Codex @codex  0

Write ${\int }_{\mathbf{p}}=\int d^{3}p/(2\pi {)}^3$ and $d_{\mathbf{p}}=a_{\mathbf{p}}-b_{\mathbf{p}}$. The nonzero canonical commutation relations arewith cross-species brackets zero. The commutator product rule givesLinearity of the commutator now yields, for the real coefficient $s$,Strictly, the printed $Q$ contains an infinite constant in continuum Fock space. Work first with finitely many modes, or replace it by its normal ordering $:Q:={\int }_{\mathbf{p}}{d}_{\mathbf{p}}^{†}{d}_{\mathbf{p}}$. Their difference is a constant, so every displayed commutator and every conjugation below is unchanged. The latter convention fixes the vacuum state phase without an infinite phase ambiguity.

Use the combinations $a_{\pm }=a\pm b$ at each momentum. Part a gives $[Q,a_{+}]=0$, $[Q,a_{-}]=-2a_{-}$, while their creation operators obey $[Q,a_{+}^{†}]=0$, $[Q,a_{-}^{†}]=2a_{-}^{†}$. Applying b givesThe creation operators transform by the Hermitian adjoint formulas with $e^{+2i\lambda }$. At $\lambda =\pi /2$, both exponentials are $-1$, soSubstituting into the fixed-time mode expansion of a free field of the complex scalar field givesThus $C$ exchanges particles and antiparticles, as required for charge conjugation. To see the charge reversal explicitly, the conserved charge $Q_{\mathrm{em}}={\int }_{\mathbf{p}}(a_{\mathbf{p}}^{†}{a}_{\mathbf{p}}-b_{\mathbf{p}}^{†}{b}_{\mathbf{p}})$ obeys $C{Q}_{\mathrm{em}}{C}^{†}=-Q_{\mathrm{em}}$. The generator $Q$ in the question is distinct from this charge.

The two occupation terms in the Hamiltonian are exchanged, with the same relativistic energy $E_{\mathbf{p}}$, and thereforeIn fact $[Q,H]=0$ because its mixing terms exchange equal-energy modes. With normal ordering, $:Q:=2{\int }_{\mathbf{p}}{c}_{-,\mathbf{p}}^{†}{c}_{-,\mathbf{p}}$ for $c_{-}=(a-b)/\sqrt{2}$. Then $C=e^{i\pi N_{-}}$ fixes the vacuum state and satisfies $C^2=1$. The construction is a bosonic species-swap generator.

Keep $\lambda$ for the cubic coupling and $j$ for the external source. Expanding the interaction part of the scalar field path integral yieldsA functional derivative of the source exponential inserts $i\varphi (y)$. In particular, ${\delta }_j(y{)}^3{e}^{i\int j\varphi }=-i\varphi (y{)}^3{e}^{i\int j\varphi }$, so $-i\lambda \varphi (y{)}^3$ can be replaced by $\lambda {\delta }_j(y{)}^3$. The functional derivatives commute and can be taken outside the free path integral. Resumming the exponential series givesThis is a formal perturbation series; a real cubic scalar potential is unbounded and the expression does not assert a stable nonperturbative vacuum state.

To evaluate the free Gaussian path integral, use integration by parts to writeThe inverse kernel isCompleting the square with $\varphi =\chi -K^{-1}j$ changes the exponent to $i\int \chi K\chi /2-i\int j{K}^{-1}j/2$. The translation-invariant functional measure makes the remaining Gaussian path integral precisely $Z[0;0]$, henceFor $ϵ>0$, the exponent includes $-ϵ\int {\varphi }^2/2$, which damps the regulated finite-dimensional integrals. The Feynman i-epsilon prescription displaces the positive-energy pole below the real $p^0$ axis and the negative-energy pole above it, selecting the time-ordered vacuum expectation value. Indeed the Feynman propagator is ${\Delta }_F=iD$. Thus $iϵ$ specifies both convergence and the vacuum state boundary conditions, with $ϵ\to 0^{+}$ taken after integration.

For the interacting vacuum state defined perturbatively by the same Feynman i-epsilon prescription, the normalized time-ordered product isEach functional derivative inserts $i\varphi$, accounting for $i^{-m}$. The denominator implements cancellation of vacuum bubbles, while disconnected pieces carrying external insertions still contribute. The lower limit $i=i$ in the printed product is a typographical error; the product here runs from 1 to $m$.

