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The equal-time canonical anticommutation relations implyApply $-i{\alpha }_{bc}^k{\partial }_{z^k}+m{\beta }_{bc}$, set $\mathbf{z}=\mathbf{y}$, and integrate over $\mathbf{y}$. The mass term cancels, leaving $i\int d^{3}y{\psi }^{†}{\alpha }^k\psi {\partial }_{y^k}{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$. Integration by parts proves the title formula as a distribution identity. The Heisenberg equation of motion then yields ${\partial }_{t}n=-\nabla \cdot \mathbf{j}$ for the physical Dirac current $j^k=\bar{\psi }{\gamma }^k\psi$. The factor $-i$ is required because the commutator of Hermitian operators is anti-Hermitian.

Treat $\varphi ,{\varphi }^{†}$ as independent variables when taking the Legendre transform. Their canonical momenta areThe canonical quantization of a complex scalar field imposes, at equal times,with $[\varphi ,{\varphi }^{†}]=[\varphi ,\varphi ]=[\pi ,{\pi }^{†}]=[\pi ,\pi ]=0$, together with their adjoints, and $[\varphi ,{\pi }^{†}]=[{\varphi }^{†},\pi ]=0$. These are canonical commutation relations for two real scalar degrees of freedom, written in a complex basis. Local fields are operator-valued distributions, so these identities are understood after smearing or with a regulator.

The Hamiltonian density is $\pi \dot{\varphi }+{\pi }^{†}{\dot{\varphi }}^{†}-\mathcal{L}$, givingThe Heisenberg equation of motion is $\dot{O}=i[H,O]$. For example, $i[\int {\pi }^{†}\pi ,\varphi ]={\pi }^{†}$, while commuting $\pi$ with the gradient term and integrating its derivative of a Dirac delta distribution gives ${\nabla }^2{\varphi }^{†}$. ThusBoth fields satisfy the Klein-Gordon equation. Their spatial Fourier modes therefore have frequencies $\pm E_p$, $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. A complex scalar field has independent coefficients for these two frequency sectors, so, defining the invariant measure $d{\Pi }_p=d^{3}p/[(2\pi {)}^{3}2E_p]$, its mode expansion isOne can recover the mode operators at any fixed time using complex scalar mode inversion:Substitution of the expansion verifies these inverses: the spatial integral selects the appropriate momentum, while the two frequency sectors acquire factors $E_p\pm E_q$. The unwanted sector vanishes when the delta function sets $E_p=E_q$. The canonical commutation relations then give, for example,Likewise $[b(p),b^{†}(q)]$ has the same value. The cross commutators vanish; their coefficients contain $E_p-E_q$ on the corresponding momentum delta function. Conversely, these oscillator brackets reproduce the equal-time field brackets, confirming the normalization of the measure.

Insert the mode expansion into $H$ and integrate over space. The terms $a^{†}{b}^{†}$ and $ba$ enforce opposite spatial momenta; their coefficient is $-E_p^2+{\mathbf{p}}^2+m^2=0$. For the diagonal terms the coefficient is $E_p^2+{\mathbf{p}}^2+m^2=2E_p^2$. Combining this with the two invariant measures leavesIn a box, $E_0={\sum }_{\mathbf{p}}{E}_p$, one half-quantum for each of the two real oscillator species. It diverges as the cutoff is removed. Normal ordering removes this vacuum energy, yielding the normal-ordered Hamiltonian of a free complex scalar field:In this nongravitational free theory an additive constant in $H$ changes only an overall state phase, not Heisenberg equations of motion, energy differences or scattering probabilities. It is therefore consistent to choose the Fock vacuum energy as zero. Both creation operators raise the energy: $[H,a^{†}(p)]=E_p{a}^{†}(p)$ and $[H,b^{†}(p)]=E_p{b}^{†}(p)$. Repeated creation builds a bosonic Fock space of spin-zero particles and antiparticles, with the same positive mass and energy. The negative-frequency part of the field does not describe negative-energy physical particles.

The global phase symmetry of a complex scalar field, with orientation $\varphi \mapsto e^{-i\alpha }\varphi$, has the indicated Noether current. Taking its divergence, the two first-derivative terms cancel, givingThe same calculation holds for its normal-ordered version. Integrating its temporal component gives a conserved charge when spatial boundary flux vanishes.

