Lecture 2 Exercise

(1) Please finish the integral

$$\begin{array}{}\text{(35)}& \int \frac{d^3\overrightarrow{k}}{(2\pi {)}^3}\frac{e^{+i\overrightarrow{k}\cdot \overrightarrow{x}}}{{\overrightarrow{k}}^2+m^2}\end{array}$$

The integral equals

$$\int \frac{d^3\overrightarrow{k}}{(2\pi {)}^3}\frac{e^{+i\overrightarrow{k}\cdot \overrightarrow{x}}}{{\overrightarrow{k}}^2+m^2}=\frac{1}{(2\pi {)}^2}\int \frac{k^2{\mathrm{d}}^{}k}{k^2+m^2}\int \sin \theta {\mathrm{d}}^{}\theta e^{ikx\cos \theta }=\frac{1}{(2\pi {)}^{2}x}{\int }_0^{\mathrm{\infty }}\frac{2u\sin u{\mathrm{d}}^{}u}{u^2+m^2{x}^2}$$

We’ll use the residue theorem to treat wirth the radial part

$$\begin{array}{c}\frac{1}{(2\pi {)}^{2}x}{\int }_0^{\mathrm{\infty }}\frac{2u\sin u{\mathrm{d}}^{}u}{u^2+m^2{x}^2}=\frac{1}{(2\pi {)}^{2}x}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{1}{2i}\frac{2u{e}^{iu}{\mathrm{d}}^{}u}{u^2+m^2{x}^2}\\ =\frac{1}{2\pi x}\mathrm{Res}[\frac{u{e}^{iu}}{u^2+m^2{x}^2},u=imx]=\frac{e^{-mx}}{4\pi x}\end{array}$$

(2) Starting from the action of the photon field (without source), please derive its gauge invariant energy-momentum tensor $T_{\mu \nu }$ and show that it is traceless.

The Lagrangian of the photon field is

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

And the energy-momentum tensor is defined as

$$T^{\mu \nu }=\frac{1}{{\mu }_0}(F^{\mu \alpha }{F}^{\nu }{}_{\alpha }-\frac{1}{4}{\eta }^{\mu \nu }{F}^{\rho \sigma }{F}_{\rho \sigma })$$

Contract the tensor, and we can find that

$${\eta }_{\mu \nu }{T}^{\mu \nu }=\frac{1}{{\mu }_0}({\eta }_{\mu \nu }{F}^{\mu \alpha }{F}^{\nu }{}_{\alpha }-\frac{1}{4}{\eta }^{\mu \nu }{\eta }_{\mu \nu }{F}^{\rho \sigma }{F}_{\rho \sigma })=\frac{1}{{\mu }_0}(F^{\mu \alpha }{F}_{\mu \alpha }-F^{\rho \sigma }{F}_{\rho \sigma })=0$$

is traceless.

(3) Please derive (26).

But a direct calculation gives

$$\begin{array}{}\text{(26)}& R_y^{-1}(\theta )L_x({\eta }_2)L_z({\eta }_1)e_{\pm }^{\mu }(k)=e_{\pm }^{\mu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}{k}^{\mu }\end{array}$$

Write the matrix for the transformation directly

$$R_y^{-1}(\theta )L_x({\eta }_2)L_z({\eta }_1)=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& \cos \theta & 0& -\sin \theta \\ 0& 0& 1& 0\\ 0& \sin \theta & 0& \cos \theta \end{array}\right)\left(\begin{array}{cccc}\cosh {\eta }_2& \sinh {\eta }_2& 0& 0\\ \sinh {\eta }_2& \cosh {\eta }_2& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)\left(\begin{array}{cccc}\cosh {\eta }_1& 0& 0& \sinh {\eta }_1\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ \sinh {\eta }_1& 0& 0& \cosh {\eta }_1\end{array}\right)$$

The calculation gives out that this product equals

$$\left(\begin{array}{cccc}\cosh {\eta }_2\cosh {\eta }_1& \sinh {\eta }_2& 0& \cosh {\eta }_2\sinh {\eta }_1\\ \cos \theta \sinh {\eta }_2\cosh {\eta }_1-\sin \theta \sinh {\eta }_1& \cos \theta \cosh {\eta }_2& 0& \cos \theta \sinh {\eta }_2\sinh {\eta }_1-\sin \theta \cosh {\eta }_1\\ 0& 0& 1& 0\\ \sin \theta \sinh {\eta }_2\cosh {\eta }_1+\cos \theta \sinh {\eta }_1& \sin \theta \cosh {\eta }_2& 0& \sin \theta \sinh {\eta }_2\sinh {\eta }_1+\cos \theta \cosh {\eta }_1\end{array}\right)$$

