by

(Λ−1)αν = Λν

α , (1.34)

which means that one simply has to change the sign of the components for whichonly one of the indices is zero (namely, Λ0

i and Λi0) and then transpose it:

Λ =

Λ00 Λ0

1 Λ02 Λ0

3

Λ10 Λ1

1 Λ12 Λ1

3

Λ20 Λ2

1 Λ22 Λ2

3

Λ30 Λ3

1 Λ32 Λ3

3

, −→ Λ−1 =

Λ00 −Λ1

0 −Λ20 −Λ3

0

−Λ01 Λ1

1 Λ21 Λ3

1

−Λ02 Λ1

2 Λ22 Λ3

2

−Λ03 Λ1

3 Λ23 Λ3

3

. (1.35)

Thus, the set of matrices that keep the inner product of 4-vectors invariant is made ofmatrices that become their own inverse when the signs of components with one timeindex are flipped and then transposed. As we will see below, such set of matricesforms a group, called the Lorentz group, and any such transformation [namely, onethat keeps the 4-vector inner product invariant, or equivalently that satisfies thecondition (1.31)] is defined as a Lorentz transformation.

To show that such set of matrices forms a group, it is convenient to write thecondition (1.31) in matrix form. Noting that when written in terms of components,we can change the ordering of product in any way we want, the condition can bewritten as

ΛµαgµνΛ

νβ = gαβ, or ΛT GΛ = G . (1.36)

A set forms a group when for any two elements of the set x1 and x2, a ‘product’x1x2 can be defined such that

1. (Closure) The product x1x2 also belongs to the set.

2. (Associativity) For any elements x1, x2 and x3, (x1x2)x3 = x1(x2x3).

1.3. LORENTZ GROUP 13

3. (Identity) There exists an element I in the set that satisfies Ix = xI = x forany element x.

4. (Inverse) For any element x, there exists an element x−1 in the set that satisfiesx−1x = xx−1 = I.

In our case at hand, the set is all 4 × 4 matrices that satisfy ΛT GΛ = G, andwe take the ordinary matrix multiplication as the ‘product’ which defines the group.The proof is straightforward:

1. Suppose Λ1 and Λ2 belong to the set (i.e. ΛT1 GΛ1 = G and ΛT

2 GΛ2 = G). Then,

(Λ1Λ2)T G(Λ1Λ2) = ΛT

2 ΛT1 GΛ1︸︷︷︸G

Λ2 = ΛT2 GΛ2 = G. (1.37)

Thus, the product Λ1Λ2 also belongs to the set.

2. The matrix multiplication is of course associative: (Λ1Λ2)Λ3 = Λ1(Λ2Λ3).

3. The identity matrix I (Iµν = δµν) belongs to the set (IT GI = G), and satisfies

IΛ = ΛI = Λ for any element.

4. We have already seen that if a 4 × 4 matrix Λ satisfies ΛT GΛ = G, then itsinverse exists as given by (1.34). It is instructive, however, to prove it moreformally. Taking the determinant of ΛT GΛ = G,

det ΛT︸︷︷︸det Λ

det G︸︷︷︸−1

det Λ = det G︸︷︷︸−1

→ (det Λ)2 = 1 , (1.38)

where we have used the property of determinant

det(MN) = det M det N (1.39)

with M and N being square matrices of same rank. Thus, det Λ �= 0 andtherefore its inverse Λ−1 exists. Also, it belongs to the set: multiplying ΛT GΛ =G by (Λ−1)T from the left and by Λ−1 from the right,

(Λ−1)T ΛT︸︷︷︸(ΛΛ−1︸︷︷︸

I

)T

G ΛΛ−1︸︷︷︸I

= (Λ−1)T GΛ−1 → (Λ−1)T GΛ−1 = G . (1.40)

This completes the proof that Λ’s that satisfy (1.36) form a group.


Since the inverse of a Lorentz transformation is also a Lorentz transformation asjust proven above, it should satisfy the condition (1.31)

gαβ = gµν(Λ−1)

µ

α(Λ−1)ν

β = gµνΛαµΛβ

ν → gµνΛαµΛβ

ν = gαβ , (1.41)

where we have used the inversion rule (1.34). The formulas (1.31), (1.41), and theirvariations are then summarized as follows: on the left hand side of the form g Λ Λ = g,an index of g (call it µ) is contracted with an index of a Λ and the other index ofg (call it ν) with an index of the other Λ. As long as µ and ν are both first orboth second indices on the Λ’s, and as long as the rest of the indices are the same(including superscript/subscript) on both sides of the equality, any possible way ofindexing gives a correct formula. Similarly, on the left hand side of the form Λ Λ = g,an index of a Λ and an index of the other Λ are contracted. As long as the contractedindices are both first or both second indices on Λ’s, and as long as the rest of theindices are the same on both sides of the equality, any possible way of indexing givesa correct formula.

A natural question at this point is whether the Lorentz group defined in thisway is any larger than the set of Lorentz boosts defined by (1.9). The answer is

yes. Clearly, any rotation in the 3-dimensional space keeps �A · �B invariant while itdoes not change the time components A0 and B0. Thus, it keeps the 4-vector innerproduct A · B = A0B0 − �A · �B invariant, and as a result it belongs to the Lorentzgroup by definition. On the other hand, the only way the boost (1.9) does not changethe time component is to set β = 0 in which case the transformation is the identitytransformation. Thus, any finite rotation in the 3-dimensional space is not a boostwhile it is a Lorentz transformation.

Furthermore, the time reversal T and the space inversion P defined by

Tdef≡ {T µ

ν}def≡

−1

1

1

1

, P

def≡ {P µν}

def≡

1

−1

−1

−1

(1.42)

satisfy

T T GT = G, P T GP = G , (1.43)

and thus belong to the Lorentz group. Even though the matrix P has the samenumerical form as G, it should be noted that P is a Lorentz transformation but Gis not (it is a metric). The difference is also reflected in the fact that the matrixP is defined by the first index being superscript and the second subscript (becauseit is a Lorentz transformation), while the matrix G is defined by both indices beingsubscript (or both superscript).

1.4. CLASSIFICATION OF LORENTZ TRANSFORMATIONS 15

As we will see later, boosts and rotations can be formed by consecutive infinitesi-mal transformations starting from identity I (they are ‘continuously connected’ to I),while T and P cannot (they are ‘disconnected’ from I, or said to be ‘discrete’ trans-formations). Any product of boosts, rotation, T , and P belongs to the Lorentz group,and it turns out that they saturate the Lorentz group. Thus, we write symbolically

Lorentz group = boost + rotation + T + P . (1.44)

Later, we will see that any Lorentz transformation continuously connected to I is aboost, a rotation, or a combination thereof.

If the origins of the inertial frames K and K ′ touch at t = t′ = 0 and �x = �x′ = 0,the coordinate xµ = (t, �x) of any event transforms in the same way as P µ:

xµ = Λµνx

ν . (1.45)

This can be extended to include space-time translation between the two frames:

xµ = Λµνx

ν + aµ , (1.46)

where aµ is a constant 4-vector. The transformation of energy-momentum is notaffected by the space-time translation, and is still given by P ′ = ΛP . Such transfor-mations that include space-time translation also form a group and called the ‘inho-mogeneous Lorentz group’ or the ‘Poincare group’. The group formed by the trans-formations with aµ = 0 is sometimes called the homogeneous Lorentz group. Unlessotherwise stated, we will deal with the homogeneous Lorentz group; namely withoutspace-time translation.

1.4 Classification of Lorentz transformations

Up to this point, we have not specified that Lorentz transformations are real (namely,all the elements are real). In fact, Lorentz transformations as defined by (1.31) ingeneral can be complex and the complex Lorentz transformations plays an impor-tant role in a formal proof of an important symmetry theorem called CPT theoremwhich states that the laws of physics are invariant under the combination of particle-antiparticle exchange (C), mirror inversion (P), and time reversal (T) under certainnatural assumptions. In this book, however, we will assume that Lorentz transfor-mations are real.

As seen in (1.38), all Lorentz transformation satisfy (det Λ)2 = 1, or equivalently,det Λ = +1 or −1. We define ‘proper’ and ‘improper’ Lorentz transformations as{

det Λ = +1 : proper

det Λ = −1 : improper. (1.47)


Since det(Λ1Λ2) = det Λ1 det Λ2, the product of two proper transformations or twoimproper transformations is proper, while the product of a proper transformation anda improper transformation is improper.

Next, look at the (α, β) = (0, 0) component of the defining condition gµνΛµ

αΛνβ =

gαβ:

gµνΛµ0Λ

ν0 = g00 = 1 , → (Λ0

0)2 −

3∑i=1

(Λi0)

2 = 1 (1.48)

or

(Λ00)

2 = 1 +3∑

i=1

(Λi0)

2 ≥ 1 , (1.49)

which means Λ00 ≥ 1 or Λ0

0 ≤ −1, and this defines the ‘orthochronous’ and ‘non-orthochronous’ Lorentz transformations:{

Λ00 ≥ 1 : orthochronous

Λ00 ≤ −1 : non-orthochronous

. (1.50)

It is easy to show that the product of two orthochronous transformations or two non-orthochronous transformations is orthochronous, and the product of an orthochronoustransformation and a non-orthochronous transformation is non-orthochronous.

From the definitions (1.42) and Iµν = δµν , we have

det I = det(TP ) = +1, det T = det P = −1,

I00 = P 0

0 = +1, T 00 = (TP )0

0 = −1(1.51)

Thus, the identity I is proper and orthochronous, P is improper and orthochronous,T is improper and non-orthochronous, and TP is proper and non-orthochronous. Ac-cordingly, we can multiply any proper and orthochronous transformations by each ofthese to form four sets of transformations of given properness and orthochronousnessas shown in Table 1.1. Any Lorentz transformation is proper or improper (i.e. det Λ =±1) and orthochronous or non-orthochronous (i.e. |Λ0

0|2 ≥ 1). Since any transforma-tion that is not proper and orthochronous can be made proper and orthochronous bymultiplying T , P or TP , the four forms of transformations in Table 1.1 saturate the

Lorentz group. For example, if Λ is improper and orthochronous, then PΛdef≡ Λ(po)

is proper and orthochronous, and Λ can be written as Λ = PPΛ = PΛ(po).It is straightforward to show that the set of proper transformations and the set

of orthochronous transformations separately form a group, and that proper and or-thochronous transformations by themselves form a group. Also, the set of properand orthochronous transformations and the set of improper and non-orthochronoustransformations together form a group.