Here the ordering convention is important. Direct substitution into the literally ordered product $i\int ({\varphi }^{†}\dot{\varphi }-{\dot{\varphi }}^{†}\varphi )d^{3}x$ yieldsThe pair terms again vanish, now because their coefficient is a difference of equal energies. Reordering the $b$ term produces a divergent vacuum constant: with a finite-mode regulator $Q_{\mathrm{bare}}=N_a-N_b-N_{\mathrm{modes}}$. Thus the requested finite charge formula uses zero vacuum charge, or equivalently the vacuum subtraction of a complex scalar charge. With that conventional subtraction,The subtraction changes neither current conservation nor the action of $Q$ on fields; without it, the unrenormalized product does not literally equal the displayed normal-ordered operator.

Finally the oscillator commutators giveor $\boxed{Q{a}^{†}=a^{†}(Q+1),Q{b}^{†}=b^{†}(Q-1)}$. Acting on a charge eigenstate, the first creator raises its charge by one and the second lowers it by one. Starting from the neutral Fock vacuum, their one-particle states therefore have charges $+1$ and $-1$ respectively. This is the complex scalar charge operator $N_a-N_b$, whereas the energy counts the sum of the two occupations. A multiparticle state has additive charge equal to the number of particles minus the number of antiparticles.

Use units $\hslash =c=1$ and the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$. The canonical momentum of the real scalar field is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$. Canonical quantization replaces the classical fields by operator-valued distributions and imposes the equal-time canonical commutation relationsThese relations are imposed on an initial slice, with smeared fields understood. In the Heisenberg picture, simultaneous unitary time evolution of both operators preserves their commutators. Since the momentum is $\dot{\varphi }$, the requested equal-time commutator is $[\varphi (t,\mathbf{x}),\dot{\varphi }(t,\mathbf{y})]=i{\delta }^{(3)}(\mathbf{x}-\mathbf{y})$.

The Legendre transform in mechanics, applied pointwise, gives the canonical Hamiltonian density of a real scalar field. ThusTo obtain its dynamics rather than assume them, use the Heisenberg equation of motion $\dot{O}=i[H,O]$. The momentum-square term gives $[H,\varphi ]=-i\pi$. The field-square term gives $[H,\pi ]=i{m}^2\varphi$, while differentiating the Dirac delta function in the gradient term and integrating by parts gives $[H,\pi ]=-i{\nabla }^2\varphi +i{m}^2\varphi$. ConsequentlyBoundary terms vanish for suitably decaying smeared fields. This is the Klein-Gordon equation as an operator identity.

A spatial Fourier transform reduces the Klein-Gordon equation to ${\ddot{\varphi }}_{\mathbf{p}}+E_p^2{\varphi }_{\mathbf{p}}=0$, where $E_p=\sqrt{{\mathbf{p}}^2+m^2}$. Its two frequency branches and the Hermitian adjoint condition giveThe invariant-normalized scalar mode extraction can be written particularly simply asSubstituting the mode expansion of a free field proves these inversion formulas: the opposite-frequency coefficient is proportional to $E_p-E_q$ and vanishes after spatial Fourier inversion. Alternatively they are the corresponding Klein-Gordon inner products. Taking their commutators using the equal-time fields gives $(E_p+E_q)(2\pi {)}^3{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$ in the mixed case. In the two equal-type cases the delta function enforces $\mathbf{q}=-\mathbf{p}$, multiplying $E_p-E_q=0$. ThereforeConversely these mode commutators reproduce $[\varphi ,\dot{\varphi }]=i{\delta }^{(3)}$: the two frequency contributions each supply half the spatial Dirac delta function.

For the covariant oscillator Hamiltonian of a real scalar field, insert the mode expansion of a free field into $H$. Spatial integration sets the two momenta equal in the mixed terms and opposite in the two-creator or two-annihilator terms. The latter have coefficient $-E_p^2+{\mathbf{p}}^2+m^2=0$. The mixed terms giveThe constant $E_0$ is the divergent zero-point energy; a finite box and a momentum cutoff make its meaning explicit. Normal ordering sets the free vacuum energy to zero and leaves all commutators unchanged. Directly applying the mode commutators to the mixed expression yieldsChoose the Fock vacuum by $a(p)|0⟩=0$. Each creation operator adds a boson of energy $E_p$ and four-momentum $p$, while each annihilation operator removes one. Products of commuting creation operators give bosonic statistics from commuting creation operators and unrestricted occupation numbers. The real scalar field has one spin-zero species, identical to its antiparticle. Its one-particle states obey $⟨q|p⟩=(2\pi {)}^{3}2E_p{\delta }^{(3)}(\mathbf{p}-\mathbf{q})$, precisely the relativistic normalization of a one-particle state.

