Number Operators and Conserved Quantities

The total number operator counts the particles or quanta represented by a chosen set of modes:

$$\widehat N = \sum_i a_i^\dagger a_i.$$

Its many-body role is broader than counting. It generates a global $U(1)$ transformation, decomposes Fock space into fixed-number sectors, supplies selection rules, and gives a one-line test for whether a Hamiltonian preserves particle number:

$$[H,\widehat N] = 0.$$

The spectra and elementary ladder commutators are developed in Number Operators. Here we use them to find symmetry sectors, test conversion and pairing terms, and distinguish conservation in a closed system from particle exchange with an environment.

Helpful background. Number operators supplies the ladder commutators; many-body Hilbert spaces explains sectors and mode cutoffs.

Total number and its sectors

Let $a_i$ be bosonic or fermionic annihilation operators in orthonormal modes of the chosen one-particle space. The mode occupations have eigenvalues $n_i=0,1,2,\ldots$ for bosons and $0,1$ for fermions. Total number is the one-body lift $\widehat N=d\Gamma(I)$: it acts as the scalar $N$ on the $N$-particle sector.

We use $U(\theta)=e^{-i\theta\widehat N}$ for global phase rotations and $\Pi_N$ for the number-sector projector. Hamiltonians are self-adjoint. The finite-dimensional calculations below are exact matrix identities; with unbounded operators, conservation means that the spectral projectors of $\widehat N$ reduce $H$ and its unitary evolution. A merely formal zero commutator on an unspecified domain is insufficient.

Fock-space decomposition

Bosonic or fermionic Fock space decomposes as

$$\mathcal F = \bigoplus_{N\in\mathcal S} \mathcal H_N,$$

where $\mathcal S=\{0,1,2,\ldots\}$ for an untruncated bosonic Fock space. For $M$ fermionic modes,

$$\mathcal S = \{0,1,\ldots,M\}.$$

The spectral resolution of total number is

$$\widehat N = \sum_{N\in\mathcal S} N\Pi_N,$$

with

$$\Pi_N\Pi_M = \delta_{NM}\Pi_N, \qquad \sum_N\Pi_N = I.$$

A vector can have components in several sectors even though the sectors are mutually orthogonal.

Projecting onto a chosen number

Because number eigenvalues are integers, the projector onto $\mathcal H_N$ has the Fourier representation

$$\Pi_N = \frac{1}{2\pi} \int_0^{2\pi} d\theta\, e^{i\theta(\widehat N-N)}.$$

Acting on a number eigenstate $\lvert M,\alpha\rangle$ gives

$$\Pi_N \lvert M,\alpha\rangle = \delta_{NM} \lvert M,\alpha\rangle.$$

This group-averaging formula is useful for number projection of variational states and for deriving sector-resolved partition functions.

Global U(1) Transformation

Define

$$U(\theta) = e^{-i\theta\widehat N}.$$

Using

$$[\widehat N,a_i] = -a_i, \qquad [\widehat N,a_i^\dagger] = a_i^\dagger,$$

one obtains

$$\begin{aligned} U(\theta)a_iU^\dagger(\theta) &= e^{i\theta}a_i, \\ U(\theta)a_i^\dagger U^\dagger(\theta) &= e^{-i\theta}a_i^\dagger. \end{aligned}$$

All modes receive the same phase. This is a global internal transformation, not a position-dependent gauge transformation.

Symmetry and Conservation

The Hamiltonian is invariant under this global $U(1)$ family when

$$U(\theta)HU^\dagger(\theta) = H$$

for every $\theta$. Differentiating at $\theta=0$ gives

$$[\widehat N,H] = 0.$$

Conversely, strong commutation makes $H$ invariant under this unitary group. In finite matrices, or with the spectral/domain condition stated above,

$$\text{global }U(1)\text{ invariance} \quad\Longleftrightarrow\quad [H,\widehat N]=0.$$

On a fixed-$N$ state, $U(\theta)$ produces only the ray phase $e^{-iN\theta}$. Across several number sectors it changes relative phases, and it acts nontrivially on charged operators.

