sidney coleman - lectures on qft

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arXiv:1110.50

13v2

[physics.ed-ph]9Jan2012

Notes from Sidney Colemans Physics 253a

Sidney Coleman

Harvard, Fall 1986

Edit and typeset by Bryan Gin-ge Chen and Yuan-Sen Ting from scans of the handwritten notesof Brian Hill. Please contact Bryan Gin-ge Chen ([email protected]) and Yuan-Sen Ting([email protected]) if you have any questions or comments.

LATEX version of October 23, 2011

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0. Preface Notes from Sidney Colemans Physics 253a 2

0 Preface

Its unexpected and heart-warming to be asked by Bryan Chen to write something aboutthese notes, 25 years after taking them. I was the teaching assistant for Sidneys quantumfield theory course for three years. In the first year, I sat in, because frankly, I hadnt learnedquantum field theory well enough the first time that I took it.

When I have the good fortune to hear a really good lecturer, I often re-copy my notes,preferably the evening on the day that I took them.

Once in a while, students would miss a class, and then ask me if they could look at mynotes. At some point, the requests started happening enough that it was suggested that acopy be put on reserve in the Harvard physics library. From there, copies of the notes justkept spreading.

Sidney once expressed disappointment about the spread of the notes. For one thing, I

even wrote down some of his anecdotes and jokes, and that made it less fun for him to re-tellthem. For another, he wrote Aspects of Symmetry which shared a lot of material with whathe taught in Physics 253b. He may have had in mind that he would write a field theorybook as a companion volume.

Of course, he never did write a field theory book, or youd be reading that, and he nevertried to rein the copies in. Now that he is gone, we are lucky that his clarity lives on.

Thanks to Bryan Chen and Yuan-Sen Ting for creating this lovely LATEX version. Some-times transcription can seem tedious, but I hope it was as valuable for them as the firstre-copying was for me, and that for you fellow student of quantum field theory theexistence of these notes is similarly valuable.

Brian Hill, www.lingerhere.org, March 10, 2011

Typesetters notes

The great field theorist Sidney Coleman for many years taught the course Physics 253 atHarvard on Quantum Field Theory. The notes you are reading were typeset from a scannedversion of handwritten lecture notes by Brian Hill from the Fall 1986 of the first half of thecourse: Physics 253a. The Harvard Physics Department has made films of the lectures fromthe 1975-1976 version of the course available on their website as well.

The typesetting for lectures 1-11 was done by Bryan Gin-ge Chen and for lectures 12-28by Yuan-Sen Ting, who also recreated most of the figures. We have attempted to stay asfaithful as possible to the scanned notes, correcting only obvious typos and changing someof the in-text references. We have noted some of these changes in comments in the LATEXsource, though this was not done systematically.

We thank Richard Sohn and Yevgeny Katz for pointing out typos in the previous version(v1). The list of typos can be found in YST and BCGs homepage.
http://www.lingerhere.org/http://www.lingerhere.org/


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Contents

0 Preface 21 September 23 7

Units; Translational invariance, Rotational invariance and Lorentz invariance for asingle free particle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2 September 25 14Position operator; Violation of Causality; Pair Production; Fock Space; Occupation

number representation; SHO; Oscillator-like formalism for Fock space. . . . . 14

3 September 30 21Causality, observables and quantum fields; Constructing the quantum field from

Fock space operators axiomatically; Translational invariance; Lorentz invari-ance and Relativistic normalization; Field constructed satisfies Klein-Gordonequation. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

4 October 2 28Constructing Fock space from the quantum field axiomatically; The Method of

the Missing Box, Classical Particle Mechanics, Quantum Particle Mechanics,Classical Field Theory, Quantum Field Theory; Quantum field from free scalartheory. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

5 October 7 38Hamiltonian recovered in free scalar theory up to infinite constant; Normal ordering;

Symmetries and conservation laws, Noethers theorem; Noethers theorem infield theory, conserved currents; ambiguity in currents; Energy-momentumtensor. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

6 October 9 49Lorentz transformations; Angular momentum conservation; Internal symmetries;

SO(2) internal symmetry; Charged field; SO(n) internal symmetry. . . . . . 49

7 October 14 58Lorentz transformation, properties of conserved quantities; Discrete symmetries;

; Charge conjugation; Parity; Ambiguity of choice of parity; Timereversal; Unitary and anti-unitary operators, angular momentum; Dilatations. 58

8 October 16 70


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Scattering theory overview; Low budget scattering theory; Turning on and off func-tion; Schrodinger picture, Heisenberg picture and interaction picture; Evolu-

tion operator; Time ordered product; Three models; Wicks theorem. . . . . 70

9 October 21 83Diagrammatic perturbation theory in Model 3; Vertex in model 1; Connected di-

agrams; Thm:

all Wick diagrams = : e

connected Wick diagrams :; Model 1solved; Model 2 begun. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

10 October 23 93Model 2 finished; Vacuum energy c.t.; S matrix is 1; Ground state energy; Yukawa

potential; Ground state wave function; Model 3 and Mass renormalization;Renormalized perturbation theory. . . . . . . . . . . . . . . . . . . . . . . . 93

11 October 28 103Feynman diagrams in Model 3; Feynman rules in model 3; A catalog of all Feynman

diagrams in model 3 to O(g2); Scattering amplitude at O(g2); Direct andexchange Yukawa potentials. . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

12 October 30 112Nucleon-antinucleon scattering at O(g2); Energy eigenstate probe; Meson-

nucleon scattering; Nucleon-antinucleon annihilation; Assembling theamplitudes for various processes into one amplitude; Mandelstam variables;Mandelstam-Kibble plot; crossing symmetry; CPT; Phase space and the S

matrix;Differential tran. prob.

unit time . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

13 November 4 131Applications of Differential tran. prob.unit time ; Decay; Cross sections, flux; Final state phase

space simplified for two bodies; dd

; Optical theorem; Final state phase spacefor three bodies; Feynman diagrams with external lines off the mass shell;they could be an internal part of a larger diagram. . . . . . . . . . . . . . . . 131

14 November 6 144Fourier transform of the new blob; A second interpretation of Feynman diagrams

with lines off the mass shell; they are the coefficients of n in 0|S|0; Athird interpretation of the blob; they are the Fourier transform of the VEVof a string of Heisenberg fields; Reformulation of scattering theory; S matrixelements without the turning on and off function; LSZ formula stated. . . . . 144

15 November 13 163


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LSZ formula proved; A second look at Model 3 and its renormalization; Renormal-ization conditions. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 163

16 November 18 171Perturbative determination of a c.t.; Problems with derivative couplings; Rephras-

ing renormalization conditions in terms of Greens functions; Lehmann-Kallenspectral representation for the propagator; Rephrasing renormalization con-ditions in terms of 1PI functions. . . . . . . . . . . . . . . . . . . . . . . . . 171

17 November 20 182Perturbative determination of c.t.; Corrections to external lines in the computation

of S matrix elements; One loop correction to meson self energy; Feynmanstrick for combining 2 denominators; Shift to make denominator SO(3, 1) in-

variant; Wick notation to make denominator O(4) invariant; Integral tablesfor convergent combinations; Self-energy at one loop studied; Combining lotsof denominators; The shift in the general case to reduce any multi-loop inte-gral to an integral over Feynman parameters. . . . . . . . . . . . . . . . . . . 182

18 November 25 196Rephrasing coupling constant renormalization in terms of a 1PI function; Exper-

imental significance of the definition; Renormalization versus the infinities;Renormalizable Lagrangians; Unstable particles, Decay products. . . . . . . 196

19 December 2 207

Unstable particles, lifetime, method of stationary phase; Where it begins again;Lorentz transformation laws of fields; Equivalent representations; Reduciblereps; The finite dimensional inequivalent irreducible representations of SO(3);Unitarity; Complex conjugation; Direct product; Projection operators andreducibility. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207

20 December 4 226Parametrizing the Lorentz group; Commutation relations for the generators; de-

composition into two sets obeying SO(3) commutation relations; The cata-log; Complex conjugation properties; Tensor product properties; Restrictionto SO(3); The vector; Rank 2 tensors; Spinors. . . . . . . . . . . . . . . . . . 226

21 December 9 240Lagrangian made of two component spinors; Solution of Weyl equations of motion;

Weyl particles; Dirac Lagrangian; Four-component spinors; Weyl basis, Diracbasis; Plane wave solutions of the Dirac equation. . . . . . . . . . . . . . . . 240


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22 December 11 249Plane wave solutions of the Dirac equation; Paulis theorem; Dirac adjoint; Pauli-

Feynman notation; Parity; Bilinears; Orthogonality; Completeness; Summary. 249

23 December 16 267Canonical quantization of the Dirac Lagrangian. . . . . . . . . . . . . . . . . . . . 267

24 December 18 273Perturbation theory for spinors; Time ordered product; Wicks theorem; Calcu-

lation of the contraction (propagator); Wick diagrams; Feynman diagrams;Matrix multiplication; Spin averages and spin sums. . . . . . . . . . . . . . . 273

25 January 6 289

Parity for spinors; Fermion and antifermion have opposite intrinsic parity; Chargeconjugation; Majorana basis; Charge conjugation properties of fermion bilin-ears; Decay of ortho and para positronium; UPUC = UCUP(1)NF; PT. . . . 289

26 January 8 301Effect of PT on states; Proof of CPT theorem in perturbation theory; Renormal-

ization of spinor theories; Propagator. . . . . . . . . . . . . . . . . . . . . . . 301

27 January 13 3131PI part of propagator; Spectral representation for propagator;

(p) to O(g2) inmeson-nucleon theory; Coupling constant renormalization; Is renormalization

sufficient to eliminate s. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31328 January 15 323

Regularization; Regulator fields; Diml regularization; Minimal subtraction; BPHZrenormalizability; Renormalization and symmetry; Renormalization of com-posite operators. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

Class: Why not use Feynmans lecture notes?Gell-Mann: Because Feynman uses a different method than we do.Class: What is Feynmans method?Gell-Mann: You write down the problem. Then you look at it and you think.Then you write down the answer.


