Lecture 7: Fock Space and Occupation States

The excitation language just introduced becomes unwieldy if indistinguishable particles are still tracked by long coordinate-space wave functions. Fock space reorganizes the problem around occupied modes and number sectors, making creation, removal, exchange signs, and conservation laws calculable without assigning identities to particles. Nearly every later interacting Hamiltonian—from Hubbard and pairing models to Green functions and transport channels—uses this change of representation. The operator algebra does more than shorten notation: it encodes the particle statistics and sector structure that determine which many-body processes are physically possible.

Imagine a gate-defined double quantum dot tuned so that zero, one, or two electrons can occupy two localized modes. A nearby charge sensor can distinguish the patterns ${(0,0)}$ , ${(1,0)}$ , ${(0,1)}$ , and ${(1,1)}$ , but it does not attach a persistent identity to either electron. When a gate pulse permits tunneling, the amplitude for a final occupation pattern depends on fermionic signs set by the ordered mode basis. An optical-lattice double well realizes the same occupation-based description on a different experimental platform: site-resolved imaging can sample occupations and correlations directly over repeated preparations.

This raises three laboratory questions. Given an initial occupation pattern and a tunneling operation, what final patterns are allowed, with what amplitudes, and what would a local measurement actually distinguish? Fock space answers the first two questions. The measurement basis answers the third.

Lecture 6’s addition-and-removal bookkeeping described an electron excitation as adding or removing an occupation from a one-particle mode and a phonon as changing a normal-mode occupation. For one mode, that language was intuitive. For hundreds of electronic orbitals or lattice modes, it becomes a computational problem: a fermionic sign depends on which other modes are occupied, a local interaction acts only on particular occupation patterns, and a change from sites to bands must transform states and observables consistently. Creation and annihilation maps address both tasks: their products construct states from the vacuum and generate number, transfer, hopping, and interaction operators. Exchange signs belong to the mode calculus; physical matrix elements supply the coefficients.

One could return to an explicitly symmetrized or antisymmetrized wave function and attach particle coordinates ${\mathbf r_1,\ldots,\mathbf r_N}$ . That is formally correct, but the labels are bookkeeping devices for identical particles, not physical identities. Every basis change and interaction would require the same exchange constraints to be rebuilt. Fock space instead makes the one-particle modes primary and stores how many indistinguishable quanta occupy each mode.

Occupations of an ordered mode basis retain the exchange constraints without assigning identities to particles. The two-site hopping calculation will test the resulting sign rule and connect it to basis-dependent readout.

This language is sometimes called “second quantization.” We will not use that name after this paragraph because it misleadingly suggests a second physical quantization. The theory is the same many-particle quantum mechanics expressed in an occupation-number basis. The term Fock-space formulation identifies the basis and operator algebra used to represent many-particle states.

The governing question for this lecture is

Once a one-particle mode basis is chosen, how do exchange symmetry and sector-changing maps determine normalized many-particle states, fermionic signs, and the operators for occupation measurements and dynamics?

The construction must preserve physical predictions under a change of basis: changing modes changes occupation coordinates and operator components, but not the physical state or consistently transformed predictions.

Before proceeding, apply a harmonic-oscillator raising operator to ${|n\rangle}$ , state the exchange symmetry of two identical fermions, and write the normalized discrete Fourier transform of a sequence on ${N}$ sites. Use these three operations to check the prerequisites before introducing multimode signs. The worked examples use declared normalized pure states. The later comparison with equilibrium Fermi occupations instead refers explicitly to an ensemble average.

From exchange symmetry to occupation sectors

Fock space is the graded state space constructed from a chosen one-particle space. Creation and annihilation operators are maps between its number sectors, and their signs and normalization follow from exchange symmetry. The finite-mode construction below is exact. Choosing only a subset of physical modes is a separate effective-model step whose control comes from spectral separation or an error estimate, not from the canonical algebra itself.

Exchange-symmetric sectors

The construction below separates structure from coordinates. Fock space and the canonical (anti)commutation relations are basis-independent structures; an occupation string is the coordinate label supplied by one ordered one- particle basis. Rotating that basis changes the strings and the corresponding creation operators but not the many-particle state or its physical matrix elements. A creation operator maps between number sectors while preserving the exchange symmetry of indistinguishable particles.

Let ${\mathcal H_1}$ be an ${M}$ -dimensional one-particle Hilbert space with an ordered orthonormal basis ${(|1\rangle,\ldots,|M\rangle)}$ . The label ${i}$ identifies a mode: it may combine site, orbital, spin, band, or momentum information. It does not label a persistent particle.

