Example 2.4 (Poincaré symmetries of the real Klein-Gordon field). Recall our conventions for Poincaré transformations \(x^{\prime \mu } =\Lambda ^{\mu }_{~~\nu } \,x^\nu + b^\mu \) from Section 1.4. A real scalar field transforms under Poincaré transformations as \(\Phi ^\prime (x^\prime ) = \Phi (x)\), i.e. the value of the transformed field \(\Phi ^\prime \) at the transformed point \(x^\prime \) is equal to the value of \(\Phi \) at \(x\). This point of view on Poincaré transformations is called passive since the field \(\Phi \) does not get transformed to a new field, but it is only expressed in a different set of coordinates \(x^\prime \). The active point of view on Poincaré transformations consists of evaluating the transformed field \(\Phi ^\prime \) at the old point \(x\) and interpreting this as a transformation \(T : \Phi (x) \mapsto T\Phi (x) = \Phi ^\prime (x)\) on the set of fields. Inverting the Poincaré transformation \(x^\prime =\Lambda x + b\) gives \(x = \Lambda ^{-1}(x^\prime - b)\), which when combined with the transformation law \(\Phi ^\prime (x^\prime ) = \Phi (x)\) yields

\begin{flalign} \label {eqn:Poincarescalarfield} T : \Phi (x) \longmapsto T\Phi (x) = \Phi ^\prime (x) = \Phi \big (\Lambda ^{-1}(x - b)\big )\quad . \end{flalign} In index notation, the argument of \(\Phi \) reads as \((\Lambda ^{-1}(x - b))^\nu = (x^\rho - b^\rho )\,\Lambda _\rho ^{~~\nu }\). Using the chain rule for partial differentiation, we find that the partial derivatives of the transformed field \(T\Phi \) are given by

\begin{flalign} \nn (\partial _\mu T\Phi )(x) &= \frac {\partial }{\partial x^\mu } \Phi \big (\Lambda ^{-1}(x - b)\big )\\ \nn &= \frac {\partial \big ((x^\rho - b^\rho )\,\Lambda _\rho ^{~~\nu }\big )}{\partial x^\mu }~~(\partial _\nu \Phi )\big (\Lambda ^{-1}(x - b)\big )\\ &=\Lambda _{\mu }^{~~\nu }\,(\partial _\nu \Phi )\big (\Lambda ^{-1}(x - b)\big )\quad , \end{flalign} which we recognize as the transformation law of a covector field under (active) Poincaré transformations. With these preparations, we can now confirm Poincaré invariance of the Klein-Gordon action (2.8) by a simple calculation

\begin{flalign} \nn S_{\mathrm {KG}}[T\Phi ] &= \int _{\bbR ^d}-\frac {1}{2}\,\bigg (\eta ^{\mu \nu }\,(\partial _\mu T\Phi )(x) \, (\partial _{\nu }T\Phi )(x) + m^2\,(T\Phi )^2(x)\bigg )\,\dd x\\ \nn &= \int _{\bbR ^d}-\frac {1}{2}\,\bigg (\eta ^{\mu \nu }\,\Lambda _{\mu }^{~~\rho }\,\Lambda _\nu ^{~~\sigma }\, (\partial _\rho \Phi )\big (\Lambda ^{-1}(x - b)\big ) \,(\partial _\sigma \Phi )\big (\Lambda ^{-1}(x - b)\big ) + m^2\,\Phi ^2\big (\Lambda ^{-1}(x - b)\big )\bigg )\,\dd x\\ \nn &= \int _{\bbR ^d}-\frac {1}{2}\,\bigg (\eta ^{\rho \sigma }\, (\partial _\rho \Phi )\big (\Lambda ^{-1}(x - b)\big ) \,(\partial _\sigma \Phi )\big (\Lambda ^{-1}(x - b)\big ) + m^2\,\Phi ^2\big (\Lambda ^{-1}(x - b)\big )\bigg )\,\dd x\\ &= \int _{\bbR ^d}-\frac {1}{2}\,\bigg (\eta ^{\rho \sigma }\, (\partial _\rho \Phi )(y) \,(\partial _\sigma \Phi )(y) + m^2\,\Phi ^2(y)\bigg )\,\dd y = S_{\mathrm {KG}}[\Phi ]\quad . \end{flalign} The third step uses the property (1.19) of the inverse Minkowski metric. In the fourth step we have changed the integration variables to \(y:= \Lambda ^{-1}(x- b)\) and used the change of variables formula \(\dd y = \vert \det (\Lambda ^{-1})\vert \, \dd x = \dd x\) for the volume element, together with the fact that \(\det (\Lambda ) = \pm 1\) which follows by taking determinants on both sides of (1.17). (Since as the default we consider proper and orthochronous Poincaré transformations, we even have \(\det (\Lambda ) = + 1\).)