Exercise 1.1 Classification of Lorentz transformations.

1.4. CLASSIFICATION OF LORENTZ TRANSFORMATIONS 17

Λ00≥1

orthochronousΛ0

0≤−1

non-orthochronous

detΛ=+1proper Λ(po) TPΛ(po)

detΛ=−1improper PΛ(po) TΛ(po)

Table 1.1: Classification of the Lorentz group. Λ(po) is any proper and orthochronousLorentz transformation.

(a) Suppose Λ = AB where Λ, A, and B are Lorentz transformations. Prove thatΛ is orthochronous if A and B are both orthochronous or both non-orthochronous,and that Λ is non-orthochronous if one of A and B is orthochronous and the other isnon-orthochronous.[hint: Note that we can write Λ0

0 = A00B

00 + �a ·�b with �a ≡ (A0

1, A02, A

03) and

�b ≡ (B10, B

20, B

30). Then use |�a ·�b| ≤ |�a||�b|. Also, one can derive �a2 = A0

02 − 1 and

�b2 = B002 − 1. ]

(b) Show that the following sets of Lorentz transformations each form a group:

1. proper transformations

2. orthochronous transformations

3. proper and orthochronous transformations

4. proper and orthochronous transformations plus improper and non-orthochronoustransformations

As mentioned earlier (and as will be shown later) boosts and rotations are con-tinuously connected to the identity. Are they then proper and orthochronous? Toshow that this is the case, it suffices to prove that an infinitesimal transformationcan change det Λ and Λ0

0 only infinitesimally, since then multiplying an infinitesimaltransformation cannot jump across the gap between det Λ = +1 and det Λ = −1 orthe gap between Λ0

0 ≥ 1 and Λ00 ≤ −1.

An infinitesimal transformation is a transformation that is very close to the iden-tity I and any such transformation λ can be written as

λ = I + dH (1.52)

where d is a small number and H is a 4×4 matrix of order unity meaning the maximumof the absolute values of its elements is about 1. To be specific, we could define it such


that maxα,β |Hαβ| = 1 and d ≥ 0, which uniquely defines the decomposition above.

We want to show that for any Lorentz transformation Λ, multiplying I + dH changesthe determinant or the (0, 0) component only infinitesimally; namely, the differencesvanish as we take d to zero.

The determinant of a n × n matrix A is defined by

det Adef≡

∑permutations

si1,i2,...inAi11Ai22 . . . Ainn (1.53)

where the sum is taken over (i1, i2, . . . in) which is any permutation of (1, 2, . . . n),and si1,i2,...in is 1(−1) if (i1, i2, . . . in) is an even(odd) permutation. When applied to4 × 4 Lorentz transformations, this can be written as

det Adef≡ εαβγδA

α0A

β1A

γ2A

δ3 , (1.54)

where the implicit sum is over α, β, γ, δ = 0, 1, 2, 3 and εαβγδ is the totally anti-symmetric 4-th rank tensor defined by

εαβγδdef≡

=

{+1

−1

}if (αβγδ) is an

{even

odd

}permutation of (0, 1, 2, 3)

= 0 if any of αβγδ are equal

(1.55)

The standard superscript/subscript rule applies to the indices of εαβγδ; namely, ε0123 =−ε0123 = 1, etc. Then, it is easy to show that

det(I + dH) = 1 + d TrH + (higher orders in d) , (1.56)

where the ‘trace’ of a matrix A is defined as the sum of the diagonal elements:

TrAdef≡

3∑α=0

Aαα . (1.57)

Exercise 1.2 Determinant and trace.Determinant of a n × n matrix is defined by

det Adef≡ si1i2,...inAi11Ai22 . . . Ainn

where sum over i1, i2 . . . in is implied (each taking values 1 through n ) and s(i1, i2 . . . in)is the totally asymmetric n-th rank tensor:

si1i2...in ≡{

+1(−1) if (i1, i2 . . . in) is an even (odd ) permutation of (1, 2, . . . n).

0 if any of i1, i2 . . . in are equal.

1.5. TENSORS 19

Show that to first order of a small number d, the determinant of a matrix that isinfinitesimally close to the identity matrix I is given by

det(I + dH) = 1 + dTrH + (higher orders in d) ,

where H is a certain matrix whose size is of order 1, and the trace (Tr) of a matrixis defined by

TrH ≡n∑

i=1

Hii .

Since all diagonal elements of H are of order unity or smaller, (1.56) tells us thatdet λ → 1 as we take d → 0. In fact, the infinitesimal transformation λ is a Lorentztransformation, so we know that detλ = ±1. Thus, we see that the determinant of aninfinitesimal transformation is strictly +1. It then follows from det(λΛ) = det λ det Λthat multiplying an infinitesimal transformation λ to any transformation Λ does notchange the determinant of the transformation.

The (0, 0) component of λΛ is

(λΛ)00 = [(I + dH)Λ]00 = [Λ + dHΛ]00 = Λ0

0 + d (HΛ)00 . (1.58)

Since (HΛ)00 is a finite number for a finite Λ, the change in the (0, 0) component tends

to zero as we take d → 0. Thus, no matter how many infinitesimal transformationsare multiplied to Λ, the (0, 0) component cannot jump across the gap between +1and −1.

Thus, continuously connected Lorentz transformations have the same ‘properness’and ‘orthochronousness. Therefore, boosts and rotations, which are continuously con-nected to the identity, are proper and orthochronous.

Do Lorentz boosts form a group?A natural question is whether Lorentz boosts form a group by themselves. The answeris no, and this is because two consecutive boosts in different directions turn out tobe a boost plus a rotation as we will see when we study the generators of the Lorentzgroup. Thus, boosts and rotations have to be combined to form a group. On theother hand, rotations form a group by themselves.

1.5 Tensors

Suppose Aµ and Bµ are 4-vectors. Each is a set of 4 numbers that transform undera Lorentz transformation Λ as

A′µ = ΛµαAα, B′ν = Λν

βBβ . (1.59)


Then, the set of 16 numbers AµBν (µ, ν = 0, 1, 2, 3) transforms as

A′µB′ν = ΛµαΛν

βAαBβ . (1.60)

Anything that has 2 Lorentz indices, which is a set of 16 numbers, and transforms as

( )′µν = ΛµαΛν

β( )αβ (1.61)

is called a second rank tensor (or simply a ‘tensor’). It may be real, complex, or evena set of operators. Similarly, a quantity that has 3 indices and transforms as

( )′µνσ = ΛµαΛν

βΛσγ( )αβγ (1.62)

is called a third-rank tensor, and so on. A 4-vector (or simply a ‘vector’) is a first-rank tensor. A Lorentz-invariant quantity, sometimes called a ‘scalar’, has no Lorentzindex, and thus it is a zero-th rank tensor:

( )′ = ( ) (scalar) . (1.63)

Contracted indices do not count in deciding the rank of a tensor. For example,

AµBµ : (scalar), AµTµν : (vector), F µνGµσ : (tensor), etc. (1.64)

The metrix gµν has two Lorentz indices and thus can be considered a second-ranktensor (thus, the metric tensor), then it should transform as

g′µν = ΛµαΛν

βgαβ = gµν (1.65)

where the second equality is due to (1.31). Namely, the metric tensor is invariantunder Lorentz transformations.

In order for some equation to be Lorentz-invariant, the Lorentz indices have to bethe same on both sides of the equality, including the superscript/subscript distinction.By ‘Lorentz-invariant’, we mean that if an equation holds in one frame, then it holdsin any other frame after all the quantities that appear in the equation are evaluatedin the new frame. In the literature, such equations are sometimes called Lorentzcovariant: both sides of the equality change values but the form stays the same. Forexample, if an equation Aµν = Bµν (which is actually a set of 16 equations) holds ina frame, then it also holds in any other frame:

A′µν = ΛµαΛν

βAαβ = ΛµαΛν

βBαβ = B′µν . (1.66)

Thus, equations such asm2 = P µPµ ,

P µ = Aµ + Bµ ,

F µν = AµBν

(1.67)

are all Lorentz-invariant, assuming of course that the quantities transform in thewell-defined ways as described above.

1.6. FIELDS (CLASSICAL) 21

1.6 Fields (classical)

A field is a quantity that is a function of space-time point xµ = (t, �x) (or ‘event’).A scalar quantity that is a function of space time is called a scalar field, a vectorquantity that is a function of space time is called a vector field, etc. The rank ofa field and the Lorentz transformation properties (scalar, vector, tensor, etc.) aredefined in the same way as before, provided that the quantities are evaluated at thesame event point before and after a Lorentz transformation; namely,

Scalar field : φ′(x′) = φ(x)

Vector field : A′µ(x′) = ΛµαAα

Tensor field : T ′µν(x′) = ΛµαΛν

βTαβ(x)

(1.68)

where x′ and x are related byx′µ = Λµ

αxα . (1.69)

For example, a vector field associates a set of 4 numbers Aµ(x) to an event pointx, say when an ant sneezes. In another frame, there are a set of 4 numbers A′µ(x′)associated with the same event x′, namely, when the ant sneezes in that frame, andthey are related to the 4 numbers Aµ(x) in the original frame by the matrix Λ. Thefunctional shape of a primed field is in general different from that of the correspondingunprimed field. Namely, if one plots φ(x) as a function of x and φ′(x′) as a functionof x′, they will look different.

When a quantity is a function of x, we naturally encounter space-time derivativesof such quantity. Then a question arises as to how they transform under a Lorentztransformation. Take a scalar field f(x), and form a set of 4 numbers (fields) bytaking space-time derivatives:

∂f

∂xµ(x) =

(∂f

∂x0(x),

∂f

∂x1(x),

∂f

∂x2(x),

∂f

∂x3(x)

)

=

(∂f

∂t(x),

∂f

∂x(x),

∂f

∂y(x),

∂f

∂z(x)

).