For ${N\ge1}$ particles, first form the tensor product ${\mathcal H_1^{\otimes N}}$ . Its positions are bookkeeping slots: exchanging slots must not be interpreted as assigning persistent identities to particles. We now define the two exchange-symmetry sectors used in this lecture.

Start with two slots. Write ${|i,j\rangle:=|i\rangle\otimes|j\rangle}$ . The swap acts as ${\hat U_{(12)}|i,j\rangle=|j,i\rangle}$ , and swapping twice is the identity ${\hat I}$ . The symmetric and antisymmetric components of a tensor vector are selected by ${\hat P_\pm^{(2)}=(\hat I\pm\hat U_{(12)})/2}$ .

For two distinct orthonormal modes ${i\ne j}$ , the only slot permutations are the identity and the swap. Projection and normalization then give

$\hat P_\pm^{(2)}|i,j\rangle =\frac{|i,j\rangle\pm|j,i\rangle}{2}, \qquad \frac{\hat P_\pm^{(2)}|i,j\rangle} {\|\hat P_\pm^{(2)}|i,j\rangle\|} =\frac{|i,j\rangle\pm|j,i\rangle}{\sqrt2}.$

If both modes are ${i}$ , symmetrization instead returns the already normalized ${|i,i\rangle}$ , whereas antisymmetrization gives zero: equal terms cancel pairwise. This is why repeating a resolved mode is allowed for bosons and excluded for fermions. For arbitrary particle number, we will average over all tensor-slot permutations to construct the corresponding sectors.

For ${N\ge1}$ , let ${\mathfrak S_N}$ be the group of permutations of the ${N}$ tensor slots. For ${\pi\in\mathfrak S_N}$ , the permutation operator ${\hat U_\pi:\mathcal H_1^{\otimes N}\to\mathcal H_1^{\otimes N}}$ is the linear map defined on product vectors by

$\hat U_\pi\bigl(|v_1\rangle\otimes\cdots\otimes|v_N\rangle\bigr) =|v_{\pi^{-1}(1)}\rangle\otimes\cdots\otimes|v_{\pi^{-1}(N)}\rangle, \qquad |v_r\rangle\in\mathcal H_1.$

Its domain is the full tensor product. It is unitary, with ${\hat U_\pi^\dagger=\hat U_{\pi^{-1}}}$ , and the inverse-index convention gives ${\hat U_\pi\hat U_\sigma=\hat U_{\pi\sigma}}$ , where ${(\pi\sigma)(r)=\pi(\sigma(r))}$ . The parity ${\operatorname{sgn}(\pi)}$ is ${+1}$ for an even number of pair exchanges and ${-1}$ for an odd number; it is independent of the chosen decomposition. Concatenating two exchange sequences gives ${\operatorname{sgn}(\pi\sigma)=\operatorname{sgn}(\pi)\operatorname{sgn}(\sigma)}$ .

On ${\mathcal H_1^{\otimes N}}$ , for ${N\ge1}$ , the symmetrization and antisymmetrization operators are the orthogonal projectors

$\hat P_+^{(N)}:=\frac{1}{N!}\sum_{\pi\in\mathfrak S_N}\hat U_\pi, \qquad \hat P_-^{(N)}:=\frac{1}{N!}\sum_{\pi\in\mathfrak S_N} \operatorname{sgn}(\pi)\hat U_\pi.$

Their domains are the full tensor product. Their ranges define the sectors ${\mathcal H_+^{(N)}:=\operatorname{Ran}\hat P_+^{(N)} =\operatorname{Sym}^N\mathcal H_1}$ and ${\mathcal H_-^{(N)}:=\operatorname{Ran}\hat P_-^{(N)} =\bigwedge^N\mathcal H_1}$ . Equivalently, every vector ${\psi}$ in the symmetric sector obeys ${\hat U_\pi\psi=\psi}$ , and every vector in the antisymmetric sector obeys ${\hat U_\pi\psi=\operatorname{sgn}(\pi)\psi}$ , for every permutation. Symmetrizing or antisymmetrizing a vector means applying the corresponding projector; a nonzero result must then be divided by its norm to obtain a unit state. The zero vector cannot be normalized.