Example 2.5 (Internal symmetries of the real and complex Klein-Gordon field). For a real scalar field \(\Phi (x)\in \bbR \), we can consider the transformation \(T : \Phi (x) \mapsto T\Phi (x) = -\Phi (x)\) that flips the sign. This transformation is a symmetry of the Klein-Gordon action (2.8), as one can easily check

\begin{flalign} \nn S_{\mathrm {KG}}[T\Phi ] &= \int _{\bbR ^d}-\frac {1}{2}\,\bigg (\eta ^{\mu \nu }\,(\partial _\mu T\Phi )(x) \, (\partial _{\nu }T\Phi )(x) + m^2\,(T\Phi )^2(x)\bigg )\,\dd x\\ &=(-1)^2 \int _{\bbR ^d}-\frac {1}{2}\,\bigg (\eta ^{\mu \nu }\,(\partial _\mu \Phi )(x) \, (\partial _{\nu }\Phi )(x) + m^2\,\Phi ^2(x)\bigg )\,\dd x = S_{\mathrm {KG}}[\Phi ]\quad . \end{flalign} Note that this transformation is also a symmetry of the action functional (2.10) with interaction terms, provided that the potential \(V(\Phi )\) is an even function, i.e. \(V(-\Phi ) = V(\Phi )\). Since \(T\Phi (x)\) at \(x\) is determined from \(\Phi (x)\) at the same point \(x\), this is an example of an internal symmetry.

The case of a complex scalar field \(\Phi (x)\in \bbC \) is richer, in the sense that there is not only one but an infinite family of internal symmetries that is labeled by a continuous parameter. Consider the transformation \(T : \Phi (x)\mapsto T\Phi (x) = e^{-\ii \,\alpha }\,\Phi (x)\) that rotates the field \(\Phi \) by any constant complex phase \(e^{-\ii \, \alpha }\in \mathsf {U}(1)\subseteq \bbC \), i.e. \(\alpha \) is taken to be independent of \(x\). The transformation of the complex conjugate field \(\Phi ^\ast \) is then given by \(T: \Phi ^\ast (x)\mapsto e^{\ii \,\alpha }\,\Phi ^\ast (x)\). The action (2.17) and its special case (2.19) are clearly invariant under this transformation, because every summand involves an equal power of \(\Phi \) and \(\Phi ^\ast \), such that the complex phases cancel out. For example, for the quadratic terms we have that \(T\Phi ^\ast (x) \, T\Phi (x) = e^{\ii \,\alpha }\, \Phi ^\ast (x)\, e^{-\ii \,\alpha }\, \Phi (x) = \Phi ^\ast (x)\,\Phi (x)\). Since there are infinitely many complex phases \(e^{-\ii \,\alpha }\in \mathsf {U}(1)\subseteq \bbC \), corresponding to choosing \(\alpha \in [0,2\pi )\), we found a continuous family of internal symmetries for the complex scalar field. It is worthwhile to note that constancy of the phase \(e^{-\ii \,\alpha }\) is crucial for having a symmetry: If \(e^{-\ii \,\alpha (x)}\) would depend on \(x\), then the partial derivatives of the transformed field receive an extra term \((\partial _\mu T\Phi )(x) = \partial _\mu (e^{-\ii \,\alpha (x)}\,\Phi (x)) = e^{-\ii \,\alpha (x)}\, (\partial _\mu \Phi )(x) -\ii \,(\partial _\mu \alpha )(x)\, e^{\ii \,\alpha (x)}\,\Phi (x)\), which would spoil the invariance of the kinetic term in the actions (2.17) and (2.19). Such local, in the sense of \(x\)-dependent, internal transformations will play an important role later in this module when we will discuss gauge theories.