(1.70)

Then pick two space-time points x1 and x2 which are close in space and in time. Theargument below is based on the observation that the difference between the valuesof the scalar field at the two event points is the same in any frame. Since f(x) isa scalar field, the values at a given event is the same before and after a Lorentztransformation:

f ′(x′1) = f(x1), f ′(x′

2) = f(x2) , (1.71)

orf ′(x′

1) − f ′(x′2) = f(x1) − f(x2) . (1.72)


Since x1 and x2 are close, this can be written as

dx′µ ∂f ′

∂x′µ (x′1) = dxµ ∂f

∂xµ(x1) , (1.73)

where summation over µ is implied, and

dx′µ def≡ x′µ1 − x′µ

2 , dxµ def≡ xµ1 − xµ

2 . (1.74)

which tells us that the quantity dxµ(∂f/∂xµ) is Lorentz-invariant. Since dxµ = xµ1−xµ

2

is a superscripted 4-vector, it follows that ∂f/∂xµ should transform as a subscripted4-vector (which transforms as Aµ = Λµ

αAα):

∂f ′

∂x′µ (x′) = Λµα ∂f

∂xα(x) . (1.75)

In fact, together with dx′µ = Λµβdxβ, we have

dx′µ ∂f ′

∂x′µ (x′) =(Λµ

βdxβ) (

Λµα ∂f

∂xα(x)

)

= ΛµαΛµ

β︸︷︷︸gα

β by (1.32)

dxβ ∂f

∂xα(x)

= dxα ∂f

∂xα(x) ,

(1.76)

showing that it is indeed Lorentz-invariant.Thus, the index µ in the differential operator ∂/∂xµ acts as a subscript even

though it is a superscript on x. To make this point clear, ∂/∂xµ is often written usinga subscript as

∂µdef≡ ∂

∂xµ=

(∂

∂t,

∂

∂x,

∂

∂y,

∂

∂z

)=

(∂

∂t, �∇

). (1.77)

Once ∂µ is defined, the standard subscript/superscript rule applies; namely, ∂µ =∂/∂xµ, etc. Symbolically, the operator ∂µ then transforms as a superscripted 4-vector:

∂′µ = Λµν∂

ν , (1.78)

with ∂′µ ≡ ∂/∂x′µ.

Example: Consider the 4-component charge current density jµ(x) = (ρ(x),�j(x)).We can see that this is indeed a Lorentz 4-vector as follows: Suppose the charge iscarried by some medium, such as gas of ions, then pick a space-time point x and let

1.7. GENERATORS OF THE LORENTZ GROUP 23

ρ0 be the charge density in the rest frame of the medium and �β be the velocity of themedium at that point. Then the charge density ρ in the frame in question is largerthan ρ0 by the factor γ = 1/

√1 − β2 due to Lorentz contraction

ρ = ρ0γ . (1.79)

Since �j = ρ�β, jµ can be written as

jµ = (ρ,�j) = (ρ0γ, ρ0γ�β) = ρ0(γ, �η) = ρ0ηµ , (1.80)

where we have defined the ‘4-velocity’ ηµ by

ηµ def≡ (γ, γ�β) . (1.81)

On the other hand, the 4-momentum of a particle with mass m can be written as

P µ = (mγ, mγ�β) = mηµ (1.82)

which means that the 4-velocity ηµ is a Lorentz 4-vector, and therefore so is jµ. Whenthe charge is carried by more than one different media, unique rest frame of the mediawhere ρ0 is defined does not exist. The total jµ, however, is the sum of jµ for eachmedium. Since jµ for each medium is a 4-vector, the sum is also a 4-vector.

Then ∂µjµ = 0 is a Lorentz-invariant equation; namely, if it is true in one frame,

then it is true in any frame. Using (1.77), we can write ∂µjµ = 0 as

∂µjµ = ∂0j

0 + ∂1j1 + ∂2j

2 + ∂3j3

=∂

∂tρ +

∂

∂xjx +

∂

∂yjy +

∂

∂zjz

=∂

∂tρ +

↑note the sign!

�∇ ·�j = 0 ,

(1.83)

which is nothing but the charge conservation equation. Thus, we see that if charge isconserved in one frame it is conserved in any frame.

1.7 Generators of the Lorentz group

In this section, we will focus on the proper and orthochronous Lorentz group. Otherelements of the Lorentz group can be obtained by multiplying T , P , and TP to the


elements of this group. The goal is to show that any element Λ that is continuouslyconnected to the identity can be written as2

Λ = eξiKi+θiLi , (i = 1, 2, 3) (1.84)

where ξi and θi are real numbers and Ki and Li are 4×4 matrices. Such group whoseelements can be parametrized by a set of continuous real numbers (in our case theyare ξi and θi) is called a Lie group. The operators Ki and Li are called the generatorsof the Lie group.

Any element of the proper and orthochronous Lorentz group is continuously con-nected to the identity. Actually we have not proven this, but we will at least showthat all boosts, rotations and combinations thereof are continuously connected to theidentity (and vice versa).

1.7.1 Infinitesimal transformations

Let’s start by looking at a Lorentz transformation which is infinitesimally close to theidentity:

Λµν = gµ

ν + ωµν (1.85)

where ωµν is a set of small (real) numbers. Inserting this to the defining condition

(1.31) or equivalently ΛναΛνβ = gαβ (1.32), we get

gαβ = ΛναΛνβ

= (gνα + ωνα)(gνβ + ων

β)

= gναgνβ + ωναgν

β + gναωνβ + ωναων

β

= gαβ + ωβα + ωαβ + ωναωνβ .

(1.86)

Keeping terms to the first order in ω, we then obtain

ωβα = −ωαβ . (1.87)

Namely, ωαβ is anti-symmetric (which is true when the indices are both subscript orboth superscript; in fact, ωα

β is not anti-symmetric under α ↔ β), and thus it has 6independent parameters:

{ωαβ} =

β −→

α

↓

0 ω01 ω02 ω03

−ω01 0 ω12 ω13

−ω02 −ω12 0 ω23

−ω03 −ω13 −ω23 0

(1.88)

2In the literature, it is often defined as exp i(ξiKi + θiLi), which would make the operatorshermitian if the transformation were unitary (e.g. representations of the Lorentz group in theHilbert space). The Lorentz transformation matrices in space-time are in general not unitary, andfor now, we will define without the ‘i’ so that the expressions become simpler.


This can be conveniently parametrized using 6 anti-symmetric matrices as

{ωαβ} = ω01{(M01)αβ} + ω02{(M02)αβ} + ω03{(M03)αβ}+ ω23{(M23)αβ} + ω13{(M13)αβ} + ω12{(M12)αβ} (1.89)

=∑µ<ν

ωµν{(Mµν)αβ}

with

{(M01)αβ} =

0 1 0 0

−1 0 0 0

0 0 0 0

0 0 0 0

, {(M23)αβ} =

0 0 0 0

0 0 0 0

0 0 0 1

0 0 −1 0

,

{(M02)αβ} =

0 0 1 0

0 0 0 0

−1 0 0 0

0 0 0 0

, {(M13)αβ} =

0 0 0 0

0 0 0 1

0 0 0 0

0 −1 0 0

,

{(M03)αβ} =

0 0 0 1

0 0 0 0

0 0 0 0

−1 0 0 0

, {(M12)αβ} =

0 0 0 0

0 0 1 0

0 −1 0 0

0 0 0 0

(1.90)

Note that for a given pair of µ and ν, {(Mµν)αβ} is a 4× 4 matrix, while ωµν is a realnumber. The elements (Mµν)αβ can be written in a concise form as follows: first, wenote that in the upper right half of each matrix (i.e. for α < β), the element with(α, β) = (µ, ν) is 1 and all else are zero, which can be written as gµ

α gνβ. For the

lower half, all we have to do is to flip α and β and add a minus sign. Combining thetwo halves, we get

(Mµν)αβ = gµα gν

β − gµβ gν

α . (1.91)

This is defined only for µ < ν so far. For µ > ν, we will use this same expression asthe definition; then, (Mµν)αβ is anti-symmetric with respect to (µ ↔ ν):

(Mµν)αβ = −(M νµ)αβ , (1.92)

which also means (Mµν)αβ = 0 if µ = ν. Together with ωµν = −ωνµ, (1.89) becomes

ωαβ =∑µ<ν

ωµν(Mµν)αβ =

∑µ>ν

ωµν(Mµν)αβ =

1

2ωµν(M

µν)αβ , (1.93)

where in the last expression, sum over all values of µ and ν is implied. The infinites-imal transformation (1.85) can then be written as

Λαβ = gα

β +1

2ωµν(M

µν)αβ , (1.94)


or in matrix form,

Λ = I +1

2ωµνM

µν . (1.95)

where the first indices of Mµν , which is a 4 × 4 matrix for given µ and ν, is takento be superscript and the second subscript; namely, in the same way as Lorentztransformation. Namely, when no explicit indexes for elements are given, the 4 × 4matrix Mµν is defined as

Mµν def≡ {(Mµν)αβ} . (1.96)

It is convenient to divide the six matrices to two groups as

Kidef≡ M0i, Li

def≡ M jk (i, j, k : cyclic) . (1.97)

We always use subscripts for Ki and Li since only possible values are i = 1, 2, 3, andsimilarly to Mµν , elements of the matrices Ki’s and Li’s are defined by taking thefirst Lorentz index to be superscript and the second subscript:

Kidef≡ {(Ki)

αβ} , Li

def≡ {(Li)α

β} . (1.98)

Later, we will see that K’s generate boosts and L’s generate rotations. Explicitly,they can be obtained by raising the index α in (1.90) (note also the the minus signin L2 = −M13):

K1 =

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

, K2 =

0 0 1 0

0 0 0 0

1 0 0 0

0 0 0 0

, K3 =

0 0 0 1

0 0 0 0

0 0 0 0

1 0 0 0

(1.99)

L1 =

0 0 0 0

0 0 0 0

0 0 0 −1

0 0 1 0

, L2 =

0 0 0 0

0 0 0 1

0 0 0 0

0 −1 0 0

, L3 =

0 0 0 0

0 0 −1 0

0 1 0 0

0 0 0 0

(1.100)

By inspection, we see that the elements of K’s and L’s can be written as

(Ki)jk = 0 , (Ki)

0µ = (Ki)

µ0 = gi

µ ,

(Li)jk = −εijk , (Li)

0µ = (Li)

µ0 = 0 ,

(i, j, k = 1, 2, 3; µ = 0, 1, 2, 3) (1.101)

where εijk is a totally anti-symmetric quantity defined for i, j, k = 1, 2, 3:

εijkdef≡

=

{+1

−1

}if (i, j, k) is an

{even

odd

}permutation of (1, 2, 3),

= 0 if any of i, j, k are equal.