Multiplying either permutation average by ${\hat U_\sigma}$ relabels the sum. Parity multiplication gives the factor in the antisymmetric case:

$\hat U_\sigma\hat P_+^{(N)}=\hat P_+^{(N)}, \qquad \hat U_\sigma\hat P_-^{(N)} =\operatorname{sgn}(\sigma)\hat P_-^{(N)}.$

Thus the projected vector has the stated exchange symmetry. On a vector already having that exchange symmetry, each signed summand returns the same vector, so the average acts as the identity. Applying the average a second time therefore leaves the result unchanged. Taking the adjoint merely relabels permutations by their inverses, so the averages are self-adjoint as well as idempotent. The physical choice of the symmetric sector for bosons or the antisymmetric sector for fermions is an input about the particle species; constructing projectors does not derive that statistics postulate.

For two slots, ${\hat P_\pm^{(2)}=(\hat I\pm\hat U_{(12)})/2}$ , where ${\hat I}$ is the identity on ${\mathcal H_1^{\otimes2}}$ . Since swapping twice gives ${\hat U_{(12)}^2=\hat I}$ , the two projectors obey ${\hat P_+^{(2)}\hat P_-^{(2)}=0}$ and ${\hat P_+^{(2)}+\hat P_-^{(2)}=\hat I}$ . Thus every tensor vector ${\psi\in\mathcal H_1^{\otimes2}}$ splits uniquely into its symmetric and antisymmetric components, as shown in Figure 7.1. This decomposition concerns the ambient tensor space; the particle species still selects the physical sector.

Figure 7.1. Symmetrization and antisymmetrization extract orthogonal components of a two-particle tensor vector, shown for at least two one-particle modes. The blue horizontal axis represents the symmetric subspace and the vermillion vertical axis the antisymmetric subspace. The black arrow is the input tensor vector; dashed lines guide its projections. Each axis stands for a whole subspace, not a single dimension or a spatial direction; lengths are schematic. The two sectors exhaust the two-particle tensor space. For three or more particles, they need not exhaust the tensor product. Project-original schematic.

For ${N\ge1}$ and one-particle vectors ${|f_1\rangle,\ldots,|f_N\rangle\in\mathcal H_1}$ , define their exterior product, or wedge product, by

$|f_1\rangle\wedge\cdots\wedge|f_N\rangle :=\sqrt{N!}\,\hat P_-^{(N)} \bigl(|f_1\rangle\otimes\cdots\otimes|f_N\rangle\bigr) \in\mathcal H_-^{(N)}.$

For two factors this gives

$|f\rangle\wedge|g\rangle =\frac{|f\rangle\otimes|g\rangle-|g\rangle\otimes|f\rangle}{\sqrt2}.$

The map is linear in each factor. Antisymmetrization reverses the sign when two factors are exchanged, so a repeated factor gives zero. In particular, ${|g\rangle\wedge|f\rangle=-|f\rangle\wedge|g\rangle}$ and ${|f\rangle\wedge|f\rangle=0}$ . The factor ${\sqrt{N!}}$ gives unit norm for orthonormal factors; arbitrary factors need not produce a unit vector.

For a complex one-particle Hilbert space ${\mathcal H_1}$ , its bosonic ( ${+}$ ) or fermionic ( ${-}$ ) Fock space is the Hilbert direct sum

$\mathcal F_\pm(\mathcal H_1) :=\bigoplus_{N=0}^{\infty}\mathcal H_\pm^{(N)} =\left\{ \boldsymbol\psi=(\psi^{(0)},\psi^{(1)},\ldots): \sum_{N=0}^{\infty}{\lVert\psi^{(N)}\rVert}^2<\infty \right\},$

where ${\psi^{(N)}\in\mathcal H_\pm^{(N)}}$ for every ${N\ge0}$ . The inner product is ${\langle\boldsymbol\phi,\boldsymbol\psi\rangle =\sum_N\langle\phi^{(N)},\psi^{(N)}\rangle}$ . The vacuum sector is ${\mathcal H_\pm^{(0)}\cong\mathbb C|0\rangle}$ . The algebraic Fock space ${\mathcal F_{\pm,\mathrm{alg}}}$ is the dense subspace of sequences with only finitely many nonzero sector components; the square-summability condition specifies the Hilbert completion of that algebraic sum. For ${\dim\mathcal H_1=M<\infty}$ , fermionic sectors with ${N>M}$ are the zero space.

For the calculations, retain the sector structure ${\mathcal F_\pm=\bigoplus_{N\ge0}\mathcal H_\pm^{(N)}}$ , including the vacuum ${\mathcal H_\pm^{(0)}=\mathbb C|0\rangle}$ . The algebraic space ${\mathcal F_{\pm,\mathrm{alg}}}$ consists of finite sums of sector vectors. For our finite set of ${M}$ modes, each sector is finite-dimensional; fermionic sectors with ${N>M}$ vanish. The full Hilbert space also includes norm-convergent sums of such vectors.