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1. September 23 Notes from Sidney Colemans Physics 253a 7

1 September 23

In NRQM, rotational invariance simplifies scattering problems. Why does the addition ofrelativity, the addition of L.I., complicate quantum mechanics?The addition of relativity is necessary at energies E mc2. At these energies

p + p p + p + 0

is possible. At slightly higher energies

p + p p + p + p + pcan occur. The exact solution of a high energy scattering problem necessarily involves manyparticle processes.

You might think that for a given E, only a finite number, even a small number of processesactually contribute, but you already know from NRQM that that isnt true.

H H+ V E 0 = 0|V|0 +

n

|0|V|n|2E0 En +

Intermediate states of all energies contribute, suppressed by energy denominators.

For calculations of high accuracy effects at low energy, relativistic effects of order

vc

2can be included. Intermediate states with extra particles will contribute corrections of or-

der Emc2

Typicalenergies in problem Typical

energy

denominator

mv2mc2 =

vc

2. As a general conclusion: the corrections of relativistic

kinematics and the corrections from multiparticle intermediate states are comparable; theaddition of relativity forces you to consider many-body problems. We cant even solve thezero-body problem. (It is a phenomenal fluke that relativistic kinematic corrections for theHydrogen atom work. If the Dirac equation is used, without considering multi-particle in-termediate states, corrections of O vc can be obtained. This is a fluke caused by someunusually low electrodynamic matrix elements.)

We will see that you cannot have a consistent relativistic picture without pair production.

Units

= c = 1 [m] = [E] = [T1] = [L1]

Because were doingrelativistic (c)

quantum mechanics ()

Sometimes 1 = 1 = 2and 12 = 1=one-bar

(1 fermi)1 197 GeV


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Lorentz Invariance

Every Lorentz transformation is the product of an element of the connected Lorentz group,

SO(3, 1), and 1, P (reflects all three space components), T (time reversal), or P T. ByLorentz invariance we mean SO(3, 1).

Metric convention: + .

Theory of a single free spinless particle of mass

The components of momentum form a complete set of commuting variables.

Momentum operator P| k

State of a spinlessparticle is completely

specified by its momentum.

= k|k

Normalization k|k = (3)(k k)The statement that this is a complete set of states, that there are no others, is

1 =

d3k|kk|

| =

d3k(k)|k (k) k|

(If we were doing NRQM, wed finish describing the theory by giving the Hamiltonian,

and thus the time evolution: H|k = |k

|2

2 |k.)We take H|k =

|k|2 + 2|k k|k

Thats it, the theory of a single free spinless particle, made relativistic.How do we know this theory is L.I.? Just because it contains one relativistic formula, it

is not necessarily relativistic. The theory is not manifestly L.I..The theory is manifestly rotationally and translationally invariant. Lets be more precise

about this.

Translational Invariance

Given a four-vector, a, specifying a translation (active), there should be a linear operator,U(a), satisfying:

U(a)U(a) = 1, to preserve probability amplitudes (1.1)

U(0) = 1 (1.2)

U(a)U(b) = U(a + b) (1.3)


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The U satisfying these is U(a) = eiPa where P = (H, P).(This lecture is in pedagogical, not logical order. The logical order would be to state:

1. That we want to set up a translationally invariant theory of a spinless particle. Thetheory would contain unitary translation operators U(a).

2. Define Pi = i Uqi

||a=0, (by (1.3) [Pi, Pj] = 0, by (1.1) P = P).

3. Declare Pi to be a complete set and classify the states by momentum.

4. Define H =

p2 + 2, thus giving the time evolution.)

More translational invariance

U(a)|0 = |a U(a) = eiPa

where |0 here means state centered at zero and |a means state centered at a.

O(x + a) = U(a)O(x)U(a)

a|O(x + a)|a = 0|O(x)|0

Non-relativistic reduction

U(a) = eiPa

eiPa|q = |q + a

Something hard to digest, but correct:

qeiPa|q = (q + a)|q + a

eiPaqei

Pa|q = (q + a)|q

ei Pa

qe

i Pa=

q + aLooks like the opposite of

eiPaO(x)ei

Pa = O(x + a)

The q operator is not an operator localized at q. No reason for these (last two equations)to look alike.


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Rotational Invariance

Given an R

SO(3), there should be a U(R) satisfying

U(R)U(R) = 1 (1.4)

U(1) = 1 (1.5)

U(R1)U(R2) = U(R1R2) (1.6)

Furthermore denote | = U(R)|| P| = R| P|

for any | i.e. U(R) P U(R) = R P (1.7)and U(R)HU(R) = H (1.8)

A U(R) satisfying all these properties is given by

U(R)|k = |RkThat (1.5) and (1.6) are satisfied is trivial.Proof that (1.4) is satisfied

U(R)U(R) = U(R)

d3k|kk|U(R)

=

d3k|RkRk| k = Rk,d3k = d3k

= d3k|kk| = 1Proof that (1.7) is satisfied

U(R) P U(R) =by (1.4)

U(R)1 P(U(R)1) by (1.5) and (1.6)

= U(R1) P U(R1)

= U(R1) P

d3k|kk|U(R1)

= U(R1) d3kk

|k

k

|U(R1)

=

d3kk|R1kR1k| k = Rk, d3k = d3k

=

d3kRk|kk|

= RP


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You supply proof of (1.8).This is the template for studying L.I.

Suppose a silly physicist took

|ks =

1 + k2z |ksk|ks = (1 + k2s )(3)(k k)

1 =

d3k

1

1 + k2z|ks sk|

If he took Us(R)|ks = |Rrs his proofs of (1.4), (1.7), (1.8) would break down becaused3k

1 + k2z

= d3k

1 + k2z

i.e.d3k

1 + k2z

is not a rotationally invariant measure!

Lorentz Invariance

k|k = (3)(k k) is a silly normalization for Lorentz invariance.d3k is not a Lorentz invariant measure.We want a Lorentz invariant measure on the hyperboloid k2 = 2, k0 > 0.d4k is a Lorentz invariant measure.Restrict it to the hyperboloid by multiplying it by a Lorentz invariant d4k(k22)(k0).This yields the measure on the hyperboloid1 d

3k2k

, k =

k2 + 2, k = (k,

k).

So we take|krelativistically

normalized

= (2)3 2k|kSo factors of 2come out right in theFeynman rules a few

months from now

.

ko

k

1Think of the function as a function ofk0 and use the general formula (f(k0)) =

zeroesof f, Ki

(k0Ki)|f(Ki)|

.


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(Using d3x(x2 R2) can get R2 sin dd)Looks like factor multiplying d3k ought to get larger as |k| gets large. This is an illusion,

caused by graphing on Euclidean paper. This is the same illusion as in the twin paradox.The moving twins path looks longer, but in fact, its proper time is shorter.

Now the demonstration of Lorentz invariance: Given any Lorentz transformation define

U()|k = |k

U() satisfies

U()U() = 1 (1.9)

U(1) = 1 (1.10)

U(1)U(2) = U(12) (1.11)

U()P U() = P (1.12)

The proofs of these are exactly like the proofs of rotational invariance, using

1 =

d3k

(2)32k|kk| d

3k

(2)32k=

d3k

(2)32k

We have a fairly complete theory except we still dont know where anything is.We need a position operator, satisfying

X = X (1.13)

R X = U(R) XU(R) (1.14)

eiPa Xei

Pa = X+ a (1.15)

Take ai of (1.15) to get i[Pi, Xj] = ij .

Determination of X in position space

(k) k| k| P| = k(k)

k| X| = i

k (k)to satisfy

inhomogeneous part ofcommutation relation

+ kF(|k|2

) (k)an arbitrary vector

commuting with the Ps,a complete set, can be

written this way


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The extra arbitrary vector can be eliminated by redefining the phases of the states. Inthe momentum state basis let

|k |kN = eiG(|k|2)|k (This is a unitary transformation, call it U.In effect, U XU is our new position operator.)

