Prof. Sachindeo Vaidya (CHEP, IISc) | PDF
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\[
\varphi(x) \;=\; \int\frac{d^3k}{(2\pi)^3}\left(a_k e^{-ik\cdot x} + a^{\dagger}_k e^{ik\cdot x}\right)
\]
\[
\left[\varphi(x),\varphi(y)\right] \;=\; \int d\mu_k\left(e^{-ik\cdot(x-y)} - e^{ik\cdot(x-y)}\right)
\]
\(\downarrow\) a fun. rather than operator
\[
\int\frac{d^4k}{(2\pi)^4}\ \frac{-i}{(k^2-m^2)}\ e^{-ik\cdot(x-y)} \;\equiv\; \Delta(x-y)
\]
since \(\left[\varphi(x),\varphi(y)\right]\) obeys (Translational Invariance).
\(G(x,y)\) obeys
\[
\left(\Box_x+m^2\right)G(x,y) \;=\; -i\,\delta^4(x-y)
\]
\[
G(x,y) \;=\; \int\frac{d^4k}{(2\pi)^4}\ \frac{i}{k^2-m^2}\ e^{-ik\cdot(x-y)}
\]
\(\hookrightarrow\) \((k^0)^2 - \left(\vec{k}^2+m^2\right)\)
\[
=\; \int\frac{dk^0}{2\pi}\int\frac{d^3k}{(2\pi)^3}\ \frac{i}{(k^0)^2-\left(\vec{k}^2+m^2\right)}\ e^{-i\left(k\cdot(x-y)\right)}
\]
Choice of contour for \(k^0\) integral is important.
\[
\begin{aligned}
G_R(x,y) &= \text{retarded greens fun.}\\
G_A(x,y) &= \text{advanced greens fun.}\\
G_F(x,y) &= \text{Feynman's Green's fun. (Feynman propagator).}
\end{aligned}
\]
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\[
G_F(x,y) \;=\; \langle 0|\,T\left(\varphi(x)\varphi(y)\right)|0\rangle
\]
More generally we are interested in --
\[
G(x_1,x_2,x_3\ldots x_n) \;=\; \langle 0|\,T\left(\varphi(x_1)\varphi(x_2)\cdots\varphi(x_n)\right)|0\rangle
\]
we can write generating functional
\[
Z[J] \;=\; \sum_N \frac{1}{N!}\int d^4x_1\,d^4x_2\cdots d^4x_N\ G(x_1,x_2\cdots x_N)\ J(x_1)J(x_2)\cdots J(x_N)
\]
\[
=\; \langle 0|\,T\,e^{\int J(x)\varphi(x)}\,|0\rangle
\]
\(J\) = Source function.
\(Z\) obeys the eq\(^{n}\)
\[
\left(\partial_x^2+m^2\right)\frac{\delta Z[J]}{\delta J(x)} \;=\; \langle 0|T\left[\left(\partial_x^2+m^2\right)\varphi(x)\,e^{\int J\varphi}\right]|0\rangle \;=\; -i\,J(x)\,Z[J]
\]
\[
\left(\partial_x^2+m^2\right)\frac{\delta Z_0[J]}{\delta J(x)} \;=\; -i\,J(x)\,Z_0[J]
\]
\[
Z_0[J] \;=\; \mathcal{N}\ e^{\frac{1}{2}\int d^4x\,d^4y\ J(x)\,G(x,y)\,J(y)}
\]
Derivation (the marginal working in the notebook) :
\[
\mathcal{L} \;=\; -\tfrac{1}{2}(\partial\phi)^2 - \frac{m^2\phi^2}{2} - \lambda\phi^4 \qquad \cdots (1)
\]
\[
=\; \underbrace{-\tfrac{1}{2}(\partial\phi)^2-\frac{m^2\phi^2}{2}}_{\mathcal{L}_0} - J\phi - \lambda\phi^4\ \Big|_{J=0}
\]
\[
\langle 0|0\rangle_J \;=\; \int D\phi\ \exp i\!\int d^4x\left(\mathcal{L}_0 - J\phi - \lambda\phi^4\right)
\]
\[
\mathcal{L} \;=\; -\tfrac{1}{2}(\partial\phi)^2-\frac{m^2\phi^2}{2}-J\phi
\]
\[
=\; \int D\phi\ \exp i\!\int d^4x\left(\mathcal{L}_0-J\phi\right)\ \underbrace{\exp i\!\int d^4x\left(-\lambda\phi^4\right)}_{\text{we will expand}}
\]
\[
=\; \sum_n \frac{(-i\lambda)^n}{n!}\int D\phi\left(\exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)\right)\int\left(\phi^4\right)^n d^4x \qquad \cdots (2)
\]
let's see.
