Lecture 7

Chasse_neige

QFT Perspective

General Remarks

What is field theory: relacistic quantum theory with Hamiltonian

$$H(t)=\int {\mathrm{d}}^{3}x\mathcal{H}(t,\overrightarrow{x})$$

Where $\mathcal{H}(t,\overrightarrow{x})$ is polynomial function of local operators $\varphi (t,\overrightarrow{x})$ and $\partial \varphi$.

We need a dynamical theory with annihilation and creation operators.

Causality: $a,a^{†}\to \varphi (t,\overrightarrow{x})$

$$\varphi (x)=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}\frac{1}{\sqrt{2E_k}}[e^{ik\cdot x}{a}_{\overrightarrow{k}}+c.c.]$$

Lagrangian:

$$\mathcal{L}=-\frac{1}{2}({\partial }_{\mu }\varphi {)}^2-\frac{1}{2}{m}^2{\varphi }^2+\text{interaction}$$

In the interaction part, operators $(\partial \varphi {)}^2$ and ${\varphi }^2$ are not allowed! We normalize $\varphi$ by the kinetic term, call these canonically normalized fields bare fields.

Using $⟨\overrightarrow{p}|{\varphi }_R(0)|\mathrm{\Omega }⟩=1$ to normalize the fields

$$\mathcal{L}=\underset{\text{renormalized field}}{\underbrace{-\frac{1}{2}Z({\partial }_{\mu }{\varphi }_R{)}^2-\frac{1}{2}Z{\varphi }_R^2}}+[\underset{\text{conter term}}{\underbrace{{\delta }_Z({\partial }_{\mu }{\varphi }_R{)}^2+{\delta }_m{\varphi }_R^2}}+\cdots ]$$

Perturbation Theory

$$S=T{e}^{-i\int {\mathrm{d}}^{}t{H}_{int}(t)}$$$$|{\varphi }_0⟩=|p_1\cdots p_N⟩=\left(\prod _i\sqrt{2E_{p_i}}{a}_{p_i}^{†}\right)|0⟩$$$$G_{\varphi }(x,y)\equiv ⟨0|T\{\varphi (x)\varphi (y)\}|0⟩$$$$G_{\varphi }=\int \frac{{\mathrm{d}}^{4}p}{(2\pi {)}^4}\frac{-i}{p^2+m^2-iϵ}{e}^{ip\cdot (x-y)}$$

Photon Field

We use the polarization vector as an adapter

$$A(x)=\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}\frac{1}{\sqrt{2E_k}}\sum _{s=\pm 1}[e^{ik\cdot x}{e}_{\mu }^{(s)}{a}_{s,k}+c.c.]$$

Under LGT: $e_{\mu }\to e_{\mu }+\alpha k_{\mu }$

$$\delta A_{\mu }(x)={\partial }_{\mu }\int \frac{{\mathrm{d}}^{3}k}{(2\pi {)}^3}\frac{1}{\sqrt{2E_k}}\sum _{s=\pm 1}[-i\alpha e^{ik\cdot x}{a}_{s,k}+c.c.]$$

So the Lagrangian

$$\mathcal{L}=-\frac{1}{2}({\partial }_{\mu }{A}_{\nu })({\partial }^{\mu }{A}^{\nu })$$

is actually not Lorentz invariant. So we exploit the index contraction, and we can get the L-int Lagrangian

$$\mathcal{L}=-\frac{1}{4}{F}_{\mu \nu }{F}^{\mu \nu }$$

We can thus using $F_{\mu \nu }$ to couple with matter.

Matter Theory: $S_M$ with global U(1) symmetry

$$\delta S_M=\int {\mathrm{d}}^{4}x{J}^{\mu }{\partial }_{\mu }\alpha (x)$$

If we identify U(1) with LGT of $A_{\mu }$

$$S_M+e\int {\mathrm{d}}^{4}x{J}^{\mu }{A}_{\mu }$$

is L-inv.

