qftm-yajnik

8/7/2019 QFTM-Yajnik

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Quantum Field Theory Methods

Dirac Equationand

Perturbation Theory

U. A. Yajnik

Physics Department, Indian Institute of Technology,Bombay 400076

Lectures delivered at IUCAA graduate school 1989


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Prefatory Remarks

These are brief notes more or less faithfully reproducing the material actually dis-cussed in the lectures. I have tried to include remarks that will provide an orientationfor further study by the student. I have adopted a nonstandard presentation of thespin-1/ 2 particles because it is deductive and more generally applicable. For thederivation of the equation alone Diracs original method is more lucid, and the stu-dent can read it in any textbook. I have relied on several sources for putting togetherthis course but most of all on lecture notes of S. Weinberg which alas remain yetunpublished. I use the metric = diag(+ ), and the units h = c = 1.I want to thank the organisers for providing me the opportunity to present theseglorious topics in the stimulating atmosphere of the School, and the students fortheir participation and comments.

C O N T E N T S

Dirac Equation

1. Lorentz Group and its representations

2. The electron and the Dirac equationPerturbation Theory

3. Interaction picture and the S -matrix

4. Rates and cross-sections from the S -matrix 5. Perturbative expansion and the Feynman rules (example of Yukawa theory)

REFERENCES


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1. Lorentz Group and its representations

1.1 SO (3) and SU (2):

In order to understand relativistic particles and their spin, we need to understand theLorentz group. We begin with basic tools for handling real and complex rotations,the latter to be dened in the following. It is sufficient to work with real rotationsin three dimensions and complex rotations in two complex dimensions. The con-

nection to the Lorentz group will appear later. We are familiar with rotations in 3dimensions, e.g.,

cos sin 0sin cos 00 0 1

represents rotations about the z axis,

cos 0 sin 0 1 0

+sin 0 cos

rotations about the y axis etc., and products of such as well. These keep the mag-nitude of vectors invariant i.e., for R a rotation matrix, if

x i = R i j x j

then

ix i

2=

jx j

2

This requires that they have the property RT R =identity. Such matrices are calledorthogonal . Rotations have the further property that det R = 1. This more restrictedclass is called special orthogonal . The important algebraic property of rotations isthat they form a group. This group is designated SO (3). Rotations of n dimensionalvectors constitute groups designated by SO (n).When the angles are small, we get for example,

R z() =1 0+ 1 00 0 1

= 1 + 0 1 01 0 00 0 0


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1 is the identity matrix and is an innitesimal. It turns out that even when isnot small one can write

R z() = 1 +

1

(Lz )n

n! exp(Lz)

Where Lz is the matrix explicitly written in the previous step, and Lnz = 1 if n = 0and Lz Lz n times otherwise.

Exercise : Calculate L2z , L3zand L4z and prove the above series expansion.

Lz is referred to as the generator of z rotations, and Lx , Ly are likewise given by

Lx =0 0 00 0 10 1 0

Ly =0 0 +10 0 0

1 0 0

Note that these are the generators of active rotations. For passive rotations, wherenot the vector but the coordinate system is being rotated, the generators are negativeof these.In the representation above, the generators are real and skew symmetric. Dene thehermitian matrices J x, y, z = iL x, y, z . Then for z rotations we get

R z() = exp( iJ z)and more generally

R n (1, 2, 3) = exp( i

J

)The J s satisfy the antisymmetric algebra

[J x , J y] = iJ z

and cyclic permutations x y z Or,[J i , J j ] = i ijk J k


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It turns out that knowing the exponential forms of Rz etc. in terms of the generatorsalong with the algebra satised by the latter is sufficient for obtaining matrices forarbitrary nite rotations. This is the subject of Lie group theory.The algebra obtained above can be thought of as abstract relations. In which casewe can nd sets of matrices of dimensions larger than 3 satisfying the same algebra.These are called representations of higher dimensions. In quantum mechanics welearn about representations of SO (3), characterised by the j in the relation

iJ 2i = j ( j + 1)

Each positive integer value of j gives an independent representation of rotation ma-trices of dimension 2 j +1. The basis of the vector space on which the representationacts is labelled by the eigenvalues m of J 3. J J x iJ y act as raising and loweringoperators. All of these consequences can be derived directly from the above algebrato be satised by the generators. Details can be found in any textbook on quantummechanics.SU (2): This is the group of unitary matrices with determinant +1, that leave mag-nitudes of 2-dimensional complex vectors invariant:

Z 1Z 1 + Z 2Z 2 = Z 1Z 1 + Z 2Z 2

Amusing fact is that the generators of SU (2) are three in number, S 1, S 2, S 3 andsatisfy

[S i , S j ] = i ijk S kDirect algebraic proof is given in Ryder[1]. This is the connection between SO (3)and SU 2. The groups however differ in their global aspects. See Schiff[2]. There isa two-to-one mapping from SU (2) to SO (3). Thus one realisation of above algebrais the one already discussed previously. But this one is essentially real aside from

overall i factor. SU (2), a complex group, permits a strictly complex representationof lower dimensionality:

S 1 =12

1 S 2 =12

2 S 3 =12

3

where s are the Pauli matrices

1 = 0 11 0 2 = 0 ii 0 3 =

1 00 -1


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This representation has S 2i = s(s + 1) =12

32 . Other representations with half-

integral values of s can be found from tensor products of this representation. Onecan also be sure that there are no more representations of the above algebra. Thisfollows from the algebraic proof given in quantum mechanics where one learns thatthe eigenvalues of S 3 must be separated by integers, and for each eigenvalue + ms ,

ms has also to occur, so that s = 0 , 12 , 1, 32 . . . are all the possibilities.This short review of these two groups prepares us for the Lorentz group.

