diagrammatic representations in quantum...
TRANSCRIPT
Bachelor of Science Thesis
Uppsala University - June 13, 2013
Diagrammatic Representationsin Quantum Theories
Supervisor: Author:
Maxim Zabzine Jacob Stenberg
Abstract
Starting from a mathematical basis where one analyses and developingdifferent techniques in how to solve and represent different kinds of integralswith diagrams. Representing the integrals as n-valent vertices and introducingpropagators is a great tool that helps with the book-keeping of the solutionsand will sometimes do the calculations redundant. The symmetries of dia-grams are analysed and how one extracts the symmetry factors from lookingat a diagram by using some fairly simple combinatorics and cleverness. Intro-ducing the probability amplitude and do some analysis of the path integralformulation is the step into physics. Discussing experiments as the double-slit experiment and deriving the Schrodinger equation from the generatingfunctional. Looking at diagrams in Quantum Mechanics and Quantum FieldTheory will explore the use of the crucial understanding of our generators forthe diagrams. This thesis makes use of the so called generating functionalalmost throughout and to connect the first discussed mathematics to realphysical theories is the aim.
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Contents
0 Introduction and Purpose 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Purpose . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1 Gaussian and general classes of integrals 41.1 Introducing the Z[0] function in R and Rn . . . . . . . . . . . . . . . 41.2 A general class of integrals . . . . . . . . . . . . . . . . . . . . . . . . 5
2 Generating functions and Feynman diagrams 72.1 Z[J] - A variable generator . . . . . . . . . . . . . . . . . . . . . . . . 72.2 The Wick theorem and Feynman diagrams . . . . . . . . . . . . . . . 9
3 Amplitudes and paths 133.1 Two-slit experiment and amplitudes . . . . . . . . . . . . . . . . . . . 143.2 Actions and Kernels . . . . . . . . . . . . . . . . . . . . . . . . . . . 163.3 The Path Integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183.4 From Feynmans Path Integral formalism to the
Schrodinger equation . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
4 Units and notations 224.1 Space and products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224.2 Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.3 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.4 Dirac delta function . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23
5 Approaching fields 245.1 From mechanics to fields . . . . . . . . . . . . . . . . . . . . . . . . . 245.2 Equation of motion for free scalar fields . . . . . . . . . . . . . . . . . 245.3 Quantization and the free-field path integral . . . . . . . . . . . . . . 26
6 Interacting fields and perturbation theory 296.1 Naive description of the interaction Lagrangian . . . . . . . . . . . . 306.2 Central Identity of Quantum Field Theory . . . . . . . . . . . . . . . 306.3 Lazy Professors and ϕ3-theory . . . . . . . . . . . . . . . . . . . . . . 326.4 Diagrams in ϕ4-theory . . . . . . . . . . . . . . . . . . . . . . . . . . 35
7 Discussion 397.1 Analogies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397.2 Symmetries and topological equivalence . . . . . . . . . . . . . . . . . 407.3 A theory that cries in the vacuum . . . . . . . . . . . . . . . . . . . . 42
References 44
A Appendix 45
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0 Introduction and Purpose
0.1 Introduction
There is a story in this thesis that is about a student who claims that a professorwho is lecturing has got something wrong. The topic is the principle of superpositionin Quantum Mechanics. The realization that a particles transition amplitude fromone point to another is given by a sum of all the trajectories that can be drawnbetween these two points(which is an infinite number of ways) is truly a fundamen-tal building block in the development of Quantum Mechanics and leads us to thedescription of the path integral.
The formulation of the path integral is a generalization of the action principle inclassical mechanics. A unique trajectory in time and space
Figure 0.1.1: The classical trajectory.
described as the classical action of the particle, which is a minimal value of theparticles “effort” to get between the two points in a certain time. In Quantum Me-chanics this principle is replaced(as above stated) by the sum of all crazy trajectorieswho can be obtained between these two points
,
Figure 0.1.2: Some of the “crazy” trajectories in Quantum Mechanicsthat deviates from the classical trajectory in the previous figure.
summing over all these possible trajectories, also called histories, will at the endgive us the quantum mechanical transition amplitude. The path integral approachin Quantum Mechanics was developed by many physicists but it was completed byR.P. Feynman in the end of the 1940’s.
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The transition amplitude in quantum theories is a very interesting thing from manydifferent aspects. Representing certain events and happenings in a pictorial andvery intuitive way was a good technique invented by R.P. Feynman. To be able torepresent very abstract mathematical representations via visual, concrete tools hashelped physicists for decades. The technique helps those who practise the subjectto avoid long and quite complicated calculations for processes in quantum theories,such as propagating, scattering and decaying particles.
Figure 0.1.3: Examples of diagrammatic representationsfor different events in quantum theories.
At the very heart of these diagrams lies a representation of abstract mathemati-cal expressions. They are perturbative contributions to the transition amplitudeswhich are calculated at different orders in coupling constants to achieve the rightaccuracy. Feynman was a brilliant man and came up with what we now a day call“The Feynman rules”. He assigned different mathematical factors for each buildingblock in a diagram, and by multiplying these together one would have obtained themathematical expression for a physical event represented by a diagram. But, short-cuts very seldom comes without consequences. By using the Feynman rules oneavoids the long and messy calculations for sure, but we loose the symmetry factorin our laziness. One has to figure out with combinatorics and cleverness what kindof symmetry a diagram has and for a bit more complicated diagrams, this can be aa big labour.
0.2 Purpose
The very goal and interest in this thesis is to study the mathematics that generatesthese diagrams, i.e. the little bit more complicated way than the Feynman rules.By developing a mathematical understanding in the first chapters and slowly work-ing out different kinds of Gaussian integrals, study the generating function(al) thatis the heart of the diagrams and then move over to make physical connections tomodern theories in physics and see what these generators actually can tell us aboutdifferent situations in our world. “Why do it the hard way?”, you might ask. The
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hard way(the hard way is to not use the Feynman rules in Quantum Field Theoryfor example) will discover symmetries for us. The generating functional does nottake in mind that two diagrams can be topologically equivalent, but produces a lotof diagrams. Studying the techniques in how to solve these integrals will give us thesymmetries of the diagrams.
This thesis aims to a group that is in the beginning of studying quantum theo-ries, particle physics and other modern theories that includes diagrams who arecurious about the mathematical tools behind the diagrammatic representations.
Note:
The first two chapters of this thesis is based on an earlier 5hp project that I, theauthor, worked on in the spring semester of 2012, Uppsala University. The projectwas to start looking at the mathematical foundation of the integral analysis towardsdiagrammatic representations and this thesis continues to make connections betweentheories in physics and the mathematical basis in said project.
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1 Gaussian and general classes of integrals
1.1 Introducing the Z[0] function in R and Rn
In R, let the function Z[0] be defined as
Z[0] =
∫Rdx e−
α2x2
(1)
(α > 0), then Z[0] is the Gaussian in R. The proof that this integral has a solutionand converge can be found in Appendix A.
Now we move over to Rn. The interpretation of the function Z[0] in Rn is a bitmore tricky. We now have the symmetric matrix as a quadratic, positive formA(x, x) = Aijx
ixj (using the Einstein summation convention), and the Gaussian is
Z[0] =
∫Rndnx e−
α2Aijx
ixj (2)
This gives n Gaussian integrals in Rn that converges.
Proof:
The matrix A is a positive definite matrix. A can be diagonalized by the followingrewritings
A = QTDQ (3)
QTQ = 1 ⇒ Q−1 = QT (4)
where the matrix Q is orthogonal. The also introduced matrix D is diagonal. Nowone element Aij can be written as
Aij =∑k,l
QikDklQlj (5)
andQTQ = 1 ⇒
∑k
QkiQkj = δij (6)
Then the exponent in (2) is∑i,j
xiAijxj =
∑i,j,k,l
xiQikDklQjlxj. (7)
Now make the variable change
xi =∑j
Qijcj ⇔
∑i
Qikxi =
∑i,j
QikQijcj = ck (8)
by the same method we use the variable change for xj and by using these facts (7)now yields ∑
i,j
xiAijxj =
∑i,j,k,l
xiQikDklQjlxj =
∑k,l
ckDklcl. (9)
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The matrix D has the matrix A’s eigenvalues at the diagonal. There are onlyeigenvalues when k = l = i (because of the diagonal properties of D) and we labelthese λii = λi.
⇒∑k,l
ckDklcl =∑i
λi(ci)2. (10)
Now we can write (2) as follows∫Rndnx e−
α2Aijx
ixj =
∫Rndnc e
−α2
∑iλi(c
i)2
=n∏i
∫Rndci e−
α2λi(c
i)2
. (11)
From (11) we see that we have n Gaussian integrals. The matrix A is a positivedefinite matrix so all of the eigenvalues of A is positive. Therefore all integralsconverge and we end up with the result∫
Rndnx e−
α2Aijx
ixj =
(2π
α
)n2 1
(λ1 · · ·λn)12
=
(2π
α
)n2 1
(detA)12
= Z[0]. � (12)
1.2 A general class of integrals
Let us introduce the integral of the form
In(α) =
∫R
dx e−12αx2
xn. (13)
With a variable change the integral kan bli solved
x =
√2
αy ⇔ y2 =
1
2αx2 (14)
dx =
√2
αdy , µ =
α
2(15)
equation (13) now takes the form
In(µ) = µ−12
∫R
e−y2(µ−
12y)ndy = µ−
(n+1)2
∫R
e−y2
yndy. (16)
Let us see what we get for n = 0
I0(µ) = µ−12
∫R
e−y2
dy = µ−12√π =
√2π
α. (17)
we see that for n = 0 the integral I0(µ) is the Gaussian. The way to calculate theintegral for n = 1 is very straight forward so we now move on the the case whenn > 1. To calculate the integral for n > 1 we will se that it is easier to go back to
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the initial variable. The integral In(µ) can be written as taking a partial derivativewith respect to µ in this way
In(µ) = − ∂
∂µIn−2(µ). (18)
Let us evaluate (18) explicitly and see why
− ∂
∂µIn−2(µ) = − ∂
∂µ
∫R
dnx e−µx2
xn−2 =
∫R
dnx x2e−µx2
xn−2 = In(µ) (19)
We can now construct a recursion formula for the integral In(µ). Because if
In(µ) = − ∂
∂µIn−2(µ) (20)
then
In−2(µ) = − ∂
∂µIn−4(µ) ⇒ In(µ) = − ∂
∂µ
(− ∂
∂µIn−4(µ)
). (21)
So if n = 2k , k = 0, 1, 2, ..
I2k(µ) =
(− ∂
∂µ
)kI0(µ) = (−1)k
∂k
∂µkI0(µ) (22)
and if we substitute (17) into the last expression we arrive at
I2k(µ) =√π(−1)k
∂k
∂µkµ−
12 . (23)
The case if n = 2k+ 1 does not have to be evaluated since the function we are inte-grating becomes an odd function and the result will always be zero if we integrateover R.