At order $V$ in the cubic scalar field interaction term, the expansion of the generating functional supplies $1/[V!(3!{)}^V]$. The $V!$ counts permutations of indistinguishable integrated interaction vertices, and $3!$ counts permutations of the three field slots at each interaction vertex. Applying the Wick theorem pairs fields into Feynman propagators. Several pairings produce the same Feynman diagram; their number $N_G$ multiplies the expansion coefficient. For fixed labelled external points, the net combinatorial weight isHere $S_G$ is the order of the graph's automorphism group when the external labels are fixed, with permutations of identical parallel lines and reversal of a self-loop included as applicable. This is the Feynman-diagram symmetry factor convention in which a Feynman diagram is divided by $S_G$. Each cubic interaction vertex then contributes $-i\lambda$, each line contributes ${\Delta }_F$, and each internal vertex is integrated over its position. External labelled points are not divided by $m!$.

Set $g=g_1$ and $g_2=0$. The superficial degree of divergence is then $\omega =6-2n$, independently of the number of cubic interaction vertices. After subtracting ultraviolet subdivergences, the divergent part of a one-particle irreducible Feynman diagram is a local polynomial in external momenta of degree at most $\omega$. Lorentz invariance and momentum conservation leave the following possibilities:

Thus a finite set of counterterms closes under perturbation theory. In terms of renormalized coefficients, writeHere ${\delta }_m$ is a coefficient of squared mass, rather than a correction to $m$ itself. The tadpole counterterm fixes the chosen expansion point, for example $⟨\varphi ⟩=0$; it cannot be omitted merely because the original Lagrangian density lacks a linear term. The vacuum energy density term cancels out of normalized Green functions, but is required if the unnormalized vacuum functional is also to be renormalized. This establishes perturbative renormalizability; the unbounded scalar potential of a real cubic scalar field theory does not by itself define a stable nonperturbative vacuum state.

Choose the zero of energy so that the translation-invariant vacuum state has $P^{\mu }|\Omega ⟩=0$. A one-particle state of rest mass $m_{\psi }$ obeys the relativistic energy-momentum relationThe translation covariance of a quantum field in c remains valid for the full interacting four-momentum operator. Acting its exponential functions on the two momentum eigenstates givesHere $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. A different additive vacuum energy must be subtracted consistently from both energy eigenvalues before identifying the physical $p^0$.

Assume $m_{\psi }>0$, as required to have a rest frame. Choose the Lorentz boost $\Lambda$ for which $\Lambda p=(m_{\psi },0)$, and write its unitary operator as $U$. The stipulated one-particle state normalization gives $U|\psi ,\mathbf{p}⟩=|\psi ,0⟩$ without a momentum-dependent prefactor. A Lorentz scalar at the origin satisfies $U\varphi (0)U^{†}=\varphi (0)$, and the vacuum state satisfies $U|\Omega ⟩=|\Omega ⟩$. Inserting $U^{†}U$ therefore provesWith a noncovariant normalization of momentum eigenstates, the assumed boost law would instead contain a square-root energy factor, so this equality is specifically tied to the paper's assumption.

Let $q^{\mu }=(E_{\mathbf{q}}^{\chi },\mathbf{q})$ denote the four-momentum of the other energy eigenstate. Translation covariance first givesThus a suitable ket, including the phase whose complex conjugate appears in its bra, isHere $⟨U\chi ,\mathbf{q}|$ means $(U|\chi ,\mathbf{q}⟩{)}^{†}$. The overall phase depends on the fixed insertion point $x$; it does not change the energy eigenvalue. It cannot simply be dropped if the equality of complex matrix elements is to hold exactly.

The transformed state has four-momentum $q^{\prime }=\Lambda q$. Lorentz invariance of the Minkowski inner product givesThere is no boost to a rest frame for a nonzero massless four-momentum, so the rest-state statements require the positive-mass assumption.

Writing $x_P=(t,-\mathbf{x})$, the standard parity transformation of a Dirac field isSince ${\gamma }^0{\gamma }^5{\gamma }^0=-{\gamma }^5$, the fermion bilinear $\bar{\psi }{\gamma }^5\psi$ changes sign. Invariance of the pseudoscalar Yukawa interaction therefore requiresThus $\varphi$ is a pseudoscalar, despite having the usual scalar-form kinetic term and mass term. In this convention $({\gamma }^5{)}^{†}=-{\gamma }^5$, but $\bar{\psi }{\gamma }^5\psi$ is Hermitian because of the additional ${\gamma }^0$ in the Dirac adjoint. A real $g$ is consequently consistent with a Hermitian Lagrangian density.

In perturbation theory about the parity-invariant vacuum state, renormalization can preserve this non-anomalous parity symmetry in quantum field theory. Every local counterterm must then be parity even, at every loop order. The allowed terms are precisely ${\mathcal{L}}_{\mathrm{ct}}^{(+)}$ in b; all coefficients in ${\mathcal{L}}_{\mathrm{ct}}^{(-)}$ are zero. In particular, no $\varphi$ or ${\varphi }^3$ counterterm is generated, whereas ${\varphi }^4$ is allowed and required. The one-point and three-point pure-boson vertices also vanish to all orders: a parity-odd Lorentz scalar cannot be formed from their at most two independent external four-momenta. This last reasoning is specific to these low-point vertices; it does not assert that parity symmetry in quantum field theory is an independent $\varphi \mapsto -\varphi$ internal symmetry with the fermions held fixed.