Charge Counting for Operator Monomials

Let

$$M_{p,q} = a_{i_1}^\dagger\cdots a_{i_p}^\dagger a_{j_q}\cdots a_{j_1}.$$

Repeated use of the commutator product rule gives

$$[\widehat N,M_{p,q}] = (p-q)M_{p,q}.$$

The integer

$$r = p-q$$

is the number charge of the monomial: it maps the $N$ sector into the $N+r$ sector.

Equal numbers of creation and annihilation operators imply $r=0$. This term-by-term test is often faster and less error prone than constructing the full commutator matrix.

The number charge fixes which sectors an operator can connect. Pair creation and annihilation stay within a parity class; single-particle loss changes parity. A neutral operator acts within one sector, where it can still change the state. Arrows indicate allowed changes of $N$, not rates or amplitudes.

Examples of number charge

Term	Number charge	Consequence
$a_i^\dagger a_j$	$0$	transfers one particle; preserves total number
$a_i^\dagger a_j^\dagger a_l a_k$	$0$	pair scattering; preserves total number
$a_i^\dagger$	$+1$	creates one particle
$a_i$	$-1$	annihilates one particle
$a_i^\dagger a_j^\dagger$	$+2$	creates a pair
$a_j a_i$	$-2$	annihilates a pair
$a_i^\dagger a_j a_k$	$-1$	one creation and two annihilations

A Hermitian number-breaking Hamiltonian contains charge-changing terms in conjugate pairs. Hermiticity does not restore number conservation.

General Weighted Charges

Many conserved quantities are weighted mode occupations:

$$\widehat Q = \sum_i q_i n_i.$$

Then

$$[\widehat Q,a_i^\dagger] = q_i a_i^\dagger, \qquad [\widehat Q,a_i] = -q_i a_i.$$

For a monomial,

$$\begin{aligned} [\widehat Q,M_{p,q}] ={}& \left( \sum_{\alpha=1}^{p}q_{i_\alpha} - \sum_{\beta=1}^{q}q_{j_\beta} \right) M_{p,q}. \end{aligned}$$

This tests electric charge, species number, spin projection in an appropriate basis, or any Abelian additive quantum number. A process can violate the unweighted excitation count while conserving a physically weighted charge.

Number-Conserving Hamiltonians

A standard Hamiltonian with one- and two-body terms is

$$H ={} \sum_{i,j} h_{ij}a_i^\dagger a_j + \frac12 \sum_{i,j,k,l} V_{ij;kl} a_i^\dagger a_j^\dagger a_l a_k.$$

Every monomial is neutral, so

$$[H,\widehat N] = 0.$$

The Hamiltonian may strongly mix mode occupations and generate correlations. Conservation of total number does not imply conservation of each $n_i$ or independent-particle dynamics.

Time-Dependent Hamiltonians

Explicit time dependence of $H(t)$ does not by itself break particle-number conservation. If

$$[H(t),\widehat N] = 0$$

for every time, then

$$\frac{d\widehat N_H}{dt} = \frac{i}{\hbar} [H_H(t),\widehat N_H] = 0.$$

A time-dependent hopping amplitude, trap, or interaction can preserve number. A coherent source or pair drive can break it. Energy conservation and number conservation are separate questions.

Block Diagonalization by Number

Insert fixed-number projectors around the commutator:

$$\Pi_N [\widehat N,H] \Pi_M = (N-M) \Pi_N H\Pi_M.$$

If $[\widehat N,H]=0$, then

$$\Pi_N H\Pi_M = 0 \qquad N\neq M.$$

Therefore

$$H = \bigoplus_N H_N, \qquad H_N = \Pi_N H\Pi_N.$$

Time evolution is block diagonal as well:

$$U(t,t_0) = \bigoplus_N U_N(t,t_0).$$

This is the algebraic reason fixed-number exact diagonalization is valid for a number-conserving Hamiltonian.