1.2 Lorentz group:

The transformations (remember c = 1)

x =x + vt

1 v2t =

t + vx1 v2

and generalisations thereof are known to leave invariant the four-dimensional dis-tance or interval

t 2 i

x i2

= t2 i

xi2

Together with ordinary rotations, already discussed, the group of these transforma-tions is called SO (3, 1) rather than SO (4) due to the different signs in the metric.We can write above transformations in a suggestive form by letting

v and (1 2) 1/ 2

and introducing :

cosh and sinh

Exercise : Check that this is allowed. is called rapidity parameter. So boost in x direction looksxt =

cosh sinh sinh cosh

xt

Aside from the appearance of hyperbolic functions instead of trigonometric ones,note the absence of opposite signs. It is easy to deduce generators:


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K x =

t x y z

t 0 1 0 0x 1 0 0 0y 0 0 0 0z 0 0 0 0

,

K y =

0 0 1 00 0 0 0

1 0 0 00 0 0 0

and K z =

0 0 0 10 0 0 0

0 0 0 01 0 0 0

.

Algebra of Ks:K xK y K yK x =0 1 0 0

1 0 0 00 0 0 00 0 0 0

0 0 1 00 0 0 01 0 0 00 0 0 0

0 0 1 00 0 0 01 0 0 00 0 0 0

0 1 0 01 0 0 00 0 0 00 0 0 0

=

0 0 0 00 0 1 00 0 0 00 0 0 0

0 0 0 00 0 0 00 1 0 00 0 0 0

= iJ z Exercise : Check the rest similarly:

[J i , J j ] = i ijk J k

[K i , K j ] = i ijk J k

[K i , J j ] = i ijk K kThis algebra seems asymmetric between space and time. This is a result of how thegenerators are labelled. The notation that makes this transparent and generalises toany number of dimensions is M xy instead of J z , M zt instead of K z etc., speciyingthe plane of rotation. Axis of rotation is an idea specic to three dimensions. Ingeneral, for the real skew symmetric generators of SO (n), the reader can check that

[M ab, M cd] = 0 if a = c = d = b


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and[M ab, M bc] = M acThis means that if the planes are unrelated, the commutator vanishes and if they

share an axis, the resulting rotation is in the plane of the remaining two axes.

The Weyl trick: How do we get rep.s of the full Lorentz group? SO (3) is a subgroupbut K s get mixed in. Let

Ai =12

(J i + iK i )B i =

12

(J i iK i )

Exercise : Then check that[Ai , A j ] = i ijk Ak[B i , B j ] = i ijk Bk[Ai , B i ] = 0

Ai , B i are hermitian and so we have two disjoint SU (2)s. In symbols, we haveSO (3, 1) SU (2)SU (2). The representations of L.G. can now be classied bythe representations of SU (2)A and SU (2)B . The largest eigenvalues of A3, B3, tobe denoted sA and sA , specify the representation. Basis vectors are labelled by mAmB . The spin of a particle species can be identied by studying properties underrotation in the rest frame. Since J = A + B , spin of the system is given by sA + sB .A typical element of SO (3, 1) is now written

U (A,B ) = exp ( iAi Ai iBi B i)which can be obtained from the standard form

U (, ) = exp (

iiJ i

iK i )

by deningAi = i ii Bi = i + iiSince K i are hermition, the K term in the exponent is antihermition, or, in the pre-

vious line with appearance of i in front of the hermitian generators, the parametersare complex. Thus U is not unitary. The group theoretic reasons for this andimplications thereof are discussed in textbooks[1],[3],[4]. I will provide an examplelater.


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2. The electron and the Dirac equation

2.1 Deducing the Dirac Equation :

Suppose we didnt know the Dirac equation and wanted the equation satised byspin- 12 particles. We know that the spin of a representation, J value, is A + B(added as angular momenta). If we want J = 12 , we must have A =

12 and B = 0

or A = 0 and B = 12 . In compact notation, either the representation (12 , 0) or the

representation (0 , 12 ). For (12 , 0), A =

12 and B = 0, so Ai can be chosen to be

the pauli matrices (times half) 12i . Substituting this in the general form of U (a, b)

obtained previously, we get for ( 12 , 0),

U ( 12 ,0) = exp( i(i ii)12

i)

which for pure boosts exp(12

ii )

= cosh | |2 sinh |

|2

Similarly, for (0 , 12 ), Ai = 0 and B i =12

i so that for pure boosts,

U (0, 12 ) = cosh | |

2+ sinh | |

2

Note the sign change.Now it turns out, neither ( 12 , 0) nor (0,

12 ) is adequate to describe spin-

12 particles

if parity is a good quantum number. This is because under parity, K i K i butJ i J i so that Ai B i and as a result, ( 12 , 0) (0, 12). One therefore needsto build the reducible representation ( 12 , 0)(0,

12 ). In matrix form, the reducible

representation is written as

U =U ( 12 ,0)

... O

O

... U (0, 12 )

Let us begin with states with J 3 = 12 in the rest frame and then boost thesestandard states to required momentum. In the present representation,


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J 3 =

12

3 ... O

O ... 12 3

Its eigenvectors can be built out of those of 3. We make the following choice

1010

,

0101

,

10

10,

010

1

The rst and third have J 3 =12 and the second and fourth have J 3 =

12 and they

constitute a linearly independent set. To distinguish between the members of eachdegenerate pair we want a matrix condition. We observe that if we dene

=0

... 1

1

... 0

then

1

010

=

1

010

, and ,

1

010=

1

010And similarly for the second and the fourth. We designate the above vectors

u(+ , 0) v(+ , 0) u(, 0) v(, 0)With u, v to designate eigenvalues of and designating the eigenvalues of J 3.The argument 0 designates the rest frame.