We try the cases k = 1 and k = 2
I2(µ) =√π(−1)
∂
∂µµ−
12 =√π(−1)
(−1
2
)µ−
32 =
1
α
√2π
α=
1
αZ[0] (24)
I4(µ) =√π(−1)2 ∂
2
∂µ2µ−
12 = −
√π
2
∂
∂µµ−
32 =
3√π
4µ−
52 =
3
α2
√2π
α=
3
α2Z[0] (25)
If we analyze the two expressions we can think of a pattern that is going to repeatitself. Each time we are taking derivatives with respect to µ we are getting an oddfactor in front of the original expression and the difference between a factor and theone before will always equal two. This is not a coincidence since the exponent islowered by 2
2for each derivative. So if we take k derivatives, the factor we will get
in front of the expression is (2k − 1)!!. If we chose k = 1, the power of α became−1. If we chose k = 2, the power of α became −2. Also, we always end upp withthe Gaussian integral as a factor. From these realizations we can state that∫
R
dx x2ke−α2x2
=(2k − 1)!!
αkZ[0]. (26)
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2 Generating functions and Feynman diagrams
2.1 Z[J] - A variable generator
In Rn, we now want to look at the integral in equation (12), in a slight differentway. We add the linear term J(x) = Jix
i, also known as a source, to the exponentand let us define Z[J ], the variable generating function such as
Z[J ] =
∫Rn
dnx e−α2Aijx
ixj+Jixi
= Z[0]e1
2αJiA
ijJj . (27)
Proof:
The approach to this proof is to use the method of completing the square. Westart with the expression(
xi − 1
αAikJk
)Aij
(xj − 1
αAjlJl
)= xiAijx
j − 1
αxiAijA
jlJl + (28)
+ − 1
αAikAijx
jJk +1
α2AikAijA
jlJkJl
where AµκAκη = δµη. Now let k = j, l = i and we obtain
(xi − 1
αAikJk
)Aij
(xj − 1
αAjlJl
)= xiAijx
j − 2
αxiJi +
1
α2JjA
ijJj. (29)
If we now look at the terms in the exponent of (27) and express xiAijxj in the terms
from (29) we get
−α2Aijx
ixj + Jixi = −α
2
(Aijx
ixj − 2
αJix
i
)=
= −α2
((xi − 1
αAikJk
)Aij
(xj − 1
αAjlJl
)− 1
α2JiA
ijJj
)(30)
If we let
yi = xi − 1
αAikJk
and substitute for y in (30) we arrive at∫Rn
dny e−α2yiAijy
j+ 12αJiA
ijJj = e1
2αJiA
ijJjZ[0] = Z[J ] �. (31)
We will now see that the function we just derived and proved is very useful to usein the process to evaluate the next introduced integral
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〈xνxτ 〉 =1
Z[0]
∫Rn
dnx xνxτe−α2xiAijx
j
=1
Z[0]
∂2
∂Jν∂JτZ[J ]
∣∣∣∣J=0
. (32)
Performing the derivatives explicitly with Z[J ] substituted for the expression in (31)(remember that the matrix A is a symmetric matrix so Aij = Aji, and again we areusing the Einstein summation convention, so when taking derivatives we have toremember that the sum gives us two terms, one for each index).
∂2
∂Jν∂Jτe
12αJνAντJτ
∣∣∣∣J=0
=∂
∂Jτ
1
2αJτ (Aντ + Aτν) e
12αJνAντJτ
∣∣∣∣J=0
=
=1
αAντe
12αJνAντJτ +
1
αJνJτA
ντe1
2αJνAντJτ
∣∣∣∣J=0
=1
αAντ (33)
Now let us look at the integral
〈xkxsxpxn〉 =1
Z[0]
∫Rn
dnx xkxsxpxne−α2xiAijx
j
=1
Z[0]
∂4
∂Jk∂Js∂Jp∂JnZ[J ]
∣∣∣∣J=0
(34)
the approach to evaluate this is the same as in the previous example with the integral〈xνxτ 〉. We start by taking the derivatives
∂4
∂Js∂Jk∂Jp∂Jne
12αJiA
ijJj =∂3
∂Js∂Jk∂Jp
1
α
(JpA
np + JsAns + JkA
nk)e
12αJiA
ijJj =
∂2
∂Js∂Jk
(1
αAnp +
1
α2
(JpA
np + JsAns + JkA
nk) (JsA
ps + JksApk + JnA
pn))
e1
2αJiA
ijJj
=∂
∂Js
(1
α2Anp(JsA
ks + · · · ) +1
α2(JsA
nkAps + JsAnsApk + · · · ) +
1
α3(· · · )(· · · )
)×
× e1
2αJiA
ijJj . (35)
The terms that are being denoted by · · · , are terms that has no factor of Js, andtherefore they are not excplicitly expressed because of the fact that they will notsurvive the derivative with respect to Js. However, the terms in the last parentheseswill have factors of Js, but at least one parenthese will have all terms with somefactor of J , and when we evaluate the expression with J = 0, it will vanish and weend up with
〈xkxsxpxn〉 =1
α2
(AnpAks + AnkAps + AnsApk
). (36)
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2.2 The Wick theorem and Feynman diagrams
The expression in (36) can be written in another form with the now introduced Wicktheorem:
〈xkxsxpxn〉 =1
n!2nαn
∑P
AkP sPApPnP (37)
where we sum over all the permutations of the indices k, s, n, p(the n in the last x-term is not the same n as in the weight factor). The new factor in front of the sumis a weight factor with some symmetry properties due to the Feynman graphs. Welook at the terms xk, xs, xp, xn as points. Points that has the properties that theyonly have one so called leg to be connected with to another point. If we connect thepoint xk to xs we have to connect xp with xn because there are no other connectionsavailable.
Figure 2.2.1: Feynman diagrams to 〈xkxsxpxn〉.
The weight factor in our integral in (36) clearly has n = 2 since it is the power ofα. If n = 2 we also get the factor 8 in the denominator in (37), and this due to thesymmetry properties that we will further discuss after the next graph.
If we merge two points together so that we get a vertex with two allowed waysof connections, we can represent the same calculations but in another pictorial way.Let the points k and s merge together and create the vertex point q and in the sameway with p and n create the vertex point z.
Figure 2.2.2: Feynman diagram to the two two-vertex points qand z with connections k, s and p, n respectively.
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After these two representations we can start to evaluate the symmetry properties abit further;
1. If there are n legs running between some points or a pair of vertices we haveto include the factor of n! for each such contribution from a pair of vertices orpoints. This is related to the number of ways to contract the indices in pairs.
2. In case of identical vertices which means that we can flip the graph withoutchanging it we include the factor of 2n where n is the number of ways to flipthe graph.
3. In case of vertices with the possibility of loop connections we get a contributionwith a factor of 2 for each such loop.
Now when we have discussed the symmetry properties in a more general way we canclarify them with an example and this will also relate to the Wick theorem as wellas the Feynman diagrams
Example 2.2.1
Consider the two two-point vertices x2 and x2. Due to our formula in (26) aswe derived and applying the Wick theorem the integral 〈x2x2〉 is
〈x2x2〉 =1
2!2!2
1
Z[0]
∫R
dx x4e−12αx2
=1
2!2!2
3!!
α2(38)
where we also have the symmetry factors to the graphs in figure 2.2.3 in an explicitform
〈x2x2〉 =1
2!2!2(1 + 2)
1
α2=
(1
8+
1
4
)1
α2(39)
Figure 2.2.3: Feynman diagrams to 〈x2x2〉 withsymmetry factors 4 and 8 respectively.
End of example.
We have now looked through a couple of examples of contracting legs from dif-ferent kinds of vertices. Now we will move over to Rn and discuss and develop someuseful settings of the Wick theorem and also check them with some examples. The
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matrix Aij has not been included much in the discussion of the graphs that we haveheld. If we look at the Wick theorem we will reformulate the formula in (37) a bitso that it will become easier to understand what the theroem actually means.
The integrals that we have looked at in the form of 〈xkxsxpxn〉, can be writtenas
〈xν1xν1 · · ·xν2k−1xν2k〉 = 〈xν1xν2〉 · · · 〈xνl−1xνl〉 · · · 〈xν2k−1xν2k〉
where we have 2k points, and it is convenient to represent the l-th point as xνl .
Example 2.2.2
The integral
〈xν1xν2xν3xν4〉 =1
Z[0]
∫Rn
dnx xν1xν2xν3xν4e−α2xiAijx
j
(40)
can be calculated very quickly after using the above statement and the Wick theorem.With information from earlier calcualtions we can quickly state that
〈xν1xν2〉〈xν3xν4〉 =1
222!α2
∑P
Aν1Pν2PAν3P
ν4P (41)
and after calculating all the permutations of the indices ν1, ν2, ν3, ν4 and having inmind the properties of the positive definite and symmetric matrix A, we obtain
〈xν1xν2xν3xν4〉 = 〈xν1xν2〉〈xν3xν4〉 =1
α2(Aν1ν2Aν3ν4 + Aν1ν3Aν2ν4 + Aν1ν4Aν2ν3)
(42)
which is the exact same answer as in (36). With this calculation we showed thepower of the Wick theorem and how it saves one a lot of time and effort instead ofdoing the messy, but straight forward, calculation with the function Z[J ] as in (34).
End of example.
In a more physical interpretation the A’s are called propagators. If we now in-troduce the n-valent vertices as monomials in x. A vertex with n entries can beconnected to another vertex with k entries with the propagators A. The A’s con-tract one index with a index from the other vertex. The main concern about thevertices is what the automorphisms of these graphs are.
Let us introduce the monomials in x as
Vn(x) =1
n!Vν1···νn1
xν1 · · ·xνn (43)
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The formula for the computation of the now introduced vertices is given by
〈Vn1(x) · · ·Vnk(x)〉 =1
n1! · · ·nk!Vν1···νn1Vσ1···σnk 〈x
ν1 · · ·xσnk 〉 (44)
and the general form of the integral for these calculations os expressed as
〈Vn1(x) · · ·Vnk(x)〉 =1
Z[0]
∫Rn
dnx Vn1(x) · · ·Vnk(x)e−α2xiAijx
j
. (45)
Before we work out a few examples let us discuss the symmetry factor a bit further.Earlier there was a brief numbering of the three symmetry properties that thesegraphs have. When we now talk about these n-valent vertices Vn(x), we will widenthese symmetry properties to the automorphism of the graphs (from now on we willdenote a graph with Γ).
There have been examples where we have repeated graphs in the summation ofall the permutations of the indices due to the properties of the propagator A. If wehave two vertices with each three entries, it is easy to see that we can connect theseentries in ((3 + 3) − 1)!! number of ways. But with the repeatedness of the graphswe have to include the autmorphism of a graph Γ, |AutΓ|.
The way to evaluate |AutΓ| is just as before, simply by looking at the graph andtry to see the symmetry properties numbered in this section, so the expression forthe automorphism is
|AutΓ| = (Loop factor)× (Vertice symmetry factor)× (Edge factor).
Now we arrive to the formula of the calculations of these graphs with the includedsymmetry factor
〈Vn1(x) · · ·Vnk(x)〉 =1
α(n1+···+nk) 12
∑Γ
1
|AutΓ|W (Γ) (46)
where we now are summing over all the graphs Γ that are being generated by thevertices and the contracting of the indices with the propagators A.