Conserved Number Distribution

For a state $\rho$, define

$$P_N = \operatorname{Tr}(\rho\Pi_N).$$

When $[H,\widehat N]=0$, every $P_N$ is constant under closed-system time evolution. Consequently, all finite moments

$$\langle\widehat N^m\rangle = \sum_N N^m P_N$$

are conserved.

This is stronger than conservation of the mean alone. A state can have constant $\langle\widehat N\rangle$ while its full number distribution changes under a specially engineered nonconserving process. For closed Hamiltonian evolution, exact $U(1)$ symmetry freezes the entire distribution. The open-system qualification below is essential.

Conserved Does Not Mean Sharp

A number-conserving Hamiltonian does not force every state to have definite number. An initial state may contain several sectors:

$$\lvert\Psi(0)\rangle = \sum_N c_N\lvert\psi_N\rangle.$$

Each sector evolves independently:

$$\lvert\Psi(t)\rangle = \sum_N c_N U_N(t)\lvert\psi_N\rangle.$$

The probabilities $|c_N|^2$ are fixed, while relative sector phases and the internal states can evolve. Thus:

• conservation does not imply a sharp value;
• a sharp value does not by itself imply a superselection rule;
• an indefinite-number state does not by itself imply number-breaking dynamics.

Selection Rules from Number Charge

Suppose an operator $O_r$ satisfies

$$[\widehat N,O_r] = rO_r.$$

Between number eigenstates,

$$\langle N',\alpha'\vert [\widehat N,O_r] \vert N,\alpha\rangle ={} (N'-N) \times \langle N',\alpha'\vert O_r \vert N,\alpha\rangle.$$

Therefore a nonzero matrix element requires

$$N' = N+r.$$

In a fixed-number state, the expectation of every charged operator vanishes:

$$\langle O_r\rangle = 0 \qquad r\neq0.$$

This is why $\langle a_i\rangle$ and $\langle a_i a_j\rangle$ vanish in an exact number eigenstate even when one-body or pair coherence is physically strong.

Basis Invariance of Total Number

Let

$$b_\alpha = \sum_i U_{i\alpha}^*a_i$$

for unitary $U$. Then

$$\begin{aligned} \sum_\alpha b_\alpha^\dagger b_\alpha ={}& \sum_{i,j,\alpha} U_{i\alpha} U_{j\alpha}^* a_i^\dagger a_j \\ ={}& \sum_i a_i^\dagger a_i. \end{aligned}$$

Individual occupations transform into off-diagonal bilinears, but their sum remains $\widehat N$. This is also exact for a unitary rotation within a finite retained one-particle space. Selecting a different subspace is a change of model, not such a rotation; its projected observables must be compared separately.

Species-Resolved Numbers

For several species or internal components,

$$\widehat N_s = \sum_i a_{is}^\dagger a_{is}, \qquad \widehat N = \sum_s \widehat N_s.$$

A Hamiltonian can conserve more than one number. Spin-conserving Hubbard hopping preserves $N_\uparrow$ and $N_\downarrow$ separately. A spin-flip term such as

$$H_{\mathrm{sf}} = \Omega \sum_i \left( c_{i\uparrow}^\dagger c_{i\downarrow} + c_{i\downarrow}^\dagger c_{i\uparrow} \right)$$

with real $\Omega$ measured in energy units preserves $N_\uparrow+N_\downarrow$ but not the two component numbers separately.

The physically conserved combinations follow from all conversion terms, not from the labels chosen before writing the Hamiltonian.

Quasiparticle Number Is a Different Charge

Phonons, magnons, Bogoliubov modes, polaritons, and other quasiparticles have their own occupation operators. Their total occupation need not equal a microscopic particle number and is often not conserved.