Exercise : What is the general choice for the eigenvectors of J 3? Can you nd for the general case?)

Starting with the standard basis dene in the rest frame we prescribe the basisin arbitrary frame by Lorent boosting. In terms of the reducible representationspecied above we dene

u(, p) = N pUu(, 0) and v(, p) = N pUv(, 0)


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Where N p is normalisation to be determined later. To obtain an explicit expressionfor these, note that with the usual notation = (1 2) 1/ 2 etc., ( v/c , not tobe confused with the matrix dened above)

cosh = =p0

m m

sinh = = | p|m

so thatcosh

2

= + m2m and sinh 2 = m2mFinally, p

Exercise : Write out the required explicit forms of u, v in terms of the physicalboost parameters , pi .

The Dirac equation can now be shown to be the generalisation to arbitrary framesof the statement that in the rest frame, u, v can be characterised as eigenvectors of . We can see that

(UU 1 I )u(+ , p) = 0Consider( p) UU 1

=exp(12 ii)

... 0

0

... exp( 12 ii )

0 ... 1

1

... 0

exp( 12 i

i )... 0

0

... exp(12 ii )(recall U 1 = U )


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=exp(

12 i

i )... 0

0 ... exp( 12 ii )

0

... exp(12 ii )

exp( 12 ii )

... 0

=0

... exp(ii )

exp(ii )... 0

=0

..

.m

p

m m +

pm

... 0 Now dene

0 =

i =0

... i

i

... 0 Then our statement above becomes

( 0 p0

p

m)u(+ , p) = ( p

m)u(+ , p) = 0

It then follows that(i m) e ix pu(+ , p) = 0The expression in the rst brackets is the Dirac differential operator. The same

equation holds for u(, p). For the v(, p) we have( p + m)v(, p) = 0

so that(i

m) eix pv(

, p) = 0

It is necessary to insert exp(+ ix p) in order to get the same differential equationas the us. Thus we may conclude that a general linear combination

(x) =p,

cp,e ix pu(, p) + dp,e

ix pv(, p)

solves the Dirac equation. Here takes the values and the cs and the ds areexpansion coefficients.


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Exercise : Show that if satises the Dirac equation then

(i + m)(i m) = ( + m2) = 0 .That is, each component of the spinor satises the Klein-Gordon equation.

We have explicitly constructed the spinorial functions describing spin-1/2 particles.They are then seen to obey the Dirac equation. These basis functions labelled bymode number (and spin) are also often called mode functions to distinguish themfrom momentum eigenstates of nonrelativistic wave mechanics. The procedure also

illustrates in principle how to construct mode functions for particles of arbitraryspin. For most purposes, the equations satised by these mode functions are notneeded. As in the exercise above, even in the general case, each component inde-pendently must always satisfy the Klein-Gordon equation. The coupled rst orderequations only serve as constraints that identify the spin content of the representa-tion. Calculations of relativistic scattering amplitudes require the knowledge of thefree propagator for particle of each species and the Lorentz scalar interaction termswhich enter the Lagrangian. Both of these can be determined from the knowledge of the free mode functions[8]. In particular the interaction terms can be constructed by

proper contraction of all group representation indices between the mode functionsof the particle species concerned.

2.2 -matrices :

The Dirac equation introduces into physics the fascinating -matrices, an efficientmachinery for dealing with the massive spin-1/2 particles. The -matrices satisfythe anticommutation algebra

{

,

}= 2 1

Where {a, b} ab+ ba is the standard notation for the anticommuting bracket, and is the metric. The matrices we have obtained in the preceeding subsection havethe further property that

0

= 0 whereas i

= 0 i 0


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The right hand side of the above relation contains the spacetime metric, the lefthand side is quadratic in the s. In this sense the -matrices are a square-rootof the Minkowskian metric. Similar matrices can be dened in arbitrary number of dimensions and with Euclidean or Minkowskian signature of the metric. An examplewe already know is the Pauli matrices with respect to 3-dimensional space. Algebrasof this kind are called Clifford algebras.The -matrices are not a closed set under multiplication even after the inclusion of the identity, as for example the Pauli matrices are. An important new matrix called 5 is obtained through

5 = i 0 1 2 3 = 1 ... 0

0

... 1

It has the properties

{ , 5}= 0 ,along with 5

= 5 and ( 5)2 = 1

The generate an algebra of 16 independent matrices in all, including the identity.Products such as (with standing for one of the 16 matrices), which arebilinear in turn out to have denite Lorentz transformation properties. We haveto abandon the discussion of their uses due to lack of time.We conclude with the remark that the algebra of the s is general, not specic tothe representation given here. The matrices obtained by similarity transformationsof those above,

= A A 1where A are arbitrary complex matrices are just as good a representation. Severaldifferent representations are in use depending upon the context. The one obtainedabove is called the chiral representation.

Why was it necessary to invent the -matrices? We can deal with higher spins suchas 1 (photons for example) and 2 (gravitons in at space) using second or higherrank tensors. It happens that all the tensor representations of SO (3, 1) or of SO (n)are real and can be written as subspaces of GL(n, R ) or repeated tensor products of the same with itself. Here the symbol GL(n, R ) refers to the General Linear Groupin n Real dimensions . This is simply the set of all possible n n real matrices. Thespinor representation that has been necessitated by the existence of the fermions isintrinsically complex. It cannot be constructed out of the real representations. The


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most convenient and compact way of dealing with it turns out to be to treat it asa subspace of GL(4, C), Creferring to the complex numbers. That the dimensionof the coplex matrices also turns out to be 4 is an accident of four dimensions. Indimensions other than four, the complex dimension of the GL group is related in anintricate way with the dimension of the real space. There is no space here to go intofurther detail. Systematic accounts can be found in Georgi[4] or old classics such asCartan[5].