Here G(|k|2) = kF(|k|2)

The only formula this affects in all that we have done so far is the expression for k| X|.With the redefined states, it is

Nk| X|n =

d3kN k|k

eiG(|k|2)(3)(kk)

k| X|N

[i

k+kF(|k|2)]eiG(|k|2)(k)

=

d3keiG(|k|2)(3)(k k)

k

G(|k|2) +kF(|k|2)

eiG + eiG

k

= i

k(k)

Up to an unimportant phase definition, we have shown that the obvious definition for Xis the unique definition, and we have done it without using L.I. or the explicit form of H.


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2 September 25

It is possible to measure a particles position in our theory, x is an observable, and this leadsto a conflict with causality.Introduce position eigenstates

k|x = 1(2)3/2

eikx

We will evaluate x|eiHt |x = 0 and see that our particle has a nonzero amplitude to befound outside its forward light cone 2.

x|eiHt |x = 0 =

d3kx|kk|eiHt |x = 0 H|k = k|k

= d3k 1(2)3 eikxeikt k = k2 + 2=

0

k2dk

(2)3

0

sin d 11

d(cos )

20

d 2

eikr cos eikt r = |x|, k = |k|

=1

(2)21

ir

0

kdk(eikr eikr)eikt k =

k2 + 2

=i

(2)2r

kdkeikreikt

These steps can be applied to the F.T. of any function of |k|.

Im 0iu

-iu

kbranch cutfor

2By translational invariance and superposition we could easily get the evolution of any initial configurationfrom this calculation.


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ei

k2+2 is a growing exponential as you go up the right side of the upper branch cut,and a decreasing exponential on the left side. Given r > 0 and r > t the product eikreikt

decreases exponentially as you go up the branch cut.Given r > 0 and r > t deform the contour to:

half circleat infinity doesnot contribute

The integral becomes

i(2)2r

k=iz(z) d(iz)ezr [e

z22t e

z22t] = e

r

22r

zdze(z)r sinh(

z2 2t)

The integrand is positive definite, the integral is nonzero.We can measure a particles position in this theory. We can trap it in a box of arbitrarily

small size, and we can release it and detect it outside of its forward light cone. The particlecan travel faster than light and thus it can move backwards in time, with all the associatedparadoxes.

Admittedly, the chance that the particle is found outside the forward light cone falls offexponentially as you get further from the light cone, and that makes it extremely unlikelythat I could go back and convince my mother to have an abortion, but if it is at all possible,it is still an unacceptable contradiction.

In practice, how does this affect atomic physics? Not at all, because we never tried tolocalize particles to spaces of order their Compton wavelength when doing atomic physics.We say that the electron is in the TV picture tube and there is not much chance that it isactually out in the room with you.

In principle, the ability to localize a single particle is a disaster, how does nature get outof it?

Particle trapped in container with reflecting walls


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If the particle is localized to a space with dimensions on the order of L the uncertainty inthe particles momentum is 1L . In the relativistic regime this tells us that the uncertaintyin the particles energy is

1

L . As L gets less than1

states with more than one particleare energetically accessible. If the box contained a photon and the walls were mirrors thephoton would pick up energy as it reflected off the descending mirror, it could turn into twophotons as it reflected.

If we try to localize a particle in a box with dimensions smaller or on the order ofa Compton wavelength it is unknown whether what we have in the box is 1 particle, 3particles, 27 particles or 0 particles.

Relativistic causality is inconsistent with a single particle theory. The real world evadesthe conflict through pair production. This strongly suggests that the next thing we shoulddo is develop a multi-particle theory.

Any number of one type of free spinless mesons

The space we construct is called Fock space. This formalism is also used in thermodynamics with the grandcanonical ensemble. Particle number instead of being fixed, fluctuates around a value determined by thechemical potential.

Basis for single particle states |k

k|k = (3)(k k) H|k = k|k P|k = k|k

This is the same as last time, except now this is just part of the basis.Two particle states |k1, k2 =

indistinguishability,

Bose statistics

|k2, k1

k1, k2|k1, k2 = (3)(k1 k1)(3)(k2 k2) + (3)(k1 k2)(3)(k2 k1)H|k1, k2 = (k1 + k2)|k1, k2P|k1, k2 = (k1 + k2)|k1, k2

etc.


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Also need a no particle state |0

0 |0= 1 H|0 = 0 P|0 = 0not part of a continuum

The vacuum is unique, it must satisfy U()|0 = |0. All observers agree that the statewith no particles is the state with no particles.

Completeness relation

1 = |00| +

d3k|kk|+ 12!

d3k1d

3k2|k1k2k1, k2|to avoid double counting. Alternatively,

just check that this works on

|k,k

Now we could proceed by setting up equations for wave functions. To specify a state, a

wave function contains a number, a function of three variables, a function of six variables,etc. Interactions involving a change in particle number will connect a function of six variablesto a function of nine variables. This would be a mess.

We need a better description. As a pedagogical device, we will work in a periodic cubicalbox of side L for a while. Since a translation by L does nothing, the momenta must berestricted to allowed values

k = 2nx

L

,2ny

L

,2nz

L satisfyingk (0, 0, L) = 2nzk

(0, L, 0) = 2ny

k (L, 0, 0) = 2nxDirac deltas become Kronecker deltas and integrals become sums

(3)(k k) kk

d3k

k

k|k = kk k1, k2|k1, k2 = k1k1k2kh2 + k1k2k2k1

Occupation number representation

Each basis state corresponds to a single function

|k1 n(k) = kk1|k1, k2 n(k) = kk1 + kk2

|0 n(k) = 0


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Given a function n(k) in the occupation number description, we write the state

| n() Inner product n()|n() = k

n(k)!n(k)n(k)

no argument, to emphasizethat the state depends on the

whole function n, not just

its value for one specific k.

Define an occupation number operator

N(k)|n() = n(k)|n() H =

k

kN(k) P =

k

kN(k)

This is a better formalism, but it could still use improvement. It would be nice to have

an operator formalism that did not have any wave functions at all.Note that H for our system has the form it would have if the system we were dealing

with was actually a bunch of harmonic oscillators. The two systems are completely different.In ours the N(k) tells how many particles are present with a given momentum. In a system

of oscillators, N(k) gives the excitation level of the oscillator labelled by k.

Review of the simple harmonic oscillator

No physics course is complete without a lecture on the simple harmonic oscillator. We willreview the oscillator using the operator formalism. We will then exploit the formal similarity

to Fock space to get an operator formulation of our multi-particle theory.

H =1

2[p2 + q2 1] [p, q] = i

If [p, A] = [q, A] = 0 then A = I.This is all we need to get the spectrum.Define raising and lowering operators

a q + ip2

a =q ip

2H = aa

[H, a] = a [H, a] = a [a, a] = 1

Ha|E = aH|E + a|E= (E+ )a|E a is the raising operator

Ha|E = (E )a|E


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a

aa

t

t

E+2E+EE-

Ladder

ofstates

Because, |H| = |aa| = ||a|||2 0 the ladder of states must stop goingdown or else E becomes negative. The only way this can happen is if a|E0 = 0. ThenH|E0 = 0.

The lowest state of the ladder, having E0 = 0 is denoted |0. The higher states are madeby

(a)n

|0

|n

H

|n

= n

|n

Get normalizations right a|n = cn|n + 1|cn|2 = n|aa|n = n + 1 cn =

n + 1

a|n = dn|n 1 |dn|2 = n dn =

n

Now we use [p, A] = [q, A] = 0 A = I to show that this ladder built from a statewith E0 = 0 is in fact the whole space. We do this by considering the projector, P, onto thestates in the ladder. Since a and a keep you within the ladder

[a, P] = [a, P] = 0 [p, P] = [q, P] = 0

P= I

The projector onto the ladder is proportional to the identity. There is nothing besidesthe states we have found.

Now we will apply this to Fock space.Define creation and annihilation operators for each momentum, ak, a

k

, satisfying

[ak, ak

] = kk [ak, ak] = 0 [ak

, ak] = 0

Hilbert space built by acting on |0 with strings of creation operators.

|k

= ak|

0

ak|

0

= 0

ak1ak2

ak3 |0 = |k1, k2, k3H =

k

kak

akP =

k

kakak


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Ifk 0 = [ak, A] = [ak, A] = A = I

This tells us there

are no other degrees

of freedom.

Weve laid out a compact formalism for Fock space. Lets drop the box normalizationand see if it is working.