The direct sum is a Hilbert-space construction. It does not assert that a closed sample fluctuates in particle number. A state with fixed ${N}$ lies entirely in one summand; the larger space lets one use maps that add or remove a mode occupation and lets number-conserving and number-changing models share one notation. Fock’s 1932 paper is an original source for representing Bose and Fermi operator calculus on a sequence of ${n}$ -particle configuration spaces [Foc32, printed pp. 622–633, especially the introductory summary and Eqs. (17)–(39)]. The completed direct-sum definition, domain language, and modern term Fock space used here are a bounded modern restatement, not a verbatim claim about Fock’s notation.

Bosonic occupation states and occupation labels

Fix an ordered orthonormal one-particle basis ${\mathcal E=(|1\rangle,\ldots,|M\rangle)}$ . For an occupation tuple ${\mathbf n=(n_1,\ldots,n_M)\in\mathbb N_0^M}$ with ${N=\sum_i n_i\ge1}$ , define the ordered product tensor ${\Phi_{\mathbf n}:={|1\rangle}^{\otimes n_1}\otimes\cdots \otimes{|M\rangle}^{\otimes n_M}}$ , omitting factors with zero occupation. The corresponding bosonic occupation state is its normalized symmetrization,

${|\mathbf n\rangle}_{\mathrm B} :=\sqrt{\frac{N!}{\prod_i n_i!}}\, \hat P_+^{(N)}\Phi_{\mathbf n}.$

Together with the normalized vacuum ${{|\mathbf 0\rangle}_{\mathrm B}:=|0\rangle}$ , these states form the bosonic occupation-number basis of ${\mathcal F_+(\mathcal H_1)}$ , the orthonormal family

$\left\{{|\mathbf n\rangle}_{\mathrm B}:\ \mathbf n=(n_1,\ldots,n_M)\in\mathbb N_0^M\right\}.$

The sum ${\sum_i n_i=N}$ selects the ${N}$ -particle sector. The index ${i}$ and entry ${n_i}$ label a one-particle mode and its occupation, not a persistent particle identity. Fermionic occupation vectors will be defined by ordered creation after deriving the fermionic operator algebra.

The bosonic normalization accounts for the ${\prod_i n_i!}$ permutations that exchange identical factors. Distinct occupation patterns have disjoint product-basis supports, and symmetrizing the product basis spans each symmetric sector. A full projector calculation below checks both statistics after the fermionic family has been constructed.

From creation maps to canonical algebra

Constructing the mode maps

The mode maps serve two related purposes: they construct many-particle states from the vacuum, and their products construct other operators. Removing an occupation and then adding one in another mode describes a transfer; adding it back in the same mode gives a number operator. Such products can preserve particle number even though their individual factors change it. We first construct the maps from the exchange projectors, without assuming their commutation relations.

Let ${\mathcal D_\pm:=\mathcal F_{\pm,\mathrm{alg}}}$ . For a one-particle state ${|f\rangle\in\mathcal H_1}$ and ${\Psi_N\in\mathcal H_\pm^{(N)}}$ , define creation by

$\begin{aligned} \hat a^\dagger(f)\Psi_N &:=\sqrt{N+1}\,\hat P_+^{(N+1)}\bigl(|f\rangle\otimes\Psi_N\bigr),\\[0.4em] \hat c^\dagger(f)\Psi_N &:=\sqrt{N+1}\,\hat P_-^{(N+1)}\bigl(|f\rangle\otimes\Psi_N\bigr). \end{aligned}$

Extend these maps linearly to ${\mathcal D_+}$ and ${\mathcal D_-}$ , respectively. For a basis mode ${|i\rangle}$ , write ${\hat a_i^\dagger:=\hat a^\dagger(|i\rangle)}$ and ${\hat c_i^\dagger:=\hat c^\dagger(|i\rangle)}$ . Creation sends the ${N}$ -particle sector into the ${N+1}$ sector, with ${\hat d^\dagger(f)|0\rangle=|f\rangle}$ , where ${\hat d=\hat a}$ or ${\hat c}$ . The annihilation map is its algebraic adjoint on the corresponding core: it lowers particle number by one, sends the vacuum to zero, and satisfies ${\langle\phi,\hat d(f)\psi\rangle =\langle\hat d^\dagger(f)\phi,\psi\rangle}$ .