(1.102)


An explicit calculation shows that K’s and L’s satisfy the following commutationrelations:

[Ki, Kj] = −εijkLk

[Li , Lj ] = εijkLk (1.103)

[Li , Kj] = εijkKk ,

where sum over k = 1, 2, 3 is implied, and the commutator of two operators A, B isdefined as

[A, B]def≡ AB − BA . (1.104)

Note that the relation [Ki, Kj] = −εijkLk can also be written as [Ki, Kj] = −Lk

(i, j, k: cyclic), etc.

Exercise 1.3 Verify the commutation relations (1.103 ). You may numerically verifythem, or you may try proving generally by using the general formula for the elementsof the matrixes.

Exercise 1.4 Boost in a general direction.Start from the formula for boost (1.9 ) where P‖ is the component of �P parallel to �β,

and �P⊥ is the component perpendicular to �β; namely,

P‖ = �P ·�n, and �P⊥ = �P − P‖�n

with �n = �β/β (and similarly for �P ′ ). Note that �β is well-defined in the primed framealso by the particular relative orientation of the two frames chosen.(a) Show that the corresponding Lorentz transformation matrix is given by

Λ =

γ γβx γβy γβz

γβx 1 + ρβ2x ρβxβy ρβxβz

γβy ρβxβy 1 + ρβ2y ρβyβz

γβz ρβxβz ρβyβz 1 + ρβ2z

, with ρ ≡ γ − 1

β2.

(b) Show that when β is small, the Lorentz transformation matrix for a boost is givento the first order in β by

Λ = 1 + βiKi . (summed over i = 1, 2, 3)

(c) In the explicit expression of Λ given above, one notes that the top row [Λ0µ (µ =

0, 1, 2, 3)] and the left-most column [Λµ0 (µ = 0, 1, 2, 3)] are nothing but the velocity

4-vector ηµ = (γ, �βγ). Let’s see how it works for general Lorentz transformations(proper and orthochronous). Suppose the relative orientation of the two frames K

and K ′ is not given by �β′K = −�βK′, where �β′

K is the velocity of the origin of K

measured in K ′, and �βK′ is the velocity of the origin of K ′ measured in K. Let Λ bethe corresponding Lorentz transformation. Express Λ0

µ and Λµ0 in terms of �β′

K and�βK′. (hint: Place a mass m at the origin of K and view it from K ′, and place a massat the origin of K ′ and view it from K.)


1.7.2 Finite transformations

Now we will show that any finite (namely, not infinitesimal) rotation can be written aseθiLi , and any finite boost can be written as eξiKi , where θi and ξi (i = 1, 2, 3) are somefinite real numbers. First, however, let us review some mathematical background:

Matrix exponentiationsThe exponential of a m × m matrix A is also a m × m matrix defined by

eA def≡ limn→∞

(I +

A

n

)n

, (1.105)

which can be expanded on the right hand side as

eA = limn→∞

n∑k=0

n(n − 1) . . . (n − k + 1)

k !

Ak

nk. (1.106)

Since the sum is a rapidly converging series, one can sum only the terms with k � nfor which n(n − 1) . . . (n − k + 1) ≈ nk. It then leads to

eA =∞∑

k=0

Ak

k !, (1.107)

which can also be regarded as a definition of eA.Using the definition (1.105) or (1.107), we see that

(eA

)†= eA†

, (1.108)

where the hermitian conjugate of a matrix A is defined by (A†)ij ≡ A∗ji. The deter-

minant of eA can be written using (1.105) as

det eA = limn→∞

[det

(I +

A

n

)]n

= limn→∞

(1 +

TrA

n+ . . . +

ck

nk+ . . .

)n

,

(1.109)

where we have used (1.56). This does not depend on ck (k > 1) since the derivativewith respect to ck vanishes as can be readily verified. Thus, ck (k > 1) can be set tozero and we have

det eA = eTrA . (1.110)


The derivative of exA (x is a number, while A is a constant matrix) with respect to xcan be obtained using (1.107),

d

dxexA =

∞∑k=1

(k xk−1)Ak

k != A

∞∑k=1

xk−1Ak−1

(k − 1) !; (1.111)

thus,

d

dxexA = AexA . (1.112)

There is an important theorem that expresses a product of two exponentials interms of single exponential, called the Campbell-Baker-Hausdorff (CBH) theorem(presented here without proof):

eAeB = eA+B+ 12[A,B]+ ··· , (1.113)

where ‘· · ·’ denotes the higher-order commutators of A and B such as [A, [A, B]],[A, [[A, B], B]] etc. with known coefficients. Note that the innermost commutator isalways [A, B] since otherwise it is zero ([A, A] = [B, B] = 0), and thus if [A, B] is acommuting quantity (a c-number), then ‘· · ·’ is zero. Applying (1.113) to B = −A,we get

eAe−A = eA−A = I , (1.114)

or(eA)−1 = e−A . (1.115)

RotationAn infinitesimal rotation around the z-axis by δθ [Figure 1.3(a)] can be written as(

x′

y′

)=

(x − δθyy + δθx

)=

(1 −δθδθ 1

) (xy

)= (I + δθLz)

(xy

), (1.116)

with

Lz =

( x y

x 0 −1y 1 0

). (1.117)

Then, a rotation by a finite angle θ is constructed as n consecutive rotations by θ/neach and taking the limit n → ∞. Using (1.116), it can be written as(

x′

y′

)= lim

n→∞

(I +

θ

nLz

)n (xy

)

= eθLz

(xy

), (1.118)


where we have used the definition (1.105).From the explicit expression of Lz (1.117), we have L2

z = −I, L3z = −Lz, L4

z = I,etc. In general,

L4nz = I, L4n+1

z = Lz, L4n+2z = −I, L4n+3

z = −Lz, (1.119)

where n is an integer. Using the second definition of eA (1.107), the rotation matrixeθLz can then be written in terms of the trigonometirc functions as

eθLz = I + θLz +θ2

2 !L2

z︸︷︷︸−I

+θ3

3 !L3

z︸︷︷︸−Lz

+ . . . (1.120)

=

(1 − θ2

2 !+ . . .

)︸︷︷︸

cos θ

I +

(θ − θ3

3 !+ . . .

)︸︷︷︸

sin θ

Lz (1.121)

=(

cos θ − sin θsin θ cos θ

), (1.122)

which is probably a more familiar form of a rotation around the z-axis by an angle θ.Similarly, rotations around x and y axes are generated by Lx and Ly as obtained

by cyclic permutations of (x, y, z) in the derivation above. Switching to numericalindices [(Lx, Ly, Lz) ≡ (L1, L2, L3)],

L1 =

( 2 3

2 0 −13 1 0

), L2 =

( 3 1

3 0 −11 1 0

), L3 =

( 1 2

1 0 −12 1 0

). (1.123)

Are these identical to the definition (1.100) which was given in 4 × 4 matrix form,or equivalently (1.101)? Since δθLi is the change of coordinates by the rotation, the

x

θθn

θn xx

θ

δθ(x,y)

(x',y')

x

y

δθ

δθx

y

(a) (b)

Figure 1.3: Infinitesimal rotation around the z-axis by an angle δθ (a), and around a

general direction �θ by an angle θ/n (b).


elements of a 4 × 4 matrix corresponding to unchanged coordinates should be zero.We then see that the L’s given above are indeed identical to (1.100).

A general rotation is then given by

eθiLi = eθ·L , (1.124)

where�θ

def≡ (θ1, θ2, θ3), �Ldef≡ (L1, L2, L3) . (1.125)

As we will see below, this is a rotation around the direction �θ by an angle θ ≡ |�θ|. Tosee this, first we write eθiLi using the definition (1.105):

eθiLi = limn→∞

(I +

θiLi

n

)n

, (1.126)

which shows that it is a series of small rotations each given by I + θiLi/n. Theaction of such an infinitesimal transformation [Figure 1.3(b)] on �x is (writing thespace components only)

x′j =

(I +

θiLi

n

)j

kxk

= gjkx

k +1

nθi (Li)

jk︸︷︷︸

−εijk by (1.101)

xk

= xj − 1

nεijk θix

k

= xj +1

n(�θ × �x)j (1.127)

where we have used the definition of the three-dimensional cross product

(�a ×�b)i = εijk ajbk . (1.128)

Thus, I+θiLi/n is nothing but a small rotation around �θ by an angle θ/n (Figure 1.3).Then n such rotations applied successively will result in a rotation by an angle θaround the same axis �θ.

BoostsA boost in x direction by a velocity β is given by (1.26):

Λ =

( t x

t γ ηx η γ

),

(γ =

1√1 − β2

, η = βγ

). (1.129)


When β is small (= δ), γ ≈ 1 and η ≈ δ to the first order in δ; then, the infinitesimalboost can be written as

Λ =(

1 δδ 1

)= I + δKx , (1.130)

with

Kx =

( t x

t 0 1x 1 0

). (1.131)

Suppose we apply n such boosts consecutively, where we take n to infinity while nδis fixed to a certain value ξ:

nδ = ξ . (1.132)

Then the resulting transformation is

Λ = limn→∞

(I +

ξ

nKx

)n

= eξKx , (1.133)

where we have used the definition (1.105). Is ξ the velocity of this boost? The answeris no, even though it is a function of the velocity. Let’s expand the exponential aboveby the second definition (1.105) and use K2

x = I:

Λ = eξKx (1.134)

= I + ξKx +ξ2

2 !K2

x︸︷︷︸I

+ξ3

3 !K3

x︸︷︷︸Kx

+ . . .

=

(1 +

ξ2

2 !+ . . .

)︸︷︷︸

cosh ξ

I +

(ξ +

ξ2

2 !+ . . .