Remark 2.7. The requirement that \(\phi \) satisfies the Euler-Lagrange equations (2.6) is in general necessary for the Noether current \(J^\mu \) to be conserved. Satisfying the Euler-Lagrange equations is often called on-shell in physics slang, which is why you will probably hear the terminology that \(J^\mu \) is an on-shell conserved current.

Remark 2.8. The reason why a conserved current \(J^\mu \) is so useful is that it has an associated conserved charge, which is defined by the integral

\begin{flalign} Q(t) := \int _{\bbR ^{d-1}} J^0(t,\mathbf {x})\, \dd \mathbf {x} \end{flalign} of the \(0\)-component of \(J^\mu \) over the space coordinates \(\mathbf {x}\in \bbR ^{d-1}\). The charge is time-independent, hence the name conserved, as shown by the following calculation

\begin{flalign} \frac {\dd Q(t)}{\dd t} = \int _{\bbR ^{d-1}} \frac {\partial J^0(t,\mathbf {x})}{\partial t}\, \dd \mathbf {x} = - \int _{\bbR ^{d-1}} \nabla \cdot \mathbf {J}(t,\mathbf {x})\, \dd \mathbf {x} = 0\quad , \end{flalign} where in the second step we have used current conservation \(0=\partial _\mu J^\mu = \frac {\partial J^0}{\partial t} + \nabla \cdot \mathbf {J}\) and in the third step Gauss’ integration theorem. (As usual, it is assumed that the fields fall off sufficiently fast at infinity so that there are no boundary terms.)

Example 2.9 (Relativistic momentum of the real Klein-Gordon field). Recall from Example 2.4 that translations of spacetime define a continuous family of symmetries \(T_b : \Phi (x)\mapsto T_b\Phi (x) = \Phi (x - b)\) for the real Klein-Gordon field. Taking the first order Taylor expansion as in (2.26) gives

\begin{flalign} \delta _b\Phi = - b^\mu \,\partial _\mu \Phi \quad . \end{flalign} For the transformation of the Lagrangian density (2.27) one finds

\begin{flalign} \delta _b\LL = \partial _\mu \big (-b^\mu \, \LL \big )\quad \Longrightarrow \quad F^\mu = -b^\mu \, \LL \quad . \end{flalign} The associated Noether current from Theorem 2.6 then reads as

\begin{flalign} \nn J^\mu &= \partial ^\mu \Phi \, b^\nu \,\partial _\nu \Phi + b^\mu \LL \\ &= b_\nu ~\bigg (\partial ^\mu \Phi \,\partial ^\nu \Phi - \frac {1}{2}\,\eta ^{\mu \nu }\Big (\partial ^\rho \Phi \,\partial _\rho \Phi + m^2\,\Phi ^2\Big )\bigg )\quad . \end{flalign} The term in the parenthesis is called the energy-momentum tensor and it is often denoted by

\begin{flalign} \label {eqn:EMtensorKG} T^{\mu \nu } := \partial ^\mu \Phi \,\partial ^\nu \Phi - \frac {1}{2}\,\eta ^{\mu \nu }\Big (\partial ^\rho \Phi \,\partial _\rho \Phi + m^2\,\Phi ^2\Big )\quad . \end{flalign} Note that current conservation \(\partial _\mu J^\mu = 0\), for all translation vectors \(b^\nu \), is equivalent to the conservation law \(\partial _\mu T^{\mu \nu } =0\) of the energy-momentum tensor that you have probably seen before in your relativity module. The conserved charges associated with the energy-momentum tensor define the relativistic momentum

