Past exam of the quantum field theory course of the University of Oxford / 2016 / 1

For the real scalar field, the two derivatives entering the Euler-Lagrange field equation areConsequently the equation of motion isThe quartic interaction therefore adds a local cubic term to the free Klein-Gordon equation. Below, write $F=(\square +m^2)\varphi +\lambda {\varphi }^3/6$; thus $F=0$ means that the field is on shell.

Take a constant infinitesimal translation with fixed-coordinate field variation $\delta \varphi ={ϵ}^{\nu }{\partial }_{\nu }\varphi$. The associated change in the Lagrangian density is $\delta \mathcal{L}={ϵ}^{\nu }{\partial }_{\nu }\mathcal{L}={\partial }_{\mu }({ϵ}^{\nu }{\delta }^{\mu }{}_{\nu }\mathcal{L})$. Substituting this total derivative into the Noether current gives $J^{\mu }={ϵ}^{\nu }{T}^{\mu }{}_{\nu }$, whereThis is the canonical stress-energy tensor. The chosen sign of the translation parameter produces precisely this sign of the Noether current; reversing the parameter reverses the current without changing its conservation law.

To check current conservation directly, differentiate the tensor. The terms containing ${\partial }_{\nu }{\partial }_{\rho }\varphi$ cancel against the derivative of the kinetic term, leavingThus ${\partial }_{\mu }{T}^{\mu }{}_{\nu }=0$ on shell. Raising the second Lorentz index gives $T^{\mu \nu }=T^{\nu \mu }$: this scalar field already has a symmetric canonical stress-energy tensor.

First take the trace in four spacetime dimensions:The product rule gives $\square {\varphi }^2=2(\partial \varphi {)}^2+2\varphi \square \varphi$. For the massless equation of motion, $\varphi \square \varphi =-\lambda {\varphi }^4/6$, soBoth coefficients vanish forThe minus sign follows from the order ${\partial }^{\mu }{\partial }^{\nu }-{\eta }^{\mu \nu }\square$ used in the question's stress-energy tensor improvement. Commuting partial derivatives shows that this improvement has identically zero divergence, so it preserves the conservation law.

The precise trace identity, without using the equation of motion, isHence the massless tensor is traceless on shell. It is not traceless for arbitrary off shell configurations: a nonzero constant $\varphi =A$ with $\lambda \ne 0$ has ${\Omega }^{\mu }{}_{\mu }=\lambda A^4/6$. This is a classical result; it does not assert absence of a quantum trace anomaly.

Under the finite dilation, the volume element transforms as $d^4{x}^{\prime }=e^{4\alpha }{d}^{4}x$, whereas ${\partial }_{\mu }^{\prime }{\varphi }^{\prime }(x^{\prime })=e^{-2\alpha }{\partial }_{\mu }\varphi (x)$. Thus the kinetic term and quartic interaction both acquire the factor $e^{-4\alpha }$. Their factors cancel the Jacobian determinant, provingA mass term would instead acquire a net factor $e^{2\alpha }$ in the action and break this scale invariance.

For the Noether current, use the variation at the same coordinate, rather than comparing values at different coordinates:At $m=0$, the corresponding Lagrangian density variation is ${\delta }_0\mathcal{L}=-\alpha (4+x\cdot \partial )\mathcal{L}=-\alpha {\partial }_{\mu }(x^{\mu }\mathcal{L})$. Thus $J_0^{\mu }=-x^{\mu }\mathcal{L}$, and the stated Noether current convention givesThe frequently used dilatation current $D^{\mu }=x_{\nu }{T}^{\mu \nu }+\varphi {\partial }^{\mu }\varphi$ is $-j^{\mu }$; this is solely an overall parameter convention. In the convention $D^{\mu }=x_{\nu }{T}^{\mu \nu }-V^{\mu }$, the virial current is $V^{\mu }=-\varphi {\partial }^{\mu }\varphi$.

Using solution and the trace computed in solution gives the useful identityIt follows that $\boxed{{\partial }_{\mu }{j}^{\mu }=0}$ in the massless theory on shell. Equivalently, $x_{\nu }{\Omega }^{\mu \nu }$ is a conserved current, since its divergence is the trace of the improved stress-energy tensor on shell. These are classical scale invariance statements; the nonzero beta function in solution explains why the interacting quantum theory need not retain them.