)︸︷︷︸

sinh ξ

Kx

=(

cosh ξ sinh ξsinh ξ cosh ξ

). (1.135)

Comparing with (1.130), we see that this is a boost of a velocity β given by

γ = cosh ξ, η = sinh ξ (1.136)

orβ =

η

γ= tanh ξ . (1.137)

Note that the relation γ2 − η2 = 1 (1.6) is automatically satisfied since cosh ξ2 −sinh ξ2 = 1.

Thus, n consecutive boosts by a velocity ξ/n each did not result in a boost of avelocity ξ; rather, it was a boost of a velocity β = tanh ξ. This breakdown of the


simple addition rule of velocity is well known: the relativistic rule of velocity additionstates that two consecutive boosts, by β1 and by β2, do not result in a boost of β1+β2,but in a boost of a velocity β0 given by

β0 =β1 + β2

1 + β1β2

. (1.138)

Due to the identity tanh(ξ1 + ξ2) = (tanh ξ1 + tanh ξ1)/(1 + tanh ξ1 tanh ξ2), however,it becomes additive when velocities are transformed by βi = tanh ξi (i = 0, 1, 2);namely, ξ0 = ξ1 + ξ2 holds.

The matrix Kx(≡ K1) given in (1.131) is identical to the 4×4 form given in (1.99)when all other elements that correspond to unchanged coordinates are set to zero.The boosts along y and z directions are obtained by simply replacing x with y or zin the derivation above. Thus, we see that K2 and K3 given in (1.99) indeed generateboosts in y and z directions, respectively.

A boost in a general direction would then be given by

Λ = eξiKi , (1.139)

where �ξ ≡ (ξ1, ξ2, ξ3) are the parameters of the boost. In order to see what kind oftransformation this represents, let’s write it as a series of infinitesimal transformationsusing (1.105):

eξiKi = limn→∞

(I +

ξi

nKi

)n

. (1.140)

From the explicit forms of Ki (1.99), we can write the infinitesimal transformation as

I +ξi

nKi = I +

1

n

0 ξ1 ξ2 ξ3

ξ1

ξ2

ξ3

0

. (1.141)

On the other hand, a boost in a general direction by a small velocity �δ is given by(1.9) with γ ≈ 1, η ≈ δ and δ ≡ |�δ|:

(E ′

P ′‖

)=

(1 δ

δ 1

) (E

P‖

), �P ′

⊥ = �P⊥ , (1.142)

or

E ′ = E + δP‖ → E ′ = E + �δ · �P

P ′‖ = P‖ + δE

�P ′⊥ = �P⊥

}→ �P ′ = �P + E �δ

(1.143)


where we have used �P (′) = P(′)‖ δ + �P

(′)⊥ (δ ≡ �δ/δ). This can be written in 4× 4 matrix

form as

E ′

P ′x

P ′y

P ′z

=

I +

0 δ1 δ2 δ3

δ1

δ2

δ3

0

EPx

Py

Pz

. (1.144)

Comparing this with (1.141), we can identify that I +ξiKi/n as a boost in �ξ direction

by a velocity ξ/n (ξ ≡ |�ξ|). Then n consecutive such boosts will result in a boostin the same direction. Since the rule of addition of velocity (1.138) is valid in anydirection as long as the boosts are in the same direction, the n boosts by velocityξ/n each will result in a single boost of velocity β = tanh ξ as before. Thus, eξiKi

represents a boost in �ξ direction by a velocity β = tanh ξ.

Boost + rotationFirst, we show that a rotation followed by a rotation is a rotation, but a boost followedby a boost is not in general a boost. Consider a rotation eθiLi followed by anotherrotation eφiLi where �θ and �φ are arbitrary vectors. Using the CBH theorem (1.113),we can write the product of the two transformations as

eφiLieθjLj = eφiLi+θjLj+12[φiLi,θjLj ] + ..., (1.145)

where ‘. . . ’ represents terms with higher-order commutators such as [φiLi, [φjLj, θkLk]]etc. Now we can use the commutation relations (1.103) to remove all commutatorsin the exponent on the right hand side. The result will be a linear combination of L’swith well-defined coefficients (call them αi) since the coefficients in ‘. . . ’ in the CBHtheorem are known. Here, there will be no K’s appearing in the linear combinationbecause of the commutation relation [Li, Lj] = εijkLk. Thus, the product is writtenas

eφiLieθjLj = eαiLi (�α : a function of �φ, �θ), (1.146)

which is just another rotation.Next, consider a boost eξiKi followed by another boost eξ′iKi :

eξ′iKieξjKj = eξ′iKi+ξjKj+12[ξ′iKi,ξjKj ] + ... . (1.147)

Again, the brackets can be removed by the commutation relations (1.103) reducingthe exponent to a linear combination of K’s and L’s. This time, there will be L’sappearing through the relation [Ki, Kj] = −εijkLk which are in general not cancelledamong different terms. Thus, a boost followed by a boost is not in general anotherboost; rather, it is a combination of boost and rotation:

eξ′iKieξjKj = eαiKi+βiLi (�α, �β : functions of �ξ, �ξ′). (1.148)


It is easy to show, however, that if two boosts are in the same direction, then theproduct is also a boost.

Any combinations of boosts and rotations can then be written as

Λ = eξiKi+θiLi = e12aµνMµν

, (1.149)

where we have defined the anti-symmetric tensor aµν by

a0idef≡ ξi, aij

def≡ θk (i, j, k : cyclic), aµν = −aνµ , (1.150)

and the factor 1/2 arises since terms with µ > ν as well as µ < ν are included inthe sum. The expression of an infinitesimal transformation (1.95) is nothing but thisexpression in the limit of small aµν . Since we now know that any product of suchtransformations can also be written as (1.149) by the CBH theorem, we see that theset of Lorentz transformations connected to the identity is saturated by boosts androtations.

We have seen that the generators K’s and L’s and their commutation relations(called the Lie algebra) play critical roles in understanding the Lorentz group. Infact, generators and their commutation relations completely determine the structureof the Lie group, as described briefly below.

Structure constantsWhen the commutators of generators of a Lie group are expressed as linear com-binations of the generators themselves, the coefficients of the linear expressions arecalled the structure constants of the Lie group. For example, the coeffients ±εijk in(1.103) are the structure constants of the Lorentz group. We will now show that thestructure constants completely define the structure of a Lie group. To see this, wehave to define what we mean by ‘same structure’. Two sets F(� f) and G(� g) aresaid to have the same structure if there is a mapping between F and G such that iff1, f2 ∈ F and g1, g2 ∈ G are mapped to each other:

f1 ↔ g1, f2 ↔ g2 (1.151)

then, the products f1f2 and g1g1 are also mapped to each other by the same mapping:

f1f2 ↔ g1g2 ; (1.152)

namely, the mapping preserves the product rule.Suppose the sets F and G are Lie groups with the same number of generators Fi

and Gi and that they have the same set of structure constants cijk

[Fi, Fj] = cijkFk, [Gi, Gj] = cijkGk . (1.153)


Any element of F and G can be expressed in exponential form using the correspondinggenerators and a set of real parameters. We define the mapping between F and G bythe same set of the real parameters:

f = eαiFi ↔ g = eαiGi . (1.154)

If f1 ↔ g1 and f2 ↔ g2, then they can be written as

f1 = eαiFi ↔ g1 = eαiGi (1.155)

f2 = eβiFi ↔ g2 = eβiGi , (1.156)

where αi and βi are certain sets of real parameters. Then the question is whetherthe products f1f2 and g1g2 are mapped to each other by the same mapping. Theproducts f1f2 and g1g2 can be written using the CBH theorem as

f1f2 = eαiFieβjFj = eαiFi+βjFj+12[αiFi,βjFj ] + ... = eφiFi , (1.157)

g1g2 = eαiGieβjGj = eαiGi+βjGj+12[αiGi,βjGj ] + ... = eγiGi , (1.158)

The numbers φi and γi are obtained by removing the commutators using the commu-tation relations (1.153), and thus completely determined by αi, βi and cijk; namely,φi = γi, and thus f1f2 and g1g2 are mapped to each other by (1.154). Thus, if two Liegroups have the same set of structure constants, then they have the same structure.


Problems

1.1 Boost and invariant mass.A particle with energy E and momentum P is moving in the x-y plane at an angle θanti clockwise from the x axis.

y

x

(E,P)

θ

(a) Boost the particle in +x direction with velocity β (or look at this particle from aLorentz frame moving in −x direction with velocity β), and write down the resulting4-momentum.(b) Calculate explicitly the invariant mass of the boosted 4-momentum and verify thatit is the same as in the original frame; i.e. E2 − P 2.(c) Express the tan θ′ in terms of β, θ and β0, where θ′ is the angle of the directionof the particle with respect to the x axis after the boost and β0 is the velocity of theparticle in the original frame.

1.2 Two-body decay.Consider the decay of a particle of mass M to two particles of masses m1 and m2.(a) Show that the momentum of the daughter particles in the rest frame of M is givenby

p =

√λ(M2, m2

1, m22)

2M

withλ(x, y, z) ≡ x2 + y2 + z2 − 2xy − 2yz − 2zx .

(b) Suppose M is moving with velocity β in the lab frame, and it decays uniformlyin 4π steradian in its own rest frame (no spin polarization). What is the maximumand minimum energies of daughter particle 1 in the lab frame? Find the energydistribution f(E1)dE1 of daughter particle 1 in the lab frame. Normalize f(E1) suchthat ∫ ∞

0f(E1)dE1 = 1 .

(hint: The uniform decay means that cos θ is distributed uniformly from −1 to 1,where θ is the polar angle of the particle with respect to some axis.)


1.3 Show that a finite rotation by �ξ and a finite boost by �θ commute when �ξ and �θare in a same direction, and that a finite boost by �ξ and another boost by �ξ′ commuteif �ξ and �ξ′ are in the same direction:[

eξiKi , eθjLj

]= 0 if �θ = c �ξ ,

[eξiKi , eξ′jKj

]= 0 if �ξ′ = c �ξ ,

where summations over i, j are implicit and c is a constant. (hint: use the CBHtheorem.)

1.4 Generators of SU(n) group.The Lie group formed by a set of complex n × n matrices which are unitary:

U †U = I

and whose determinants are unity:

det U = 1

is called the special unitary group in n dimensions or SU(n). Show that the generatorsGk of a SU(n) group are traceless hermitian matrixes and that there are n2 − 1 ofthose:

{Gk (k = 1, . . . n2 − 1)}, with G†k = Gk , TrGk = 0 .

