Self Energy

8/13/2019 Self Energy

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Feynman parameter integrals

We often deal with products of many propagator factors in loop integrals. The trick is tocombine many propagators into a single fraction so that the four-momentum integration canbe done easily. This is done commonly using so-called Feynman parameters.

We rewrite the product of propagators

1

(A1+i)(A2+i) (An+i), (1)

whereAi has the form ofp2 m2. The sign ofAi is not fixed, but the imaginary part has the

fixed sign because ofi. This turns out to be useful.A single propagator can be rewritten using a simple integral

1

A= i

0dteit(A+i). (2)

The surface term att vanishes because ofiterm. Using this multiple times, we find

1

(A1+i)(A2+i) (An+i)= (i)n

0

dt1 dtnein

i ti(Ai+i). (3)

Here comes another trick. We use the identity

1 =0

d

1

1

ni

ti

. (4)

This can be shown using the general formula (f(x)) =(x x0)/|f(x0)| wherex0is the zero

off(x), i.e., f(x0) = 0. The delta function in the above integral therefore can be rewrittenas

1

1

ni

ti

=

( n

i ti)12n

i ti=(

ni

ti), (5)

and the integration can be done trivially to yield unity. We used the fact that all of ti inEq. (3) are positive to limit the integration above zero.

Inserting the unity Eq. (4) into Eq. (3), we find

1

(A1+i)(A2+i) (An+i)= (i)n

0

dt1 dtnd

1

1

ni

ti

ein

i ti(Ai+i). (6)

Now we change the variables fromti to ti= xi, and find

= (i)n0

dx1 dxndn1

1

ni

xi

ein

i ti(Ai+i). (7)

Now the integral can be done by using the integral representation of the gamma function

(n) = (n 1)! =0

dt tn1et. (8)

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This expression can be generalized to the following one

(n 1)! 1

Xn =

0dt tn1etX (9)

as long as Re(X)> 0 so that the integral converges. Because Re(i(A + i)) = >0, we find

1

(A1+i)(A2+i) (An+i)= (i)n

0

dx1 dxn(n 1)!

(in

i ti(Ai+i))n

1

ni

xi

.

(10)Finally, note that all xi are positive while the sum ofxi must be unity. Therefore the inte-gration region can be limited to 0 < xi< 1:

1

(A1+i)(A2+i) (An+i)= (n 1)!

10

dx1 dxn1

(ni ti(Ai+i))n

1

n

ixi

.

(11)The simple cases ofn = 2, 3 are

1

(A1+i)(A2+i)=

10

dx

(x(A1+i) + (1 x)(A2+i))2, (12)

1

(A1+i)(A2+i)(A3+i)= 2

10

dx dy dz (1 x y z)

(x(A1+i) +y(A2+i) +z(A3+i))3. (13)

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General Ideas on the Self-energy Diagrams

We calculate the two-point function G2(x y) = |(x)(y)| in perturbation theory.

An important point is to realize is that the full two-point function can be obtained once onehas all the 1PI (one particle irreducible) diagrams i(p) computed because

G2(p)

d4xG2(x y)eip(xy)

= i

p m0+

i

p m0(i(p))

i

p m0+

i

p m0(i(p))

i

p m0(i(p))

i

p m0+

= i

p m0 (p). (14)

Here,m0 is the bare mass in the Lagrangian which is different from the physical (kinetic)mass.

To correctly identify the mass of the particle, we look for the zero of the denominator inthe two-point function. We define the physical mass mof the particle by the equation

p m0 (p)|p=m= m m0 (m) = 0. (15)

Then we expand the self-energy diagram (p) aroundp= m as

(p) =m (Z12 1)(p m) +Z12 R(p), (16)

with m = (m), (Z12 1) = (p)/p|p=m, and R(p) behaves as O(p m)2 when

p m. Then the two-point function becomes

G2(p) = i

p m0 m+ (Z12 1)(p m) Z

12 R(p)

= iZ2

p m R(p), (17)

with m = m0 + m. Therefore, m is interpreted as the correction to the mass of theparticle due to interactions, and this is why these diagrams are called self-energy diagrams.The factorZ2 is called the wave-function renormalization factor which describes the strength(square root of the probability) of the field operator creating the one-particle state. Atthe lowest order in perturbation theory Z2 = 1, but is different from unity once higher ordercorrections are taken into account. This is the factor that appears in the LSZ reductionformula.

