Past exam of the quantum field theory course of the University of Oxford / 2017 / 1

Treat $\varphi$ and ${\varphi }^{\ast }$ as independent variables in the Euler-Lagrange equations. Write $s={\varphi }^{\ast }\varphi$, and assume the scalar potential $f(s)$ is real and differentiable. Variation with respect to ${\varphi }^{\ast }$ givesConsequently the equations for the complex scalar field and its complex conjugate areNo polynomial assumption is needed: only the first derivative of the scalar potential enters. For $f(s)=m^{2}s$, both reduce to the free Klein-Gordon equation.

For a constant real parameter, ${\partial }_{\mu }(e^{i\alpha }\varphi )=e^{i\alpha }{\partial }_{\mu }\varphi$. The phases therefore cancel separately in the kinetic term and in ${\varphi }^{\ast }\varphi$, leaving any $f({\varphi }^{\ast }\varphi )$ unchanged. This is the global phase symmetry of a complex scalar field, with circle group $U(1)$.

Use infinitesimal variations per unit parameter, $\delta \varphi =i\varphi$, $\delta {\varphi }^{\ast }=-i{\varphi }^{\ast }$. The Lagrangian density is exactly invariant, so no total-derivative correction is needed in Noether's theorem. The Noether current isIt depends only on derivatives of the Lagrangian density with respect to field derivatives, so the derivative-free scalar potential cannot alter its form. Direct differentiation cancels the mixed first-derivative terms:Thus it is conserved on solutions of the Euler-Lagrange equations. Its spatial integral is a conserved Noether charge when the flux through the boundary vanishes. Its overall sign is fixed here by $\delta \varphi =+i\varphi$; reversing the orientation of the phase transformation reverses the current.

Let ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$. The conjugate momentum fields are ${\pi }_{\varphi }={\dot{\varphi }}^{†}$ and ${\pi }_{{\varphi }^{†}}=\dot{\varphi }$. Inserting the given mode expansion of a free field into $i\int d^{3}x({\dot{\varphi }}^{†}\varphi -{\varphi }^{†}\dot{\varphi })$ produces spatial Dirac delta factors. The $b^{†}{c}^{†}$ and $cb$ terms vanish after integration: they contain ${\omega }_{\mathbf{p}}-{\omega }_{\mathbf{q}}$ on ${\delta }^3(\mathbf{p}+\mathbf{q})$, and ${\omega }_{-\mathbf{p}}={\omega }_{\mathbf{p}}$. The number-conserving terms instead add their two frequency factors, cancelling the mode normalization. The un-subtracted ordered expression isThe last term is an infinite vacuum constant, not a finite physical vacuum charge. Choosing the Fock vacuum to have zero Noether charge, equivalently using normal ordering, givesA state containing $n_c$ quanta created by $c^{†}$ and $n_b$ by $b^{†}$ has Noether charge $n_c-n_b$. Thus the two families describe opposite unit Noether charges; with this source's current the $c^{†}$ quantum has positive Noether charge and the $b^{†}$ quantum negative Noether charge. This is an internal Noether charge; identifying it with electric charge requires a specified electromagnetic coupling. The vacuum subtraction of a complex scalar charge leaves all commutators unchanged.^[ref]

Scope of the expansion: a fixed-frequency expansion with time-independent oscillator coefficients is the free-field or asymptotic-field form. For arbitrary interacting $f$, it can be used as canonical oscillator coordinates at a chosen time, rather than as the exact interacting time dependence. For example, $f(s)=m^{2}s+\lambda s^2$ gives $(\square +m^2)\varphi +2\lambda {\varphi }^{\ast }\varphi \varphi =0$, which a generic free plane-wave superposition does not satisfy. The Noether charge and its generator action remain valid in the interacting theory; the symmetry does not depend on using a free Hamiltonian.

The two oscillator families commute with each other. Their canonical commutation relations giveBoth terms in the mode expansion of a free field therefore acquire the same factor, givingThe Noether charge is the generator of the internal symmetry. In the active-operator convention $U(\alpha )=e^{i\alpha Q}$, repeated commutators giveThe sign agrees with part b. A negative sign for $[Q,\varphi ]$ would correspond to the opposite current convention, not to the current used here. Subtracting a multiple of the identity from $Q$ to fix the vacuum Noether charge has no effect on this generator relation.