Here, any element of the group is written in terms of the generators as

U = ei γkGk ,

where γk is a real number and the index k is summed over.

2.3 Quunturn Lorentz Transfirmadions 5 5

does mean that the U(T(B)) are uniquely determined in at least a finite neighborhood of the coordinates 8" = 0 of the identity, in such a way that Eq. (2.2.15) is satisfied if U, a, and f (8,8) are in this neighborhood- The extension to all 0" i s discussed in Section 2.7.

There is a special case of some importance, that we will encounter again and again. Suppose that, the function f ( O , @ (perhaps just for same subset of the coordinates H a ) is simply additive

This i s the case for instance for translations in spacetime, or for rotations about any one fixed axis (though not for both together). Then the coefficients fab, in Eq. (2.2.19) vanish, and so do the structure constants (2.2.23). The generators then all commute

Such a group is called Abelian. In this case, it is easy to calculate U(T{O)) for all 6)'. From Eqs. (2.2.18) and (2.2.241, we have for any integer iV

Letting N + yr;, and keeping only the first-order term in U ( T ( B / N ) ) , we have then

and hence

2.3 Quan turn Lorentz Transformations

Einstein's principle of relativity states the equivalence of certain 'inertial' frames of reference. It is distinguished from the Galilean principle of relativity, obeyed by Newtonian mechanics, by the transformation connecting coordinate systems in different inertial frames. If xp are the coordinates in

2 3 one inertial frame (with xl,x , x Cartesian space coordinates, and x0 = t a time coordinate, the speed of light being set equal to unity) then in any other inertial frame, the coordinates i~ ' must satisfy

or equivalently

56 2 Relutiz;istic Quantum Mechanics

Here rl,, is the diagonal matrix, with elements

and the summation convention is in force: we sum over any index like p and v in Eq. (2.3.21, which appears twice in the same term, once upstairs and once downstairs. These transformations have the special property that the speed of light is the same (in our units, equal to unity) in all inertial frames;' a light wave travelling at unit speed satisfies Jdx/dtJ = 1, or in other words q,,dxfidrv = d x 2 - dr2 = 0, from which it follows that also q,,,d~'Pdx'~ = 0, and hence Idxl/dt'l = 1.

Any coordinate transformation xti + x'p that satisfies Eq. (2.3.2) is linear3"

with up arbitrary constants, and a constant matrix satisfying the conditions

For some purposes, it is useful to write the Lorentz transformation condition in a different way. The matrix g,, has an inverse, written qp", which happens to have the same components: it is diagonal, with

1 1 - 2 2 - 3 3 = = -1, q - q - 9 +l . Multiplying Eq, (2.3.5) with f 7 A K , and inserting parentheses judiciously, we have

Multiplying with the inverse of the matrix q,,Ap, then gives

These transformations form a group. If we first perform a Lorentz transformation (2.3.41, and then a second Lorentz transformation x'fi + x"P, with

then the effect is the same as the Lorentz transformation xp + x"P, with

(Note that if APv and A" both satisfy Eq. (2.331, then so does rT",M,, SO this is a Lorentz transformation. The bar is used here just to distinguish

' There is a larger class of cmrdinate transformations, known as crrnfirmal transformations, for which tji,,, dx'fidr'~ is proportional though generally not equal to q,, dxPdxv, and which therefore also leave the speed or light invariant. Conformal invarianoe in two dimensions has provcd enormously important in string theory and statistical mechanics, but the physical relevance of the= cmformal transformations in four spacetime dimensions is not yet clear.

one Lorentz transformation from the other.) The transformations T(A, a) induced on physical states therefore satisfy the composition rule

r[k,a)~(A,a) - T(AA,Au + a ) . (2.3.8) Taking the determinant of Eq. (2.3.5) gives

so As, has an inverse, (A- ' )~ , which we see from Eq. (2.3.5) takes the form

- 1 ~ ) Y - A P = - v - ~ v p ~ p f l ~ p ~ (2.3.10)

The inverse of the transformation T(A,a) is seen from Eq. (2.3.8) to be T(A-', A - ' a ) , and, of course, the identity transformation is Tj1,O).

In accordance with the discussion in the previous section, the transformations T(A, a) induce a unitary linear transformation on vectors in the physical Hilber t space

The operators U satisfy a composition rule

(As already mentioned, to avoid the appearance of a phase factor on the right-hand side of Eq. (2.3.11), it is, in general, necessary to enlarge the Lorentz group. The appropriate enlargement is described in Section 2.7.)

The whole group of transformations T(A,a ) is properly known as the inhumogeneows Lvrentz g r w p , or Poinc~11"i group. It has a number of important subgroups. First, those transformations with u p = 0 obviously form a subgroup, with

T(A, 0) 7 (A, 0) = T(AA, o), (2.3.12) known as the homogeneous Lorentz group. Also, we note from Eq. (2.3.9) that either DetA = +1 or DetA = -1; those transformations with DetA = +1 obviously form a subgroup of either the homogeneous or the inhomogeneous Lorentz groups. Further, from the 00-components of Eqs. (2.3.5) and (2.3.61, we have

with l' summed over the values 1,2, and 3. We see that either Aoo 2 +1 or noo 5 -1. Those transformations with hao 2 +I form a subgroup. Note that if AiL, and A!', are two such As, then

But Eq. (2.3.13) shows that the three-vector ( A I ~ , hZo, has length d m , and similarly the three-vector (no1, hoz, no3) has length

5 8 2 Relathistic Quantum Mechanics

\ / ( A O ~ ) Z - 1, SO the scalar product of these two three-vectors is bounded by

and so

The subgroup of Lorentz transformations with DetA = +1 and hoo 2 +1 is known as the proper orthochruraous Lorentz group. Since it is not possible by a continuous change of parameters to jump from Det A = t.1 to DetA = -1, or from noo 2 +1 to Aoo -1, any Lorentz transformation that can be obtained from the identity by a continuous change of parameters must have DetA and Aoo of the same sign as for the identity, and hence must belong to the proper orthochronous Lorentz group.

Any Lorentz transformation is either proper and orthochronous, or may be written as the product of an element of the proper orthochronous Lorentz group with one of the discrete transformations r9 or ,F or PF, where P is the space inversion, whose non-zero elements are

and 9- is the time-reversal matrix, whose non-zero elements are

Thus the study of the whole Lorentz group reduces to the study of its proper orthochronous subgroup, plus space inversion and time-reversal. We will consider space inversion and time-reversal separately in Section 2.6, Until then, we will deal only with the homogeneous or inhomogeneous proper ort hochronous Lorentz group.

2.4 The Poincari! Algebra

As we saw in Section 2.2, much of the information about any Lie symmetry group is contained in properties of the group elements near the identity. For the inhomogeneous Lorentz group, the identity is the transformation A/', = P,, d': = 0, so we want to study those transformations with

both w", and &"being taken infinitesimal. The Lorentz condition (2.3.5)

2.4 The Poinuark Algebra

reads here

We are here using the convention, to be used throughout this book, that indices may be lowered or raised by contraction with g , , , or q p '

Keeping only the terms of first order in UI in the Lorentz condition (2-3.51, we see that this condition now reduces to the antisymmetry of w,,,,

A n antisymmetric second-rank tensor in four dimensions has (4 x 3) /2 = 6 independent components, so including the four components of ep, an inhomogeneous Lorcntz transformation js described by 6 + 4 = 10 parameters.

Since U{l,O) carries any ray into itself, it must be proportional to the unit operator,* and by a choice of phase may be made equal to it. For an infinitesimal Lorentz transformation (2.4.11, UjI + w, F;-) must then equal 1 plus terms linear in w,,, and 6,. We write this as

Here Jf'" and PP are o- and €-independent operators. and the dots denote terms of higher order in rfl and/or 6. In order for U(l +o, E ) to be unitary, the operators Jiafl and PP must be Hermitian

Since m,,, is antisymmetric, we can take its coefficient Jfn to be antisymmetric also

As we shall see, P ', P I , and ~ b r e the components of the momentum operators, J*" p3', and J ' ~ are the components of the angular momentum vector, and P O is the energy operator, or Hanailtonian.*'

* In thc absence of supersclcction rules. the possibility that the proportionality factor may depend on the slate on which 1;(1,0) acts can be ruled out by thc same reasuning that we used in Section 2.2 Lo rule out the possibility that the phase5 in projective represcntations of symmetry groups may depend on the states on which the symmetries acL. Whcrc superselection rules apply, i t may be necessar) to rcdefine U(1.0) by phase factors that dcpend on the sector on which it acts.

" We will see that lhis identification uf the angular-rnomenlurn generators is forced on us by the ccrmmu~.atiim relations d the J"? On the other hand. thc cornmulation rclat~ons do not allow us to distinguish between PI' and -Yv. su the sign for the epPrl tcrm i n (2.4.3) is a matler of convention. The consistency of the choice in (2.4.3) with the usual definition of the Ilarnilhnian P O is shown in Seclion 3.1.

60 2 Relativistic Quantum Mechanics

We now examine the Lorentz transformation properties of JPG and PP. We consider the product

where Apt, and aF are here the parameters of a new transformation, unrelated to w and F. According to Eq. 2 . 3 , the product U ( A - I , -Apia) UjA, a ) equals U(1, O), so u(A1', -h-'a) is the inverse of U(A, a). It follows then from (2.3.1 1) that

To first order in CL) and F, we have then

Equating coefficients of w,, and F, on both sides of this equation (and using (2.3.10)), we find

For homogeneous Lorentz transformations (with & = O), these transformation rules simply say that JPv is a tensor and P p is a vector. For pure translations (with Ap, = P,), they tell us that PP is translation-invariant, but J P u is not. In particular, the change of the space-space components of JPu under a spatial translation is just the usual change of the angular momentum under a change of the origm relative to which the angular momentum is calculated.

Next, let's apply rules (2.4.81, (2.4.9) to a transformation that is itself infinitesimal, i.e., Apt, = P, +wPy and a p = .+, with infinitesimals up, and ehnrelated to the previous w and e. Using Eq. (2.4.31, and keeping only terms of first order in c d l , and FP, Eqs. (2.4.8) and (2.4.9) now become

Equating coefficients of (up, and F , on both sides of these equations, we find the commutation rules

This is the Lie algebra of the Poincark group.