[ak, ak

] = (3)(k k) [ak, ak] = 0 = [ak, ak

]

H =

d3kkak

akP =

d3kkakak

Check energy and normalization of one particle states.

k

|k

=0|a

kak|

0

=0|[a

k, ak

]|0

= (3)(k k)0|0 = (3)(k k)[H, ak] =

d3kk[ak

ak, ak] = kak

H|k = Hak|0 = [H, ak

]|0 = k|k[ P, ak] =

kak P|k = k|k

Check normalization of two-particle states |k1, k2 = ak1ak2

|0. Using commutation rela-tions check

k1, k2|k1, k2 = (3)(k1 k1)(3)(k2 k2) + (3)(k2 k1)(3)(k2 k1)

Mathematical Footnote:

Weve been calling the ak and ak

operators. An operator takes any normalizable vector in

Hilbert space to another normalizable vector: A| is normalizable whenever | is normal-izable. Even x in 1-d QM is not an operator.

dx|f(x)|2 <

dx|f(x)|2x < . x is

an unbounded operator. An unbounded operator has A| normalizable for a dense set3of |. ak and ak are even more awful than unbounded operators. An extra meson in aplane wave state added to any state is enough to make it nonnormalizable. a

k

is an operatorvalued distribution. You only get something a mathematician would be happy with afterintegration.

d3kf(k)ak is acceptable with a sufficiently smooth function.

3Any | is the limit of a sequence in the dense set.


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3 September 30

In ordinary QM, any hermitian operator is observable. This cant be true in relativisticquantum mechanics. Imagine two experiments that are at space-like separation. Ifx1 R1and x2 R2 then (x1 x2)2 < 0.

space-time regionaccessible to observer 1

space-time regionaccessible to observer 1

R1 R2

Suppose observer 2 has an electron in his lab and she measures k. If observer 1 canmeasure y of that electron it will foul up observer 2s experiment. Half the time whenshe remeasures x it will have been flipped. This tells her that observer 1 has made ameasurement. This is faster than light communication, an impossibility. It is a little hardto mathematically state the obvious experimental fact that I cant measure the spin of anelectron in the Andromeda galaxy. We dont have any way of localizing particles, any positionoperator, yet. We can make a mathematical statement in terms of observables:

If O1 is an observable that can be measured in R1 and O2 is an observable that can bemeasured in R2 and R1 and R2 are space-like separated then they commute:

[O2, O2] = 0

Observables are attached to space time points. A given observer cannot measure allobservables, only the ones associated with his or her region of space-time.

It is not possible, even in principle for everyone to measure everything. Out of the hordesof observables, only a restricted set can be measured at a space-time region. Localization ofmeasurements is going to substitute for localization of particles.

The attachment of observables to space-time points has no analog in NRQM, nor inthe classical theory of a single particle, relativistic or nonrelativistic, but it does have ananalog in classical field theory. In electromagnetism, there are six observables at each point:

Ex(x) = Ex(x, t), Ey(x), Ez(x), Bx(x), By(x), Bz(x). We cant design an apparatus herethat measures the Ex field now in the Andromeda galaxy.

In classical field theory, these observables are numbers. In quantum mechanics, observ-ables are given by operators. The fields will become quantum fields, an operator for eachspacetime point. We can see in another way that the electric field is going to have to becomea quantum field: How would you measure an electric field? You mount a charged ball, pith


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ball, to some springs and see how much the springs stretch. The location of the ball is givenby an equation like:

qx = Ex(x) might be

d4xf(x)Ex(x) where f(x)gives some suitable average over the pith ball.

The amount the ball moves is related to the E field, and if the world is quantum me-chanical, x must be an operator, and so Ex must be an operator.

We dont have a proof, but what is strongly suggested is that QM and relativistic causalityforce us to introduce quantum fields. In fact, relativistic QM is practically synonymous withquantum field theory.

We will try to build our observables from a complete commuting set of quantum fields.

a(x)a=1,...,N

operator valued functions of space-time.

Observables in a region R will be built out ofa(x) with x R. Observables in space-likeseparated regions will be guaranteed to commute if

1. [a(x), b(y)] = 0 whenever (x y)2 < 0.

We are going to construct our fields out of the creation and annihilation ops. Thesefive conditions will determine them:

2. a(x) = a(x) hermitian, observable

and that they have proper translation and Lorentz transformation properties

3. eiPaa(x)eiPa = a(x a)4. U()a(x)U() = a(1x)

If this were not a scalarfield, there would be extra

factors here reflecting achange of basis; as well as

a change of argument

5. a(x) =

d3k[Fak (x)ak + Gak(x)ak

]

We can think of our unitary transformations in two ways; as transformations on the

states | U|, or as transformations on the operators A UAU. NOT BOTH!Whats embodied in assumption (3):

Given U(a) the unitary operator of space translation by a (U(a) = ei Pa) the translationof a state | is a state | = U(a)|. Suppose the value of some observable, like chargedensity is

f(x) = |(x)|


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then it should be that|(x)| = f(x a).

x

xf( )

xa

f( - )ax

Rewrite the first equation with x x af(x a) = |(x a)|

Equate |(x a)| = |(x)|= |ei Pa(x)ei Pa|

A hermitian operator is determined by its expectation values

eiPa(x)ei

Pa = (x a)

(3) is just the full relativistic form of this equation. The equation with a = (t,0) is justthe time evolution for Heisenberg fields.

For the exact same reason as x a appears in the RHS of (3), 1x appears in the RHSof (4). (4) gives the Lorentz transformation properties of a

scalar field. This is not much ofan assumption. We can get fields transforming as vectors or tensors by taking derivatives ofthe a. Out of vector or tensor fields, we could make scalars.

In order to apply condition (4), it is nice to have the discussion phrased in terms ofrelativistically normalized creation and annihilation operators.

Recall the relativistically normalized one particle states

|k = (2)3/22k|kIntroduce (k) = (2)3/2

2kak, (k)|0 = |k.

Multiparticle states are made by

(k1) (kn)|0 = |k1, . . . , kn


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The Lorentz transformation properties of the states are

U()|0

= 0

U()|k1, . . . , kn = |k1, . . . , knand U(a) = eiPa is found from

P|0 = 0P|k1, . . . , kn = (k1 + + kn)|k1, . . . , kn

In Eqs. (1.9)-(1.12) and Eqs. (1.1)-(1.3) of the Sept. 23 lecture we set up the criteria thatU() and U(a) must satisfy. At the time our Hilbert space consisted only of the one particlepart of the whole Fock space we have now. You should check that the criteria are satisfiedin Fock space.

We can determine Lorentz transformation and translation properties of the (k). Con-sider,

U()(k)U()|k1, . . . , kn = U()(k)|1k1, . . . , 1kn= U()|k, 1k1, . . . , 1kn= |k, k1 . . . , kn

That is U()(k)U()|k1, . . . , kn = (k)|k1, . . . , kn|k1, . . . , kn is an arbitrary state in our complete basis so we have determined its action

completelyU()(k)U() = (k)

Similarly, or by taking the adjoint of this equation

U()(k)U() = (k)

An analogous derivation shows that

eiPx(k)eiPx = eikx(k)

eiPx(k)eiPx = eikx(k)

Now to construct the field (if there is more than one well label them when weve foundthem) satisfying all 5 conditions. First well satisfy condition (5) except well write the linear

combination of ak and ak

in terms of our new (k) and (k)

(x) =

It would be stupid

not to use theL.I. measure

d3k

(2)32k[fk(x)(k) + gk(x)

(k)]


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By (3), (x) = eiPx(0)eiPx, that is,

(x) = d3k(2)32k [fk(0)eiPx(k)eiPx + gk(0)eiPx(k)eiPx]=

d3k

(2)32k[fk(0)eikx(k) + gk(0)e

ikx(k)]

We have found the x dependence of fk(x) and gk(x) now we will use (4) to get their kdependence. A special case of (4) is

(0) = U()(0)U()

d3k

(2)32k[fk(0)(k) + gk(0)

(k)] =

d3k(2)32k

[fk(0) U()(k)U()

(k)

+gk(0) U()(k)U()

(k)

]

change variablesmeasure is unchanged k 1k

=

d3k

(2)32k[f1k(0)(k) + g1k(0)

(k)]

The coefficients of (k) and (k) must be unchanged fk(0) = f1k(0) and gk(0) =g1k(0).

k ranges all over the mass hyperboloid (k0 > 0 sheet), but a Lorentz transformation can

turn any of these ks into any other. So fk(0) and gk(0) are constants, independent of k.

(x) =

d3k

(2)32k[f eikx(k) + geikx(k)]

We have two linearly independent solutions of conditions (3), (4) and (5), the coefficientsof the complex constants f and g. Well name them. (Switching back to our old creationand annihilation ops.)

+(x) =

d3k

(2)3/22kakeikx (x) =

d3k

(2)3/22kake

ikx

Note (x) = +(x) convention is bananas, but itwas estd by Heisenberg and Pauli 50 years ago.

Now well apply hermiticity. Two independent combinations satisfying (2) are

(x) = +(x) + (x) and (x) =1

i[+(x) (x)]


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These are two independent cases of the most general choice satisfying (2):

(x) = ei+(x) + ei(x)

Now to satisfy (1). There are three possible outcomes of trying to satisfy (1).

Possibility A: Both of the above combinations are OK. We have two fields 1 and2 commuting with themselves and each other at spacelike separation. In thispossibility +(x) and (x) commute with each other at spacelike separation.

Possibility B: Only one combination is acceptable. It is of the form = ei++ei.While may be arbitrary, only one is acceptable.

Possibility C: The program crashes, and we could weaken (5) or think harder.