For finitely many modes the fermionic maps extend to bounded operators on ${\mathcal F_-}$ , and those extensions are Hilbert-space adjoints. The bosonic maps for nonzero ${f}$ are unbounded: their Hilbert-space adjoints require their maximal operator domains, so the displayed core may not be replaced by all of ${\mathcal F_+}$ .

Creation depends linearly on the added one-particle state, as well as on the input many-particle vector. The same formulas extend to unnormalized vectors. If ${|f\rangle=\sum_i f_i|i\rangle}$ , the projector construction gives

$\hat d^\dagger(f)=\sum_i f_i\hat d_i^\dagger, \qquad \hat d(f)=\sum_i f_i^*\hat d_i, \qquad \hat d=\hat a\ \text{or}\ \hat c.$

Thus a superposition of one-particle states determines the same superposition of creation maps on every sector. For fermions, the factor ${\sqrt{N+1}}$ matches the normalization convention for exterior products; it does not normalize every output. Dividing each nonzero output by its norm would destroy linearity. The exterior product in Definition 7.3 is already defined for arbitrary one-particle vectors and is linear in each factor. Its output has unit norm for orthonormal factors, but need not have unit norm for general one-particle states.

Annihilation as contraction of an antisymmetric product

For the fermionic algebra, work with the declared finite ${M}$ -dimensional one-particle space. We first find the annihilator on arbitrary exterior products, without introducing occupation coordinates. Write ${W_N:=|f_1\rangle\wedge\cdots\wedge|f_N\rangle}$ for arbitrary ${|f_r\rangle\in\mathcal H_1}$ ; orthogonality and unit norm are not assumed. Set ${W_0:=|0\rangle}$ for the empty product.

Antisymmetrizing all slots absorbs an antisymmetrizer on the last ${N}$ slots:

$\hat P_-^{(N+1)}(\hat I\otimes\hat P_-^{(N)}) =\hat P_-^{(N+1)} \qquad(N\ge1).$

Each permutation of those slots contributes its parity in the inner average and the same parity under the full projector. Their product is one. Hence ${\sqrt{N+1}\sqrt{N!}=\sqrt{(N+1)!}}$ in the creation definition gives

$\hat c^\dagger(f)W_N =|f\rangle\wedge|f_1\rangle\wedge\cdots\wedge|f_N\rangle.$

For ${N=0}$ , this is the defining rule ${\hat c^\dagger(f)|0\rangle=|f\rangle}$ . Creation therefore prepends one factor with the exterior-product normalization already fixed.

To obtain annihilation, let ${|\Psi_N\rangle\in\mathcal H_-^{(N)}}$ , with ${N\ge1}$ , and test the adjoint relation against an arbitrary ${|\Phi_{N-1}\rangle\in\mathcal H_-^{(N-1)}}$ . The projector is self-adjoint and fixes the input, ${\hat P_-^{(N)}|\Psi_N\rangle=|\Psi_N\rangle}$ , so

$\begin{aligned} \langle\Phi_{N-1}|\hat c_i|\Psi_N\rangle &={\bigl(\hat c_i^\dagger|\Phi_{N-1}\rangle\bigr)}^\dagger |\Psi_N\rangle\\ &=\sqrt N\,{\Bigl[ \hat P_-^{(N)}\bigl(|i\rangle\otimes|\Phi_{N-1}\rangle\bigr) \Bigr]}^\dagger|\Psi_N\rangle\\ &=\sqrt N\, \bigl(\langle i|\otimes\langle\Phi_{N-1}|\bigr) \hat P_-^{(N)}|\Psi_N\rangle\\ &=\sqrt N\, \bigl(\langle i|\otimes\langle\Phi_{N-1}|\bigr)|\Psi_N\rangle\\ &=\langle\Phi_{N-1}|\Bigl[ \sqrt N(\langle i|\otimes\hat I^{\otimes(N-1)})|\Psi_N\rangle\Bigr]. \end{aligned}$

The partial inner product on the right remains antisymmetric in its ${N-1}$ remaining slots. Adjointness therefore identifies it with ${\hat c_i|\Psi_N\rangle}$ . For ${N=1}$ , its scalar value multiplies the vacuum: ${\hat c_i|f\rangle=\langle i|f\rangle|0\rangle}$ .