2.4 The Poincari Ajgebra 61

In quantum mechanics a special role is played by those operators that are conserved, i.e., that commute with the energy operator H = PO. Inspection of Eqs. (2.4.13) and (2.4.14) shows that these are the momentum three-vector

P = { P I , ~ 2 , P} (2.4.1 5 )

and the angular-momentum three-vector 23 31 12 J = { J , J , J }

and, of course, the energy P O itself, The remaining generators form what is called the 'boost' three-vector

K = ( J ' ~ , J ~ ~ , J ' ~ ) , (2.4.17) These are not conserved, which is why we do not use the eigenvalues of K to label physical states. In a three-dimensional notation, the commutation relations (2.4.l2), (2.4.13), (2.4. t 4) may be written

where i, j , k , etc. run over the values 1, 2, and 3, and F i j k is the totally antisymmetric quantity with €123 = + l . The commutation relation (2.4.18) will be recognized as that of the angular-momentum operator.

The pure translations T(1,a) form a subgroup of the inhomogeneous Lorentz group with a group multiplication rule given by (2.3.7) as

~ ( l , a ) T ( l , a ) = T(1,iif a). (2.4.25) This is additive in the same sense as (2.2.24), so by using (2.4.3) and repeating the same arguments that led to (2.2.26), we find that finite translations are represented on the physical Hilbert space by

In exactly the same way, we can show that a rotation Ro by an angle (81 around the direction of 8 is represented on the physical Hilbert space by

It is interesting to compare the Poincare algebra with the Lie algebra of the symmetry group of Newtonian mechanics, the Galilean group. We

62 2 Relatiuistic Quantum Mechanics

could derive this algebra by starting with the transformation rules of the Galilean group and then following the same procedure that was used here to derive the Poincare algebra. However, since we already have Eqs. (2.4.18)-(2.4.241, it is easier to obtain the Galilean algebra as a low-velocity limit of the Poincark algebra, by what is known as an InKm- Wgner contraction4J. For a system of particles of typical mass m and typical velocity tl, the momentum and the angular-momentum operators are expected to be of order J - 1, P - mu. On the other hand, the energy operator is H = M -t W with a total mass M and non-mass energy W

? (kinetic plus potential) of order M -- rn, W - ma-. inspection of Eqs. (2.4.18)<2.4.24) shows that these commutation relations have a limit for v << 1 of the form

with K of order l / v . Note that the product of a translation x 4 x + a and a 'boost' x + x + vt should be the transformation x -, x + vt + a, but this is not true for the action of these operators on Hilbert space:

exp(iK - v ) exp(-iP . a) = exp{iMa . v/2) exp (l(K v - P a)) . The appearance of the phase factor exp(iMa , v/2) shows that this is a projective representation, with a superselection rule forbidding the superposition of states of different mass. In this respect, the mathematics of the Poincark group is simpler than that of the Galilean group. However, there is nothing to prevent us from formally enlarging the Galilean group, by adding one more generator to its Lie algebra, which commutes with all the other generators, and whose eigenvahes are the masses of the various states. In this case physical states provide an ordinary rather than a projective representation of the expanded symmetry group. The difference appears to be a mere matter of notation, except that with this reinterpre- tation of the Galilean group there is no need for a mass superselection rule.

2.5 One-Particle States

We now consider the ~Iassification of one-particle states according to their transformation under the inhomogeneous Loren tz group.

The components of the energy-momentum four-vector all commute with each other, so it is natural to express physical state-vectors in terms of

2.5 One-Particle States 6 3

eigenvectors of the four-momentum. Introducing a label rr to denote all other degrees of freedom, we thus consider state-vectors Y,-, with

For general states, consisting for instance of several unbound particles, the label 17 would have to be allowed to include continuous as well as discrete labels. We take as part of the definition of a one-particle state, that the label o is purely discrete, and will limit ourselves here to that case. {However, a specific bound state of two or more particles, such as the lowest state of the hydrogen atom, is to be considered as a one- particle state. It is not an elementary particle, but the distinction between composite and elementary particles is of no relevance here.)

Eqs. (2.5.1) and (2.4.26) tell us how the states Y p , transform under translations:

We must now consider how these states transform under homogeneous Lorentz transformations.

Using (2.4-91, we see that the effect of operating on Y , , with a quantum homogeneous Lorentz transformation U(A,O) = U(A) is to produce an eigenvector of the four-momentum with eiganvalue Ap

Hence U ( A ) Y p , must be a linear combination of the state-vectors

~ ~ w ' , ~ = x Cnro(A, ~ ) ' b p , d . (2.5.3) a'

In general, it may be possible by using suitable linear combinations of the Y,, to choose the 0 labels in such a way that the matrix C,r,(A, p) is block-diagonal; i n other words, so that the .VP, with a within any one block by themselves furnish a representation of the inhomogeneous Lorentz group. It is natural to identify the states of a specific particle type with the components of a representation of the inhamogeneous Lorentz group which is irreducible, in the sense that it cannot be further decomposed in this way." Our task now is to work out the structure of the

Of course, different particle species may correspond to representations that are isomorphic, i.e., thal hare matrices Cu~,(A.p) that are either idenlical, or identical up lo a similarity transrorma- Lion. In some cases i t may hc convenienl to definc particle types as irreducible representations of larger groups that contain the inhomogeneous proper urthochronous lo rent^ group as a subgroup; for instance, as we shall sec, for massless particies whose interactions respect the symmetry of space inversion i t is customary to treat all the components of an irreducible representation d the inhomugeneous Lorentz group including space inversion as a single particle typc.

64 2 Relativistic Qualaturn Mechanics

coefficients C,), (A, p) in irreducible representations of the inhomogeneous Lorentz group.

For this purpose, note that the only functions of pp that are left invariant by a 1 proper orthclchronous Lorentz transformations A" are the invariant square I qpvp'pv, and for p2 5 0, also the sign of Hence, for each value of p2, and (for p2 5 0) each sign of we can choose a 'standard' four-momentum, say kp, and express any p p of this class as

where L p , is some standard Lorentz transformation that depends on pp, and also implicitly on our choice of the standard k p . We can then defZne the states V , , of momentum p by

where N ( p ) is a numerical normaIiratian factor, to be chosen later. Up to this point, we have said nothing about how the o labeIs are related for different momenta; Eq. (2.5.5) now fills that gap.

Operating on (2.5.5) with an arbitrary homogeneous Lorentz transformation U(A), we now find

The point of this last step is that the Lorentz transformation L - ~ ( A ~ ) A L ( ~ ) takes k to L(p)k = p, and then to Ap, and then back to k , so it belongs to the subgroup of the homogeneous Lorentz group consisting of Lorentz transformations W p , that leave k)' invariant:

This subgroup is called the little For any W satisfying Eq. (2.5,7), we have

u( W ) y t . ~ Ddr(W)vk,d . (2.5.8) of

The coefficients D ( W ) furnish a representation of the little group; that is, for any elements W, W we have

and so

2.5 One- Particle Srates 6 5

In particular, we may apply Eq. (2.5.8) to the little-group transformation

and then Eq. (2.5.6) takes the form

Apart from the question of normalization, the problem of determining the coefficients C,!, in the transformation rule (2.5.3) has been reduced to the problem of finding the representations of the little group. This approach, of deriving representations of a group like the inhomogeneous Lorentz group from the representations of a little group, is called the method of indu~ed representatiom6

Table 2.1 gves a convenient choice of the standard momentum k p and the corresponding little group for the various classes of four-momenta.

Of these six classes o f four-momenta, only (a), (c) , and (f) have any known interpretations in terms of physical states. Not much needs to be said here about case If) - p p = 0; it describes the vacuum, which is simply left invariant by U(A). In what follows we will consider only cases (a) and (c), which cover particles of mass M > 0 and mass zero, respectively.

This is a good place to pause, and say something about the normalization of these states. By the usual orthonormalization procedure of quantum mechanics, we may choose the states with standard momentum k P to be orthonormal, in the sense that

(The delta function appears here because Y k , and Yk~,a~ are eigenstates of a Hemitian operator with eigenvalues k and kf, respectively.) This has the immediate consequence that the representation of the little group in Eqs. (2.5.8) and (2.5.1 1) must be unitary**

Now, what about scalar products for arbitrary momenta? Using the unitarity of the operator UjA) in Eqs. (2.5.5) and (2.5.1 I), we find for the

** The little groups SU(2.1) and S O ( 3 , 1) for p2 > O and $' = 0 have no non-trivial finite- dimensimal unitary representations, so if there were any states with a given momentum @' with p2 > O or p" = O that transform non-trivially under the little group, there would have to be an infinite number of them.


Table 2.1. Standard momenta and the corresponding little group for various classes of four-momenta* Here K is an arbitrary positive energy, say I eV. The little groups are mostly pretty obvious: S U ( 3 ) is the ordinary rotation group in three dimensions (excluding space inversions), because rotations are the only proper orthochronous Lorentz transformations that leave at rest a particle with zero momentum, while S0(2,1) and SO(3,l) are the Lorentz groups in (2 + 1)- and (3 + 1)-dimensions, respectively. The group l S O ( 2 ) is the group of Euclidean geometry, consisting of rotations and translations in two dimensions. Its appearance as the little group for p2 - O is explained below.

Standard kp Little Group

scalar product :

where kr = L - ' ( ~ ) ~ ' . Since also k = L - ' ( ~ ) ~ , the delta function S3(k - k'j is proportional to 63(p - p'). For p' = p, the little-group transformation here is trivial, w ( L - ~ ( ~ ) , p ) = I , and so the scalar product is

It remains to work out the proportionality factor relating h3(k - kr) and 4Tp - p'). Note that the Lorentz-invariant integral of an arbitrary scalar function f I p ) over four-momenta with -p2 -- M~ L 0 and > 0 (i.e.,

Slates

cases (a) or (c)) may be written

(8(# j is the step function: O(x) = 1 for x 2 0, B(x) = 0 for x: < 0.) We see that when integrating on the 'mass shdl' p2 + M' = 0, the invariant volume element is

d3p/dp2 + M~ . The delta function is defined by

so we see that the invariant delta function is

Since p' and p are related to kt and k respectively by a Lorentz transformation, L(p) , we have then

$63(p' - p) = k067(k' - k)

and therefore

The normalization factor N ( p ) is sometimes chosen to be just N ( p ) = 1, but then we would need to keep track of the f ' /k0 factor in scalar products. Instead, I will here adopt the more usual convention that

We now consider the two cases of physical interest: particles of mass M > 0, and particles of zero mass.