Example 2.2.3
We have the vertices V4(x) and V2(x), the first vertex has four entries and the secondvertex has two. Let us perform the calculations we have discussed above using thesetwo vertices
〈V4(x)V2(x)〉 =1
Z[0]4!2!
∫Rn
dnxVν1ν2ν3ν4Vσ1σ2xν1xν2xν3xν4xσ1xσ2 (47)
Using the Wick theorem we obtain
〈V4(x)V2(x)〉 =1
4!2!α3(12W (Γ1) + 3W (Γ2)) =
1
α3
(1
4W (Γ1) +
1
16W (Γ2)
). (48)
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By some fairly simple combinatorics the expected number of graphs is 15. This alsocoincides with the expected value ((4+2)-1)!!. One can come to the realization thatwith these two vertices there are two graphs that can be constructed, also shownfrom the result above, but only Γ1 will have any purpose to show because Γ2 is a socalled vacuum graph with really no results of happenings, just as the second graph inFigure 2.2.3.
Figure 2.2.4: Graph to 〈V4(x)V2(x)〉 with symmetry factor 4.
This result also coincides with the general formula in (46). One can easy see thatthere is always one loop in the graph and two edges running between the vertices,therefore, |AutΓ| = 4, and with that (46) would exactly have the same result as (48).
End of example.
3 Amplitudes and paths
In previous chapters we have been presenting a lot of mathematics without so muchof physical interpretations. This is not without thought because the mathemat-ics above is very useful in the underlying understanding of physical events in bothclassical and quantum mechanics. So in the coming section a lot more physics andthought experiments will be presented.
A nice story
In the beginning of the 20th century, a professor had a lecture in quantum me-chanics about amplitudes for an electron that is fired from a source S at a screen
Figure 3.1: A schematic picture of the thought experiment.
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A with two holes in it, and detected at another screen B behind A. The professorsaid that the amplitude of a particle to be emitted from the source S and then travelthrough a hole in the screen and then detected at a position x in a time t at screenB is concluded by a postulate of quantum mechanics, the superposition principle.He claimed that the amplitude is the sum of the amplitudes that the particle goesthrough from S to hole one, and then to B, and the amplitude for the particle togo from S to hole number two, and then to B. A student raised his hand and askedthe preofessor:
”What if we put in another screen with two holes in it, what is then the proba-bility for the particle to arrive at position x at the time t?”
The professor looked at the student and replied that the probability is then thesum over all amplitudes for each path that the particle can take. Just as the pro-fessor was about to start, the very same student interrupts him and says:
”But if we then put in another screen, and another after that and so on, and alsodrills an infinite ammount of holes in the screens so that the screens are barely there.What is the amplitude then?”
The professor started to loose his patience with the student, sighed heavily andreplied that he thought it was obvious that you just sum over all the holes andpaths. He proceeded the lecture in a hurry to cover the rest of the material.
End of story
The story tells us that is not restricted to a finite set of examples of the exper-iment with a number of screens and holes. Because that the fact if we drill aninfinite amount holes in an infinite amount of screens, the screens and holes wouldcover the whole space between the source and the detector, and a new formulationwas born1.
3.1 Two-slit experiment and amplitudes
So what was it that the student had in mind? He thought of a new way to describethe transition amplitudes for a particle, and this time, just moving in free spacefrom one point to another. The amplitude for a certain path has to be determined,and then sum over all the amplitudes to obtain the total amplitude. As we know, weget different kinds of patterns on the detecting screen in the double slit experiment,i.e. if we have one hole open, both holes open or measure through which hole theparticle passes. We introduce the complex numbers to represent the amplitude totake a path through hole one in the screen A as φ1. The amplitude for the otherpath through hole number two is then φ2. Recall the thought experiment that the
1The formulation is called the path integral and we will come backto this later in the chapter of path integrals in quantum mechanincs.
14
professor lectured about2, if we sum the amplitudes, we get the total amplitde, thusφ = φ1 + φ2. Hence one expects that the probability that a particle arrives at apoint x at the screen B is the sum of the probability for each path, i.e P = P1 +P2,where we now represent the probability pf passing through hole one with hole twoclosed, and vice versa, as
P1 = |φ1|2 , P2 = |φ2|2. (49)
By closing one hole, and calculate the probability for the particle to arrive at x whenit takes this particular path, and vice versa for the other one. One expects that thetotal probability should be
P = P1 + P2 = |φ|2 (50)
when both of the holes are open. By direct realizations from the experiment weimmediately obtain that
P = P1 + P2 6= |φ1 + φ2|2. (51)
When both holes are open we obtain the familiar probability distribution of theinterference phenomena from particles that have wave-like behaviour. But in someway, we destroy this feature by knowing which hole the particles passes through.One can perform the same experiment with two holes open, but now have lightemitted from a source behind the screen A so that light is scattered behind hole oneor two. This also interrupts the behaviour of the particles thus the result would notcoincide with the probability distribution in figure a, thus the distribution is alteredby knowing through which hole the particle passes.
Figure 3.1.1: Schematic pictures (a, b and c) of the three discussed variations ofthe experiment with P = |φ|2, P1 = |φ1|2 and P2 = |φ2|2 respectively.
From elementary probability theory, if two outcomes are independent, the sum ofthe both probabilities is the total probability. Yet, we conclude that this is false inthis case. It can also easily be seen by the probability graphs P (x), sum b and cand it would look like a bell curve, which is no way near the outcome in a3. Thus,
2It is a thought experiment because this experimental setup is extremelysimplified. Experiments like these demands a lot of advanced hardware.
3The graphs is far from exact. It is just a pictorial way to show how theprobabilities distribute
15
if we make an observation of the particle to determine which hole it passes through,we exclude important information of the system. If we do not look, we cannot saywhich hole the particle passed through, but it sure did pass through one of them,we just do not know through which. When both holes are open, the two alternativepaths are not exclusive alternatives, and the laws of probability has to change inorder to coincide with the reality. These are now interfering alternatives and arefundamental in quantum mechanics.
3.2 Actions and Kernels
Now the discussed amplitudes needs some more formal and mathematical discussionsto be fully defined and understood. We now constrain ourselves to one dimension.The particles trajectory can be described as at the time t1 the particles position isgiven by x(t1) = x1. The function x(t) therefore generates the trajectory of the par-ticles path. In classical mechanics the constraint is that there is only one trajectoryfrom x1 to x2, the so called classical trajectory.
The principle of least action is a way to describe a particular path out all possi-ble paths. The classical path x will be a minimum and the quantity S will be aextremum. The trick is to vary the path x by an ammount of δx but still have theendpoints fixed. S is given by
S =
t2∫t1
L(x, x, t)dt (52)
where L is the Lagrangian of the system. Fixing the ends gives the conditions
δx(t1) = 0 , δx(t2) = 0 (53)
and the condition of S to be an extremum we interpret as
δS = S(x+ δx)− S(x) = 0. (54)
By means of expressing S(x+ δx) from (52) we obatin
S(x+ δx) =
t2∫t1
L(x+ δx, x+ δx, t)dt =
t2∫t1
(L(x, x, t) + δx
∂L
∂x+ δx
∂L
∂x
)dt =
= S(x) +
t2∫t1
(δx∂L
∂x+ δx
∂L
∂x
)dt (55)
integrating the first term in the integral above by parts gives
S(x+ δx) = S(x) + δx∂L
∂x
∣∣∣∣t2t1
−t2∫t1
δx
(d
dt
∂L
∂x− ∂L
∂x
)dt (56)
16
inserting this result in (54) we obtain
δS = δx∂L
∂x
∣∣∣∣t2t1
−t2∫t1
δx
(d
dt
∂L
∂x− ∂L
∂x
)dt = 0 (57)
the firs term on the right hand side is equal to zero by the conditions in (53), hencewe obtain
d
dt
∂L
∂x− ∂L
∂x= 0 (58)
this equation is the Euler-Lagrange equation of classical motion and is one of themost fundamental equations in classical mechanics.
Example 3.2.14
For a free particle the Lagrangian is L = mx2
2. We put this Lagrangian into the
action S and get
S =
t2∫t1
L(x)dt =
t2∫t1
mx2
2dt
the velocity of the particle is constant in time and by replacing x = x2−x1
t2−t1 we obtainthe minimal action on classical solutions as
S =m
2
(x2 − x1
t2 − t1
)2t2∫t1
dt =m
2
(x2 − x1)2
t2 − t1.
End of example.
To every path a particle can take from two points there is an amplitude. Thisis often called the Kernel. The Kernel will be the sum over all trajectories from thepoint x1 to x2 and is written K(2, 1). The amplitude is a sum of all the contributionsφ(x(t)) from each path,
K(b, a) =∑
all paths from a to b
φ(x(t)). (59)
We must take into account that is not only the extreme path which goes from a tob that contributes to the amplitude, every possible path contributes. Actually, theycontribute equal amounts to the amplitude, but does so at different phases. Thephase of the contribution is proportional to the action S in units of ~
φ(x(t)) = const× ei~S(x(t)). (60)
4This is actually problem 2-1 from R.P.Feynman, A.R.Hibbs,”Quantum mechanics and path integrals”.
17
At a first glance it is not clear how one of the paths becomes the most important,since they all contribute with an equal amount. In the classical limit, i.e. S becomesvery big in comparison to ~, therefore the phase-contribution from is a very largeangle. When the angle becomes very big, the sine or the cosine becomes as likelyto be a plus as a minus sign. When alternating the path with δx which is small inthe classical scale, the change in the action is also small in the classical scale. Butmeasured in units of ~ the change is not as small any more. The change in paths willgenerate an enormous change in phase and the sines and cosines will oscillate veryfrequently, and therefore the net contribution with respect to quantum mechanicswill be zero since the oscillating functions vary between plus and minus in an equalamount. This leads to that a small change in x(t) when S is an extremum will notchange S, at least not in the first order of correction.
3.3 The Path Integral
Since a path is some way defined by the two endpoints it should be reasonable tothink that the path can be divided in to sub-paths. If we have the endpoints xaand xb then define a point xc lying on the path between these two. The techniqueof summing the action holds, e.g.
S(b, a) = S(b, c) + S(c, a). (61)
The Kernel is proportional to exp( i~S(b, a)), substituting (61) gives the rule of mul-tiplication of Kernels
K(b, a) ∼∫xc
ei~ (S(b,c)+S(c,a))dxc =
∫xc
ei~S(b,c)e
i~S(c,a)dxc =
∫xc
K(b, c)K(c, a)dxc.
(62)To construct a path from xa to xb, consider the time interval between these twopoints as tb − ta. By defining the whole interval as Nε = tb − ta, where ε is a smalltime interval, i.e. ε = tj+1 − tj. This procedure will generate a multiple integralover all paths5 due to all values of xj between 1 and N −1 steps(since the endpointsxb = xN and xa = x0 are fixed there are no integrations over these coordinates).The resulting expression for the Kernel is then obtained as
K(b, a) ∼∫ ∫
· · ·∫φ(x(t))dx1dx2 · · · dxN−1. (63)
It can now be tempting to introduce a limit where we make ε smaller, but actually,the limit does not yet exist. A normalizing factor which depends on ε must beintroduced to make the the whole thing converge. We then can write (63) as
K(b, a) = limε→0
1
A
∫ ∫· · ·∫
ei~S(b,a)dx1
A
dx2
A· · · dxN−1
A. (64)
5For a more detailed discussion on how to construct the sum and the paths see R.P.Feynman,A.R.Hibbs, ”Quantum mechanics and path integrals”, McGraw-Hill, New York, 1965.