A Hamiltonian diagonal in quasiparticle operators,

$$H_{\mathrm{qp}} = E_0 + \sum_\alpha E_\alpha \beta_\alpha^\dagger\beta_\alpha,$$

conserves $\sum_\alpha\beta_\alpha^\dagger\beta_\alpha$ only within that quadratic effective model. Interactions among quasiparticles can create and annihilate them while preserving an underlying atomic, electric, or spin charge.

Always name the counted degrees of freedom. Bogoliubov Quasiparticles works this distinction through explicitly for bosonic and fermionic transformations, including expected microscopic number, exact parity, and number-conserving formulations.

Local Number Versus Total Number

For a spatial region $\Omega$,

$$\widehat N_\Omega = \int_\Omega d^dr\, n(\mathbf r).$$

Even when total number is conserved, $\widehat N_\Omega$ generally is not. Particles can cross the boundary:

$$\frac{d\widehat N_\Omega}{dt} = - \int_{\partial\Omega} d\mathbf S\cdot \mathbf j_N + \widehat S_\Omega.$$

The source $\widehat S_\Omega$ vanishes for a locally number-conserving closed model. Taking $\Omega$ to be the complete closed system removes boundary flux and recovers global conservation.

The global commutator $[H,\widehat N]=0$ does not by itself identify a unique local current. Locality and the density decomposition are additional inputs.

Chemical Potential and the Grand Hamiltonian

When particle number is conserved, define

$$K = H-\mu\widehat N.$$

Then

$$[K,\widehat N] = 0.$$

On one fixed-$N$ sector,

$$K_N = H_N-\mu N I_N.$$

The chemical-potential term is a constant energy shift inside that sector. It does not change fixed-$N$ eigenvectors or closed-system expectation values. Its thermodynamic role is to change the relative weights assigned to different number sectors. In general, $H$ generates laboratory real-time evolution; using $K$ instead is a rotating-frame choice. Correlations containing number-changing fields acquire a chemical-potential phase even if the initial state has fixed number.

An equilibrium chemical potential normally couples to a conserved charge. If the proposed charge is not conserved by the microscopic isolated dynamics, the interpretation of its multiplier requires additional justification.

Grand-Canonical Mixtures Are Not Coherent Superpositions

For $[H,\widehat N]=0$, the grand-canonical density operator is block diagonal:

$$\rho_{\beta,\mu} = \bigoplus_N P_N\rho_{N,\beta}.$$

It is a statistical mixture over number sectors, not a coherent vector with an observable phase between sectors. Number fluctuations in this ensemble therefore do not by themselves violate a particle-number superselection restriction.

Conservation Is Not Superselection

Three statements must remain distinct:

1. $[H,\widehat N]=0$: the chosen closed-system dynamics preserves number sectors;
2. $\rho=\sum_N\Pi_N\rho\Pi_N$: the state has no inter-sector coherences;
3. $[A,\widehat N]=0$ for every allowed observable $A$: accessible operations cannot compare number-sector phases.

Only the third is an operational superselection restriction. A direct-sum decomposition or one commuting Hamiltonian is not enough to establish it. For particle-number restrictions, a phase reference or reservoir can make relative phases accessible in an enlarged number-conserving description; see particle-number superselection. This does not authorize odd fermionic observables or lift fundamental fermion-parity restrictions.

Fermion Parity

Define fermion parity spectrally by

$$P_F = (-1)^{\widehat N} = e^{i\pi\widehat N}.$$

It acts as

$$P_Fa_iP_F^\dagger = -a_i, \qquad P_Fa_i^\dagger P_F^\dagger = -a_i^\dagger.$$

An operator with even total ladder degree preserves fermion parity. In particular, pair creation and annihilation change $N$ by two, so they break continuous $U(1)$ number symmetry but preserve the discrete $\mathbb Z_2$ parity symmetry.

Bosonic models can also possess number parity when every process changes $N$ by an even integer. That model symmetry should not be conflated with the broader role of fermion parity in fermionic observable algebras.