And the relation between ${\eta }_1$, ${\eta }_2$ and $\theta$ satisfies

$$\cos \theta =\cosh {\eta }_1-\sinh {\eta }_1$$

and

$$\tan \theta =\sinh {\eta }_2$$

So we can get $\cosh {\eta }_2=\frac{1}{\cos \theta }$, and

$$\sinh {\eta }_1=\frac{{\sin }^2\theta }{2\cos \theta }\cosh {\eta }_1=\frac{1+{\cos }^2\theta }{2\cos \theta }$$

So the whole matrix, using $\theta$ to depict, is

$$\left(\begin{array}{cccc}\frac{1+{\cos }^2\theta }{2{\cos }^2\theta }& \tan \theta & 0& 2{\tan }^2\theta \\ \sin \theta \cos \theta & 1& 0& -\sin \theta \cos \theta \\ 0& 0& 1& 0\\ \frac{{\sin }^2\theta (1+2{\cos }^2\theta )}{2{\cos }^2\theta }& \tan \theta & 0& \frac{2{\cos }^4\theta -{\cos }^2\theta +1}{2{\cos }^2\theta }\end{array}\right)=I+\left(\begin{array}{cccc}\frac{1-{\cos }^2\theta }{2{\cos }^2\theta }& \tan \theta & 0& 2{\tan }^2\theta \\ \sin \theta \cos \theta & 0& 0& -\sin \theta \cos \theta \\ 0& 0& 0& 0\\ \frac{{\sin }^2\theta (1+2{\cos }^2\theta )}{2{\cos }^2\theta }& \tan \theta & 0& \frac{2{\cos }^4\theta -3{\cos }^2\theta +1}{2{\cos }^2\theta }\end{array}\right)$$

Therefore, put the total transformation on the vectors

$$e_{\pm }^{\mu }(k)=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\ 1\\ \pm i\\ 0\end{array}\right)$$

we’ll get

$$I{e}_{\pm }^{\mu }(k)+\left(\begin{array}{cccc}\frac{1-{\cos }^2\theta }{2{\cos }^2\theta }& \tan \theta & 0& 2{\tan }^2\theta \\ \sin \theta \cos \theta & 0& 0& -\sin \theta \cos \theta \\ 0& 0& 0& 0\\ \frac{{\sin }^2\theta (1+2{\cos }^2\theta )}{2{\cos }^2\theta }& \tan \theta & 0& \frac{2{\cos }^4\theta -3{\cos }^2\theta +1}{2{\cos }^2\theta }\end{array}\right)e_{\pm }^{\mu }(k)=e_{\pm }^{\mu }(k)+\frac{1}{\sqrt{2}}\left(\begin{array}{c}\tan \theta \\ 0\\ 0\\ \tan \theta \end{array}\right)$$$$e_{\pm }^{\mu }(k)+\frac{1}{\sqrt{2}}\left(\begin{array}{c}\tan \theta \\ 0\\ 0\\ \tan \theta \end{array}\right)=e_{\pm }^{\mu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}{k}^{\mu }$$

(4) Please prove (29).

What are they? $R_z(\theta )$ is clearly an SO(2) generator; For 2 $R\times L$-type transform, taking $\theta \to 0$ we get generators:

$$\begin{array}{}\text{(27)}& R_z(\theta )=1-i\theta \underset{J_z}{\underbrace{\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& i& 0\\ 0& -i& 0& 0\\ 0& 0& 0& 0\end{array}\right)}},R_y^{-1}{L}_x{L}_z=1-i\theta \underset{T_x}{\underbrace{\left(\begin{array}{cccc}0& i& 0& 0\\ i& 0& 0& -i\\ 0& 0& 0& 0\\ 0& i& 0& 0\end{array}\right)}}\end{array}$$$$\begin{array}{}\text{(28)}& R_x^{-1}{L}_y{L}_z=1-i\theta \underset{T_y}{\underbrace{\left(\begin{array}{cccc}0& 0& i& 0\\ 0& 0& 0& 0\\ i& 0& 0& -i\\ 0& 0& i& 0\end{array}\right)}}\end{array}$$$$\begin{array}{}\text{(29)}& [J_z,T_x]=i{T}_y,[J_z,T_y]=-i{T}_x,[T_x,T_y]=0\end{array}$$

The little group is ISO(2).