18
Feynman introduced a convenient way to write this whole expression in a morecompact way than in (64), it is often written as
K(b, a) =
∫ xb
xa
Dx(t)ei~S(b,a) (65)
where the D implies a product of the integrals and all paths. In quantum mechanics,the wave function ψ(x, t) is an amplitude and in this field it is more useful to usethis notation. We can write a wave function ψ(x2, t2) by using the definitions of theKernels stated above
ψ(x2, t2) =
∫ ∞−∞
K(x2, t2;x1, t1)ψ(x1, t1)dx1, (66)
this expression makes perfect sense in the physical interpretation. The amplitudeto arrive at the point (x2, t2) is the sum of all possible values of x1 of the amplitudeto be at x1, the time has of course to fixed to t1, multiplied by the amplitude to gofrom x1 to x2. In this case the Kernel.
While we are at the topic of quantum mechanics, Paul Dirac actually made thisvery same interpretation of the path-integral earlier than Feynman. Using theDirac bracket notation we will now obtain the same result as in (65) in the quan-tum mechanical regime. By denoting a particles position in the position space as|qi, ti〉H = |qi〉H in the Heisenberg picture. The amplitude of finding a particle atposition |qj〉 (tj > ti), is denoted by the inner product
〈qj|qi〉H = 〈qi|e−i~ H(tj−ti)|qi〉S (67)
where we have used the time-evolution operator to define the jth state as |qj〉 =
e−i~ H(tj−ti)|qi〉S. From here we are going to denote the initial state at time ti = t0 = 0
as |qi〉 and the final state at time tf as |qf〉. Define the time-interval between finaland initial state as τ . We now divide the interval in N infinitesimal intervals ∆t,such that N∆t = τ . Now we can write the amplitude as
〈qf |qf〉H = 〈qf |e−i~ Hτ |qi〉S = 〈qf |e−
i~ H∆te−
i~ H∆t · · · e−
i~ H∆t|qi〉S. (68)
We have constructed our states such that they are normalized, i.e. 〈qj|qj〉 = 1 and〈qj+1|qj〉 = δ(qj+1 − qj). Recall the unit operator in position-space∫
dq |q〉〈q| =
(N−1∏k=1
∫dqk
)|qN−1〉〈qN−1| · · · |q1〉〈q1| = 1, (69)
the unit operator in momentum-space is similar, and recalling that 〈p|q〉 = 1√2e−ipx,
which means that the momentum state projected on the coordinate space is a planewave, we get ∫
dp〈qj+1|p〉〈p|qj〉 =
∫dp
2πei~p(qj+1−qj) = δ(qj+1 − qj) (70)
19
After inserting (69) in (68) we take out one of the terms out and start focusing on
rewriting the expression with a Hamiltonian H = p2
2m+ V (q)∫
dp
2π〈qk+1|e−
i~p2
2m∆t|p〉〈p|e−
i~V (q)∆t|qk〉, (71)
since the momentum and the potential operators are acting on eigenstates, we cansimply remove the operator-hat and insert their respective eigenvalue
e−i~V (qk)∆t
∫dp
2πe−
i~p2
2m∆t〈qk+1|p〉〈p|qk〉 = e−
i~V (qk)∆t
∫dp
2πe− i
~
(p2
2m∆t−p(qk+1−qk)
),
(72)by completing the square in the exponent as
p2
2m∆t− p(qk+1 − qk) =
∆t
2m
((p− m
∆t(qk+1 − qk)
)2
− m2
∆t2(qk+1 − qk)2
), (73)
we write the whole thing as
ei~( m
2∆t(qk+1−qk)2−V (qk)∆t)
∫dp
2πe−
i∆t2m~(p− m
∆t(qk+1−qk))
2
, (74)
the integral over p is a shifted Gaussian. Solving the integral we obtain(−im
2π~∆t
) 12
ei~( m
2∆t(qk+1−qk)2−V (qk)∆t). (75)
Plugging this solution in (68) combined with (69), and moving over to the continuumlimit where ∆t→ 0 and therefore N →∞, i.e.
N−1∑k=0
i
~∆t
(m
2
(qk+1 − qk
∆t
)2
− V (qk)
)→ i
~
τ∫0
dt(m
2q2 − V (q)
), (76)
thus the final expression is
〈qf |e−i~ Hτ |qi〉 =
( m
i2π~∆t
)N2
(N−1∏j=1
∫dqj
)ei~
τ∫0
dt (m2 q2−V (q)). (77)
We now see that the Lagrangian has appeared in the exponent, and that the totalintegral in the exponent is the action, which is in agreement with (65). The constantin front of the integral and the product of integrals is the same as in the notation∫Dx(t), where we have all paths that is determined by all the trajectory producing
functions q(ti) from the initial point to the final point, corresponding to the stateevolution in the Quantum Mechanical interpretation.
20
3.4 From Feynmans Path Integral formalism to theSchrodinger equation
We look at the small time interval ε between t2 and t1. In first order of correctionin ε the action is the short time interval ε times the Lagrangian for the system. Weare looking at a system which have the Lagrangian L = 1
2mx2 − V (x, t). We have
ψ(x, t+ ε) =
∞∫−∞
dx1
Aexp
[εi
~L
(x2 − x1
ε,x1 + x2
2
)]ψ(x1, t)
=
∞∫−∞
dx1
A
[exp
(i
~m(x2 − x1)2
ε
)]×[exp
(− i~εV
(x1 + x2
2, εt
))]ψ(x1, t), (78)
only if x1 is near x2 we get important contributions to the amplitude. When x1
is appreciably different from x2 the exponent in the first factor becomes large andoscillates very rapidly. So, we make the substitution x1 = x2 + ς. We then obtain
ψ(x, t+ ε) =
∞∫−∞
dς1
Aexp
(imς2
2~ε
)exp
[−iε~V
(x+ ς
2, t
)]ψ(x2 + ς, t), (79)
we can now expand the expression to the first order in ε. Since we made a substi-tution with the small parameter ς in the process to the solution we must expand tothe second order in ς. Also the term V
(x+ς
2, t)
can be replaced by εV (x, t) due tothe error which will be of order higher than ε. Expanding and we obtain
ψ(x, t) + ε∂ψ
∂t=
∞∫−∞
dς1
Aexp
(imς2
2~ε
)×[1− iε
~V (x, t)
]
×[ψ(x, t) + ς
∂ψ
∂x+
1
2ς2∂
2ψ
∂x2
], (80)
equating the first term from the left hand side with the first term in the integralabove yields
ψ(x, t) =
∞∫−∞
dς1
Aexp
(imς2
2~ε
)ψ(x, t), (81)
if both sides agree in the limit, then we are free to choose the constant A such thatthe integral in (81) is equal to one
⇒∞∫
−∞
dς1
Aexp
(imς2
2~ε
)= 1⇔ A =
∞∫−∞
dς exp
(imς2
2~ε
)=
(2πi~εm
) 12
, (82)
and look, we now actually determined the constant A that was former discussedwhen we introduced it in (64). We look at the right hand side of (69), writing the
21
terms out with the determined coefficient A (take a look back at chapter 1.2 for arecap of the evaluations of the integrals6) we have
ψ(x, t) + ε∂ψ
∂t= ψ(x, t) +
ε
i~V (x, t)ψ(x, t) +
i~ε2m
∂2ψ
∂x2
⇔ i~ε∂ψ
∂t= −ε ~
2
2m
∂2ψ
∂x2+ εV (x, t)ψ(x, t), (83)
this expression will be true to the order of ε iff ψ(x, t) satisfies
i~∂ψ
∂t=
(− ~2
2m
∂2
∂x2+ V (x, t)
)ψ(x, t), (84)
and we have obtained the Schrodinger equation from the Path-Integral formalism.
4 Units and notations
In almost all modern theories the constants c- the speed of light in vacuum, and~-Planck’s modified constant(or the Dirac symbol) shows up very frequently andphysicist developed a “new language” to simplify and make algebraic expressionslook more elegant. Therefore we will now introduce the natural units
c = ~ = 1. (85)
As a remark one also have the Boltzmann constant kB as a natural unit in particlephysics, but this constant will not show up anywhere in this paper. If one expressesa speed, let us say v = 4
5, this is directly understood such as the speed is measured
in fractions of the speed of light, c. Earlier in the paper we discussed different sizesof actions. As one can see, the actions is measured in units of ~ hence the ~ in thedenominator of the exponent.
4.1 Space and products
A four-vector in space-time is denoted as7
xµ = (x0, x1, x2, x3) , µ = 0, 1, 2, 3 (86)
where x0 is the time-component and often written as ct hence it has the dimensionof length. The last three components is often denoted as xi where i = 1, 2, 3. Thesecomponents are the regular spacial components. A scalar product in space-time isdefined by
aµbµ = gµνa
νbµ, (87)
6For more discussion about these kind of integrals see Appendix 2 in A.Zee, “Quantum fieldtheory in a nutshell”, Princeton University Press 41, 2003.
7See W.Rindler, “Introduction to Special Relativity”, chapter 7, for very thorough discussionsabout four-vectors and space-time properties.
22
where we have introduced the 4× 4 matrix, the metric tensor
gµν = diag(−1, 1, 1, 1)µν . (88)
The scalar product can be performed by the method of “contracting indices”, andsince gµν is diagonal, µ = ν will give the only non-zero terms in the product. Hence
aµbµ = gµνa
νbµ = −a0b0 + a1b1 + a2b2 + a3b3. (89)
4.2 Integrals
Integrals will appear in almost every calculation from now on, and they will bemore than one dimension in most cases, therefore it is convenient to write a three-dimensional integral as ∫∫∫
dxdydz :=
∫d3x, (90)
and the four-dimensional(space-time) version∫dt
∫∫∫dxdydz :=
∫d4x. (91)
4.3 Differentials
In the Lagrangians of fields, the four-vector of differential operators ∂µ, shows upvery frequently. There are different ways to write ou this four-vector of operatorsbut in this paper we will stick to
∂µ :=∂
∂xµ=
(− ∂
∂t, ~∇)
, ∂µ :=∂
∂xµ=
(∂
∂t, ~∇)
(92)
and the squared operator
∂2 = ∂µ∂µ = gµν∂
ν∂µ = − ∂2
∂t2+ ~∇2. (93)
The functional derivative generates a delta function via
δf(~y)
δf(~x)= δ3(~y − ~x)
even= δ3(~x− ~y) (94)
4.4 Dirac delta function
The equations of motion for the fields is very fundamental in how to know how thefields works and how(if they do) interact with other fields. In the derivation oneuses the Dirac delta function which simplifies integrals over many dimensions a lot.The function has the properties
δ3(~x− ~y) =
{0 , ~x 6= ~y
∞, ~x = ~y, (95)
23
and ∫d3x δ3(~x− ~y) =
{0, ~x 6= ~y
1, ~x = ~y. (96)
With these new notations, tools and units we are ready to look into some fieldtheory.
