Past exam of the theoretical physics major option course of the University of Oxford / 2022 / 6

The real scalar field action has spacetime translations and Lorentz invariance, together giving Poincaré symmetry. With a real scalar transformation law it also has parity and time reversal. A generic polynomial potential has no continuous internal symmetry. The additional field reversal $\varphi \mapsto -\varphi$ is a symmetry precisely when every nonzero odd-power coupling vanishes. The free massless special case also admits a constant shift of $\varphi$; generic interactions remove it.

In mass units, $[{\partial }_{\mu }]=1$ and the dimensionless action requires $[\mathcal{L}]=D$. Normalizing the kinetic term fixes the classical mass dimension of the scalar field:For each interaction, $[{\lambda }_n]+n[\varphi ]=D$, soThese are classical or engineering scaling dimensions; quantum corrections can change the scaling of operators and couplings. The scale transformation prescribed later has field weight one, which equals this canonical weight only at $D=4$.

Put $V(\varphi )=m^2{\varphi }^2/2+{\sum }_{n>2}{\lambda }_n{\varphi }^n/n!$. The Euler-Lagrange equation for the scalar field is $\square \varphi +V^{\prime }(\varphi )=0$, or explicitlyThe derivative of ${\varphi }^n/n!$ is ${\varphi }^{n-1}/(n-1)!$; all interaction terms therefore enter with the displayed plus sign for the chosen Minkowski metric.

The finite passive dilatation is $x^{\prime }=e^{-\alpha }x$, ${\varphi }^{\prime }(x^{\prime })=e^{\alpha }\varphi (x)$. To compare at the same argument, invert the coordinate change:Consequently the fixed-argument infinitesimal variation isThis also follows from $\delta \varphi =\alpha \varphi$ and $\delta x^{\mu }=-\alpha x^{\mu }$. The plus sign on the derivative term is a consequence of this specified contraction of coordinates.

Under the finite dilatation, $d^D{x}^{\prime }=e^{-D\alpha }{d}^{D}x$, ${\partial }_{\mu }^{\prime }{\varphi }^{\prime }(x^{\prime })=e^{2\alpha }{\partial }_{\mu }\varphi (x)$ and ${\varphi }^{\prime }(x^{\prime }{)}^n=e^{n\alpha }\varphi (x{)}^n$. Keeping the couplings fixed givesThus the full finite change isFor invariance for arbitrary fields and all constant $\alpha$, the fixed nonzero kinetic term requires $D=4$. Then the mass term requires $m=0$, and every nonzero coupling must have $n=4$. Henceincluding the free choice ${\lambda }_4=0$. At $D=4$ the canonical mass dimension is $[\varphi ]=1$ and $[{\lambda }_n]=4-n$, so only the quartic interaction has a dimensionless coupling. A mass introduces a scale. This recovers the same result from solution.

The fixed field weight in the question matters: if it were replaced by the general canonical weight $(D-2)/2$, the kinetic term could be invariant in other dimensions, and a nonzero monomial interaction would require $n=2D/(D-2)$ for $D\ne 2$. That is a different transformation from the one specified here. The result above is classical; quantum running generally breaks the quartic theory's scale invariance.

In the classically scale-invariant case from solution, use $\partial \mathcal{L}/\partial ({\partial }_{\mu }\varphi )={\partial }^{\mu }\varphi$, $\bar{\delta }\varphi =\alpha (\varphi +x^{\nu }{\partial }_{\nu }\varphi )$ and $\delta x^{\mu }=-\alpha x^{\mu }$. Dividing the supplied Noether current expression by $\alpha$ givesSince ${\theta }^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$, this isThe first term in this dilatation current comes from the intrinsic weight of the scalar field, while the second comes from moving its spacetime argument. Using the canonical energy-momentum tensor alone without the first term would not give the stated conserved current.

For a translation-invariant potential, the on-shell energy-momentum tensor obeys ${\partial }_{\mu }{\theta }^{\mu \nu }=0$. The divergence of the dilatation current is thereforeUse the equation of motion $\square \varphi =-V^{\prime }(\varphi )$ to obtain the useful general identityFor $D=4$, $m=0$ and only ${\lambda }_4$ nonzero, every term vanishes. More explicitly, $\varphi \square \varphi =-{\lambda }_4{\varphi }^4/6$ while ${\theta }^{\mu }{}_{\mu }=-(\partial \varphi {)}^2+{\lambda }_4{\varphi }^4/6$; adding the first kinetic term cancels both remaining contributions. Thus ${\partial }_{\mu }{j}^{\mu }=0$, and $\int d^{3}x{j}^0$ is conserved when the boundary flux vanishes. Outside the parameter conditions of solution, the boxed formula displays the classical sources of scale breaking; the same current need not be conserved.

For a free real scalar field in four spacetime dimensions, $\mathcal{L}=(\partial \varphi {)}^2/2-m^2{\varphi }^2/2$ and $\pi =\dot{\varphi }$. Canonical quantization promotes these canonical fields to Hermitian operator-valued distributions with equal-time canonical commutation relationsExpand the solution of the Klein-Gordon equation into positive- and negative-frequency modes:Hermiticity pairs the two coefficients as adjoints. The field commutators are equivalent toFor example, inserting the expansion in $[\varphi ,\dot{\varphi }]$ yields two equal Fourier integrals, each with coefficient $i/2$, and hence $i{\delta }^{(3)}$. The vacuum obeys $a(\mathbf{k})|0⟩=0$, and repeated creation operators generate the symmetric bosonic Fock space. Normal ordering of the quadratic Hamiltonian removes its vacuum constant and gives $H=\int d{\Pi }_k{\omega }_k{a}^{†}a$; the same computation for the spatial momentum gives the stipulated four-momentum $P^{\mu }=\int d{\Pi }_k{k}^{\mu }{a}^{†}a$.

The oscillator algebra now impliesConsequentlyIndeed, ${\partial }^{\mu }{e}^{\mp ikx}=\mp i{k}^{\mu }{e}^{\mp ikx}$. For $\mu =0$, the result is $[H,\varphi ]=-i\dot{\varphi }$, which is the Heisenberg equation of motion with the correct sign. As usual, the field identities are interpreted after smearing the operator-valued distributions, or on a suitable common finite-particle domain.