\[
\frac{\delta}{\delta J(y)}\left[\exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)\right] = i\!\int d^4x\ \frac{\delta}{\delta J(y)}\left[\mathcal{L}_0-J(x)\phi(x)\right]\times \exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)
\]
\[
=\; \left[i\!\int d^4x\ -\delta(x-y)\,\phi(x)\right]\times\exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)
\]
\[
=\; -i\,\phi(y)\,\exp i\!\int d^4x\left(\mathcal{L}_0-J\phi\right) \qquad \cdots (3)
\]
On Generalizing.
\[
\frac{\delta^{4n}}{\delta J(y)^{4n}}\left[\exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)\right] \;=\; (-i)^{4n}\,\phi^{4n}(y)\ \exp i\!\int d^4x\left[\mathcal{L}_0-J\phi\right]
\]
we got
\[
\phi^{4n}(y)\,\exp i\!\int d^4x\,(\mathcal{L}_0-J\phi) \;=\; \frac{\delta^{4n}}{\delta J(y)^{4n}}\left[\exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)\right] \qquad \cdots (4)
\]
use (4) in (2)
\[
=\; \sum_n \frac{(-i\lambda)^n}{n!}\int D\phi \int \frac{\delta^{4n}}{\delta J(y)^{4n}}\left[\exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)\right]
\]
\[
=\; \sum_n \frac{(-i\lambda)^n}{n!}\int\left(\frac{\delta}{\delta J(y)}\right)^{4n}\int D\phi\ \exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)
\]
\[
=\; \sum_n \frac{\left[\int -i\lambda\left(\frac{\delta}{\delta J(y)}\right)^4\right]^n}{n!}\ \underbrace{\int D\phi\ \exp i\!\int d^4x\,(\mathcal{L}_0-J\phi)}_{Z_0(J)}
\]
\[
=\; \left[\exp\int\left(-i\lambda\,\frac{\delta^4}{\delta J(y)^4}\right)\right] Z_0(J)
\]
\[
\mathcal{L} \;=\; \tfrac{1}{2}(\partial\varphi)^2 - \tfrac{1}{2}m^2\varphi^2 - \lambda\varphi^4
\]
E.O.M :
\[
\left(\partial_x^2+m^2\right)\varphi(x) + 4\lambda\varphi^3(x) \;=\; 0
\]
\[
Z[J] \;=\; \mathcal{N}\ e^{-i\lambda\int d^4x\left(\frac{\delta}{\delta J(x)}\right)^4}\ Z_0[J]
\]
\[
=\; e^{\frac{1}{2}\int J(x)G(x,y)J(y)}\ e^{\int J(x)G(x,y)\frac{\delta}{\delta\varphi(y)}}\ e^{\frac{1}{2}\int G(x,y)\frac{\delta}{\delta\varphi(x)}\frac{\delta}{\delta\varphi(y)}}\ e^{i\lambda\int\varphi^4(x)\,d^4x} \qquad \text{[?]}
\]
where
\[
V \;=\; \frac{\delta}{\delta\varphi(z_1)}\cdots\frac{\delta}{\delta\varphi(z_n)}\qquad \mathcal{F}[\varphi] \qquad \text{[?]}
\]
\[
\mathcal{F}[\varphi] \;=\; \mathcal{N}\ \exp\left(\tfrac{1}{2}\int G\,\frac{\delta}{\delta\varphi(x)}\frac{\delta}{\delta\varphi(y)}\right)\ \exp\left(S_{\text{int}}\right) \ , \qquad \left(S_{\text{int}}=\int\lambda\varphi^4\right)
\]
\[
V(x_1 x_2\cdots x_n) \;=\; \prod_j \int i\left(\partial_{x_j}^2+m^2\right) G(x_1,x_2\cdots x_n)
\]
Scattering amplitude \((S)\) :
\[
S\left(k_1\cdots k_n \longrightarrow k_{n+1}\cdots k_N\right)
\]
\[
=\; \prod_{j=1}^{n}\int u_{k_j}(x_j)\ i\left(\partial_{x_j}^2+m^2\right)\ \prod_{r=n+1}^{N} u^{*}_{k_r}(x_r)\ i\left(\partial_{x_r}^2+m^2\right)\ G(x_1x_2\cdots x_N)
\]
where \(x_j=(x_j^0,\vec{x}_j)\).