5 Approaching fields
In “Quantum Field Theory in a nutshell”, A.Zee, there is a beautiful descriptionof how one can start constructing a field of a plane with harmonic oscillators ondiscrete space-time points, and then take the limit of the distance between them tozero. The mattress is a rather naive way to describe field theory. But it is a goodexample to start with to get some intuitive feeling of a field theory. The example isa toy-model and is not associated with any real physical fields.8
5.1 From mechanics to fields
The Lagrangian for n-coupled harmonic oscillators in 1-dimension at a distance lfrom each other is
L =n∑i=1
(1
2Al
(∂qi∂t
)2
− 1
2Alm2q2
i −1
2A
(qi+1 − qi)2
l
), (97)
with some constant A. Considering the limit l→ 0⇔ n→∞, and also introducingthe field ϕ =
√Aq, one obtains
L =
∫dx
(1
2
(∂ϕ
∂t
)2
− 1
2m2ϕ2 − 1
2
(∂ϕ
∂x
)2)
=
∫dxL, (98)
we have obtained the Lagrangian density L as a functional of the field ϕ(x, t). Theexact same reasoning as above and a bit more work with the analysis one can quiteeasily obtain the analogous expression for a field in 3-dimensions, ϕ(~x, t). A veryimportant thing to stress is that the argument ~x is not a dynamical variable, it isjust a label(it works in the same way as the index i did in the sum for the oscillators)to keep track on which field we are talking about, e.g. ϕ( ~x1, t) is the field ϕ1 at thespace-time point ( ~x1, t). A general space-time point like this is often just written asx1, e.g. ϕ( ~x1, t)→ ϕ(x1), in the Heisenberg picture.
5.2 Equation of motion for free scalar fields
The action for a 3-dimensional free scalar field is
S(ϕ(x)) =
∫d4xL(ϕ(x)) =
∫d4x
(−1
2∂µϕ∂
µϕ− 1
2m2ϕ2
). (99)
8See page 4 in “Quantum Field Theory in a nutshell”, A.Zee, to read more about on how toconstruct the mattress and how to find its Lagrangian.
24
From the principle of least action, S should stay unchanged with respect to S →S + δS. This requires that δS = 0 if we vary the Lagrangian with respect to ϕ.Hence L(∂µϕ, ϕ) → L(∂µϕ + ∂µδϕ, ϕ + δϕ) ≈ L(∂µϕ, ϕ) + δL(∂µϕ, ϕ). The firstterm is simply the normal action and inserting this in the variation gives us thecondition δS = 0. We have
δS =
∫d4xδL chain rule
=
∫d4x
(∂L∂ϕ
δϕ(x) +∂L
∂(∂µϕ)∂µδϕ(x)
), (100)
varying with respect to ϕ yields
δS
δϕ(y)=
∫d4x
(∂L∂ϕ
δ4(x− y) +∂L
∂(∂µϕ)∂µδ
4(x− y)
), (101)
we have to use partial integration on the last term to move the derivative so we canuse the properties of the delta function properly, after partial integration we obtain
δS
δϕ(y)=
∂L∂(∂µϕ)
δ4(x− y)
∣∣∣∣Boundary
+
∫d4x
(∂L∂ϕ
δ4(x− y)− ∂µ∂L
∂(∂µϕ)δ4(x− y)
),
(102)the first term on the right hand side will vanish. The switch between the variables~x and ~y does not matter at all, we just need a dummy variable to integrate over andthen we can relabel it back to ~x. We can now perform the integration easily withthe delta functions and we obtain
δS
δϕ=∂L∂ϕ− ∂µ
∂L∂(∂µϕ)
= 0. (103)
With the Lagrangian that we have in (99)
−m2ϕ(x) + ∂µ∂µϕ(x) = 0⇔ (−∂µ∂µ +m2)ϕ(x) = 0, (104)
equation (104) is the famous Klein-Gordon equation. The equation is the equationof motion for a free scalar field. It is important to stress that this is the equationof a real scalar field, and has nothing to do with a quantized field. Thus there arenot any factors of ~ in this equation hence the parameter m cannot be thought ofas the mass of the field. The general solution9 to the Klein-Gordon equation is
ϕ(x) =
∫d3k
(2π)32ω
(a(~k)eikx + a∗(~k)e−ikx
)(105)
for a real physical scalar field and the constants in the denominator are chosen byconvenience. a(~k) and its corresponding conjugate are some arbitrary functions of
the vector ~k. And
x = (t, ~x) , k = (ω,~k) , kx = kµxµ = −ωt+ ~k · ~x. (106)
9In “Quantum Field Theory”, M.Srednicki derives the properties of the field and the constraintsit must have to be real in p.24-26.
25
5.3 Quantization and the free-field path integral
By means of looking back to some Quantum Mechanics we can develop the pathintegral for our free-field theory by looking at the path integral for the harmonicoscillator. Remember that the free-fields Lagrangian always includes a term thatcorresponds quantum mechanically to a harmonic oscillator potential.
To be thorough, we start by look into the Heisenberg picture of Quantum Me-chanics. The main difference between these two is that in the Schrodinger picturethe state-vectors are time dependent and the operators are not, which means thatthe observables themselves is time dependent. In the Heisenberg picture the statesare time independent but the operators are not. This is obtained by the unitarytransformation(for some time independent Hamiltonian)
QH(t) = eiHtQSe−iHt (107)
for some operator Q. The subscript H denotes the Heisenberg picture and S theSchrodinger picture. A state is represented by
|ψ〉H = eiHt|ψ, t〉S. (108)
If the state-vector in the Heisenberg picture truly is time independent then
∂
∂t|ψ〉H = 0. (109)
Proof:
We begin by taking the derivative from the left on both sides
∂
∂t|ψ〉H =
∂
∂t
(eiHt|ψ, t〉S
)= iHeiHt|ψ, t〉S + eiHt
∂
∂t|ψ, t〉S, (110)
multiplying both sides with i
i× (110) = −HeiHt|ψ, t〉S + eiHti∂
∂t|ψ, t〉S, (111)
identifying the coefficient in front of the last term as the Hamiltonian in the Schrodingerpicture(with ~ = 1), we obtain the commutator
i∂
∂t|ψ〉H = [eiHt, H]|ψ, t〉S
expansion= 0. (112)
�
As a remark, one can obtain the time evolution of the operator QH(t) by taking aderivative of (107) and one will find the Heisenberg equation of motion
∂
∂tQH(t) = i[H, QH(t)]. (113)
26
Heisenberg introduced operators to the position q and momenta p and their corre-sponding operators are Q(t) and P (t) and also demanding the quantizing conditionsfor the two operators
[Q, P ] = i. (114)
Acting on a ket-vector with Q(t)
Q(t)|q〉H = q|q, t〉H ,
which gives us an instantaneous eigenvalue for the operator. The instantaneouseigenstate can be expressed as |qi, ti〉H = eiHti |q, t = 0〉S where the state is timeevolved from t = 0 in the Schrodinger picture.. We can now use these notations towrite a probability amplitude of a particle at position qi at the time ti to be found ata later position qf at time tf . As we did before in the derivation of the path integral,we use the same tools but in the Heisenberg picture the amplitude is written
〈qf , tf |qi, ti〉H = 〈qf |e−iH(tf−ti)|qi〉S. (115)
From here on, we just continue as we did in chapter 3.3 with the harmonic oscillatorpotential10. Why? Because we want to mimic the description of a quantized free-fieldtheory, but first there is one confession to make. In the evaluation of exponents withoperators there was a bit sloppy notation. It is not general true that eA+B = eAeB.In fact, one has to use the Baker-Campbell-Haussdorf formula
eA+B = eAeB × e12
[A,B]+ 12
[A,[A,B]]+···, (116)
but since we are dealing with small intervals of time δt, the δt will appear inhigher powers for each commutator hence it can be justified to use approximationeA+B ≈ eAeB.
Since we are trying the describe the quantized version of the field-theory, we alsohave to have quantization conditions for our fields. The earlier classical scalar ϕ(x)is now being associated to be an operator. The conjugate momenta to ϕ(x) is of-ten denoted as π(x). Mimicing Quantum Mechanics we impose some quantizationconditions
[π(x), ϕ(x′)] = −iδ3(x− x′) , [ϕ(x), ϕ(x′)] = 0 , [π(x), π(x′)] = 0 (117)
With theˆnotation to denote operators and the since we have a continuous indexthe Kronecker delta has been replaced by the Dirac delta function. The solution tothe Klein-Gordon equation is now similar, but we have to promote operators thatcan create and annihilate particles in our space. If one looks at (105), the solutionto the Klein-Gordon equation for the classical scalar field, this turns into
ϕ(x) =
∫d3k
(2π)32ω
(a(~k)eikx + a†(~k)e−ikx
)(118)
10In chapter 7, “Quantum Field Theory”, M.Srednicki, there is a very rigorous derivation of theharmonic oscillator path integral. There are a few more mathematical tricks than the derivationin chapter 3.3 in this theses, but the logic and steps are fairly the same.
27
where a(~k) and a†(~k) are creation and annihilation operators. Expressing the corre-sponding conjugate momenta in terms of these operators, we get the commutationrelation of the operators
[a(~k), a†(~k′)] = (2π3)2ωδ3(~k − ~k′) , [a†(~k), a†(~k′)] = [a(~k), a(~k′)] = 0, (119)
quite similar to the creation and annihilation operators in Quantum Mechanics forthe harmonic oscillator. Now we are starting to get somewhere in our quantizedfree-field theory. We are now allowed to create and annihilate particles at somepositions and this is very useful in the calculations of amplitudes.