2.3 Orthogonality and normalisation :

A consequence of the fact that boosts U are not unitary is that the complex mag-nitude uu of the vectors is not left invariant. It turns out however that u 0u doesremain invariant. Because of this, one denes a Dirac conjugate u u 0, a dualvector more handy than the hermitian conjugate u. Consider

u( , p)u(, p) = u( , 0)U 0Uu(, 0)N p N p

Now

0U =0

... 1

1 ... 0 e

12 i

i ... 0

0 ... e12 i

i

=e

12 i

i ... 0

0

... e12 i

i

0 ... 1

1

... 0

= U 1 0So

u( , p)u(, p) = N p N pu( , 0)u(, 0)

= N 2 p So this is a Lorentz scalar. Similarly

v( , p)v(, p) = N 2

p

There are two conventions for normalisation

u( , p)u(, p) = OR u( , p)u(, p) =m

The rst is Lorentz invariant. But the second has the advantage of providing amore physical normalisation for creation and destruction operators in the quantumtheory, to which we pass next.


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2.4 Quantization:

Klein and Jordan had showed that a consistent quantum theory of many identicalbosons could be obtained by canonical quantisation applied to the Klein-Gordoneld. (see Pauli[6] and Wentzel[7] for early accounts). Jordan and Wigner showedthat to get the same for a system of Dirac particles, one had to use anticommutatorsinstead of commutators[6][7]. The exclusion principle is then automatically built in.Accordingly, one needs to require

{a ( x, t),

b( x , t )

}= 3( x, x )ab

a, b are spinor indices. Imposing this condition makes the coefficients cp,, dp,introduced in sec. 2.2 into operators. The reader can verify on his or her own orconsult a textbook to verify that the conditions on these coefficients become

{cp,, cp } = 3( p, p )

{dp,, dp } = 3( p, p )and all brackets of the following form vanish

{c, d}= {c, d}= 0To prove this one needs the following result

Exercise : Prove the spin sums

ua (, p)ub(, p) = (

p + m2

)ab

va (, p)vb(, p) = ( p m

2)ab

Exercise : Check that had N p been different, anticommutators of the cs and theds would have to be normalised differently to retain the canonical quantizationcondition.

Exercise : Write a Lagrangian that will give the Dirac equation and make theconjugate variable.


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3.5 Physical interpretation :

(i

x m) = 0

( i

x m) = 0

(i

x + m) = 0

From which we can show ( ) = 0

Hence by Gauss theorem

A d3x 0 = d3x is conserved in every Lorentz frame. Also, the density 0 is positive denite.When Dirac invented the equation, he thought this fact was its prime advantage.

This was because was meant to be a single-particle probability amplitude wave.The rst relativistic equation was written by Schr odinger who proposed

( 2

t 2+ 2 m2)( x, t) = 0

This is more commonly known as Klein-Gordan equation. This was in keeping with p02 p2 m2 = 0 and the implementation p0 i/t , pi i/x i . But therewere two problems: 1) solutions with negative energy (i.e., eigenvalue of i/t exist,and 2) the density whose integral is conserved is not positive denite:

= i( )Thus the Dirac theory appeared to be an advance. In Diracs theory negative en-ergies persisted but he proposed stability against indenite transitions to negativeenergy states by using fermi statistics, Dirac sea and hole theory.The fact that has subsequently become clear is the interpretation rst advanced byHeisenberg and Pauli, wherein , are operator valued spacetime elds. i is not


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the p operator. p, x are only c-members. i/x give spacetime-translations,brought about by the Hamiltonian H , or the momenta P i , distinct from mode labels p.There are several approaches to relativistic quantum theory, but if one uses space-time elds, the many-particle theory with quantized elds and not the wave me-chanics is the correct Relativistic Quantum Mechanics.


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3. Interaction picture and the S -matrix

3.1 Interactions among (otherwise) free particles :

Interactions among particles result in a change in their momenta, and at high ener-gies, change in the species of the particles, e.g.,

4He++ + Au 4He++ + Au (Rutherford expt .) p + e n + e (inverse decay)

or decays : + + + 0 2 It is convenient to think of otherwise free particles that interact briey and result

in the nal product that is again free. Such transitions can be represented byinteraction terms of the form

a p2 2 b p2 2 c p3 3 d p4 4

where the letters of the alphabet designate different particle species, and the sub-scripts label momentum and spin of single particle states. Acting on a state withfree c and d particles it connects the latter with a state of free a and b providedthe various momenta and spins match. On physical grounds we need to constructterms such that 1) energy and momentum are conserved and 2) various charges areconserved and discrete symmetries are obeyed. A further requirement is Lorentzcovariance of the nal answer, since different observers must agree about the pro-cess. A skeletal expression such as the one above can not guarantee all this. Whichis where the construct of spacetime eld operators with denite Lorentz covariantproperties helps. Let the interaction piece of H be

Hd3x. As we shall see later,

for Lorentz invariance, we need that

[H(x), H(y)] = 0 for (x y)(x y) spacelikeIt turns out that this can be achieved if His a product of local elds and theirderivatives and the Lorentz indices on the elds are properly contracted to make Ha Lorentz scalar. Typical Hare : (g, etc. are real constants)


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g3, 4, 6

with a scalar , real for 3 possibly complex for others

g, a fermion

g, g

eA A a vector

If we expand individual elds in terms of their modes, we get a summation of termswhose operator parts look like the skeletal example written above. We now proceedto calculating the effects of such Htreated as perturbation to the free hamiltonian.There are many good texts on the subject. Good references for beginners are Lee[9]and Lurie[10] to name but two.