6 8 2 Relativistic Quan turn Mechanics

The little group here is the three-dimensional rotation group. Its unitary representations can be broken up into a direct sum of irreducible unitary representations7 D $ ~ ( R ) of dimensionality 2 j+ 1, with j = 0, f , 1, . - -. These can be built up from the standard matrices for infinitesimal rotations Kk = dik 4- Oik, with Oik = -Oki infinitesimal:

with t~ running over the values j , j - I, , . . , - j . For a particle of mass M > 0 and spin j, Eq. (2.5.1 1 ) now becomes

with the little-group element W ( A , p ) (the Wigner rotation5) given by Eq. (2.5.10) :

To calculate this rotation, we need to choose a 'standard boost' L(p) which carries the four-momentum from kt' = (O,O,O, M ) to pp. This is conveniently chosen as

It is very important that when Ag, is an arbitrary three-dimensional rotation .%, the Wigner rotation W ( A , p) is the same as 2 for all p. To see this, note that the boost (2.5.24) may be expressed as

where R($) is a rotation (to be defined in a standard way below, in Eq.

2.5 One-Particle States

(2.547)) that takes the three-axis into the direction of p, and

Then for an arbitrary rotation 2

But the rotation ~ - ' ( 8 $ ) 9 R ( f i ) takes the three-axis into the direction A and then into the direction &$, and then back to the three-axis, so it must be just a rotation by some angle 0 around the three-axis

Since R(0) commutes with B(lpl), this now gives

and hence

as was to be shown. Thus states of a moving massive particle (and, by extension, multi-particle states) have the same transformation under rotations as in non-relativistic quantum mechanics. This is another piece of good news - the whole apparatus of spherical harmonics, Clebsch4ordan coefficients, etc. can be carried over wholesale from non-relativistic to relativistic quantum mechanics.

Mass Zero

First, we have to work out the structure of the little group, Consider an arbitrary little-group element WP,, with WP,,kv = k p , where k p is the standard four-momentum for this case, k p = (Q0, I, 1). Acting on a time-like four-vector $ = (0,0,0,1), such a Lorentz transformation must yield a four-vector W t whose length and scalar product with Wk = k are the same as those of t :

Any four-vector that satisfies the second condition may be written


and the first condition then yields the relation

It follows that the effect of W p , on tv is the same as that of the Lorentz transformation

This does not mean that W equals S(cc,P), but. it does mean that S -I (a, p ) W is a Lorentz transformation that leaves the time-like four- vector (0,0,0,1) invariant, and is therefore a pure rotation. Also, S", like Wit, leaves the light-like four-vector (0,0,1,1) invariant, so s-'(u, b) W must be a rotation by some angle 0 around the three-axis

S-' (a, p ) W = ~ ( 0 ) , (2.5.27) where

The most general element of the little group is therefore of the form

What group is this? We note that the transformations with B = 0 or with cc = 0 = 0 form subgroups :

These subgroups are Abelian - that is, their elements all commute with each other. Furthermore, the subgroup with 8 = 0 is invariant, in the sense that its elements are transformed into other dements of the same subgroup by any member of the group

From Eqs. (2.5.29)42.5.31) we can work out the product of any group elements. The reader will recognize these multiplication rules as those of the group I S 0 ( 2 ) , consisting of translations (by a vector (a, P ) ) and rotations (by an angle 0) in two dimensions.

Groups that do not have invariant Abelian subgroups have certain simple properties, and for this reason are called semi-simple. As we have

2.5 One-Parrick States 7 1

seen, the little group iSO(2) like the inhomogeneous Lorentz group is not semi-simple, and this leads to interesting complications. First, let's take a look at the Lie algebra of ISO(2) . For O,a,P infinitesimal, the general group element is

From (2.4.31, we see then that the corresponding Hilbert space operator is

where A and 3 are the Hermitian operators

and, as before, J3 = J12, Either from (2.4.1 8)-(2.4.20), or directly from Eqs. (2.5.29) - (2.5.311, we see that these generators have the commutators

[JLA] = +iB , [ J 3 , B ] = -iA , [A, B ] = 0 .

Since A and 3 are commuting Hermitian operators they (like the momentum generators of the inhomogeneous Lnrentz group) can be simultane- ously diagonalized by states Yk,,,

The problem is that if we find one such set of non-zero eigenvalues of A,B , then we find a whole continuum. From Eq. (2.5.321, we have

U[R(O)]A U-' ti)] = A cos 0 - B sin O , U[R(I) )]B UP' [R(O)] = A sin 0 + B cos B ,

and so, for arbitrary 8,

where


Massless particles are not observed to have any continuous degree of freedom like 0 ; to avoid such a continuum of states, we must require that physical states (now called YA,~ ) are eigenvectors of A and B with a = b = O :

These states are then distinguished by the eigenvalue of the remaining generator

Since the momentum k is in the three-direction, a gives the component of angular momentum in the direction of motion, or helicity.

We are now in a position to calculate the Lorentz transformation properties of general massless particle states. First note that by use of the general arguments of Section 2.2, Eq. (2.5.32) generalizes for finite .% and D to

U(S(M, f i ) ) = exp(ictA + ipB) (2.5.40) and for finite 9 to

An arbitrary element W of the little group can be put in the form (2.5281, so that

U( W)Yk, , = exp(iaA + $3) e ~ p ( i 8 ~ ~ ) ~ ~ , , = e ~ p j i O a ) Y ~ , ~ and therefore Eq. (2.5.8) gives

Dalu( W ) e ~ p ( i 0 0 ) J ~ l ~ , where B is the angle defined by expressing W as in Eq. (2.5.28). The Lorentz transformation rule for a massless particle of arbitrary helicity is now given by Eqs. (2.5.11) and (2.5.18) as

with O(A, p) defined by

We shall see in Section 5.9 that electromagnetic gauge invariance arises from the part of the little group parameterized by a and B.

At this point we have not yet encountered any reason that would forbid the helicity of a massless particle from being an arbitrary real, number. As we shall see in Section 2.7, there are topolqycal considerations that restrict the allowed values of 0 to integers and half-integers, just as for massive particles.

To calculate the little-group element (2.5.43) for a given A and p, (and also to enable us to calculate the effect of space or time inversion on these states in the next section) we need to fix a convention for the standard Lorentz transformation that takes us from kil = (O ,O , IC, PC) to pi1. This may conveniently be chosen to have the form

where B ( M ) is a pure boost along the three-direction:

and R@) is a pure rotation that carries the three-axis into the direction of the unit vector fi. For instance, suppose we take fi to have polar and azimuthal angles B and 4:

p = (sin 8 cos 4, sin0 sin @, cos 8 ) . (2.5.46) Then we can take R@) as a rotation by angle O around the two-axis, which takes (0,0,1) into (sin 8, 0, cos 91, followed by a rotation by angle # around the three-axis:

where 0 5 8 n, 0 < 4 < 271. (We give U(R(p)) rather than RIP), together with a specification of the range of 4 and 8, because shifting 0 or 4 by 27c would give the same rotation R(p), but a different sign for U ( R ( 6 ) ) when acting on half-integer spin states.) Since (2.5.47) is a rotation, and does take the three-axis into the direction (2.5.46), any other choice of such an R(p) would differ from this one by at most an initial rotation around the three-axis, corresponding to a mere redefinition of the phase of the one-particle states.

Note that the helicity is Lorentz-invariant; a massless particle of a given helicity a looks the same (aside from its momentum) in all inertial. frames. Indeed, we would be justified in thinking of massless particles of each different helicity as different species of particles. However, as we shall see in the next section, particles of opposite helicity are related by the symmetry of space inversion. Thus, because electromagnetic and gravitational forces obey space inversion symmetry, the massless particles of helicity f 1 associated with electromagnetic phenomena are both called photons, and the massless particles of helicity 5 2 that are believed to be associated with gravitation are both called gruoitons. On the other hand, the supposedly massless particles of helicity +1/2 that are emitted in nuclear beta decay have no interactions (apart from gravitation) that


respect the symmetry of space inversion, so these particles are given different names: neutrinos for helicity +1/2, and antr'nezdtriplos for helicity -1/2.

Even though the helicity of a massless particle is Lorentz-invariant, the state itself is not. In particular, because of the helicity-dependent phase factor expjirr H ) in Eq. (2.5.42), a state formed as a linear superposition of one-particle states with opposite helicities will be changed by a Lorentz transformation into a different superposition. For instance, a general one-photon state of bur-momenta may be written

where

The generic case is one of elliptic polarizatb'on, with I r+ - I both non-zero and unequal. Circular polarization is the limiting case where either .r+ or r- vanishes, and linear polarizativla is the opposite extreme, with 182+1 = 12-1. The overall phase of a+ and a- has no physical significance, and for linear polarization may be adjusted so that a- = a;, but the relative phase is still important. Indeed, for linear polarizations with a- = a;, the phase of a+ may be identified as the angle between the plane of polarization and some fixed reference direction perpendicular to p. Eq. (2.5.42) shows that under a Lorentz transformation A@,, (his angle rotates by an amount %(A,p). Plane polarized gravitons can be defined in a similar way, and here Eq. (2.542) has the consequence that a Lorentz transformation A rotates the plane of polarization by an angle 20(A,p).

2.6 Space Inversion and Time-Reversal

We saw in Section 2.3 that any homogeneous Lorentz transformation is either proper and orthochronous (ix., DetA = +1 and 2 +1) or else equal to a proper orthochronous transformation times either .9 or 9 or 99, where 9 and .F are the space inversion and time-reversal transformations

I t used to be thought self-evident that the fundamental multiplication rule of the Poincari: group

Chapter 1 Lorentz Group and Lorentz Invariance where one second is deﬁned (1967) to be 9192631770...

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Transcript of Chapter 1 Lorentz Group and Lorentz Invariance where one second is deﬁned (1967) to be 9192631770...