Now, the path integral for a quantized free-field theory in the presence of sourcescan be stated
Z0[J ] = 〈0|0〉J =
∫Dϕ exp
(i
∫d4x [L0 + J(x)ϕ(x)]
)= exp
(i
2
∫d4xd4yJ(x)D(x− y)J(y)
), (120)
where D(x − y) is the free-field propagator. As one can see the condition for thevacuum expectation value Z0[0] = 〈0|0〉 evaluated at J = 0 is equal to 1, as it shouldbe. Let us consider a very simple example of how to find an amplitude for a freepropagating particle
Example 5.3.1
We are dealing with free-field theory at the moment, there are no interactions anda propagating particle is not disturbed by anything. Consider a particle that propa-gates in space from a point x1. The amplitude to find the particle at some point x2
is written as
〈x2|x1〉 =
(1
i
)2δ
δJ(x2)
δ
δJ(x1)〈0|T[ϕ(x1)ϕ(x2)]|0〉, (121)
where the T is the time ordering operator which orders operators such as the opera-tors with earliest time argument is to the left and the latest to the right, and in casewith multiple operators they will be ordered from left to right in the same way. Weuse our knowledge of perturbed Gaussian integrals from the first two chapters in thisthesis and obtain
〈0|T[ϕ(x1)ϕ(x2)]|0〉 =
∫Dϕϕ(x1)ϕ(x2) exp
(i
∫d4xL0
)=
1
i
δ
δJ(x1)
1
i
δ
δJ(x2)
∫Dϕ exp
(i
∫d4x [L0 + J(x)ϕ(x)]
)=
1
i
δ
δJ(x1)
1
i
δ
δJ(x2)exp
(i
2
∫d4xd4yJ(x)D(x− y)J(y)
), (122)
28
denote the free-field path integral with Z0[J ] and taking the functional derivatives weobtain
=1
2
1
i
δ
δJ(x1)
[∫d4xd4y δ3(x− x2)D(x− y)J(y) +
∫d4xd4y J(x)D(x− y)δ3(y − x2)
]×Z0[J ] =
1
2i
[∫d4y D(x2 − y)δ3(y − x1) +
∫d4x δ(x− x1)D(x− x2)
]Z0[J ]
=1
2i(D(x2 − x1) +D(x1 − x2))Z0[J ] + (a bunch off terms with J’s)Z0[J ],
as usual in the end, set J = 0 and the last terms with J’s will vanish and we obtain
〈0|T[ϕ(x1)ϕ(x2)]|0〉 =1
iD(x1 − x2) (123)
x1 x2
Figure 5.3.1: Graph to 〈0|T[ϕ(x1)ϕ(x2)]|0〉.
End of example.
We actually start to see why we went through all the mathematics in chapter 1and chapter 2, here comes the pay-off: If we wanted to calculate the amplitudes andthe diagrams for
〈0|T[ϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4)]|0〉, (124)
one can quite quickly see that the way of doing it and by construction of the pathintegral, we would get the exact same graphs that is corresponding the graphs incalculation (36) and the graphs who are displayed in the beginning of chapter 2.1.So, there was a purpose to develop the understanding of a little bit more generalmathematics at first, then switch over and see how the physics is constructed and lastbut not least, i.e. the very goal of everything in this thesis is to see the connection,see the mathematics in use for real physical calculations.
6 Interacting fields and perturbation theory
In the beginning of this paper there are sections were we talked about differentGaussian integrals and how to solve the different types with mathematical tricks andtechniques. The general purpose of those sections was to build up a mathematicalunderstanding of how to generate so called n-point functions and their correspondingdiagrams. In field theory, the perturbation theory from the path-integral approach isthe key to generate Feynman diagrams for different kinds of events, such as scatteringof particles. One have to make a few adjustments on our former development of thegenerating function so that it now is a generating functional and has different kindsof Lagrangians for fields. The following calculations and examples are for interactingscalar fields.
29
6.1 Naive description of the interaction Lagrangian
We begin with the path integral for our free-field theory
Z0[J ] = 〈0|0〉J =
∫Dϕ ei
∫d4x(L0+J(x)ϕ(x)), (125)
where L0 is the free-field Lagrangian density in (99) and J is the source we introducedin chapter 2.1. If we look at the integral as it is and evaluates it with J = 0, itis normalized so it equals 1 with the absorbed constants in the notation Dϕ. Theintegral evaluated with no additional terms in the Lagrangian is being denoted asZ0[J ]. With almost the same methods(but similar logic) one can obtain an analogousexpression for the generating functional11 as we have in chapter 2.1
Z0[J ] = exp
(i
2
∫d4x d4y J(x)∆(x− y)J(y)
), (126)
Earlier we had the matrix Aij that played the role of our defining propagator be-tween the points xi and xj, the analogy here is the ∆(x− y)(actually 1
i∆(x− y) is
the propagator) that describes a propagator from the space-time point x to y. Fromhere we have a very good tool to describe propagations in fields.
Consider the Lagrangian density for a field
L = L0 + Lint , Lint =∞∑n=3
1
n!gnϕ
n + (counterterms), (127)
where we now have all the general interacting terms for a field. In most cases, onerestricts the interacting terms to only one type. This theory involves interactionswith n-point vertices, i.e. each vertex contracts n sources to it. The generalizationlike this considers a ϕn theory. But let us proceed to a very important identityin QFT and perturbation theory, the very important tool we have to generate thediagrams that we are looking for.
6.2 Central Identity of Quantum Field Theory
If we consider the Lagrangian in (127), but without focusing on any of the countert-erms for the moment, and inserting it in the path-integral for our free-field theorywe have
〈x′1 · · · , x′k|x1, · · · , xk〉J = 〈0|T[ϕ(x′1) · · ·ϕ(x′k)ϕ(x1) · · ·ϕ(xk)]|0〉, (128)
we know that we can reproduce the amplitude by the regular methods in how wetreat these integrals. Starting by rewriting the first part of the integral in ourcommon way by introducing our sources as
(128) =
(1
i
)2k k∏i=1
δ
δJ(xi)
k∏j=1
δ
δJ(x′j)
∫Dϕ ei
∫d4x(L0+Lint+J(x)ϕ(x))
∣∣∣∣J=0
, (129)
11In M.Srednicki “Quantum Field Theory”, chapter 8-9, there is a very good derivation in howto obtain this and more information about the Feynman propagator ∆(x− y).
30
and now to a clever mathematical trick to rewrite the interaction part of the La-grangian. We take out the interaction part outside of the integral in the exponent,but we cannot do it just right away. We have to use or sources such as we replacethe argument of the interaction Lagrangian with a functional derivative operator,e.g. Lint(ϕ(x))→ Lint(
1i
δδJ(x)
) hence we obtain
(1
i
)2k k∏i=1
δ
δJ(xi)
k∏j=1
δ
δJ(x′j)× e
[i∫d4x
2k∑n=3
gnn! (
1i
δδJ(x))
n
] ∫Dϕ ei
∫d4x(L0+J(x)ϕ(x))
∣∣∣∣J=0
=
(1
i
)2k k∏i=1
δ
δJ(xi)
k∏j=1
δ
δJ(x′j)exp
(i
∫d4x
2k∑n=3
gnn!
(1
i
δ
δJ(x)
)n)
×exp
(i
2
∫d4x d4y J(x)∆(x− y)J(y)
) ∣∣∣∣J=0
(130)
Now, we can Taylor expand the function in front of the integral, sum up all thefunctional derivatives, perform them and we get the desired power of ϕ(x) in eachterm in the sum. Then we can do the expansion backwards and we have arrivedat were we started from in (129). Here is the key, we can now choose in whichpower of the coupling constant gn we want to do our calculations in. We choose atheory that describes by n′, put all the coupling constants in the sum with n 6= n′
to zero and then we expand the whole expression. If we want to O(g2), we expandthe coefficient with functional derivatives in front of the integral to the order of g2.We do not have to worry about the terms we get from O(g) since the calculationin O(g2) is more accurate than the first, and so on. It is common to define the twointegrals in (130) as a functional Z1[J ], which means that the hole expression canbe written in avery compact and nice form as
〈x′1, · · · , x′k|x1, · · · , xk〉 = 〈0|T [ϕ(x′1) · · ·ϕ(x′k)ϕ(x1) · · ·ϕ(xk)] |0〉
=
(1
i
)2k k∏i=1
δ
δJ(xi)
k∏j=1
δ
δJ(x′j)Z1[J ]. (131)
The double perturbative expansion means that we can also expand the last term(thefree-field theory path integral in the presence of sources) in powers of J . Expandingboth terms we obtain
Z1[J ] ∼∞∑V=0
1
V !
[i∞∑n=3
gnn!
∫d4x
(1
i
δ
δJ(x)
)n]V
×∞∑P=0
1
P !
[i
2
∫d4y d4z J(y)∆(y − z)J(z)
]P, (132)
the proportional sign showed up because there is a constant of proportionality thathas to be fixed when evaluating everything with J = 0, bet we will not worry moreabout that because this constant will just multiply each diagram, thus we do notreally have to present it. The labellings V and P is a convenient way to label the
31
expansion. A calculation to O(g2) will involve two vertices, and the parameter thatdecides which order of g we will have is the expansion parameter V . The number ofpropagators we will have in a diagram will be determined by how many powers ofthe Feynman propagator in the last integral we have, hence the label P for the lastexpansion parameter.
6.3 Lazy Professors and ϕ3-theory
In ϕ3-theory we are dealing with interactions that has 3-valent vertices. Some ϕ3-theories are not physical at all, but are good examples to study to get some insightin how it works. Why is some ϕ3-theories physically important to deal with? Arestriction is that the Hamiltonian must be bounded from below. If we include aninteraction term as
Lint =1
3!gϕ3 + (counterterms), (133)
the corresponding Hamiltonian will have the tendency to be arbitrarily negative forlarge values of ϕ. Anyway, let us study an entertaining example12
The lazy Professor
There was a student at the university who was very eager to study Quantum FieldTheory. At the moment, the student sat down with ϕ3-theory with the given La-grangian
L = −1
2∂µϕ∂µϕ−
1
2m2ϕ2 +
1
3!gϕ3 + (counterterms), (134)
and tried to evaluate a Feynman diagram. The diagram that the student wanted toevaluate was
.
No matter for how long the student studied the diagram, he could not in any wayunderstand how one extracted the symmetry factor from just looking at a diagram.He started to loose his patience and finally he gave up. He raised his hand and askedthe professor:
”Professor, how does one extract the symmetry factor from a diagram? I’m get-ting really confused with all these factorials and descriptions in the book of how onerearranges all the building blocks in the diagram.”
The Professor came up to the student, glimpsed at the diagram and mumbled some-thing about loops, rearranging propagators and so on and sad that the symmetry
12This story is made up.
32
factor was probably 4. The student stared at the diagram and was even more con-fused than before while the discouraging Professor walked away.
The student knew that if he started from the beginning from the mathematicalformulation for the diagram and calculated it, he would extract the symmetry fac-tor when the calculation was finished.
The students calculation
Putting V = 1 and P = 2 in the double perturbative expansion we get
ig
3!2!1!
(i
2
)2(1
i
)3 ∫d4x
δ3
δJ(x)3
∫d4yd4y′d4zd4z′J(y)∆(y−z)J(z)J(y′)∆(y′−z′)J(z′),
(135)without writing out all the constant terms in front of the integrals, he started by tak-ing one functional derivative. Recall that one functional derivative generates a deltafunction, so the student performed the integration directly over the correspondingcoordinate. Taking one functional derivative∫
d4y′d4zd4z′∆(x− y)J(z)J(y′)∆(y′ − z′)J(z′) + (136)∫d4y′d4zd4z′J(y)∆(x− y)J(y′)∆(y′ − z′)J(z′) + (137)∫d4yd4y′d4z′J(y)∆(y − z)J(z)∆(x− z′)J(z′) + (138)∫d4yd4y′d4zJ(y)∆(y − z)J(z)J(y′)∆(y′ − x), (139)
at this step, one can see that it is allowed to shift integration variables back andforth since the integration variables only acts as dummy variables. In (137), lety → z, in (138) let y → y′, z → z′, z′ → z and in (139) let y → y′, z → z′. Thiswill give him four identical integrals,
(135) = 4× ig
3!2!1!