3.2 In, out states and the interaction picture :

Calculations in quantum mechanics are done in either of the equivalent pictures:

Schrodinger: States undergo time evolution, operators are time independent

i

t |, t

S = H S

|, t

S

iddt

H S = 0

Here labels the state.

Heisenberg: States are time dependent, operators undergo dynamical evolution

i t | H = 0i ddt OH = [H H , OH (t)]

for any operator O. Choose

|, t = 0 S = | H and OH (0) = OS Then the two pictures are related through


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| S = e iHt

| H

OH = eiHt OS e iHtand H S same as H H .In situations where interactions come on for short times, a hybrid picture is prefer-able. Write

H = H 0 + V where eigenvalues and eigenstates of H 0 should be known exactly. Consider aSchrodinger picture state

| = n cn (t)|n, twhere H 0|n, t = E n |n, t ,

i.e., |n, t = e iE n t |n, 0then i

t | = ( H 0 + V )|

cn (t) =m

cm (t) n, 0|V |m, 0 ei(E n E m )t

=m

cm (t) n, 0|eiH 0 t V e iH 0 t |m, 0

DeneV I (t) = eiH 0 t V S e iH 0 t

I stands for interaction picture. The interpretation is that V I is the operator whosematrix elements give the instantaneous rate at which the amplitude of occupationof a particular state n is changing. Dene correspondingly

| I = eiH 0 t | S And for a general operator

O,

ddt OI = i[H 0, OI ]

The important property of the I -picture can then be proved by the reader:

Exercise : Show that i t | I = V I | I


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Interaction picture proves to be very convenient in eld theory which is done inthe Heisenberg picture. In the absence of interaction, V = 0, I -picture reducesto the H -picture. However, the relation of I -picture directly to H -picture lookscomplicated:

| I = eiH 0 t e iHt | H Of course, matrix elements have to be the same, calculated in any picture.3.3 Evolution operator :

For a scattering process, consider the H -picture states in the Fock space of the

all the particles involved in the interaction. We distinguish the basis beforethe interaction, called the in basis from that after the interaction, called theout basis. We denote the basis vectors |in (t) and |out (t) . denotes allpossible quantum numbers required to specify the basis. Although the bases areisomorphic the two are not identical at all nite times due to interaction. Thebases {|out () }and {|in () }differ at most by an overall phase, whichwould be unimportant to observations. But in general the two are mappedone onto the other by a unitary operator, involving the interaction hamiltonianwhich is assumed to be operative for a brief period. For further discussion of theassumptions involved and their implications see for instance Lee[9] and Itzyksonand Zuber[3].

We seek amplitudes of the form

out (t)|in (t)The H -picture |in is just as well an I -picture state at t = . We introduce anoperator that will evolve it as an I -picture state

|in (t) I U (t, )|inwhere U is the appropriate operator. From the preceeding exercise it follows that

iddt

U (t, ) = V I (t)U (t, )To conserve probability U ought to be unitary, which can be checked; since V I ishermitian, from the preceeding equation we get


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iU = U V I

iddt

(UU ) = 0

So UU is a constant, and U (t, ) = 1 ; hence the result follows.The desired amplitude is then

out (t) |in (t)= out ()|U (t, )U (t, )|in ()= |U (, )|

where both and are taken to be eigenstates of the free hamiltonian ignoringthe overall phase difference between in and out.

Exercise : Prove the last equality involving the U s.The S -matrix is dened to be

S limt t

U (t, t )

And

S = |S |which is seen to be the required matrix element.Using the differential equation satised by U (t, ) we can obtain an expansionfor U (, ). First convert to the integral equation

U (t, t 0) = 1 i t

t 0V I (t ) U (t , t0) dt

which is equivalent and takes account of the initial condition. Then, upon iterating,

U (0)

(t, t 0) = 1

U (1) (t, t 0) = i t

t 0dt1V I (t1)

U (2) (t, t 0) = ( i)2 t

t 0dt1

t 1

t 0dt2V I (t1)V I (t2)

U (n )(t, t 0) = ( i)n

t

t 0dt1 . . .

tn 1

t 0dtn V I (t1) . . . V I (tn )


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where n U (n ) gives n th order approximation to U . Recast in a more elegant form:

First,

t

t 0dt1

t 1

t 0dt2V I (t1)V I (t2) =

t

t 0dt2

t 2


by a change of variables. Now,

t

t 0dt2

t 2

t 0dt1 =

t

t 0dt2

t

t 0dt1(t2 t1)

= t

t 0dt1

t

t 1dt2

in the second step exchanging dt1 and dt2 and using up the step function . So

t

t 0dt1

t 1

t 0dt2V I (t1)V I (t2) =

12

t

t 0dt1

t 1


+12

t

t 0dt1

t


Both t1 and t2 integrations range from t0 to t but the order of the operators isdifferent depending on t1 < or > t 2. Hence dene

T {V I (t1)V I (t2)} (t1 t2)V I (t1)V I (t2)+ (t2 t1)V I (t2)V I (t1)called the time ordered product, introduced by Dyson. Then rewrite

t

t 0dt1

t 1

t 0dt2V I (t1)V I (t2) =

12

t

t 0dt1

t

t 0dt2T {V I (t1)V I (t2)}

It can be shown similarly that

t

t 0dt1 . . .

tn 1

t 0dtn V I (t1) . . . V I (tn )

=1n!

t

t 0dt1 . . .

t

t 0dtn T {V I (t1) . . . V I (tn )}

SoU (t, t 0) = 1 +

n=1

(i)nn!

t

t 0dt1 . . .

t

t 0dtn T {V I (t1) . . . V I (tn )}

T exp { i t

t 0V I (t )dt }


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The second is a convenient notation for the full expansion above it. This allows usto calculate the S -matrix to any desired order in the interaction. While evaluatingthe matrix elements, a further fact needs to be kept in mind. We know that the freehamiltonian needs to be normal ordered to avoid innite energy of the ground state.The same prescription needs to be applied to the interacting part of the hamiltonianV . If this is not done we once again encounter spurious innities, as can be checkeddirectly.