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Explicit Calculation of the One-loop Self-energy Diagram

At the one-loop level, the self-energy diagram is given by

i(p) = (ie)

d4k

(2)4i

p+ k m0+i

igk2 +i

(ie). (18)

This integral, however, is divergent both in the ultraviolet k and the infraredk 0. Todeal with these divergences, we regulate the integral, i.e., introduce parameters which makethe integral formally convergent, but the parameters must be taken to zero or infinity at theend of calculations. There are many ways of regularizing the integrals, but we adopt thefollowing prescription for this purpose:

1

k2 +i

1

k2 2 +i

1

k2 2 +i=

2 2

(k2 2)(k2 2). (19)

In the end we take the limit 0 and to recover the original expression, but theintegral becomes finite as long as we keep both and finite. They are called infrared orultraviolet cutoffs. Below, we often dropi terms but it is understood that they are alwaysthere.

The regularized form is then

i(p) = e2

d4k

(2)4(p+ k+m0)

(p+k)2 m20

2 2

(k2 2)(k2 2). (20)

The next step is to use Feynman parameters to combine three propagator factors and simplify

the numerator by the identity a= 2a,

i(p) = e22 10

dxdydz(1 x y z)

d4k

(2)4(2(p+ k) + 4m0)(

2 2)

[x((p+k)2 m20) +y(k2 2) +z(k2 2)]3

. (21)

The denominator can be simplified using x+y+z= 1 to

denom = [k2 + 2xk p+xp2 xm20 y2 z2]3

= [(k+xp)2 +x(1 x)p2 xm20 y2 z2]3. (22)

By shifting the integration variable k k xp, we find

i(p) = e22 10

dxdydz(1 x y z)

d4k

(2)4(2((1 x)p+ k) + 4m0)(

2 2)

[k2 +x(1 x)p2 xm20 y2 z2]3. (23)

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The kterm in the numerator can be dropped because it is an odd function ofk and vanishesupond4k integration.

The next step is to perform d4k integration. We need to work out a general formula

d4k

(2)41

(k2 M2 +i)3 =

i

(4)21

2

1

M2 i. (24)

Here we recovered i term because it is important. To show this, we first focus on k0 inte-

gration. The (triple) poles are located at k0 =

k2 +M2 i and hence the pole with +sign is below the real axis, and that with sign is above the real axis. Since the integrandgoes to zero sufficiently fast at the infinity |k0| on the complexk0 plane, the integrationcontour can be rotated from k0 (, ) to k0 (i, i) without hitting the poles.This is called Wick rotation. Then we can change the integration variable k0 =ik4 such thatthe new variable ranges in (, ). Then the above integral is rewritten as

d4k

(2)41

(k2 M2 +i)3 =i

d4kE(2)4

1

(k2E M2 +i)3

. (25)

Here, d4k = dk0dk, d4kE = dk4dk, and k2E = (k

4)2 +k2 = (k0)2 +k2 = k2. Now thatthe denominator is positive definite (no poles along the integration contour), we do not haveto worry about i in performing integration. We then use the polar coordinates for four-dimensionalkEand the integration volume is d

4kE= 22k3EdkE=

2k2Edk2Eafter integrating

over three angle variables. (In general, dnx= 2n/2

(n/2)xn1dx.) Then the integral becomes quite

simple,

=

i2

(2)20 dk

2Ek

2E

1

k2E+ M2 i. (26)

Thek2Eintegration can be done in parts and we find Eq. (24).Now going back to Eq. (23), we can perform d4k integration using Eq. (24) and find

i(p) = e2 10

dxdydz(1 x y z) i

(4)2(2(1 x)p+ 4m0)(

2 2)

xm20+y2 +z2 x(1 x)p2 i. (27)

The integration overzcan be done trivially using the delta function,

i(p) = e2 1

0

dx 1x

0

dy i

(4)2

(2(1 x)p+ 4m0)(2 2)

xm

2

0+y2

+ (1 x y)2

x(1 x)p2

i

. (28)

Note that the integration region ofy is now limited as [0, 1 x] because of the delta functionconstraintx + y+ z= 1 andz >0. The integration over y is also a simple logarithm and wefind

i(p) = ie2

(4)2

10

dx(2(1 x)p+ 4m0)logxm20+ (1 x)

2 x(1 x)p2 i

xm20+ (1 x)2 x(1 x)p2 i

. (29)

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Even the integration over x can be done with elementary functions only, but the expressionbecomes lengthy and not very inspiring and therefore we keep x integration. Finally using= e2/4,

(p) =

4

10

dx(2(1 x)p+ 4m0)logxm20+ (1 x)

2 x(1 x)p2 i

xm20+ (1 x)2 x(1 x)p2 i

. (30)

Following the general discussions, we now identify m and Z2. Since (p) and hencem= m m0 are O(e

2), I can replace m0 in (p) bym by neglecting O(e4) corrections.