For the Feynman propagator, put $z=x-y$ and contract the two fields in the Fock vacuum. Only $a{a}^{†}$ survives, giving the Wightman functionThe stipulated time-ordered product convention is $i{\Delta }_F(z)=\theta (z^0)W(z)+\theta (-z^0)W(-z)$. In a contour integral over $p^0$, the Feynman i-epsilon prescription places the positive-energy pole just below the real axis and the negative-energy pole just above it. For $z^0>0$ close below, clockwise, obtaining $-i{e}^{-i{E}_p{z}^0}/(2E_p)$; for $z^0<0$ close above, obtaining $-i{e}^{i{E}_p{z}^0}/(2E_p)$. Changing $\mathbf{p}\mapsto -\mathbf{p}$ in the second contribution therefore provesActing with $\square +m^2$ multiplies the integrand by $-(p^2-m^2)$. The remaining factor tends to $-1$ as a distribution, so Fourier inversion givesThe sign also follows from the derivative jump of a free scalar time-ordered two-point function: the first time derivative of $i{\Delta }_F$ jumps by $⟨[\dot{\varphi },\varphi ]⟩=-i{\delta }^{(3)}$, so that of ${\Delta }_F$ jumps by $-{\delta }^{(3)}$.

For the real scalar field, the mostly-plus Minkowski metric gives a positive kinetic energy. Its canonical momentum and Hamiltonian operator areThe Euler-Lagrange field equation is $(\square -m^2)\varphi =0$, or $({\partial }_t^2-{\nabla }^2+m^2)\varphi =0$. Canonical quantization of a real scalar field promotes the fields to Hermitian operators and replaces the equal-time Poisson brackets by canonical commutation relations:The Heisenberg equation of motion then gives $\dot{\varphi }=\pi$ and $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, so the classical field equation remains an operator identity. Products at coincident points need a regulator; equivalently, the field is an operator-valued distribution.

Resolve the free field into positive- and negative-frequency plane waves. With $E_{\mathbf{k}}=\sqrt{|\mathbf{k}{|}^2+m^2}$ and $k\cdot x=-E_{\mathbf{k}}t+\mathbf{k}\cdot \mathbf{x}$,Reality pairs the two terms by Hermitian conjugation. The equal-time canonical commutation relations are equivalent toFor example, the two mixed terms in $[\varphi ,\pi ]$ each supply half the Dirac delta function. Thus the normalization $1/\sqrt{2E_{\mathbf{k}}}$ is fixed, not optional. Each independent mode is a quantum harmonic oscillator, with $a$ and $a^{†}$ its annihilation operator and creation operator.

Choose the Fock vacuum by $a(\mathbf{k})|0⟩=0$. Repeated creation operators construct the bosonic Fock space. Substitution into the Hamiltonian operator yieldswhere a finite volume $V$ can be used as a regulator. Normal ordering sets the flat-space vacuum reference to zero. The spatial momentum operator isIn particular, $[H,a^{†}(\mathbf{k})]=E_{\mathbf{k}}{a}^{†}(\mathbf{k})$ and $[\mathbf{P},a^{†}(\mathbf{k})]=\mathbf{k}{a}^{†}(\mathbf{k})$. A one-particle state therefore has positive energy, momentum $\mathbf{k}$, and invariant mass $m$. A scalar field transforms in the trivial spin representation, so its particles have spin zero. Because the field is real there is only one set of oscillators: the particle is its own antiparticle, with no distinct conserved particle-minus-antiparticle charge. The free theory does conserve its occupation-number sum, though generic scalar interactions need not.

A momentum eigenstate extends throughout space. A localized particle is described by a superposition such asIts wave packet evolves through the phase $e^{-i{E}_{\mathbf{k}}t}$; a narrow packet has group velocity $\mathbf{v}=\mathbf{k}/E_{\mathbf{k}}$, whose magnitude is at most one. Localization and dispersion concern these superpositions, rather than classical trajectories attached to individual field modes.

Relativistic spacetime behavior is encoded by microcausality. Directly from the oscillator commutators,This is a Lorentz-invariant distribution and vanishes at equal time. Any spacelike separation can be transformed to an equal-time separation, so $[\varphi (x),\varphi (y)]=0$ for spacelike $x-y$. Thus local operations at spacelike-separated points commute. The vacuum Wightman function need not vanish outside the light cone, but that correlation does not transmit a controllable signal. The Feynman propagator iswhich propagates positive energy forward and negative energy backward in the time-ordered product. For causal response one uses the retarded Green function, whose support lies in or on the future light cone.