$$\begin{gathered} [J_z,T_x]=-\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& i& 0\\ 0& -i& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{cccc}0& i& 0& 0\\ i& 0& 0& -i\\ 0& 0& 0& 0\\ 0& i& 0& 0\end{array}\right)+\left(\begin{array}{cccc}0& i& 0& 0\\ i& 0& 0& -i\\ 0& 0& 0& 0\\ 0& i& 0& 0\end{array}\right)\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& i& 0\\ 0& -i& 0& 0\\ 0& 0& 0& 0\end{array}\right) \\ =-\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& -1\\ 0& 0& 0& 0\end{array}\right)+\left(\begin{array}{cccc}0& 0& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& -1& 0\end{array}\right)=\left(\begin{array}{cccc}0& 0& -1& 0\\ 0& 0& 0& 0\\ -1& 0& 0& 1\\ 0& 0& -1& 0\end{array}\right)=i{T}_y \end{gathered}$$$$\begin{gathered} [J_z,T_y]=-\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& i& 0\\ 0& -i& 0& 0\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{cccc}0& 0& i& 0\\ 0& 0& 0& 0\\ i& 0& 0& -i\\ 0& 0& i& 0\end{array}\right)+\left(\begin{array}{cccc}0& 0& i& 0\\ 0& 0& 0& 0\\ i& 0& 0& -i\\ 0& 0& i& 0\end{array}\right)\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& i& 0\\ 0& -i& 0& 0\\ 0& 0& 0& 0\end{array}\right) \\ =-\left(\begin{array}{cccc}0& 0& 0& 0\\ -1& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)+\left(\begin{array}{cccc}0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right)=\left(\begin{array}{cccc}0& 1& 0& 0\\ 1& 0& 0& -1\\ 0& 0& 0& 0\\ 0& 1& 0& 0\end{array}\right)=-i{T}_x \end{gathered}$$$$\begin{gathered} [T_x,T_y]=-\left(\begin{array}{cccc}0& i& 0& 0\\ i& 0& 0& -i\\ 0& 0& 0& 0\\ 0& i& 0& 0\end{array}\right)\left(\begin{array}{cccc}0& 0& i& 0\\ 0& 0& 0& 0\\ i& 0& 0& -i\\ 0& 0& i& 0\end{array}\right)+\left(\begin{array}{cccc}0& 0& i& 0\\ 0& 0& 0& 0\\ i& 0& 0& -i\\ 0& 0& i& 0\end{array}\right)\left(\begin{array}{cccc}0& i& 0& 0\\ i& 0& 0& -i\\ 0& 0& 0& 0\\ 0& i& 0& 0\end{array}\right) \\ =\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)+\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)=0 \end{gathered}$$

(5) Please derive (33).

The polarization tensor $e_{\pm }^{\mu \nu }$ is symmetric, traceless, and transverse. Once again, $e_{\pm }^{\mu \nu }$ is not a Lorentz tensor. Under the two “Abelian” LGTs:

$$\begin{array}{}\text{(33)}& e_{\pm }^{\mu \nu }(k)\to e_{\pm }^{\mu \nu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}[e_{\pm }^{\mu }(k)+\frac{\tan \theta }{2\sqrt{2}{k}^0}{k}^{\mu }]k^{\nu }+(\mu \leftrightarrow \nu )\end{array}$$

We’ve already known that $e_{\pm }^{\mu \nu }(k)=e_{\pm }^{\mu }(k)e_{\pm }^{\nu }(k)$, so transformation rule for this tensor is

$$\begin{array}{c}{e}_{\pm }^{\mu \nu }(k)\to (R_y^{-1}(\theta )L_x({\eta }_2)L_z({\eta }_1)e_{\pm }^{\mu }(k))(R_y^{-1}(\theta )L_x({\eta }_2)L_z({\eta }_1)e_{\pm }^{\nu }(k))\\ =(e_{\pm }^{\mu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}{k}^{\mu })(e_{\pm }^{\nu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}{k}^{\nu })\\ =e_{\pm }^{\mu }(k)e_{\pm }^{\nu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}(e_{\pm }^{\mu }(k)k^{\nu }+k^{\mu }{e}_{\pm }^{\nu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}{k}^{\mu }{k}^{\nu })\\ =e_{\pm }^{\mu \nu }(k)+\frac{\tan \theta }{\sqrt{2}{k}^0}[e_{\pm }^{\mu }(k)+\frac{\tan \theta }{2\sqrt{2}{k}^0}{k}^{\mu }]k^{\nu }+(\mu \leftrightarrow \nu )\end{array}$$