(i
2
)2(1
i
)3 ∫d4x
δ2
δJ(x)2
∫d4y′d4zd4z′∆(x−z)J(z)J(y′)∆(y′−z′)J(z′).
(140)Another functional derivative
δ
δJ(x)
∫d4y′d4zd4z′∆(x− z)J(z)J(y′)∆(y′ − z′)J(z′) = (141)∫d4y′d4z′∆(x− x)J(y′)J(z′)∆(y′ − z′)J(z′) + (142)∫
d4z′d4z∆(x− z)J(z)∆(x− z′)J(z′) + (143)∫d4y′d4z∆(x− z)J(z)J(y′)∆(y′ − x), (144)
33
in (143) let z → y′ and in (144) let z → z′, this will give∫d4y′d4z′ (∆(x− x)J(y′)∆(y′ − z′) + 2×∆(x− z′)J(y′)∆(x− z′)J(z′)) , (145)
and the last functional derivative
δ
δJ(x)(145) = ∆(x− x)
(∫d4z′∆(x− z′)J(z′) +
∫d4y′J(y′)∆(x− y′)
)+
2×∆(x− x)
(∫d4y′J(y′)∆(x− y′) +
∫d4z′∆(x− z′)J(z′)
)= [relabelling] = 6×
∫d4y′J(y′)∆(y′ − x)∆(x− x). (146)
Plugging this into (135) we obtain the final expression for the diagram
(135) =1
2ig
∫d4xd4y′ iJ(y′)
1
i∆(y′ − x)
1
i∆(x− x). (147)
And the calculation is finished.
The student is sure that he performed the calculations right, and he knew thatthe symmetry factor always shows up as a rational term in the front of the integral.
End of calculation.
The student proved the Professor wrong. The symmetry factor is 2 since 4×63!2!1!×4
= 12.
The generating functional will generate two topologically equivalent diagrams whichwe can add up,and the factor of 1
2will cancel the symmetries.
Feynman developed rules in how one extracts the mathematical expression froma diagram, i.e. beginning in the other end, and then has to evaluate the symme-try factor by combinatorics and cleverness. Starting from a general diagram, heassigned a certain factor for each of the components in a diagram. Label the pointswhere the building blocks lies. When this is done, the only thing one has to do isto multiply them together. The building blocks for a general diagram are
A line segment from x to y −→ 1i∆(x− y)
Vertex −→ (ig)∫d4x
Source at z −→ i∫d4z J(z)
Symmetry factor −→ 1S
.
As we can see this extraction backwards works for the student diagram. Label thepoint where the source is with y′, the point where the vertex lies with x. Thus wehave our building blocks
1
i∆(y′ − x),
1
i∆(x− x), ig
∫d4x, i
∫d4y′J(y′),
1
Sand multiplying these together with S = 2 we will have the diagram as in theexample of the lazy Professor.
34
6.4 Diagrams in ϕ4-theory
Consider the Lagrangian
L = −1
2∂µϕ∂µϕ−
1
2m2ϕ2 − 1
4!λϕ4 + (counterterms). (148)
A baby example
No matter which theory we choose, it should be possible for a particle to propa-gate from one point to another. This means that we have no vertices(V = 0) andone propagator(P = 1). We get our partition function Z1[J ] to be
Z1[J ] =i
2
∫d4y d4z J(y)∆(y − z)J(z), (149)
truly trivial. The diagram is also trivial
.
Figure 6.4.1: Diagram for V = 0 and P = 1with symmetry factor 2.
End of the baby example.
The two filled black dots represents sources, which we can use two create or anni-hilate particles with. As in example 5.3.1, we used 〈0|T[ϕ(x2)ϕ(x1)]|0〉 to generatea particle that propagates from x1 to x2. It is now that our functional derivativescome in hand. What the derivatives actually do is when acting on the diagram, theystrip of one source each, and labels the end that is left with the argument that is inthe functional derivative, e.g. if you act with δ
δJ(xi)on the diagram, it strips off one
source and labels the end with xi. So the two functional derivatives actually justremoved the sources, labelled them x1 and x2.
A very nice thing to notice is that the symmetry factor is 2. We saw in Exam-ple 5.3.1 that the generating functional produced two diagrams that we could addup and cancel the symmetry factor of 1
2. Things really starts to coincide.
So, for higher powers of the coupling constant, the more vertices we get. Thehigher powers of Feynman propagators, the more propagators we get. One can seethat the number of functional derivatives for V > 1 becomes large very quick sincewe have the exponent 4 for the derivatives in ϕ4 theory. Feynman was a smartman and developed rules so one can work backwards from first have drawn all thetopologically in-equivalent13, assigning certain factors for the building blocks in the
13With topologically in-equivalent means that one cannot construct another diagram by twisting,stretching and rotating another diagram without tearing our gluing things together.
35
diagrams and then multiply them together. Although, the symmetry factors doesnot come out without effort by this method, if you are not sure about what you do.So, let us try something harder.
In ϕ4-theory there is a relation that is very handy to have in mind when one istrying to draw different diagrams. The relation is
S = 2P − 4V, (150)
where S stands for number of external sources, P for number of internal propagatorsand V for number of vertices in a diagram. For the baby example, S = 2, V = 0 thusP = 1. When drawing the diagrams, one has to consider all the topologically in-equivalent diagrams when can imagine during the process, this is were some routineand cleverness comes into the picture. We will only be interested in the connecteddiagrams.
Figure 6.4.2: All connected diagrams at O(λ) withsymmetry factors 4! and 4 respectively.
Figure 6.4.3: All connected diagrams at O(λ2) with symmetryfactors 8, 16, 12, 8 and 12 respectively.
It is of course possible two calculate all these diagrams from scratch using the doubleperturbative expansion, but as one will see, it gets rather messy and there is a lotof terms to keep track on. But it is of course irresistible two do at least one of thesecalculations.
Example 6.4.1
We choose to calculate the diagram to the right in Figure 6.4.2, e.g. this dia-gram
,
36
starting the set up with counting the vertices in the diagram and we see that V = 1.There are two sources in the diagram which from (150) we conclude that there arethree propagators, so P = 3. The vertex factor in this theory is −iλ. Plugging thisin to our favourite tool, the double perturbative expansion, we then have
Z[J ] =−iλ4!3!
(i
2
)3 ∫d4x
δ4
δJ(x)4
∫d4x1 · · · d4x6 (J1∆12J2)(J3∆34J4)(J5∆56J6),
(151)the parenthesis is only for notational usefulness when considering the derivatives.We have introduced the abbreviations
Ji := J(xi) , ∆ij := ∆(xi − xj), ∆xi := ∆(x− xi) (152)
Skipping the numerical factors in front of the integral for now and taking one func-tional derivative, and performing the integration over the corresponding delta func-tion directly
1st derivative: ∫d4x2d
4x3d4x4d
4x5d4x6(∆x2J2)(J3∆34J4)(J5∆56J6)
+
∫d4x1d
4x3d4x4d
4x5d4x6(J1∆1x)(J3∆34J4)(J5∆56J6)
+
∫d4x1d
4x2d4x4d
4x5d4x6(J1∆12J2)(∆x4J4)(J5∆56J6)
+
∫d4x1d
4x2d4x3d
4x5d4x6(J1∆12J2)(J3∆3x)(J5∆56J6)
+
∫d4x1d
4x2d4x3d
4x4d4x6(J1∆12J2)(J3∆34J4)(∆x6J6)
+
∫d4x1d
4x2d4x3d
4x4d4x5(J1∆12J2)(J3∆34J4)(J5∆5x)
relabelling= 6×
∫d4x2 · · · d4x6(∆x2J2)(J3∆34J4)(J5∆56J6) (153)
2nd derivative:
δ
δJ(x)(153) = 6×
∫d4x3d
4x4d4x5d
4x66×∆xx(J3∆34J4)(J5∆56J6)
6×∫d4x2d
4x4d4x5d
4x6(∆x2J2)(∆x4J4)(J5∆56J6) + 6× (3 terms)
relabelling= 6×
∫d4x2d
4x3d4x4d
4x5d4x6∆xx(J3∆34J4)(J5∆56J6)
+24×∫d4x2d
4d4x4d4x5d
4x6(∆x2J2)(∆x4J4)(J5∆56J6), (154)
We have got two terms with different structure. This means that we will generatetwo diagrams. Every time a derivative is hitting the integrals, one integration is per-formed directly to the corresponding coordinate as explained in the beginning. But
37
every time, we can relabel the dummy variables so that have we can add up the termsthat have the same structure. This will not really be displayed any more since thesteps are being easier and easier after every derivative.
If a derivative strikes an expression like ∆xiJi(Jk∆klJl), it will for quite obviousreasons generate three terms, but the last two terms(if we work from left to rightwith the derivative) we can add up because they have the same structure. Let uskeep taking derivatives
3rd derivative:
δ
δJ(x)(154) = 6× 4
∫d4x4d
4x5d4x6∆xx(∆x4J4)(J5∆56J6)
+24× 2
∫d4x4d
4x5d4x6∆xx(∆x4J4)(J5∆56J6)
+24× 2
∫d4x4d
4x5d4x6(∆x4J4)(J5∆5x)(∆x6J6), (155)
the coloured number in front of the integrals is just to keep track on that this numbercame from the new derivative after relabelling. The two first terms have the samestructure so we add them up
(155) = (6× 4 + 24× 2)
∫d4x4d
4x5d4x6∆xx(∆x4J4)(J5∆56J6)
+24× 2
∫d4x4d
4x5d4x6(∆x4J4)(J5∆5x)(∆x6J6) (156)
4th derivative:
δ
δJ(x)(156) = (6× 4 + 24× 2)× 1
∫d4x5d
4x6∆xx∆xx(J5∆56J6)
+(6× 4 + 24× 2)× 2
∫d4x5d
4x6∆xx(J5∆5x)(∆x6J6)
+24× 2× 3
∫d4x5d
4x6∆xx(J5∆5x)(∆x6J6) (157)
the last two terms have been relabelled so they have the same indices and they arealso equal structure wise. Adding them up we get the results
(157) = 72×∫d4x5d
4x6∆xx∆xx(J5∆56J6)
+288×∫d4x5d
4x6∆xx(J5∆5x)(∆x6J6). (158)
Let us just analyse the two different terms. We are only interested in the connectedtype of diagrams and if we would draw the diagram to the first term it would be
,
38
which is obviously a disconnected diagram. So we omit this term from now on.Plugging the remaining term into (151) we arrive at
Z[J ] = (−iλ)1
4
∫d4xd4x5d
4x6iJ(x5)1
i∆(x5 − x)
1
i∆(x− x)
1
i∆(x− x6)iJ(x6),(159)
and we for sure did the calculation correct since the symmetry factor S = 4(asclaimed in Figure 6.4.2) dropped out. The mathematical expression we arrivedat makes perfect sense. A particle is created at point x5 by J(x5), propagates tothe point x(where the loop connects to the line by the vertex) via ∆(x5 − x), per-forms a loop from x to x via ∆(x−x) then propagates to the point x6 via ∆(x−x6)and gets annihilated by the source J(x6). This can also happen in the reversed order.
End of example.