\begin{flalign} P^\mu := \int _{\bbR ^{d-1}} T^{0\mu }\,\dd \mathbf {x} \end{flalign} of the Klein-Gordon field. Explicitly, the \(\mu =0\)-component reads as

\begin{flalign} E:= P^0 = \int _{\bbR ^{d-1}} T^{00}\,\dd \mathbf {x} = \int _{\bbR ^{d-1}} \bigg (\frac {1}{2}\dot {\Phi }^2 + \frac {1}{2}(\nabla \Phi )^2 +\frac {m^2}{2}\,\Phi ^2\bigg )\,\dd \mathbf {x}\quad , \end{flalign} where \(\dot {\Phi } := \frac {\partial \Phi }{\partial t}\) denotes the time derivative, and it describes the energy. (As we will see in Section 2.3, this agrees with the Hamiltonian.) The spatial components

\begin{flalign} P^i = -\int _{\bbR ^{d-1}} \dot {\Phi }\,\partial ^i \Phi \,\dd \mathbf {x}\quad , \end{flalign} for \(i=1,\dots , d-1\), describe the momentum.

Example 2.10 (Relativistic angular momentum of the real Klein-Gordon field). Recalling again Example 2.4, we consider now the case of a Lorentz transformation \(x^\prime = \Lambda x\). To obtain a suitable continuous family of Lorentz transformations, we write \(\Lambda = e^{\omega }= \sum _{n=0}^\infty \frac {\omega ^n}{n!}\) as the exponential of a matrix \(\omega \). Expanding the condition (1.17) to first order in \(\omega \) yields

\begin{flalign} \nn \eta _{\mu \nu }\,\Lambda ^{\mu }_{~~\rho }\,\Lambda ^\nu _{~~\sigma } &= \eta _{\mu \nu }\,\big (\delta ^{\mu }_{~~\rho } + \omega ^\mu _{~~\rho }\big )\,\big (\delta ^\nu _{~~\sigma } + \omega ^\nu _{~~\sigma }\big )+\mathcal {O}(\omega ^2) \\ &= \eta _{\rho \sigma } + \omega _{\rho \sigma } + \omega _{\sigma \rho } + \mathcal {O}(\omega ^2) = \eta _{\rho \sigma }\quad , \end{flalign} hence the index lowering \(\omega _{\mu \nu } := \eta _{\mu \rho }\,\omega ^{\rho }_{~~\nu }\) must be antisymmetric

\begin{flalign} \label {eqn:omegaantisym} \omega _{\mu \nu } = - \omega _{\nu \mu }\quad . \end{flalign} One can check that the latter condition implies that \(\Lambda = e^{\omega } \) satisfies (1.17) to all orders in \(\omega \). Hence, we obtain a continuous family of symmetries \(T_\omega : \Phi (x)\mapsto T_\omega \Phi (x) = \Phi (e^{-\omega } x)\) for the real Klein-Gordon field. Taking the first order Taylor expansion as in (2.26) gives

\begin{flalign} \delta _\omega \Phi = -\omega ^\mu _{~~\nu }\,x^\nu \,\partial _\mu \Phi \quad . \end{flalign} For the transformation of the Lagrangian density (2.27) one finds

\begin{flalign} \delta _\omega \LL = -\omega ^{\mu }_{~~\nu }\,x^\nu \,\partial _\mu \LL = \partial _\mu \Big (-\omega ^\mu _{~~\nu } \,x^\nu \,\LL \Big )\quad \Longrightarrow \quad F^\mu = -\omega ^\mu _{~~\nu } \,x^\nu \,\LL \quad , \end{flalign} where in the second equality we have used that \(\omega ^{\mu }_{~~\mu }= \eta ^{\mu \nu }\omega _{\nu \mu }=0\) as a consequence of the antisymmetry condition (2.39). The associated Noether current from Theorem 2.6 then reads as