Finally, commuting creation operators implyThe multiparticle state is symmetric under the particle exchange operator, which is Bose statistics. For a normalized single mode, the occupation states are $(a^{†}{)}^n|0⟩/\sqrt{n!}$ with $n=0,1,2,\dots$: any number of identical bosons can occupy it. There is no Pauli exclusion principle for these particles. This construction agrees with the Spin-statistics theorem relating integer spin to bosonic exchange symmetry under relativistic locality and positive-energy assumptions. It demonstrates the required scalar case without assuming that theorem as the quantization prescription.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, the corresponding coefficient is a difference of those square roots and multiplies ${\delta }^3(\mathbf{p}+\mathbf{q})$; it vanishes because $E_{\mathbf{p}}=E_{-\mathbf{p}}$. Taking the adjoint gives $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ is the vacuum expectation value of the time-ordered product of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,Here $\theta$ is the Heaviside step function, and changing $\mathbf{p}$ to $-\mathbf{p}$ in the second term gives the last line.

For the Fourier transform convention ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for $t>0$, and above for $t<0$, reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, soIt is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.

The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.

For a time-independent Hamiltonian operator, a symmetry operator $U$ satisfies $UH{U}^{†}=H$, or $[U,H]=0$. Thus applying $U$ to an energy eigenstate preserves its energy. Commuting Hermitian observables can be diagonalized within each degenerate energy eigenspace, allowing states to be labelled by symmetry quantum numbers.

A continuous unitary symmetry $U(\epsilon )=e^{-i\epsilon Q/\hslash }$ has Hermitian generator $Q$. Differentiating the invariance condition gives $[H,Q]=0$. The Heisenberg equation of motion then yieldswhen $Q$ has no explicit time dependence. spatial translations connect to momentum conservation, rotations to angular-momentum conservation, and time-translation invariance to energy conservation. A central potential energy conserves all components of angular momentum, but they cannot all be assigned sharp simultaneous values because they do not commute; $H,L^2,L_z$ are a compatible choice. Discrete reflection symmetry gives parity labels without a continuously variable generator. A quantum degeneracy can be protected by a symmetry, but a quantum degeneracy alone does not identify its physical origin.

For observables $A,B$, nonzero $[A,B]$ obstructs a complete common eigenbasis. In a finite-dimensional system, commuting Hermitian operators can be simultaneously diagonalized; quantum degeneracy may require choosing an appropriate basis inside the degenerate eigenspace. Noncommuting observables can still share particular eigenvectors, so incompatibility is not a claim that every individual state must have nonzero spreads in both quantities.

The Robertson uncertainty relation follows by applying Cauchy-Schwarz inequality to $(A-⟨A⟩)|\psi ⟩$ and $(B-⟨B⟩)|\psi ⟩$:In particular $[x,p_x]=i\hslash$ gives $\Delta x\Delta p_x\ge \hslash /2$. These are preparation spreads, not necessarily measurement errors. A zero commutator expectation can make this bound zero even when the linear operators do not commute.

Measurement order can matter: preparing $|+z⟩$, measuring $S_x$, then measuring $S_z$ gives both z outcomes with probability $1/2$ when the intermediate result is ignored. A direct z measurement would give $+\hslash /2$ certainly. Finally the Heisenberg equation of motion $dA/dt=(i/\hslash )[H,A]+\partial A/\partial t$ relates commutation to conservation: an observable with no explicit time dependence and $[H,A]=0$ is conserved. For unbounded self-adjoint operators these statements require the appropriate common domains; a formal commutator calculation alone need not justify simultaneous spectral measurements.

For a normalized state, the expectation value isFor a time-independent operator, differentiation givesThe Schrödinger equation says $|\dot{\psi }⟩=-iH|\psi ⟩/\hslash$. The self-adjointness of $H$ gives $⟨\dot{\psi }|=i⟨\psi |H/\hslash$, soIn the integral derivation, the same self-adjointness step is integration by parts with vanishing boundary terms. An explicitly time-dependent observable would also contribute $⟨\partial Q/\partial t⟩$. These are the expectation-value equations underlying the Heisenberg equation of motion.

