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

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.