Lets first calculate some commutators. Using their expansions in terms of the ak and akand the commutation relations for ak and a

k

we find

[+(x), (y)] = 0 = [(x), +(y)]

and [+(x), (y)] =

d3k

(2)32keik(xy) +(x y, 2)

or just +(xy)

also [(x), +(y)] = +(x y, 2)+ is manifestly Lorentz invariant +(x) = +(x).

Possibility A runs only if +(x y) = 0 for (x y)2

< 0. We have encountered a similarintegral when we were looking at the evolution of one particle position eigenstates. In fact,

i0+(x y) =

d3k

(2)3eik(xy)

is the very integral we studied, and we found that it did

not vanish when (x y)2 < 0.Possibility A is DEAD.On to possibility B. Take (x) = ei+(x) + ei(x) and calculate

[(x), (y)] = +(x y) +(y x) dependencedrops out

Does this vanish when (x y)2 < 0? Yes, and we can see this without any calcula-tions. A space-like vector has the property that it can be turned into minus itself by a(connected) Lorentz transformation. This and the fact that + is Lorentz invariant, tellsus that [(x), (y)] = 0 when x y is space-like. We can choose arbitrarily, but we cantchoose more than one . Well choose = 0. Any phase could be absorbed into the ak and

ak.


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Possibility B is ALIVE, and we dont have to go on to C.We have our free scalar field of mass

(x) = +(x) + (x) = d3k

(2)3/2

2k[akeikx + ake

ikx]

Our field satisfies an equation (show using k2 = 2)

(+ 2)(x) = 0 =

This is the Heisenberg equation of motion for the field. It is called the Klein-Gordonequation. If we had quantized the electromagnetic field it would have satisfied Maxwellsequations.

Actually, Schrodinger first wrote down the Klein-Gordon equation. He got it at the sametime as he got the Schrodinger equation:

i0 = 12

2

This equation is obtained by starting with E = p2

2m noting that E = (when = 1) andpi = ki and for plane waves = i0 and k

i = 1i i.Schrodinger was no dummy, he knew about relativity, so he also obtained (+2)(x) = 0

from p2 = 2.He immediately saw that something was wrong with the equation though. The equation

has both positive and negative energy solutions. For a free particle the energies of its possiblestates are unbounded below! This is a disgusting relativistic generalization of a single particlewave equation, but with 50 years hindsight we see that this is no problem for a field thatcan create and destroy particles.


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4. October 2 Notes from Sidney Colemans Physics 253a 28

4 October 2

We have constructed the quantum field. It is the object observables are built from. More thanthat though, we can reconstruct the entire theory from the quantum field. The structure webuilt in the last three lectures is rigid. We can make the top story the foundation. Supposewe started with a quantum field satisfying (as our does),

1. (x) = (x), hermiticity

2. (+ 2)(x) = 0, K.-G. equation

3. [(x), (y)] = +(x y) +(y x) =

d3k(2)32k

[eik(xy) eik(xy)]

4.U()(x)U() = (1x)

U(a)(x)U(a) = (x a) is a scalar field5. (x) is a complete set of operators, i.e. ifx [A, (x)] = 0 A = I.

then from these properties, we could reconstruct the creation and annihilation operatorsand the whole theory. Conversely, all these properties follow from the expression for (x)in terms of the creation and annihilation operators. The two beginning points are logicallyequivalent.

Defining ak and ak

and recovering their properties

Property (2) of , that is a solution of the Klein-Gordon equation, tells us that

(x) =

d3k[keikx + ke

ikx] (k0 = k =

k2 + 2)

This is because any solution of the Klein-Gordon equation can be expanded in a completeset of solutions of the K.-G. eqn., and the plane wave solutions are a complete set. Because is an operator, the coefficients in the expansion, k and k, are operators. Because of

property (1), k = k

. Just to please my little heart, lets define

These funny factors will make the

commutation relations for the ak and akcome out nice

k =ak

(2)3/2

2kk = k

=ak

(2)3/2

2k


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Then the expression is

(x) = d3k(2)3/22k [akeikx + akeikx]This defines implicitly the ak and a

k

. To find their commutation relations well first haveto solve for them. Note that

(x, 0) =

d3k

(2)3/2

2k[akeikx + ake

ikx]

While

(x) = d3k

(2)3/22k [ak(ik)eikx + ak(ik)e

ikx]

And

(x, 0) =

d3k

(2)3/2

k2

[akeikx + ake

ikx]

The coefficient of 1(2)3/2

eikx in the expansion for (x, 0) is 1

2k(ak + ak) therefore

12k

(ak + ak) =

d3x

(2)3/2(x, 0)eikx

The coefficient of 1(2)3/2

eikx in the expansion for (x, 0) is

k2

(iak + iak) therefore

k2 (iak + iak) =

d3x(2)3/2

(x, 0)eikx

Thats just the inverse Fourier transformation applied.Now I can use these two expressions to solve for ak. Take

2k times the expression for

12k

(ak + ak) and add

2

ktimes the expression for

k2 (iak + iak) and divide by 2 to

get

ak =1

2

d3x

(2)3/2

(x, 0) +

i

k(x, 0)

ei

kx

Take the hermitian conjugate to get an expression for ak.

Having solved for the ak and ak

, we can now find their commutation relations. Write out

the double integral for [ak, ak

] and use property (3) below (which is weaker than property

(3)). You should get (3)(k k).From property (4), you can derive, what U(a)(x)U(a)akU(a) and U()

akU()4 are.

You can also derive and state the analog of property (5).

4It is easier to find the action of Lorentz transformations on a(k) (2)3/22kak.


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The properties of the field (x) are actually a little overcomplete. We can weaken one ofthem, (3), without losing anything. Since obeys the K.-G. equation, (2),

(x + 2)[(x), (y)] = 0

We see that the commutator obeys the K.-G. equation for any given y. For a given y, weonly need to give the commutator and the time derivative on one initial time surface, andthe K.-G. equation determines its evolution off the surface. So we will weaken (3) to

3.[(x, t), (y, t)] = 0[0(x, t), (y, t)] = i(3)(x y)

We can easily check that this is the right specialization of property (3) by doing theintegrals, which are easy with x0 = y0. (For a given y = (y, t) we have chosen our initial

surface to be x0 = t. The equations in (3) are called equal time commutation relations.)

[(x, t), (y, t)] =

d3k

(2)32k[ei

k(xy) eik(xy)]Change variables k k

in second term

= 0

[0(x, t), (y, t)] =

d3k

(2)32k(ik)[ei

k(xy) + eik(xy)]

Again k k in second term

= d3k

(2)3

2k

(ik)2eik(xy)

= i(3)(x y) Reminds us a littleof a bad dream inwhich [pa, qb] = iab

In the remainder of this lecture we are going to develop a completely different approach

to quantum field theory. We will obtain a field satisfying properties (1), (2), (3), (4) and(5), and then well stop. We have just shown that given a field satisfying these propertieswe can recover everything we did in the first three lectures. The new approach is


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The method of the missing box

classicalparticlemechanics

classicalfield

theory

quantum

particlemechanics

quantum

fieldtheory

continuumlimit

canonicalquantization

Anyone looking at this diagram can see there is a missing box

Taking the continuum limit essentially is just letting the number of coordinates, degreesof freedom go to infinity. This is usually done in a cavalier way. It doesnt matter a wholelot if you have a discrete infinity or a continuous infinity. With a continuous infinity you canFourier transform the coordinates to obtain a discrete set of coordinates.

Canonical quantization is a turn the crank way of getting quantum mechanics, beginningwith a classical Hamiltonian.

Well combine these two standard operators to get the missing box, but first, a lightningreview of the essential principles of the three boxes we have.

Classical Particle Mechanics

We start with a Lagrangian, a function of the generalized coordinates, and their time deriva-tives

L(q1, . . . , qN, q1, . . . , qN, t) qa(t), a 1, . . . , N e.g. L = 1

2mq2 V(q)

We define the action S = t2t1 dtL and apply Hamiltons principle to get the equations ofmotion. That is, we vary S by arbitrarily varying qa(t) except at the endpoints

qa(t1) = qa(t2) = 0

and demand0 = S


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S = t2t1 dta Lqa qa + Lqa qa parts integration

=

t2t1

dt

a

L

qa d

dtpa

pa Lqa

qa + paq

a|t2t1 vanishes because ofendpoint restriction

Since the variation qa(t) is arbitrary

L

qa= pa, for all a e.g. p = mq, p = dV

dq

These are the Euler-Lagrange equations.Around 1920(!) the Hamiltonian formulation of classical particle mechanics was discov-

ered.Define H(p1, . . . , pN, q1, . . . , qN, t) =

a

paqa L

Hmust be written in terms of the ps and qs only, not the qs. (This is not always possible.The new variables must also be independent, so it is possible to vary them independently.Examples where the ps are not complete and independent are electromagnetism and aparticle on a sphere in R3. The Lagrangian for the latter system may be taken as

L =1

2

mr2 + (r2

a2)

V(r)

The equation of motion for the variable enforces the constraint. In the passage to theHamiltonian formulation p = 0;

5 p is not an independent function. In this system, oneway to et to the Hamiltonian description is to first eliminate and one more coordinate,taking perhaps , , polar coordinates for the system, rewriting the Lagrangian, and thentrying again.)