Exact Reduced BCS Hamiltonian

Define the pair operators

$$b_{\mathbf k}^\dagger = c_{\mathbf k\uparrow}^\dagger c_{-\mathbf k\downarrow}^\dagger, \qquad b_{\mathbf k} = c_{-\mathbf k\downarrow} c_{\mathbf k\uparrow}.$$

A reduced BCS grand Hamiltonian is

$$K_{\mathrm{red}} ={} \sum_{\mathbf k,\sigma} \xi_{\mathbf k} c_{\mathbf k\sigma}^\dagger c_{\mathbf k\sigma} - g \sum_{\mathbf k,\mathbf k'} b_{\mathbf k}^\dagger b_{\mathbf k'},$$

Here $\xi_{\mathbf k}=\epsilon_{\mathbf k}-\mu$, and the sums run over a specified finite pairing shell with the corresponding time-reversed modes. The positive $g$ has energy units for these discrete sums; any volume normalization is absorbed into it. Since

$$[\widehat N,b_{\mathbf k}^\dagger] = 2b_{\mathbf k}^\dagger, \qquad [\widehat N,b_{\mathbf k}] = -2b_{\mathbf k},$$

the pair-scattering product is neutral:

$$[\widehat N,b_{\mathbf k}^\dagger b_{\mathbf k'}] = 0.$$

Therefore

$$[K_{\mathrm{red}},\widehat N] = 0.$$

The exact reduced pairing Hamiltonian scatters pairs; it does not create particles from nothing.

Reduced BCS Model develops the exact fixed-number problem. The comparison here isolates what changes under mean-field decoupling.

BCS Mean-Field Hamiltonian

Introduce

$$\Delta = g \sum_{\mathbf k} \langle b_{\mathbf k}\rangle.$$

The usual unprojected mean-field grand Hamiltonian has the form

$$K_{\mathrm{MF}} ={} \sum_{\mathbf k,\sigma} \xi_{\mathbf k} c_{\mathbf k\sigma}^\dagger c_{\mathbf k\sigma} - \sum_{\mathbf k} \left( \Delta b_{\mathbf k}^\dagger + \Delta^*b_{\mathbf k} \right) + \frac{|\Delta|^2}{g}.$$

Holding the chosen c-number $\Delta$ fixed,

$$[\widehat N,K_{\mathrm{MF}}] ={} -2\sum_{\mathbf k} \Delta b_{\mathbf k}^\dagger + 2\sum_{\mathbf k} \Delta^*b_{\mathbf k},$$

which is generally nonzero. The approximation mixes sectors whose particle numbers differ by two while preserving fermion parity.

This does not mean the underlying reduced BCS model violates particle-number conservation. The number breaking belongs to the symmetry-breaking mean-field representation. Reduced BCS Model develops the exact fixed-number sectors, blocking, and Richardson solution.

The thermodynamic saddle, coherence factors, and number-projected interpretation are developed in BCS Mean-Field Theory.

Pair-field phase and number projection

The direction of a phase transformation depends on whether we transform an operator or a state. With $U(\theta)=e^{-i\theta\widehat N}$,

$$U b_{\mathbf k}U^\dagger=e^{2i\theta}b_{\mathbf k}.$$

For an actively transformed state $\rho'=U\rho U^\dagger$, however,

$$\Delta'=g\sum_{\mathbf k}\operatorname{Tr}(\rho'b_{\mathbf k}) =e^{-2i\theta}\Delta, \qquad UK_{\mathrm{MF}}(\Delta)U^\dagger =K_{\mathrm{MF}}(e^{-2i\theta}\Delta).$$

The opposite phase appears because the expectation contains $U^\dagger bU$. Alternatively, a passive redefinition $b'=e^{2i\theta}b$, $\Delta'=e^{2i\theta}\Delta$ leaves the expression $-\Delta b^\dagger-\Delta^*b$ unchanged. These are two descriptions of the same phase convention, not contradictory transformation laws.