3.4 Covariance :In the hamiltonian formulation of dynamics, the time is specic to a particular ref-

erence frame and so are the canonical momenta so that in this formulation, Lorentzinvariance, even if present is not obvious. Thus the formalism above seems to ob-scure Lorentz invariance. We shall verify that the latter is nevertheless very muchpresent. First of all V (t) can be written as

V (t) = H( x, t) d3xIn the absence of derivative coupling (involving terms) His the same as Lint ,the interaction part of the lagrangian density, and is a Lorentz scalar. Thus,

U (, ) = 1 +

1

(i)nn! d4x1 . . . d4xn T {HI (x1) . . . HI (xn )}

Which has a greater chance of being Lorentz invariant. The T -product however doesnot seem Lorentz invariant due to the presence of the -function. It is, with furtherproviso on H. His hermitian and therefore an observable. Causality requires that

[H(x), H(x )] = 0 for (x x)(x x) < 0

with this natural requirement we can show theT

-product to be L-invariant.Consider two spacetime points ( ta , xa ) and (tb, xb). Under a boost by speed v in thex direction, we nd

(ta tb) (t a t b + vxa vxb)Working directly in the primed frame, we would have written (ta tb) so thearguments of the two step functions dont match. But note that if ( xa xb) (xa xb) > 0, i.e., the interval is timelike, the sign of ta tb has to be the same


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as that before the transformation, so the meaning of both the -functions is thesame. On the other hand, for the interval spacelike, the relative sign between thearguments can be negative. But this is precisely the case when the Hs commuteand the ordering of operators is no longer of any signicance.

3.5 Existence of S :For S to exist, the double limit lim

t t

U (t, t ) must exist and V should switch off

in the same limits. The latter requirement is satised in non-relativistic scatteringif the range of the potential is limited. But in relativisitic eld theory it leads to

intricate issues and sometimes signals phenomena that cannot be described in thisformalism. In some of the cases eld renormalisation cures the problem, in othercases infrared divergences have to be handled by making explicit some additionalphysical requirements. But in the case of quarks and gluons which are in principlemassless and possess strong self and mutual coupling, no S can exist. The freeparticle basis for expressing the S -matrix is provided only by the bound states,nucleons and mesons, and not the supposed constituents.There can be other cases where the S -matrix may not exist, as in the followingexample[3] where two isomorphic Hilbert spaces are not mappable onto each other

by a unitary transformation.Consider a 1-dimensional lattice of N spins. Let states be specied by the eigenvaluesof N i

(i)3 . The ground state | may be taken to be that with all the spins down,

and characterised byN

i(i) |

12

, 12

. . . 12

= 0

s denoting the appropriate lowering operators. Make a change to new operators

2 = 2, 1 = cos 1 + sin 3, 3 = sin 1 + cos 3This change can be expressed by the unitary transformation

(i)a = [exp( i2

(i)2 )] (i)a [exp( i

2

(i)2 )]

The states are similarly related, so that

| =N

iexp(i

2

(i)2 )|


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In the limit of innite number of degrees of freedom however, the transformationdoes not exist because

| = (cos2

)N N 0and the same is true of inner product between any two states from the two differentHilbert spaces because an innite number of cos / 2 or sin/ 2 factors will appearin every case.

Exercise : Check the above calculation using 2 i2 = +

.


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4. Rates and cross-sections

The probability for the reaction isP ( ) = |S |2

But S is of the formS = 1 + (2 )4 i 4(P P )Mwhere 1 corresponds to the possibility of no interaction. The remaining piece is

imaginary and singular. We know that any reasonable way of calculating transitionswill give energy momentum conservation so we write it in the form above. In thenext section we shall see that this is indeed justied. In nding P , we ignore the 1 .But we still get ( 4(P P ))2 which needs interpreting. Roughly speaking, since wehave two s of the same argument, one of them is (0). Consider

4(i

pi) p=0

=1

(2)4 d4x e ix ( pi ) p=0 = 1(2)4 d4xSo it will be interpreted as the total spacetime volume. But this is innite. We haveto start in a box of nite volume and then let V after making sure it cancels.In a box, states are normalized differently

p1, 1| p2, 2 = 3( p1, p2)1 2While

p1, 1| p2, 2 box = p1 ,p2 1 2Since d3 p box (2)

3

V p

,

the corresponding

3( p1, p2)box

V

(2)3 p1 ,p2Thus

p1| p2 box = [V

(2)3] 1 p1| p2

SoS box = [

V (2)3

] (N + N )/ 2S Where N , N are numbers of particles in the two states.Also, we calculate the differential


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P d where d N

id3 pi

which in the box isd box = [

V (2)3

]N d So if we rst do everything in a box and re-express in the continuum language,

P ( )d = (2 )4 4( p p ) V T |M |2[V

(2)3] (N + N )

[

V

(2)3 ]

N d

Where V T represents the (4-dim) box volume. Hence the rate

d( ) = (2 )44( p p ) V [V

(2)3] N

|M |2 d N is usually 1 or 2. 3 is difficult to arrange and rare in nature. For single particledecay, we calculate rate

d( ) = (2 )44( p p ) (2)3|M |2d For a collision, we calculate rate per unit ux, also known as the differential cross-section. This gets rid of the extra V . The ux is u /V where u is relativespeed of the particles. For a xed target experiment, u = uincident . Then

d( ) =d

= (2 )4

4( p p )(2)

6

|M

|2

u d

The cross-section can be cast in a manifestly Lorentz invariant form, a demonstra-tion we shall skip. It may also be noted that the normalisation p| p = 3( p, p ) isnot Lorentz invariant. Some authors prefer p| p = (2 )32 p03( p, p ). With corre-sponding changes in d and ux expressions, nal answer should remain the same.We now need a method for calculating M.