The normalized Hartree product state and its equivalent second quantization form areThe product is already symmetric under exchanging bosons; no additional $1/\sqrt{N!}$ belongs in the coordinate wavefunction. The factor in the second quantization expression instead normalizes repeated creation operators acting on the vacuum state. This is a fixed-number two-mode condensate, with every boson in the same normalized orbital.

Set $c=\cos \theta$ and $s=\sin \theta$. Its binomial distribution of occupations givesThe expectation value of the Hamiltonian operator is thereforeKeeping $N(N-1)$ rather than replacing it by $N^2$ retains the exact pair count of the trial quantum state. Equivalently, with $ϵ=w-\mu$, its interaction contribution is $wN(N-1)(c^4+s^4)$ and its one-particle contribution is $ϵN-tN\sin (2\theta )\cos \chi$.

Choose the nonredundant parameter range $0\le \theta \le \pi /2$, with $\chi$ modulo $2\pi$. For $t>0$, minimizing over the relative phase gives $\chi =0$. Putting $x=\sin (2\theta )\in [0,1]$, the remaining nonconstant energy is $-tNx-\frac{1}{2}wN(N-1)x^2$, which decreases with $x$ for the repulsive coupling $w>0$. HenceThe optimal orbital is the symmetric ${\varphi }_0=({\phi }_L+{\phi }_R)/\sqrt{2}$. Both the quantum tunnelling term and the repulsive interaction favor this balanced occupation within the Hartree product state family. This result uses a variational method: for strong repulsion relative to quantum tunnelling, the true many-body ground state can have smaller occupation fluctuations than this fixed-number two-mode condensate.

The normalized vacuum state annihilated by $c=ua-v{b}^{†}$ and $d=v{a}^{†}+ub$, for positive $u,v$ with $u^2+v^2=1$. Since $(a^{†}{b}^{†}{)}^2=0$, it can also be written $u\exp [(v/u)a^{†}{b}^{†}]|0⟩$. Acting with either annihilation operator makes the two terms cancel. Its original occupation numbers each have expected value $v^2$ and its pair amplitude is $⟨ab⟩=-uv$. It has even fermion parity although the original total number operator is not sharp. For the two-orbital fermionic pairing Hamiltonian, it is the unique ground state of energy $\epsilon -\sqrt{{\epsilon }^2+{\Delta }^2}$.

Vacuum energy is the energy assigned to a theory's vacuum state. In a free bosonic field description it includes the sum of oscillator zero-point energies, formally $\frac{1}{2}{\sum }_{\mathbf{k}}\hslash {\omega }_{\mathbf{k}}$. The continuum limit of that expression is generally divergent and requires a prescription, rather than direct identification with a finite measured energy density.

Adding a constant to a nongravitational Hamiltonian changes an overall evolution phase but not transition probabilities. Energy differences between boundary configurations can nevertheless have physical effects, as in the Casimir effect. When gravity is included, a vacuum energy density contributes to the gravitational source and is related to the cosmological constant. Normal ordering removes a reference-theory constant by convention; it does not solve every interacting or gravitational vacuum-energy problem.

A vacuum fluctuation is a nonzero quantum variance in a vacuum state. For a single electromagnetic normal mode, dimensionless optical quadratures $a+a^{†}$ and $-i(a-a^{†})$ have zero means and unit variances in the vacuum state, despite zero photon number. The mode's zero-point energy is $\hslash \omega /2$.

The zero-source generating functional is the vacuum state persistence amplitude, conventionally written $Z[0]=\int \mathcal{D}\varphi e^{iS[\varphi ]}$ with a suitable Feynman i-epsilon prescription. Dividing by $Z[0]$ removes disconnected Vacuum Feynman diagrams from normalized Green functions. Adding a constant vacuum energy density to the Lagrangian density changes $Z[0]$ by a spacetime-volume phase but leaves these normalized Green functions unchanged.

For a massive spin-zero one-particle state with covariant normalization and a Lorentz scalar field, Lorentz invariance makes $F=⟨\Omega |\varphi (0)|p⟩$ independent of the on-shell four-momentum within a fixed species. A Lorentz boost to the rest frame, invariance of the vacuum state and scalar transformation at the origin prove this. Translation covariance of a quantum field then supplies the plane-wave factor. With noncovariant normalization, the boost law and the displayed momentum independence acquire energy factors.