\begin{flalign} J^\mu = - \omega _{\nu \rho }\,x^\nu \,T^{\mu \rho } = \frac {1}{2}\,\omega _{\nu \rho }\,\Big (x^\rho \,T^{\mu \nu }-x^\nu \,T^{\mu \rho } \Big ) \quad , \end{flalign} where \(T^{\mu \rho }\) denotes the energy-momentum tensor (2.36) and in the last step we used that \(\omega _{\nu \rho } = -\omega _{\rho \nu }\) is antisymmetric. Denoting the term in the parenthesis by

\begin{flalign} (\mathcal {J}^\mu )^{\rho \nu } := x^\rho \,T^{\mu \nu }-x^\nu \,T^{\mu \rho }\quad , \end{flalign} current conservation \(\partial _\mu J^\mu =0\), for all parameters \(\omega _{\nu \rho }\), is equivalent to the conservation law \(\partial _\mu (\mathcal {J}^\mu )^{\rho \nu }=0\). The conserved charges associated with \((\mathcal {J}^\mu )^{\rho \nu } \) define the relativistic angular momentum

\begin{flalign} \label {eqn:KGangularmomentum} L^{\rho \nu } := \int _{\bbR ^{d-1}} (\mathcal {J}^0)^{\rho \nu } ~\dd \mathbf {x} = \int _{\bbR ^{d-1}} \Big (x^\rho \,T^{0\nu } - x^{\nu }\,T^{0\rho }\Big ) ~\dd \mathbf {x} \end{flalign} of the Klein-Gordon field. Since the angular momentum (2.44) contains only contributions arising from the dynamics of the field, we can conclude that the Klein-Gordon field does not carry any internal angular momentum, a.k.a. spin.

Example 2.11 (Electric charge of the complex Klein-Gordon field). Recall from Example 2.5 that the complex Klein-Gordon field admits a continuous family of internal symmetries \(T_\alpha : \Phi (x)\mapsto T_\alpha \Phi (x) = e^{-\ii \,\alpha }\,\Phi (x)\). Taking the first order Taylor expansion as in (2.26) gives

\begin{flalign} \delta _\alpha \Phi = -\ii \,\alpha \,\Phi \quad ,\qquad \delta _{\alpha }\Phi ^\ast = \ii \,\alpha \,\Phi ^\ast \quad . \end{flalign} For the transformation of the Lagrangian density (2.27) one finds

\begin{flalign} \delta _\alpha \LL = 0 \quad \Longrightarrow \quad F^\mu = 0 \end{flalign} because the Lagrangian density (2.19) is invariant under phase rotations. The associated Noether current from Theorem 2.6 then reads as

\begin{flalign} J^\mu = \frac {\partial \LL }{\partial (\partial _\mu \Phi )}\,\delta _\alpha \Phi + \frac {\partial \LL }{\partial (\partial _\mu \Phi ^\ast )}\,\delta _\alpha \Phi ^\ast = \alpha \,\Big ( \ii \,\partial ^\mu \Phi ^\ast \, \Phi - \ii \, \partial ^\mu \Phi \, \Phi ^\ast \Big )\quad . \end{flalign} The term in the parenthesis is called the electromagnetic current of the complex Klein-Gordon field and it is often denoted by

\begin{flalign} j^\mu := \ii \,\Big ( \Phi \,\partial ^\mu \Phi ^\ast -\Phi ^\ast \, \partial ^\mu \Phi \Big )\quad . \end{flalign} The conserved charge associated with \(j^\mu \) defines the electric charge

\begin{flalign} Q:= \ii \, \int _{\bbR ^{d-1}} \Big (\Phi ^\ast \, \dot {\Phi } - \Phi \,\dot {\Phi }^\ast \Big )\,\dd \mathbf {x}\quad , \end{flalign} where \(\dot {\Phi } = \frac {\partial \Phi }{\partial t} = \partial _0 \Phi = -\partial ^0\Phi \) denotes the time derivative. (The minus sign in the last step is due to \(\eta ^{00}=-1\) in the Minkowski metric.)