Vary the coordinates and momenta

dH =

a

dpaq

a +padqa

Fortunately we didnthave to expand out dqa

Lq a

dqa L

qadqa

pa by the E-L equations

Read off Hamiltons equations

H

pa= qa

H

qa= pa

Notice: Lt

= 0 dHdt

= 0in that case H is called the energy (which

is the name reserved for the conserved quantityresulting from time translation invariance)

5How can I vary that?!?


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Quantum Particle Mechanics

We replace the classical variables pa, qa by operator valued functions of time satisfying

[qa(t), qb(t)] = 0 = [pa(t), pb(t)]

and [pa(t), qb(t)] = iba = 1

The ps and qs are hermitian observables. They are assumed to be complete. Theydetermine the Hilbert space. For example 1-d particle mechanics, the space of square inte-grable functions. A basis set can be the eigenstates of q. p or rather eipx tells you how torelate the phases of the various eigenstates and indeed even that the range of q is R (not[-1,1] or anything else).

The quantum Hamiltonian, which determines the dynamics, is just the classical Hamil-

tonian except it is now a function of the operator ps and qs. For any AdA

dt= i[H, A] +

A

t if there is explicit dependence ontime other than that implicit in the

time dependence of the ps and qs

H is the generator of infinitesimal time translations just as in classical mechanics.H suffers from ordering ambiguities. Because p and q dont commute, it is not clear, and

it may matter, whether you write p2q, qp2 or pqp. Sometimes this ambiguity can be clearedup using other criteria. For example, in central force problems, we quantize in Cartesiancoordinates and then transform to central coordinates.

Heisenberg equations of motiondqa

dt= i[H, qa]

= i

i H

pa

=

H

pa

This step depends only on the commutationrelations for the ps and qs. Fairly easy to seefor a polynomial in the ps and qs.

Similarly,dpadt

= Hqa

This is the motivation for the commutation relations. It is a way of putting the correspon-dence principle into the theory. The quantum equations resemble the classical equations, atleast up to the ordering ambiguities, and any variation due to ordering ambiguities is downby a factor of. We wouldnt expect any general procedure for turning classical theories intoquantum theories, motivated only by the correspondence principle, to be able to fix thoseambiguities. We can be a little more precise about how the classical equations are actually


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recovered. In general, the Heisenberg equations of motion for an arbitrary operator A relateone polynomial in p, q, p and q to one another. We can take the expectation value of this

equation to obtain (a quantum mechanical average of ) an equation between observables. Inthe classical limit, when fluctuations are small, expectations of products can be replaced byproducts of expectations, pn pn, pq pq, and this turns our equation amongpolynomials of quantum operators into an equation among classical variables.

The other two boxes are actually going to go quite quickly, because if you are just a littlecavalier, the continuum limit is little more than some new notation.

Classical Field Theory

We have an infinite set of generalized coordinates, real number functions of time, a(x, t)labelled by a discrete index, a, and a continuous index, x.

a(x, t) qa(t)t ta a, x

It is sometimes a handy mnemonic to think of t, x, as generalization of t, but that isnot the right way to think about it. For example we are used to giving initial value data atfixed t in CPM. In CFT we dont give initial value data at fixed t and x, that is obviouslyincomplete. We give initial value data for fixed t and all x.

With cowboy boldness, everywhere in CPM we see a sum on a, well just replace it with asum on a and an integral over x, and everywhere we see ab well replace it with

ab

(3)(x

y).

The Dirac delta function has the exact same properties in integrals that the Kronecker deltahas in sums.

The next thing to do is write down Lagrangians. Because a CPM Lagrangian can containproducts ofqa with different a, our Lagrangian could contain products of(x, t) with differentx

d3xd3yd3zfabc(x, y, z)a(x, t)b(y, t)c(z, t)

a,b,c

fabcqa(t)qb(t)qc(t)

But note that a CPM Lagrangian does not contain products at different times.With an eye to Lorentz invariance, and noticing that Lagrangians are local in time, we will

specialize to Lagrangians that are local in space. Also, you know that when taking continuum

limits, differences of neighboring variables become spatial derivatives and a combination like

1

a2[((n 1)a, t) + ((n + 1)a, t) 2(na,t)]

would become 2

x2 in the continuum. But again with an eye to Lorentz invariance, becauseonly first derivatives with respect to time appear in the Lagrangian, we will only consider


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first derivatives with respect to the xi. So L has the form

L(t) = d3kL(a(x), a(x), x)And the action S =

t2t1

dtL(t) =

d4xL

The Euler-Lagrange equations which come from varying S will be Lorentz covariant ifL is a Lorentz scalar. At three points we have used Lorentz invariance to cut down on thepossible forms for L. These have been specializations, not generalizations. Now well applyHamiltons principle

0 = S under arbitrary variations

satisfying a(x, t1) = a(x, t2) = 0=

a

d4x L

a(x, t)a(x, t) +

La a

a

a

=

a

d4x

La

a

a

Do parts integration. As usual, therestrictions on a make the surfaceterms at t1 and t2 drop out

La

= a (for all x and a)

Euler-Lagrangeequations

a should not be thought of as a four-vector generalization of pa. The correspondence is

0a(x, t) pa(t)

In fact 0a is often jut written a.We should say something about the surface terms at spatial infinity which we ignored

when we did our parts integration. One can say we are only considering field configura-tions which fall off sufficiently rapidly at spatial infinity that we can ignore surface terms.Alternatively we could work in a box with periodic boundary conditions. Anyway, well justbe slothful.

A simple example of a possible

LMost general L satisfying (3) conditions1. Build L out of one reals calar field,

= (not = , were not doing QFT yet)

2. L is a Lorentz scalar


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3. L is quadratic in and (1) and (3) are for simplicity. A motivation for (3) is that a quadratic action yields linear

equations of motion, the easiest ones to solve. Most general L is

L = 12

a[ + b2]

One of these constants is superfluous; we are always free to rescale , |a| . Lbecomes

L = 12

[ + b2]

The Euler-Lagrange equations for our example are

L

= and

L

= 0 or ( b) = 0

In general the Hamiltonian which was

apaqa L in CPM is

H =

a

d3x(0a0

a L) =

d3xH

H = a 0a0

a L is the Hamiltonian density

In our example

H =

d3x[1

2(0)2 +

1

2()2 b2]

Since each of these terms separately can be made arbitrarily large, if the energy is to bebounded below, they each better have a positive coefficient. better be + and b better be

b = 2 definition of 0

The E-L equation is now + 2 = 0

Gosh, this is looking familiar now.We are ready to fill in the last box. We are ready to canonically quantize classical fieldtheory. Since we are not worrying about the passage to an infinite number of variables, thiswill be little more than a notational change in the canonical quantization of CPM. It wouldnot even have required a notational change if Newton hadnt chosen two ways of writing Sfor sum,

and

.


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Quantum Field Theory

We replace the classical variables

a

,

0

a by quantum operators satisfying[a(x, t), b(y, t) = 0 = [0a(x, t),

0b (y, t)]

and [0a(x, t), b(y, t)] = iba(3)(x y)

The Dirac delta for the continuous index x is just the continuum generalization of theKronecker delta. H =

d3xH(0a, a, x) determines the dynamics as usual. The commu-

tation relations are set up to reproduce the Heisenberg equations of motion. Since this isjust a change of notation there is no need to redo any proofs, but lets check that things areworking in our example anyway.

H = d3x12[(0)2 + ( )2 + 22]0(y, t) = i[H, (y, t)] = i(i)

d3x0(x, t)(3)(x y)

= 0(y, t)

Similarly, 00(y, t) = 2 2. you need the commutator [0(y, t), (x, t)] which is

obtained by taking the gradient of the equal time commutation relation (ETCR).We have reproduced our quantum field satisfying (1), (2), (3), (4) and (5). This was

accomplished in one lecture rather than three because this is a mechanical method withoutphysical insight. The physical interpretation comes at the end instead of at the beginning.In our first method, the constructive method, we shook each object we introduced to makesure it made sense, and we finally obtained a local observable, the quantum field. we cantput interactions in in this method though, because in the very first steps we had to knowthe whole spectrum of the theory, and about the only theory we know the exact spectrumfor is the free theory. In our magical canonical quantization method its easy to put ininteractions: just let L L 4. At the first order in perturbation theory the part of4 that has two creation and two annihilation operators produces two-into-two scattering!At second order well get two-into-four and two-into-six 6 scattering. Looks easy. . . if therewerent any booby traps. . . but boy are there going to be booby traps.

6pair production!


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5 October 7

The simplest example obtained from applying canonical quantization to a free scalar fieldled to the same theory we got by constructing a local observable painstakingly in the theorywe had before.