7 Discussion
7.1 Analogies
The mathematical analysis in chapter 2 discusses the Wick Theorem, but have weactually used it in our calculations sometime? Actually, no. In free-field theory wehave the propagator D(x − y), which is somewhat analogous to the first represen-tation of a propagator, the inverse of the matrix Aij defined as Aij. We used theindices of the matrix to define propagations between two points and the propagatorwere defined. One restriction of the matrix A was that it actually had an inverse.What about D(x− y)? Is there an inverse?
To analyse the propagator D(x − y), we write it out in an explicit form as a four-dimensional integral
D(x− y) =
∫d4k
(2π)4
eik(x−y)
k2 +m2 − iε, (160)
the propagator is a Green’s function for the Klein-Gordon equation which meansthat it satisfies
”Inverse function× Function” = ”1”, (161)
so that the “Inverse function” is the operator (−∂x+m2) in the Klein-Gordon equa-tion and the ”1” is equal to the four-dimensional Dirac delta function δ4(x − y).This agrees in the limit where ε→ 0 which is straight forward to plug in. Thereforewe have a direct analogy between these two different, but very alike, propagators.
We saw that one could use the Wick Theorem to pair up indices with each other inchapter 2 which we used to generate the graphs in Figure 2.2.1 with the matrixpropagator Aij. Similarly, we can use the Wick Theorem for the propagator D(x−y)to pair up the indices in different combinations
〈0|Tϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4)|0〉 =1
i2
∑pairings
D(xi1 − xi2)D(xi3 − xi4), (162)
39
where we sum over the different pairings, not permutations, of the indices and wewill obtain the result
〈0|Tϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4)|0〉 =1
i2D(x1 − x2)D(x3 − x4) +
+1
i2D(x1 − x3)D(x2 − x4) +
1
i2D(x1 − x4)D(x2 − x3), (163)
diagrammatic this is the exact same result as in Figure 2.2.1. By having analysedthe pure mathematical formulation of graph generators in the beginning of the thesisone gains the crucial understanding of how these things works. It is very impor-tant to have that in mind and not just to start calculating things without knowingwhere they come from, how they work and what they actually represent in physicalsituations.
7.2 Symmetries and topological equivalence
Throughout the analysis of every diagram there has always been a remark aboutsymmetries and symmetry factors. Where does these symmetries from diagramsactually arise from? For starters, let us take a very simple example, the propagationdiagram in free-field theory
.
It is claimed that the symmetry factor for this diagram is 2. Where does this 2 arisefrom? We have the generating functional for this simple diagram
Z1[J ] =i
2
∫d4y d4z J(y)∆(y − z)J(z). (164)
We have learned that by hitting the “diagram” with functional derivatives we canproduce a propagation diagram for a particle in a free-field theory. Since the deriva-tive operator is 1
iδ
δJ(xi)and the diagram has a product of two “J − functions” it will
for obvious reasons produce a sum of two terms. The symmetry arise from that theoperator can hit one or the other “J − function” and actually it not actually can, itwill hit them both.
.
The figure is displaying how the derivatives hits the diagram. The left derivativewill first take, let us say, the blue “path” and hit the closest source and the rightderivative will take the same coloured “path” and hit the closest source. But theleft derivative will also take the red “path” and hit the source furthest away and the
40
right derivative will follow the colour that the left one decides. This will generate 2diagrams. But the diagrams will appear to be topologically equivalent, which meansthat we can create the first diagram of the second diagram without changing anylabels of the ends by twisting and turning, without tearing it apart or glue stufftogether
+ ,
and without changing any labels we can turn the second diagram by π around theaxis which is perpendicular to the diagram in the upward direction and we simplyget
2× ,
hence the symmetry factor of 2 for the diagram. This is the simplest diagram onecan imagine and here it would not be necessary to carry out the calculation for thediagram. If one has some experience and is a bit clever, the symmetry factors arenot very hard to figure out. Although, in some cases when one is not completelyconvinced of the symmetries of a diagram, the power of the double perturbativeexpansion reveals itself. The integrals themselves are not really hard to do but theytests the “bookkeeping” skills of the person. The whole calculation is just actuallybased on that the integrals are number-valued objects, not functions, e.g.
I1 =
1∫0
dx x , I2 =
1∫0
dt t ⇒ I = I1 + I2I1=I2= 2I1. (165)
Thus to calculate the many integrals, which for sure can be very many as seen inExample 6.4.1(a total of 360 integrals), is just to keep track on structure wiseequal terms and relabelling dummy integration variables. It requires some work butis not always complicated. The diagram that is explicitly calculated in Example6.4.1 is not a very complicated diagram, although, the principle and technique ofcalculating any diagram will still be the same. Let us just look at one more diagram
.
Does this diagrams symmetry factor demand a long calculation to reveal its sym-metry factor? Not really. By means on the evaluation of the simple diagram atthe beginning of this subsection we can use the same “theory” of how the func-tional derivatives hits the diagram. We need a total number of 4 derivatives tonot get the total result of zero. How can these act on the diagram? The first onehas 4 choices, the third has 3 choices, the second one has 2 choices and then thereis only one “choice” left for the last derivative. This gives a symmetry factor of4 × 3 × 2 × 1 = 4!. No matter how the dervatives rearrange themselves all thediagrams will be topologically equivalent which is quite easy to see if one looks atthe diagram. It does not matter how the ends are labelled, we can just twist and
41
turn the different legs around different axis’s and always end up with 24 identicaldiagrams.
Three derivatives would leave one of the sources untouched and when putting J = 0this would kill the diagram. This is actually a direct analogy to the integrals firstdiscussed in chapter 1. The integral∫
R
dx x2k+1e−12αx2
, k ∈ N (166)
is odd under evaluation over R and would give a result of zero. The “exact” samething happens in the calculations of the diagrams. We would end up with terms thatall have terms of J ’s in them, and at the end putting J = 0 would kill all terms.
7.3 A theory that cries in the vacuum
This subsection is nearly just a remark, although, it is worth to mention someproblems of certain theories. In chapter 6 there is some analysis of diagrams inϕ3- and ϕ4-theory. In the Lagrangian densities there are terms that have not beendiscussed much about, the counter terms. What are the purpose of the counterterms? A demand for a Quantum Field Theory is that the fields obeys the condition
〈0|ϕ(x)|0〉 = 0. (167)
Consider a ϕ3-theory. By a Legendre transformation from the Lagrangian densityin (134) would give us the Hamiltonian density
H =1
2π2 +
1
2(∇ϕ)2 +
1
2m2ϕ2 − 1
3!ϕ3. (168)
By making ϕ arbitrarily big, we can get the Hamiltonian density to be arbitrarilynegative. This leads to that the Hamiltonian density does not have a proper groundstate. How does this connect to the condition in (167)? As we saw in the exampleof the lazy professor we had this diagram
.
The condition in (167) can be translated into our favourite tool, the generatingfunctional, as
〈0|ϕ(x)|0〉 =1
i
δ
δJ(x)Z1[J ]
∣∣∣∣J=0
. (169)
But if we look at the diagram, the functional derivative will hit the diagram, strip offone source and label it and directly after that we will put J = 0. The diagram willnot vanish! A Quantum Field Theory has to have a well-defined vacuum expectationvalue(〈0|ϕ(x)|0〉 = 0) but this theory violates this condition. These kind of vacuum
42
fluctuations has to be dealt with and this is where the counter terms comes inhand. The counter terms will provide us with “cancelling” diagrams for the vacuumexpectation value. The mathematics and analysis of these types of diagram are notdiscussed because it is beyond the level of this thesis. 14
14In chapter 9, “Quantum Field Theory”, M.Srednicki, there is a lot more discussion about howthe counter term diagrams actually works and how one rescales the Lagrangian density to involvethem.
43
References
[1] J.Qiu, M.Zabzine, “Introduction to Graded Geometry, Batalin-Vilkovisky For-malism and their Applications”, [arXiv:1105.2680], 14 Dec 2011.
[2] J.Sawon, “Perturbative expansion in Chern-Simons theory”, 13 Mar 2009, [arX-ivmath/0504494].
[3] R.P.Feynman, A.R.Hibbs, ”Quantum mechanics and path integrals”, McGraw-Hill, New York, 1965.
[4] A.Zee, “Quantum field theory in a nutshell”, Princeton University Press 41, 2003.
[5] W.Rindler, “Introduction to Special Relativity”, Clarendon Press, Oxford, 1991.
[6] M.Srednicki, “Quantum Field Theory”, Cambride University Press, New York,2007.
[7] J.J. Sakurai, Jim Napolitano, ‘‘Modern Quantum Mechanics”, Pearson Educa-tion, publishing as Addison-Wesley, San Francisco, 1994.
44
A Appendix
Gaussian integral
We have the Gaussian integral
I =
∫Re−
12αx2
dx⇔ I2 =
(∫Re−
12αx2
dx
)×(∫
Re−
12αy2
dy
),
we can rewrite this as∫∫R2
e−12α(x2+y2)dxdy = [polar coordinates] =
∞∫0
2π∫0
re−12αr2
drdφ =2π
α,
therefore
I2 =2π
α⇔ I =
√2π
α.
Disconnected diagrams calculation
These two short-hand notations will be used
∂
∂Ji= ∂i , Z[J ] = Z
∂p∂nW [J ] = ∂p∂n ln Z = − 1
Z2∂pZ∂nZ +1
Z∂p∂nZ,
∂k∂s∂p∂nW [J ] =
− 6
Z4∂kZ∂sZ∂pZ∂nZ +2
Z3∂k∂sZ∂pZ∂nZ +2
Z3∂sZ∂k∂pZ∂nZ +2
Z3∂sZ∂pZ∂k∂nZ
+2
Z3∂kZ∂s∂pZ∂nZ−1
Z2∂k∂s∂pZ∂nZ−1
Z2∂s∂pZ∂k∂nZ +2
Z3∂kZ∂pZ∂s∂nZ
− 1
Z2∂k∂pZ∂s∂nZ−1
Z2∂sZ∂k∂p∂nZ +2
Z3∂kZ∂sZ∂p∂nZ−1
Z2∂k∂sZ∂p∂nZ
− 1
Z2∂sZ∂k∂p∂nZ−1
Z2∂kZ∂s∂p∂nZ +1
Z∂k∂s∂p∂nZ
and after removing the terms which is identically zero and evaluating with J = 0,we end up with
∂k∂s∂p∂nW [J ] = − 1
Z2[0]∂s∂pZ∂k∂nZ−
1
Z2[0]∂k∂pZ∂s∂nZ
− 1
Z2[0]∂k∂sZ∂p∂nZ +
1
Z[0]∂k∂s∂p∂nZ.
45
With the knowledge from section 2.1 that the three first terms can be written as
−(
1
Z[0]∂µ∂νZ
)(1
Z[0]∂ε∂ηZ
)= − 1
α2AµνAεη
and the result from the integral in (34), we obtain
∂k∂s∂p∂nW [J ] = − 1
α2
(AspAkn + AkpAsn + AksApn
)+
+1
α2
(AnpAks + AnkAps + AnsApk
)= 0,
diagrammatic this is
= 0.
46