First,m= (p)|p=m is given by

m = (p)|p=m=

4

10

dx(2(1 x)m+ 4m)logxm2 + (1 x)2 x(1 x)m2 i

xm2 + (1 x)2 x(1 x)m2 i

= m

4 1

0

dx2(1 +x)logx2m2 + (1 x)2 i

x2

m2

+ (1 x)2

i

. (31)

The argument of the logarithm is manifestly positive, and we can safely drop i. Moreover,we take the limit and 0 in the end, and we can neglect x2m2 in the numeratorand (1 x)2 in the denominator. Then the expression becomes drastically simpler and theend result is

m = m

4

3log

2

m2+

3

2

. (32)

This is the correction to the mass of the electron.The interpretation of this self-energy is quite interesting. In classical electrodynamics, we

actually have a linearly-divergent self-energy. An electron creates a Coulomb field around it,

and it feels its own Coulomb field. If, for instance, one imagines the electron to be a sphereof radius re with a uniform charge density, the total potential energy is V =

35e2

re. The total

energy of the electron is the sum of the rest energy m0c2 and the potential energy V and

hence the total mass of the electron we observe is given by

mc2 =m0c2 +

3

5

e2

4re. (33)

In the limit of re 0, the bare mass m0 needs to be sent negative to cancel the linearlydivergent Coulomb self-energy to obtain the observed mass of the electron. In the quantummechanical language, cleary this is an ultraviolet divergence as it corresponds to short-

distance physics re 0. If we imagine the electron to be as small as the Planck size re =hGN/c3 = 1.6 10

33 cm, where presumably the quantum gravity takes over physics, we

need to make the bare mass as negative as 5.34 1019 MeV which is cancelled by theself-energy for 20 digits to get 0.511 MeV. This is absurd. The classical electrodynamicstherefore breaks down at the distance scale where the rest energy and the self-energy becomecomparable, re e

2/mc2 1013 cm, and a better (deeper) theory needs to take over theclassical electrodynamics below this distance scale.

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What we have learnt here is that the situation in the QED is much better. The total massof the electron is

mc2 =m0c2

1 +

43log

2

m2

+3

2 . (34)

The ultraviolet cutoff corresponds to the inverse size of the electron in the classicallanguage. Again if we imagine the electron to be as small as the Planck size, the correctionto the electron mass is only 18.0%. The crucial difference from the classical theory is that(1) the dependence on the size of the electron is only logarithmic instead of power, and (2)the correction is proportional to the electron mass itself and hence can never be much largerthan the bare mass. Compared to the classical case Eq. (33), the quantum result Eq. (34)corresponds to the cutoffre h/mc which is nothing but the Compton wave length. Belowthis distance scale, quantum effects are essential and the classical theory does not apply anymore.

There are several lessons to be learnt from this discussion. First, the QED knows itsown limitation by the fact that it requires the ultraviolet cutoff to regulate the theory. Thetheory does not apply beyond certain energy scale. Such a theory is called an effectivefield theory, and is true to almost all quantum field theories. However, the cutoff can beextremely large unlike in the classical electrodynamics. Even though the QED should beregarded as a theory which should be taken over by a yet deeper theory at extreme highenergies, its applicability is practically infinite. Second, even though there is an ultravioletdivergence in the electron mass, what we observe is the totalof the bare mass and the selfenergy. Therefore, any calculations should be expressed in terms of the observedmass of theelectron instead of the bare mass. The same comment applies to the fine-structure constantas we will see later. It turns out that any physical quantities are finite once expressed in terms

of the observedmass and fine-structure constants despite the fact that there are dependenceon the ultraviolet cutoff as well as bare parameters at the intermediate stage of calculations.This is a general property ofrenormalizable quantum field theories: all physical quantitiesare finite once expressed in terms of observable quantities. Because of this property, one canuse the QED to do precise calculations without worrying about what physics is there at theenergy scale of the ultraviolet cutoff.

Next topic is the wave-function renormalization factor. Following the general discussions,we calculate

Z12 1 = (p)

p

p=m

. (35)

Using Eq. (30) again, and paying attention to the fact that p2

= (p)2

, we find

Z12 1 =

4

10

dx

2(1 x)log

xm2 + (1 x)2 x(1 x)p2 i

xm2 + (1 x)2 x(1 x)p2 i

+(2(1 x)p+ 4m)

x(1 x)2p

xm2 + (1 x)2 x(1 x)p2 i

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x(1 x)2p

xm2 + (1 x)2 x(1 x)p2 i

p=m

. (36)

We can drop the first term in the parenthesis because it vanishes in 0 limit, and alsom2 (2) in the numerator (denominator) in the logarithm can be set zero. After a simpleintegration, we find

Z12 1 =

4

2log

m2

2 log

2

m2

9

2

. (37)

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Self Energy