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5. Perturbation expansion and the Feynman rules

5.1 Evaluating matrix elements:

To calculate M , we need to evaluate things like

d4x1d4x2 0|

a T {: HI (x1) : : HI (x2) :}

a |0

We shall do this in a specic example and see the general rules. Consider

L=12

12m22 + (i / M ) g

SoH 0 =

12

(2 + | |2 + m22) + (i 0 + i i i + M)andV = g

In this theory a scalar can emit fermion-antifermion pair. Alternatively, a fermion(or antifermion) can emit a scalar and continue as a fermion (or antifermion). Whenthe emitted scalar is absorbed by another fermion, there is an effective interactionbetween the two fermions. This is a simplied version of the Yukawa theory of nucleonic forces.We need V in the interaction picture. i.e., as a functional,

V I = V [I , I , I ]

But the I , I undergo evolution through H 0 so their mode expansion is identicalto the free eld expansion

I ( x, t) =

d3k

(2)3/ 2 {a

k

e ikx

2wk+ a

k

eikx

2wk }I (x) = d

3k(2)3/ 2 {

b k, u( k, )e ikx + c k, v(

k, )eikx }This is the advantage of the interaction picture. Consider evaluating the secondorder term

(i)2g22 d4x1d4x2T {: (x1)(x1)(x1) : : (x2)(x2)(x2) :}


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between the states

| p; ( k, 12 ) and

| p ; ( k, 12 ) where p refers to and k to the spinor,

created by b k, . (To distinguish the antiparticle created by c

k, we need furthernotation, but we dont need it for the present example). For evaluating

0|a p b k, 12 T {: (x1)(x1)(x1) :: (x2)(x2)(x2) :}apb

k 12 |0

the strategy is to take all destruction operators towards the right so that when theyreach |0 they give zero, but along the way generate c-numbers from commutatorsor anticommutators. This process by which two eld operators are made to yielda c-number is referred to as the contraction between the two. The fact that each

HI term is normal ordered prevents contractions within the same HI . A generalformula giving the vacuum expectation value of the time ordered product of normalordered expressions in terms of c-number functions and delta-functions is the contentof Wicks theorem. Our example will provide a specic case of the general result,and will also demonstrate how the latter is proved.The contractions of the s are independent of those of s and vice versa so we canconsider each case separately. We introduce the shorthand notation 1 (x1) anddenote by des and cr the parts of containing the destruction operators and thecreation operators respectively.

0|a p T {(x1)(x2)}ap|0

= (t1 t2) 0|a p (des1 + cr1 )(des2 + cr2 )ap|0

+ (t2 t1) 0|a p (des2 + cr2 )(des1 + cr1 )ap|0

= (t1 t2){0|a p des1 cr2 ap|0 + 0|a p cr1 des2 a

p|0 }

+ (t2 t1){0|a p des2 cr1 ap|0 + 0|a p cr2 des1 a

p|0 }

= 0

|a p cr1

des2 a

p

|0 + 0

|a p cr2

des1 a

p

|0

+ 0|a p ap|0 {(t1 t2)[des1 , cr2 ] + (t2 t1)[des2 , cr1 ]}= [a p , cr1 ][

des2 , a

p] + [a p ,

cr2 ][

des1 , a

p] + p,p 0|T {(x1)(x2)}|0

Diagrammatically, this can be written as


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Figure 1 The nodes attached only to external lines

Here each dashed line represents a factor from the preceeding line. The incompletepieces of solid line represent uncontracted fermionic parts and can be ignored forthe moment. A factor such as [ des2 , a

p] represents an incoming scalar disappearing

at vertex. It has the value exp ( ipx)/ (2)3/ 222. Similarly the outgoing lines,whose value can also be easily found. In addition we have the expression

0

|T

{(x1)(x2)

}|0

represented by a line originating at a vertex and terminating at another. This inter-nal line is the famous Feynman propagator, rst proposed by E. C. G. Stueckelberg.Its value we shall record later. It can be shown that this expression is a Greenfunction of the Klein-Gordon equation with mixed boundary conditions - it propa-gates positive frequency mode functions exp(it ) forward in time and negativefrequency mode functions exp(+ it)

backward in time. In the old wavefunc-tion interpretation negative frquency solutions meant negative energy particles. Byresorting to Diracs hole theory (for fermions) Feynman guessed that negative fre-quency solutions needed to be propagated backward in time. He also argued that atheory which permitted pair creation needed such a propagator for consistency. Thesystematic derivation given here (sec. 4.3 and present) is due to Dyson and Wick.

Exercise : Prove by direct computation that the propagator above is given by theexpression listed in the Feynman rules (sec. 5.2) below.

For each of the bosonic terms obtained above, there are three possiblities for fermions


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{b k 12 ,

cr1

}0

|T

{(x1)(x2)

}|0

{des2 , b

k 12 }+ {b k 12 , cr2 }0|T {(x2)(x1)}|0 {des1 , b

k 12 }+ {b k 12 , b

k 12 }Tr 0|T {(x2)(x1)}|0 0|T {(x1)(x2)}|0

Here the vacuum expectation value expressions are matrix expressions for the fer-mionic propagator. In the rst two terms these are sandwitched between a row anda column vector, and in the third, the product is traced.