An exact number eigenstate has $\langle b_{\mathbf k}\rangle=0$. Pairing can still appear in neutral correlations $\langle b_{\mathbf k}^\dagger b_{\mathbf k'}\rangle$. Projecting a BCS variational vector with $\Pi_N$ restores a sharp number and zero anomalous mean, while preserving nontrivial neutral pair correlations. It does not make a nonzero charged expectation compatible with exact fixed number.

Explicit, Spontaneous, and Effective Breaking

These phrases describe different physics.

Explicit breaking

The Hamiltonian itself does not commute with $\widehat N$. For bosonic modes a coherent source is

$$H_{\mathrm{src}} = \sum_i \left( f_i a_i^\dagger + f_i^*a_i \right).$$

It changes subsystem number by one and fixes a phase reference. A fermionic Hamiltonian must be even in fermion parity: an ordinary complex coefficient multiplying a lone $c_i$ is not a physical closed-system source. Fermions can exchange with explicit fermionic reservoir modes through even bilinears, or with a pair field through even pairing terms.

Spontaneous breaking

The microscopic Hamiltonian preserves $U(1)$, but an infinite-system equilibrium phase need not. A standard construction takes the thermodynamic limit before removing an infinitesimal phase-selecting source. A finite variational ansatz with a chosen phase can approximate such a phase; the ansatz alone does not prove spontaneous breaking. Finite exact number eigenstates can show strong neutral pairing correlations.

Effective breaking

The modeled subsystem exchanges particles with an omitted condensate, reservoir, drive, or reaction channel. The subsystem Hamiltonian or master equation need not conserve its number even when an enlarged closed description conserves a total charge.

Using the word “broken” without identifying which of these mechanisms is intended obscures the approximation.

Number-Changing Bosonic Terms

Bosonic effective models commonly contain a linear source or a pairing drive:

$$H_{\mathrm{pair}} = \frac12 \sum_{i,j} \left( \Delta_{ij}a_i^\dagger a_j^\dagger + \Delta_{ij}^*a_j a_i \right).$$

The source changes $N$ by one. The pair drive changes $N$ by two and can preserve bosonic number parity. Stability of a bosonic pairing Hamiltonian is a separate condition; a formal parity symmetry does not guarantee a spectrum bounded below.

A mean-field replacement $a_i\to\phi_i$ is an approximation involving a phase reference or symmetry-breaking limit. It is not an identity in a finite fixed-number Hilbert space.

Open-System Number Balance

For a Lindblad–GKSL equation

$$\dot\rho = -\frac{i}{\hbar}[H,\rho] + \sum_\mu \kappa_\mu \mathcal D[L_\mu]\rho,$$

with

$$\mathcal D[L]\rho = L\rho L^\dagger - \frac12 \{L^\dagger L,\rho\},$$

the number expectation obeys

$$\frac{d}{dt} \langle\widehat N\rangle ={} \frac{i}{\hbar} \langle[H,\widehat N]\rangle + \sum_\mu \kappa_\mu \left\langle L_\mu^\dagger\widehat N L_\mu - \frac12 \{L_\mu^\dagger L_\mu,\widehat N\} \right\rangle.$$

If

$$[\widehat N,L_\mu] = -r_\mu L_\mu$$

then $L_\mu^\dagger L_\mu$ commutes with $\widehat N$ automatically, and that channel contributes

$$-r_\mu\kappa_\mu \langle L_\mu^\dagger L_\mu\rangle.$$

Single-particle loss has $r_\mu=1$, pair loss has $r_\mu=2$, and number dephasing with $L_\mu=n_i$ has $r_\mu=0$.

The dissipator is unchanged by $L\mapsto e^{i\phi}L$. Consequently a loss generator can be covariant under global $U(1)$ rotations while still reducing $\langle N\rangle$. Symmetry covariance of an open-system equation is weaker than conservation of its symmetry generator (Buča and Prosen, 2012). If $H$ and every jump commute with $N$, the sector probabilities are conserved; charged jumps instead transfer probability between sectors. This distinction prevents carrying the closed-system symmetry argument over to a dissipator without checking its adjoint action.

