Field theory

04-28

下午走錯教室了，懶得換了，就聽了一節。

1. $Z[J]=int D[psi]s^{-S[psi]+<f,psi>}$

for free theory

$=Nexp(frac{1}{2}<J,frac{1}{-partial^2+m^2}J>)$

$=N(1+frac{1}{2}<G(1,2)j(1)j(2)>_{1,2}+...+frac{1}{N!}<G(1,...,N)J(1)...J(N)>_{1,2,...,N}+...$

where

$<G(1,...,N)J(1)...J(N)>_{1,2,...,N}=int dx_1dx_2...dx_nG(x_1,...,x_n)J(x_1)...J(x_n)$

this is the Taylor expansion.we use this to define $G(x,y)$ although we know G is the green function.

$f(x)=sum_{n=0}^{infty}frac{x^n}{n!}f^{(n)}(0)$

$G(1,...,N)=frac{1}{N}frac{delta^n Z[J]}{delta J(1)...J(N)}|^{J=0}=<psi(1)...psi(N))>$

$G(1,2)=Delta_F(1,2)$

$Z[J]/N=1+frac{1}{2}<JDelta_F J>+frac{1}{8}<JDelta_F J>^2+frac{1}{48}<JDelta_F J>^3+...$

$G(1,2,3,4)=frac{1}{8}(Delta_F(1,2)Delta_F(3,4)+Delta_F(1,3)Delta_F(2,4)+Delta_F(1,4)Delta_F(2,3))$

idea is:

$J longleftrightarrow J$

4 $1 longleftrightarrow J$

$J longleftrightarrow J$

to or

4 $1 longleftrightarrow 2$ 4*2 $1 longleftrightarrow J$

$J longleftrightarrow J$ $2longleftrightarrow J$

to or or

4*2 $1 longleftrightarrow 2$ 4*2 $1 longleftrightarrow 3$ 4*2 $1 longleftrightarrow J$

$3 longleftrightarrow J$ $2 longleftrightarrow J$ $2 longleftrightarrow 3$

now calculate the third term:

it seems the third term is also equedistribution.

so it appears that $G_N=1,2,...,N$ with $Delta_F$ connecting pairs of them and summed together Rule each point is on exactly one propagator.

$Z[J]/N=exp(frac{1}{2}Jlongleftrightarrow J)=1+frac{1}{2}(Jlongleftrightarrow J)+(frac{1}{2})^2frac{1}{2}(Jlongleftrightarrow J)^2+...$

N-th terms 2N-th order in $J$ with coefficient $frac{2^{-N}}{N!}$ .

consider arbitrary

$exp()=1+()+frac{()()}{2}+...+frac{()()...()}{N!}+...$

then $ln Z=()$

take another look at $Z[J]$

it looks like Fourier-Laplace transform of $e^{-S[psi]}$

take $S[psi]=frac{1}{2}<psi M psi>$ .

$Z[J]=int D[psi]exp(-frac{1}{2}<psi Mpsi>+<Jpsi>)$

$=int D[psi]exp(-frac{1}{2}<psi^` M psi^`>+<JM^{-1}J>)$

$=N exp(<JM^{-1}J>/2)$ , $N=Z[0]$

$psi^`=psi-M^{-1}J$ means $psi^`(1)=psi(1)-M^{-1}(1,2)J(2)$ .

$M M^{-1}=1$ $M[z]=ln z=frac{1}{2}<JM^{-1}J>+constant$

Recall in C,M , $L=frac{mx^2}{2}$ , $Longrightarrow H=frac{p^2}{2m}$

Legendre transform,Do it for $S[psi]$ .

Def $J(1)=frac{delta S}{delta psi(1)}$ ,use $J$ as the new variable.

Def $T[J]=<Tpsi>-S$

Example:

$S=frac{1}{2}<psi Mpsi>$

$Longrightarrow J=Mpsi Longrightarrow psi=M^{-1}J$

$T[J]=<JM^{-1}J>-frac{1}{2}<JM^{-1}J>=frac{JM^{-1}J}{2}$ .

we differ from $T$ by a constant because $S$ is a free theory.

because $S$ is a free theory,constant here represents vacuum.

$Z[J]=int D[psi]e^{-S[psi]+<J,psi>}$

$S_J[psi]=S[psi]-<J,psi>$

classical E M

$0=frac{delta S_J[psi]}{delta psi}=frac{delta S}{delta psi}-J$ .

so $J=frac{delta S}{delta psi}$ is simply the $Ecirc M$ of $S_{J}[psi]$ .

therefore legendre transform is simply classical physics.just like the path integral/partition is quantum.

$Gamma$ :Quantum effective action.

This lead to a natural question:

can the quantum $W[J]=lnZ[J]$ be the legendre transform of a $Gamma[psi]$

then quantum $W[J]$ can be "understand" as coming an effective "classical"-looking act $Gamma[psi]$ .then $Gamma[psi]$ treated as classical,captures the information the quantized theory of $S[psi]$ .