Exercise : Prove the above expansion.

Diagramatically we can show these as

Figure 2 Generating internal lines

Putting together the bosonic and the fermionic calculations, we get the nine dia-grams shown in g. 3. The term corresponding to a given diagram, can be obtainedif we assign specic factors to each kind of line according to the Feynman rulestabulated in the next subsection.


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Figure 3 All the terms from contractions


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5.2 Feynman rules :

incoming boson1

23/ 2e ip x

2 p

outgoing boson1

23/ 2eip x

2 pincoming electron u( k, ) e

ik x

23/ 2

outgoing electron u( k , )eik x

23/ 2

electron propagator i d4q

(2)4e iq(x 1 x2 )

q + M q2

M 2 + i

Other factors not appearing in the present example are

incoming positronv( k, )e ik x

23/ 2

outgoing positron v( k , ) eik x

23/ 2

boson propagator i d4q

(2)4e iq(x 1 x 2 )

q2 M 2 + i


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In our example we can see that some of the diagrams contain distinct pieces not con-nected to each other. They are called disconnected diagrams. In practice, one doesnot draw all the diagrams including the disconnected ones, but only the connectedones. This is because the subdiagrams of a disconnected diagram either correspondto particles that undergo no interaction and hence are discarded along with the in S , or they are diagrams corresponding to a different scattering processthan the one under consideration and are also perhaps of some other order in theperturbative expansion. Also, note several diagrams get repeated, drawn differently.This results from taking all vertices in all possible orders. We need to draw onlyone representative diagram. The repetition cancles the 1

n!in front.

Momentum space rules: For a given connected diagram in x-space, we have d4xfor every H(x). Once every external line has been replaced bythe wavefunction factor and internal line with a propagator,the d4x integrals can all be trivially carried out. At a vertexwith two external and two internal lines, such as that in theadjoining diagram, we get

d4x1e ip 1 x 1 e ip 2 x1 d4q1

(2)4e iq1 (x1 x r )

q21 m2 + i d4q2(2)4

e iq2 (x 1 xs )

q22 m2 + iHere xr , xs are locations of some other vertices and we have not displayed factorsor spinors which do not involve xi . This gives

d4q1

(2)4d4q2(2)4

(2)44( p1 + p2 + q1 + q2) 1

q21 m2 + i 1

q22 m2 + iThus the d4x at every vertex simply gives conservation of energy momentum owingthrough that vertex.

The equivalent momentum space rules then say that

1. For external lines leave out factors eip x of coordinate space rules.

2. Leave out eiqx in propagators

3. Put four-momentum conserving 4 at every vertex and leave out the d4xoperations. 4. Integrate over all internal momenta.


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It is easy to check that in doing 4, one always picks up an overall 4( p

p)

balancing the total incoming and outgoing four momentum. This 4 is the one we ex-tracted in anticipation while dening M . M is then obtained after integratingthe remaining internal momenta.

Figure 4 The two independent connected diagrams at tree level

We can now return to our example and illustrate these comments. Firstly we ignorethe disconnected diagrams. Of the remaining, there are two distinct diagrams, each

of which appears twice. They are shown above. Leaving out 1 / 2!, we get for thediagram on the left

(i g)2 d4x1 d4x2 e ip x1

(2)32 2wp

u( k, 12 )e ik x 1

(2)32

i d4q

(2)4e iq(x 1 x 2 )

q + M q2 M 2 + i

eip x 2

(2)32 2wp

u( k , 12 )e ik x2

(2)32

=(ig)2(i)

(2)6 d4q

(2)4(2)44( p + k + q) (2)44( p + k + q)

u( k , 12 )(q

+ M )u( k, 12 )

2 p 2 p(q2 M 2 + i )= i(2)44( p + k p k )

(ig)2

(2)6u( k , 12 ) (

( p + k) + M ) u( k, 12 )2 p p ( ( p + k)2 M 2 + i )

From this, one contribution M(1) to Mcan be read off. The second contribution

M(2) comes from the diagram on the right. We can see that it can be easily obtained


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from

M(1) by interchanging p with p but leaving k and k where they are. There

remains now the rather forbidding task of obtaining |M(1) + M(2) |2, which alsoinvolves taking traces of products of Dirac matrices. We shall not take this up here.

REFERENCES

1. Ryder L. H., Quantum Field Theory , Cambridge University Press, (1985)2. Schiff L. I., Quantum Mechanics , 3rd Ed., McGraw-Hill Pub. Co., (1968)3. Itzytkson C. and Zuber J-B., Quantum Field Theory , McGraw-Hill Pub. Co.,(1980)4. Georgi H., Lie Algebras for Particle Physics , Benjamin-Cummings PublishingCo., (1982)5. Cartan E., Theory of Spinors (English translation), Dover Publishers

6. Pauli W., Principles of Quantum Mechanics , Translated by P. Achuthan and K.Venkatesan, Allied Publishers and Springer-Verlag India Ltd., (1980)7. Wentzel G., Quantum Theory of Fields , Interscience Publishers, (1949)8. Weinberg S., Phys. Rev. 133 , 1318 (1964); 134 , 882 (1964).9. Lee T. D., Particle Physics and Introduction to Field Theory , Harwood AcademicPublishers, (1982)10. Lurie D., Particles and Fields , Interscience Publishers, (1968)

For Lie Group theory refer also to the recently published lectures:11. N. Mukunda in Introduction to Topologfy, Differential Geometry and GroupTheory for Physicists by S. Mukhi and N. Mukunda, Wiley Eastern Ltd., (1990)

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