Exercises

Number blocks and sector probabilities

Assume $[H,\widehat N]=0$.

1. Show that $\Pi_NH\Pi_M=0$ for $N\neq M$.
2. Show that $P_N(t)=\operatorname{Tr}[\rho(t)\Pi_N]$ is constant under unitary evolution.

Solution

For the first part,

$$\Pi_N [\widehat N,H] \Pi_M = (N-M) \Pi_NH\Pi_M.$$

The left side vanishes. If $N\neq M$, division by $N-M$ gives

$$\Pi_NH\Pi_M = 0.$$

Thus $H$ commutes with every spectral projector $\Pi_N$. For the second part,

$$\begin{aligned} \frac{dP_N}{dt} &= -\frac{i}{\hbar} \operatorname{Tr} \left( [H,\rho]\Pi_N \right) \\ &= -\frac{i}{\hbar} \operatorname{Tr} \left( \rho[\Pi_N,H] \right) = 0. \end{aligned}$$

Every sector probability is conserved.

Chemical potential in a fixed-number sector

Let $K=H-\mu\widehat N$ with $[H,\widehat N]=0$. Show that on $\mathcal H_N$, evolution under $K$ differs from evolution under $H$ only by a global phase.

Solution

On the fixed-$N$ sector,

$$K_N = H_N-\mu N I_N.$$

Because the second term is proportional to the identity,

$$\begin{aligned} e^{-iK_Nt/\hbar} &= e^{-i(H_N-\mu NI_N)t/\hbar} \\ &= e^{i\mu Nt/\hbar} e^{-iH_Nt/\hbar}. \end{aligned}$$

The prefactor is a global phase on $\mathcal H_N$, so every expectation value within that sector is unchanged. The chemical potential matters when comparing or statistically weighting different $N$ sectors.

Single-particle loss

For one bosonic mode with number-conserving $H$ and master equation

$$\dot\rho = -\frac{i}{\hbar}[H,\rho] + \kappa\mathcal D[a]\rho,$$

derive the equation for $\langle n\rangle$. Is the generator nevertheless covariant under $U(\theta)=e^{-i\theta n}$?

Solution

The Hamiltonian contribution vanishes because $[H,n]=0$. For the dissipator, use

$$[n,a] = -a,$$

so the jump removes one quantum. The adjoint dissipator gives

$$\begin{aligned} \frac{d}{dt}\langle n\rangle ={}& \kappa \left\langle a^\dagger n a - \frac12 \{a^\dagger a,n\} \right\rangle \\ ={}& -\kappa \langle a^\dagger a\rangle. \end{aligned}$$

Therefore

$$\frac{d}{dt} \langle n\rangle = -\kappa\langle n\rangle,$$

with solution $\langle n(t)\rangle=e^{-\kappa t}\langle n(0)\rangle$. Number is not conserved in the subsystem because the environment receives the lost quanta. Nevertheless, $UaU^\dagger=e^{i\theta}a$ and $\mathcal D[e^{i\theta}a]=\mathcal D[a]$ establish covariance: rotating the initial state rotates its entire evolution. Covariance does not fix its number distribution.

Atom–molecule conversion

For independent bosonic atom and molecule modes $a,b$, take $H_g=g b^\dagger a^2+g^*a^{\dagger2}b$. Which is conserved: $N_a+N_b$ or $Q=N_a+2N_b$? Find the allowed basis in the sector $Q=4$.

Solution

The term $T=b^\dagger a^2$ satisfies $[N_a,T]=-2T$ and $[N_b,T]=T$. Thus $[Q,T]=0$, and the same holds for $T^\dagger$. The unweighted object count changes by one, but the count of constituent atoms is conserved. The $Q=4$ basis is $|4,0\rangle,|2,1\rangle,|0,2\rangle$. Conversion connects neighboring vectors with amplitudes $g\sqrt{12}$ and $2g$, respectively. A symmetry block based on $N_a+N_b$ would incorrectly remove both processes.

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