In the Schrödinger picture, a time-independent Hamiltonian gives $|{\psi }_S(t)⟩=U(t)|\psi (0)⟩$, where $U=e^{-iHt/\hslash }$. The Heisenberg picture transfers this evolution to each observable:Both pictures give the same probabilities and expectation values. Differentiating yields the Heisenberg equation of motionThe last term is an explicit change of the measured operator. If it is absent and $[H,A]=0$, the observable is conserved.

The corresponding classical equation is $dA/dt=\{A,H{\}}_{\mathrm{PB}}+\partial A/\partial t$. The correspondence principle relates the Poisson bracket to $[A,H]/(i\hslash )$, with suitable operator ordering and a classical limit. For $H=p^2/(2m)+V(x)$, the canonical commutation relation givesTaking means gives Ehrenfest theorem. Its second equation contains $-⟨V^{\prime }(x)⟩$, which need not equal $-V^{\prime }(⟨x⟩)$. A narrow wave packet in a slowly varying force field can follow the classical equations approximately; a quadratic potential makes the equality exact for the means.

For a quantum harmonic oscillator, direct solution gives $x_H(t)=x_H(0)\cos \omega t+p_H(0)\sin \omega t/(m\omega )$ and $p_H(t)=p_H(0)\cos \omega t-m\omega x_H(0)\sin \omega t$. These have the classical form as operator identities. A coherent state supplies a localized center moving on the classical orbit, whereas an energy eigenstate has stationary density and $⟨x⟩=0$, even at large excitation. Thus large quantum numbers alone do not create a localized classical trajectory. Classical behavior also concerns state preparation, relative fluctuations and the resolution of the measurements.

Define the constant Dirac gamma matrix combinations ${\alpha }^k={\gamma }^0{\gamma }^k$ and $\beta ={\gamma }^0$. The Hamiltonian operator isEvaluate its commutator with $n(\mathbf{x})$ using the bilocal result in a. The mass term cancels immediately when $\mathbf{z}=\mathbf{y}$, because its two Dirac delta distributions coincide. For the derivative term first keep the two points distinct:The terms in which the derivative acts on ${\psi }_c$ cancel against each other; the derivative of the second Dirac delta distribution remains. ThusThe last equality is integration by parts, or the defining derivative rule for a distribution. Consequently the literal conserved current whose divergence equals $C$ isBecause a commutator of two Hermitian operators is anti-Hermitian, this $\mathbf{J}$ includes $-i$. The physical Hermitian Dirac current is instead $j^k=\bar{\psi }{\gamma }^k\psi$, so $C=-i\nabla \cdot \mathbf{j}$. The Heisenberg equation of motion now givesThis establishes the continuity equation directly from the required commutator, without assuming the Dirac equation or invoking the Noether theorem as a substitute for the computation.

Write ${\int }_{\mathbf{p}}=\int d^3\mathbf{p}/(2\pi {)}^3$. The canonical commutation relations and the product rule for a commutator giveThe other-species terms vanish. Integrating the Dirac delta distributions in the Hamiltonian therefore giveswhere $E_{-p}=E_p$. Consequently the complex scalar field satisfiesDirect differentiation of the time factors gives the same expression, proving the Heisenberg equation of motion

For three odd fermionic operators, the identityconverts the commutator with a quadratic Hamiltonian into the given canonical anticommutation relations. Writing a mode label as $i=(\mathbf{p},s)$, this yieldsThe opposite-species number terms commute with each fermionic creation operator or fermionic annihilation operator, because two anticommutations give a positive sign. Restoring the continuous normalization, the Dirac delta distribution in the mode brackets cancels $d^3\mathbf{q}/(2\pi {)}^3$. ConsequentlySubstitute these into the mode expansion of a Dirac field. The result isSince $i{\partial }_t{e}^{-ip\cdot x}=E_{\mathbf{p}}{e}^{-ip\cdot x}$ and $i{\partial }_t{e}^{ip\cdot x}=-E_{\mathbf{p}}{e}^{ip\cdot x}$, this is exactlyThe Heisenberg equation of motion uses an ordinary commutator with the even Hamiltonian, despite the odd fermionic operators obeying anticommutators among themselves.

The printed identity has the wrong sign for the specified relative phase. Differentiating the energy found above and evaluating the operator commutator giveThus the phase derivative of a two-mode condensate energy isA concrete counterexample to the printed sign is $N=100$, $\theta =\pi /4$, $\chi =\pi /2$ and $t>0$: its left side is $100t$, whereas its printed right side is $-100t$, independently of $w$ and $\mu$.