Well double check that we get the same Hamiltonian we had in that theory. Evaluate

H =1

2

d3x[2 + ()2 + 22] =

Write in terms of its Fourier expansion

(x) =

d3k

(2)3/2

2k

[akeikx + ake

ikx]

Substitution would lead to a triple integral, but the x integration is easily done yieldinga function, which does one of the k integrals7.

H =1

2

d3k

2k

ak akk=k seefootnote 1

e2ikt(2k + |k|2 + 2) +akak(2k + |k|2 + 2)

That what multiplies e2ikt is 0 is good.H should not be time-dependent.

+ akak(2k

+ |k|2 + 2) + akake

2ikt

(2k + |k|2 + 2)

H had four types of terms: ones creating particles with momentum k and

k, ones

destroying particles with momentum k and k, ones creating a particle with momentum k,then destroying one with momentum k, and ones creating a particle with momentum k,then destroying one with momentum k. These all conserve momentum.

H =1

2

d3kk(akak

+ akak

)

Almost, but not quite what we had before.

Using [ak, ak

] = (3)(k k) gives H = d3kk akak + 12 (3)(0). (3)(0)! Can get someidea of the meaning of this by putting the system in a box. H becomes

1

2 k

k(akak + akak) = k

akak + 12In a box we see that this is just a zero point energy. (It is still infinite though.) Just

as the spectrum of the Hamiltonian H = 12 (p2 + 2q2) starts at 12 , not zero, the spectrum

7d3xei

kxeikx = (2)3(3)(k + k) k = k


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of our Hamiltonian starts at 12

k k, not zero. That this constant is infinite even in abox is because even in a box the field has an infinite number of degrees of freedom. There

can be excitations of modes with arbitrarily short wavelengths. The zero point energy isultraviolet divergent. In general, infinite systems, like an infinite crystal, also have infrareddivergences. If the energy density of an infinite crystal is changed by an infinite amount,the total energy is changed by an infinite amount. Our Hamiltonian for the system in infinitespace, would still have an infinity ((3)(0)) even if the momentum integral were cut off, say

by restricting it to |k| < K. That part of the infinity is eliminated by putting the systemin a box, eliminating long wavelengths. Our system in infinite space has a zero-point energythat is also infrared divergent.

This is no big deal for two reasons.

(A) You cant measure absolute energies, only energy differences, so its stupid to

ask what the zero point energy is. This even occurs in introductory physics. Weusually put interaction energies to be zero when particles are infinitely separated,but for some potentials you cant do that, and you have to choose your zeroanother place.

(B) This is just an ordering ambiguity. Just as the quantum Hamiltonian for theharmonic oscillator could be chosen as

H =1

2(p + iq)(p iq)

we could reorder our Hamiltonian (see below).

In general relativity the absolute value of the energy density does matter. Einsteinsequations

R 12

gR = 8GTcouple directly to the energy density T00. Indeed, introducing a change in the vacuum energydensity, in a covariant way,

T T gis just a way of changing the cosmological constant, a term introduced by Einstein andrepudiated by him 10 years later. No astronomer has ever observed a nonzero cosmologicalconstant.8 Our theory is eventually going to be applied to the strong interactions, maybeeven to some grand unified theory. Strong interactions have energies typically of 1 GeV and

a characteristic length of a fermi, 1013

cm. With a cosmic energy density of 1039

GeV/cm3

,the universe would be about 1 km long according to Einsteins equations. You couldnteven get to MIT without coming back where you started. We wont talk about why thecosmological constant is zero in this course. They dont explain it in any course given atHarvard because nobody knows why it is zero.

8[BGC note: As of 1998, measurements on supernovae have indicated that > 0.]


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Reordering the Hamiltonian:

Given a set of free fields (possibly with different masses) 1(x1), . . . , n(xn), define thenormal ordered product: 1(x1) n(xn) :

as the usual product except rearranged so that all the annihilation operators on the rightand a fortiori all creation operators on the left. No further specification is needed sinceall annihilation operators commute with one another as do all annihilation operators. Thisoperation is only defined for free fields satisfying the K.-G. equation, which tells us we canwrite the field in terms of time independent creation and annihilation operators.

Redefine H to be : H :. This gets rid of the normal ordering constant.9

This is the first infinity encountered in this course. Well encounter more ferocious ones.We ran into it because we asked a dumb question, a physically uninteresting question, aboutan unobservable quantity. Later on well have to think harder about what weve done wrongto get rid of troublesome infinities.

If we wanted to get as quickly as possible to applications of QFT, wed develop scatteringtheory and perturbation theory next. But first we are going to get some more exact resultsfrom field theory.

Symmetries and Conservation Laws

We will study the relationship between symmetries (or invariances) and conservation lawsin the classical Lagrangian framework. Our derivations will only use classical equationsof motion. Hopefully everything will go through all right when Poisson brackets becomecommutators, since the commutation relations are set up to reproduce the classical equationsof motion. In any given case you can check to see if ordering ambiguities or extra terms incommutators screw up the calculations.

Given some general Lagrangian L(qa, qa, t) and a transformation of the generalized coor-dinates

qa(t) qa(t, ) qa(t, 0) = qa(t)9 Whats wrong?

a

a = aa

1 Normal order both sides: aa : =: aa : 1

0 = 1

Answer. We dont normal order equations. Normal ordering is not derived from the ordinary productany more than the cross product is derived from the scalar product. Normal ordering cannot accurately beused as a verb.


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we can define

Dqa

qa

=0This is useful because the little transformations will be important.Some examples:

1. Space translation of point particles described by n vectors ra, a = 1, . . . , n. TheLagrangian well use is

na=1

ma2

|ra|2

ab

Vab(|ra rb|)

The transformation is ra ra + e, all particles in the system moved by e

Dra = e

2. Time translations for a general system. Given some evolution qa(t) of the system, thetransformed system is a time ahead qa(t) qa(t + )

Dqa =qa

t

3. Rotations in the system of example (1)

ra Rotation

matrixR ( eangle, axis

)ra Dra = e ra

(Ife = z, Dxa = ya, Dya = xa, Dza = 0.)Most transformations are not symmetries (invariances).

Definition: A transformation is a symmetry iffDL = dFdt for some F(qa, qa, t). This equality

must hold for arbitrary qa(t), not necessarily satisfying the equations fo motion.Why is this a good definition? Hamiltons principle is

0 = S =t2

t1dtDL =

t2t1

dFdt

= F(t2) F(t1)

Thus a symmetry transformation wont affect the equations of motion (we only considertransformations that vanish at the endpoints).

Back to our examples:


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1. DL = 0, F = 0

2. IfL has no explicit time dependence, all of its time dependence comes from its depen-dence on qa and qa, Lt = 0, then DL =

dLdt , F = L

3. F = 0

Theorem (E. Noether say Nota): For every symmetry there is a conserved quantity.The proof comes from considering two expressions for DL.

DL =

a

L

qaDqa + paDq

a

used pa Lqa

=

a( paDq

a + paDqa) used E-L equations

=d

dt

a

paDqa by equality of mixed partials

Dqa= ddt

Dqa

By assumption DL = dFdt . Subtracting these two expressions for DL we see that thequantity

Q =

a

paDqa F satisfies dQ

dt= 0

(There is no guarantee that Q = 0, or that for each independent symmetry well getanother independent Q, in fact the construction fails to produce a Q for gauge symmetries.)

We can write down the conserved quantities in our examples1. (pa = mar

a), Dra = e, F = 0.

Q =

a

mae ra = e

a

mara

For each of three independent es we get a conservation law. The momentum

P =

amar

a is conserved.

d P

dt= 0

.

Whenever we get conserved quantities from spatial translation invariance, whether ornot the system looks anything like a collection of point particles, well call the conservedquantities the momentum.


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2. Dqa = qa

t , F = L. Note: Q is identical to H.

Q = a

paqa

L is conserved when Lt = 0Whenever we get a conserved quantity from time translation invariance, well call theconserved quantity the energy.

3. Dra = e ra, F = 0.Q =

a

pa (e ra) =

a

e (ra pa)

= e a

ra pa three laws

J =

a

ra pa is conserved

d J

dt= 0

Whenever we get a conserved quantities from rotational invariance, well call them theangular momentum.

There is nothing here that was not already in the Euler-Lagrange equations. What thistheorem provides us with is a turn the crank method for obtaining conservation laws from avariety of theories. Before this theorem, the existence of conserved quantities, like the energy,had to be noticed from the equations of motion in each new theory. This theorem organizes

conservation laws. It explains, for example, why a variety of theories, including ones withvelocity dependent potentials all have a conserved Hamiltonian, or energy (example (2)).

From the conserved quantity, we can usually reconstruct the symmetry. This can bedone in classical mechanics using Poisson brackets. Well do it in quantum mechanics usingcommutators.10

Assume Dqb and F depend only on the qa, not the qa so that their expression in theHamiltonian formulation only depends on the qa, not the pa. Then

[Q, qa] =

b

pbDqb F, qa

=

b

[pb, qa]

iabDqb

= iDqa

This assumption is not at all necessary. Usually the result holds. For example the energygenerates time translations even though the assumption doesnt hold.