\[
S \;=\; -\tfrac{1}{4}\int d^4x\left(F_{\mu\nu}F^{\mu\nu} - A_\mu J^\mu\right)
\]
\[
=\; \tfrac{1}{2}\int d^4x\left[\left(\vec{E}^2-\vec{B}^2\right) - A_0 J_0 + \vec{A}\cdot\vec{J}\right]
\]
where,
\[
E_i = F_{0i}\ ,\qquad B_i = \tfrac{1}{2}\epsilon_{ijk}F_{jk}
\]
Maxwell's eq\(^{n}\) --
\[
\partial_\mu F^{\mu\nu} \;=\; J^\nu \ , \qquad \partial_\mu\left(\tfrac{1}{2}\epsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}\right) \;=\; 0
\]
\(A_\mu\) has a gauge symmetry, i.e. \(A_\mu \mapsto A_\mu + \partial_\mu\theta(t,x)\) gives same \(F_{\mu\nu}\), i.e. \((\vec{E}\) & \(\vec{B})\).
redundent eq\(^{n}\) --
\[
\partial_i E_i \;=\; J^0 \qquad (\text{Not a time evolution eq}^{n}).
\]
Fix the gauge (or getting rid of redundencies) : one choice is to choose \(A^0=0\). (\(A\) is a 4 vector in other frame \(A^0\) can take non zero value, so there is some tension b/w lorentz invariance & fixing the gauge.)
then split,
\[
A_i \;\equiv\; A_i^{T} + A^{L}_{i}
\]
\(A^L \longrightarrow \partial_i f\), and \(A^T\) obeys
\[
\partial_i A_i^{T} = 0 \quad \xrightarrow{\ \text{in }\vec{k}\text{ space}\ } \quad \left(\vec{k}\cdot\vec{A}^{T} = 0\right)
\]
\[
E_i \;=\; \partial_0 A_i - \partial_i A_0^{\,\,(=0)} \;=\; \partial_0 A_i^{T} + \partial_i\left(\partial_0 f\right)
\]
\[
\Rightarrow\quad \partial_i E_i \;=\; \partial_i\partial_i\left(\partial_0 f\right) \;=\; J_0
\]
\[
\left(\partial_0 f\right)(x^0,\vec{x}) \;=\; \int d^3y\ \boxed{G_c(\vec{x}-\vec{y})}\ J_0(x^0,\vec{y})
\]
\(\downarrow\) \(-\frac{1}{4\pi}\frac{1}{|\vec{x}-\vec{y}|}\)
\[
S \;=\; \tfrac{1}{2}\int d^4x\left[\left(\partial_0 A_j^{T}\right)\left(\partial_0 A_j^{T}\right) - \left(\partial_i A_j^{T}\right)\left(\partial_i A_j^{T}\right)\right]
\]
\[
\qquad +\ \int d^4x\int d^4y\ \tfrac{1}{2}\,J_0(x)\,G_c(x,y)\,J_0(y)\,\delta(x^0-y^0)\ +\ \int d^4x\ A_i^{T}J_i
\]
The condition \(A^0 = 0 = \vec{\nabla}\cdot\vec{A}\) is called radiation gauge. This is action for 2 massless field (the two transverse poles of \(A_i\)).
If \(J^\mu\) is zero, then we have
\[
\partial^2 A_i^{T} \;=\; 0
\]
plane wave sol\(^{n}\) \(e^{-ikx}\)
it has solutions
\[
A_i^{T}(x) \;=\; \sum_{k,\lambda}\left(a_{k\lambda}\,e_i^{(\lambda)}(k)\,u_k(x) + c.c\right)
\]
\(\downarrow\) polarisation vectors
\[
\sum_{\lambda=1,2} e_i^{(\lambda)}(k)\,e_j^{(\lambda)}(k) \;=\; \left(\delta_{ij} - \frac{k_ik_j}{\vec{k}^2}\right)
\]
There wouldn't be much necessary for the things to come!