Photon Propagator

$$\begin{array}{c}{G}_{\mu \nu }=⟨0|T\{A_{\mu }(x)A_{\nu }(y)\}|0⟩\\ =\int \frac{{\mathrm{d}}^{4}p}{(2\pi {)}^4}\frac{-i}{p-iϵ}{e}^{ip\cdot (x-y)}\times P_{\mu \nu }(p)\end{array}$$

Where $P$ is described by the “adaptors”

$$\begin{array}{c}{P}_{\mu \nu }(p)=\sum _{h=\pm 1}{e}_{\mu }^{(h)}(p)e_{\nu }^{(h)\ast }(p)\\ =A{\eta }_{\mu \nu }+B\times \frac{p_{\mu }{p}_{\nu }}{p^2}\end{array}$$

Take the photon propagating in the z direction , we can know that

$$\{\begin{array}{l}A=1\\ B\text{undefined}\end{array}$$

For a current conserved theory, the propagator can take the form

$$G_{\mu \nu }(p)=\frac{-i{\eta }_{\mu \nu }}{p^2-iϵ}$$

Or in some QFT textbooks, we can do this by calculating the inverse of the Lagrangian

$$\mathcal{L}+{\mathcal{L}}_{GF}$$

Consider the current between two sources, the amplitude is

$$iM=(ie{)}^2{J}_A^{\mu }(-p)\frac{-i{\eta }_{\mu \nu }}{p^2-iϵ}{J}_B^{\nu }(p)$$

Take $p=(E,0,0,p)$ which is off-shell, the charge conservation gives that

$$0=p_{\mu }{J}_{A,B}^{\mu }=-E{J}_{A,B}^0+p{J}_{A,B}^3$$

So

$$M=\frac{e^2(-{\rho }_A{\rho }_B+J_A^i{J}_B^i)}{-E^2+p^2-iϵ}=-\frac{e^2{\rho }_A{\rho }_B}{p^2}+\frac{J_A^1{J}_B^1+J_A^2{J}_B^2}{-E^2+p^2-iϵ}$$

Doing Fourier transform, the first term gives out

$$\delta (t_A-t_B)\frac{e^2{\rho }_A{\rho }_B}{4\pi |x_A-x_B|}$$

The second term is proportional to

$$J_A^1{J}_B^1+J_A^2{J}_B^2=\sum _{h=\pm 1}{J}_A^{(h)}{J}_B^{(h)}$$

Graviton Field

Similarly, the polarization tensor

$$e_{\mu \nu }^{(\pm 2)}\to e_{\mu \nu }^{(\pm 2)}+2{\xi }_{(\mu }{k}_{\nu )}$$

Define the brakets as

$$A_{(\mu }{B}_{\nu )}=\frac{1}{2}(A_{\mu }{B}_{\nu }+A_{\nu }{B}_{\mu })$$

We guess the Lagrangian of the graviton field

$$\mathcal{L}=-\frac{1}{4}{\partial }_{\lambda }{h}_{\mu \nu }{\partial }^{\lambda }{h}^{\mu \nu }+\cdots$$

Adding some terms to make the Lagrangian gauge invariant, with options:

1. We do the genuine gauge transform with ${\partial }_{\mu }{\xi }^{\mu }=0$ and $h_{\mu \nu }$ traceless.
2. Allow $h\ne 0$, using arbitrary ${\xi }^{\mu }$.

Try option 2 ($h\ne 0$)

$$\mathcal{L}=\frac{1}{4}{h}_{\mu \nu }◻h^{\mu \nu }+ah◻h+bh{\partial }^{\mu }{\partial }^{\nu }h+c{h}_{\mu \nu }{\partial }^{\mu }{\partial }_{\lambda }{h}^{\nu \lambda }$$

Using the gauge invariance constrain, we can get

$$\mathcal{L}=\frac{1}{4}{h}_{\mu \nu }◻h^{\mu \nu }-\frac{1}{4}h◻h+\frac{1}{2}h{\partial }^{\mu }{\partial }^{\nu }h-\frac{1}{2}{h}_{\mu \nu }{\partial }^{\mu }{\partial }_{\lambda }{h}^{\nu \lambda }$$

This is consistent with the second order term of

$$S=\frac{M_{pl}^2}{2}\int {\mathrm{d}}^{4}x\sqrt{-g}R$$

Couple $h_{\mu \nu }$ to matter, we have tensors in forms like

$${\partial }^{\mu }{\partial }^{\nu }{h}_{\mu \nu }-{\partial }^{2}h$$

but this will not work. We should choose “minimal coupling” in the form

$${\mathcal{L}}_{int}=\frac{\kappa }{2}{h}_{\mu \nu }{T}^{\mu \nu }$$

Why option 1 doesn’t work:

Choose $h=0$ and ${\partial }_{\mu }{\xi }^{\mu }=0$

$${\mathcal{L}}_{UM}=\frac{1}{4}{h}_{\mu \nu }◻h^{\mu \nu }-\frac{1}{2}{h}_{\mu \nu }{\partial }^{\mu }{\partial }_{\lambda }{h}^{\nu \lambda }$$

This is called Unimodular Gravity.

$${{\mathcal{L}}^{\prime }}_{UM}={\mathcal{L}}_{UM}+Ah\Rightarrow S_{EH}+\int {\mathrm{d}}^{4}xA(\sqrt{-g}-1)$$

How does matter couple to $h_{\mu \nu }$ in unimodular gravity

The energy momentum tensors with the constrain

$${\partial }_{\mu }{T}^{\mu \nu }={\partial }^{\nu }\mathrm{\Phi }(x)$$

can couple with $h_{\mu \nu }$. The coupling term is

$${\mathcal{L}}_{int}=\frac{\kappa }{2}{h}_{\mu \nu }(T^{\mu \nu }-\frac{1}{4}{\eta }^{\mu \nu }T)$$

Graviton Propagator

Choosing the gauge fixing term as

$${\mathcal{L}}_{GF}=-\frac{1}{2}{\left({\partial }^{\mu }{\overline{h}}_{\mu \nu }\right)}^2$$

The total Lagrangian is

$$\begin{array}{c}\mathcal{L}={\mathcal{L}}_{free}+{\mathcal{L}}_{GF}=\frac{1}{4}{h}_{\mu \nu }◻{\overline{h}}^{\mu \nu }\\ =\frac{1}{8}{h}_{\mu \nu }(2{\eta }^{\mu (\rho }{\eta }^{\sigma )\nu }-{\eta }^{\mu \nu }{\eta }^{\rho \sigma })◻h_{\rho \sigma }\end{array}$$

The propagator is described by the equation

$$\frac{1}{2}(2{\eta }^{\lambda (\mu }{\eta }^{\nu )\kappa }-{\eta }^{\lambda \kappa }{\eta }^{\mu \nu }){◻}_x{G}_{\mu \nu \rho \sigma }(x,y)=i{\delta }^{(4)}(x-y){\delta }_{(\rho }^{\lambda }{\delta }_{\sigma )}^{\kappa }$$

This gives out the propagator

$$G_{\mu \nu ,\rho \sigma }(p)=-\frac{i}{p^2-iϵ}[2{\eta }_{\mu (\rho }{\eta }_{\sigma )\nu }-{\eta }_{\mu \nu }{\eta }_{\rho \sigma }]$$

Doing the same trick between the two sources

$$\begin{array}{c}iM(p)={\left(\frac{i\kappa }{2}\right)}^2{T}_A^{\mu \nu }(-p)G_{\mu \nu ,\rho \sigma }(p)T_B^{\rho \sigma }(p)\\ =\frac{i{\kappa }^2}{4p^2}[2T_A^{\mu \nu }(-p)T_{B\mu \nu }(p)-T_A(p)T_B(p)]\end{array}$$

Using the current conservation law

$$0=p_{\mu }{T}_{A,B}^{\mu \nu }=-E{T}_{A,B}^{0\nu }+p{T}_{A,B}^{3\nu }$$

And this will give out the amplitude

$$\begin{array}{c}M(p)=\frac{{\kappa }^2}{4(-E^2+p^2)}[(T_A^{11}-T_A^{22})(T_B^{11}-T_B^{22})+4T_A^{12}{T}_B^{12}]\\ +\frac{{\kappa }^2}{4p^2}(1-\frac{E^2}{p^2})T_A^{00}{T}_B^{00}\\ +\frac{{\kappa }^2}{4p^2}[T_A^{00}(T_B^{11}+T_B^{22})+(T_A^{11}+T_A^{22})T_B^{00}-4(T_A^{01}{T}_B^{01}+T_A^{02}{T}_B^{02})]\end{array}$$

The first term is on-shell poles of the gravitons, the second term is the Newtonian potential, and the third term gives out the velocity-dependent potential.