10Show: [Qi,Qj] = iijkQk] when the Qis are generators of an SO(3) internal symmetry. For a general

internal symmetry group, the quantum Qs recreate the algebra of the generators.


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Symmetries and Conservation Laws in (Classical) FieldTheory

Field theory is a specialization of particle mechanics. There will be more that is true in fieldtheory. What is this more?

Electromagnetism possesses a conserved quantity Q, the electric charge. The charge isthe integral of the charge density p, Q =

d3x(x, t). There is also a current and there is

a much stronger statement of charge conservation than dQdt = 0. Local charge conservationsays

t+ = 0

Integrate this equation over any volume V with boundary S to get

dQVdt

= S

dAn using Gausss theorem and(Q=V d3x(x,t))This equation says you can see the charge change in any volume by watching the current

flowing out of the volume. You cant have:

t

x

simultaneously wink out of existence with nothing happening anywhere else.

This picture satisfies global charge conservation, but violates local charge conservation.You have to be able to account for the change in charge in any volume, and there would haveto be a flow of current in between the two charges. Even if there were not a current and alocal conservation law, we could invoke special relativity to show this picture is inconsistent.In another frame the charges dont disappear simultaneously, and for a moment global chargeconservation is violated.

Field theory which embodies the idea of local measurements, should have local conser-vation laws.Given some Lagrangian density L(a, a, x) and a transformation of the fields

a(x) a(x, ) a(x, 0) = a(x)

we define Da = a

=0

.


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Definition: A transformation is a symmetry iff DL = F for some F(a, a, x). Thisequality must hold for arbitrary a(x) not necessarily satisfying the equations of mo-

tion.We are not using special relativity. F is not necessarily a 4-vector (just some set of 4objects).

Note that the previous definition can be obtained. To wit,

DL = D

d3xL =

d3xF

=

d3x0F

0 =d

dTF

where F =

d3xF0.Why is this a good definition? Hamiltons principle is

0 = S = d4xDL = d4xF=

d3x[F0(x, t2) F0(x, t1)] blithe as usual aboutcontributions from

spatial infinity

Thus a symmetry transformation does not affect the equations of motion (we only considertransformations that vanish at the endpoints when deriving the equations of motion.).

Theorem (maybe this is Noethers theorem?): For every symmetry there is a conservedcurrent.

The proof comes from considering two expressions for DL.

DL = a La Da + a Da used a La=

a

(a Da + a D

a) used E-L equations

=

a

a Da by equality of mixed partials

Da=Da

By assumption DL = F. Subtracting these two expressions for DL we see that thefour quantities

J =

a

a Da satisfy J

= 0

which implies dQVdt = S dAn JThis equation justifies calling

J0 the density of stuffand Ji the current of stuff.

, S =boundary of V. The total amount of stuff, Q is

independent of time.There is an ambiguity in the definition of F and J since F can be changed by any

satisfying = 0. In the particle mechanics case, F was ambiguous, but only by a


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time independent quantity. For arbitrary antisymmetric A we can let F F + A(A

= 0). As a result of this change J J A.

Q =

d3xJ0 Q d3xiAi = Qi.e. Q is unchanged, ignoring contributions from spatial infinity as usual.

This is a lot of freedom in the definition of J. For example in a theory with three fieldswe can let

J J (3)4(12 12)This J is as good as any other. There are 500 papers arguing about which energy-

momentum tensor is the right one to use inside a dielectric medium. 490 of them are idiotic.Its like if someone passes you a plate of cookies and you start arguing about which copy

is #1 and which is #2. Theyre all edible! Sometimes one J is more useful in a givencalculation than another, for some reason. Instead of arguing that the most convenient J

is the right one, you should just be happy that you had some freedom of choice.From cranking Hamiltons principle, we can give another derivation of the relation be-

tween symmetries and conserved quantities.

S =

dtL =

a

paDqa|t2t1

for an arbitrary variation about a solution of the equations of motion. By assumption of asymmetry

S = dtDL = dt dFdtt2

t1

Subtracting we see

apaDqa F is time independent.

[Aside: Noethers Theorem derived at the level11 of

the action

If without using the equations of motion, for an arbitrary function of space time (x)parametrizing a transformation, to first order in (x)

L = (x)k + (x)j

(Equivalent to DL = k when is a constant).11i.e. using Hamiltons principle rather than mucking around with the equations of motion


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Then by Hamiltons principle, for fields satisfying the equations of motion,

0 = d4xL = d4x((x)k + (x)j)=

d4x((x)k

(x)j) Let (x) vanish at to do parts integration.

So it must be that J = k j is conserved.

Example: Space-time translation invariance

= = ()

L =

L +

L= L

L

L

is a parameter for thetype of transformation.

= L plays the same role

I read offk = gL, j = so T = gL is conserved.]

Noethers Theorem: when the variation is a hermitiantraceless matrix. (useful in theories with fields most

easily written as matrices.)

Suppose (without using the equations of motion) S =

d4x Tr matrix

F

hermitian, tracelessbut otherwise arbitrary

.

Then Hamiltons principle says

0 =

d4xTr F

Then F is a matrix of conserved currents, except there is no conservation law associatedwith the trace.

If you like, write this statement as

F 1

NId Tr F

dimension of F

= 0


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Space-time translations in a general field theory

We wont have to restrict ourselves to scalar fields because under space-time translations,an arbitrary field, vector, tensor, etc., transforms the same way. Just let the index a alsodenote vector or tensor indices.

Translations are symmetries as long as L does not have any explicit dependence on x. Itdepends on x only through its dependence on a and

a, Lx

= 0.

a(x) a(x + e) e some fixed four vectorDa = e

a DL = eL = (eL)F = eL J =

a

a e

a eL = eT

T

= a a a gLThere are four conserved currents, four local conservation laws, one for each of the four

independent directions we can point e, i.e. T = 0 since J

= 0 for arbitrary e.T0 is the current that is conserved as a result of time translation invariance, indeed,

H = T00 =

a

aa L, H =

d3xH

T00 is the energy density, Ti0 is the current of energy. Ti are the three currents fromspatial translation invariance.

Pi =

d3xT0i

T0i = the density of the ith component of momentum

Tji = the jth component of the current of the ith component of momentum

For a scalar field theory with no derivative interactions, a = so

T =

a

aa gL

Note that T is symmetric, so you dont have to remember which index is which. T

can be nonsymmetric, which can lead to problems, for example gravity and other theoriesof higher spin. Can try to symmetrize it.


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6 October 9

Lorentz transformationsUnder a Lorentz transformation all vectors transform as

a a

where specifies the Lorentz transformation. must preserve the Minkowski space inner

product, that is ifb b then ab ab

This must be true for arbitrary a and b. (The equation this condition gives is g

=

g.) Well be interested in one parameter subgroups of the group of Lorentz transformations

parametrized by . This could be rotations about some specified axis by an angle or aboost in some specified direction by a rapidity . In any case, the Lorentz transformation isgiven by a family

a a() = ()aUnder this (active) transformation (we are not thinking of this as a passive change of

coordinates) the fields a transform as (a is a scalar),

a(x) a(x, ) = a(()1x)We are restricting ourselves to scalar fields. Even though we only used scalar fields in

our examples of T, the derivation of the conservation of T from space-time translation

invariance applies to tensor, or vector, fields. With Lorentz transformations, we only considerscalars, because there are extra factors in the transformation law when the fields are tensorial.For example a vector field A(x) A(1x).

We need to get D =

=0. Well define

D defines some matrix From the invariance of ab, well derive a condition on

.

0 = D(ab) = (Da)b + a

(Db)

= ab + a

b

= ab + ab relabel dummy indices

in the second term,

,.

= ( + )ab =

since this has to hold for arbitrary a and b. and range from 0 to 3, so there are 4(41)2

= 6independent , which is good since we have to generate 3 rotations (about each axis) and 3boosts (in each direction).


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As a second confidence-boosting check well do two examples.Take 12 = 21 = 1, all other components zero.

Da1 = 12a2 = 12a2 = a2

Da2 = 21a2 = 21a1 = +a1

a

x2

x1infinitesimallyshifted a

This says a1 gets a little negative component proportional to a2 and a2 gets a littlecomponent proportional to a1. This is a rotation, in the standard sense about the positivez axis.

Take 01 = 10 = +1, all other components zero.

Da0 = 01a1 = 01a1 = a1Da1 = 10a

0 = 10a0 = a0

This says x1, which could be the first component of the position of a particle, gets a littlecontribution proportional to x0, the time, which is definitely what a boost in the x1 directiondoes. In fact, Da0 = a1, Da1 = a0 is just the infinitesimal version of

a0 cosh a0 + sinh a1a1 sinh a0 + cosh a1

Without even thinking, the great index raising and lowering machine has given us all theright signs.Now assuming L is a scalar, we are all set to get the 6 conserved currents.


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6. October 9 Notes from Sidney Colemans P

sidney coleman - lectures on qft

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