For ${N\ge2}$ , group the signed permutation average defining ${W_N}$ by the factor in the first slot. Bringing factor ${r}$ to the front takes ${r-1}$ exchanges; the remaining permutations form an exterior product. In each indexed wedge below, omit the excluded factor and take the surviving factors in increasing index order; the empty product is the vacuum:

$W_N=\frac1{\sqrt N}\sum_{r=1}^{N}{(-1)}^{r-1} |f_r\rangle\otimes \Bigl(\bigwedge_{1\le s\le N,\;s\ne r}|f_s\rangle\Bigr).$

Applying the partial inner product cancels ${1/\sqrt N}$ and gives the contraction rule, including ${N=1}$ with the empty product understood as the vacuum:

(7.1)

$\hat c_i W_N =\sum_{r=1}^{N}{(-1)}^{r-1}\langle i|f_r\rangle \bigwedge_{1\le s\le N,\;s\ne r}|f_s\rangle, \qquad N\ge1.$

On the vacuum the defining annihilation rule gives zero. Thus annihilation removes each possible factor, weights it by its one-particle overlap, and counts the exchanges needed to expose it. This conclusion used only the projector, exterior-product normalization, and adjoint relation.

Hide the displayed creation and contraction formulas for eight minutes. In the ordered three-mode space ${(1,2,3)}$ , evaluate ${\hat c_2^\dagger(|1\rangle\wedge|3\rangle)}$ and ${\hat c_3(|2\rangle\wedge|3\rangle)}$ , showing every exchange sign. Then act with both orderings in ${\hat c_i\hat c_j^\dagger+\hat c_j^\dagger\hat c_i}$ on an arbitrary exterior product and predict their sum. Use the projector definition and adjoint relation to recover a missing step; do not assume the operator anticommutation relations.

Canonical anticommutation from the projector definition

The contraction rule lets us compare adding a factor before removal with removing a factor before addition. The new first factor supplies one term; all contractions with the original factors occur in both orders.

For the declared finite-dimensional one-particle space, the fermionic maps in Definition 7.6 satisfy, on ${\mathcal D_-=\mathcal F_-(\mathcal H_1)}$ ,

$\{\hat c_i,\hat c_j^\dagger\}=\delta_{ij}\hat I, \qquad \{\hat c_i^\dagger,\hat c_j^\dagger\}=0, \qquad \{\hat c_i,\hat c_j\}=0.$

Here ${\{\hat A,\hat B\}:=\hat A\hat B+\hat B\hat A}$ . These are the canonical anticommutation relations (CAR).

For an arbitrary exterior product ${W_N}$ , prepend ${|j\rangle}$ and apply the contraction rule. The new factor contributes its one-particle overlap. Each original factor has shifted one position to the right, reversing its contraction sign. Factor the creator out of the remaining sum by linearity; the bracketed expression is the original contraction sum:

$\begin{aligned} \hat c_i\hat c_j^\dagger W_N &=\hat c_i\bigl(|j\rangle\wedge W_N\bigr)\\ &=\langle i|j\rangle W_N+ \sum_{r=1}^{N}{(-1)}^r\langle i|f_r\rangle |j\rangle\wedge \Bigl(\bigwedge_{1\le s\le N,\;s\ne r}|f_s\rangle\Bigr)\\ &=\delta_{ij}W_N-\hat c_j^\dagger\Bigl[ \sum_{r=1}^{N}{(-1)}^{r-1}\langle i|f_r\rangle \bigwedge_{1\le s\le N,\;s\ne r}|f_s\rangle\Bigr]\\ &=\delta_{ij}W_N-\hat c_j^\dagger\hat c_iW_N. \end{aligned}$

For the vacuum, the same identity reads ${\hat c_i\hat c_j^\dagger|0\rangle=\delta_{ij}|0\rangle}$ . For two creators, exchanging the two prepended factors gives

$\hat c_i^\dagger\hat c_j^\dagger W_N =|i\rangle\wedge|j\rangle\wedge|f_1\rangle\wedge\cdots\wedge|f_N\rangle =-\hat c_j^\dagger\hat c_i^\dagger W_N.$

Taking adjoints gives the two-annihilator relation. Product tensors span the finite tensor space, so their antisymmetrizations span each ${\mathcal H_-^{(N)}}$ . The identities therefore hold on every sector and hence on the finite-mode Fock space. No occupation-state action has entered the proof. In particular, ${2{(\hat c_i^\dagger)}^2=0}$ : the algebra already enforces repeated-mode exclusion before occupation labels are assigned.

Fermionic occupation states from ordered creation

CAR now lets us construct and normalize a basis by adding distinct modes to the vacuum. The written mode order fixes the sign convention.

Fix the ordered orthonormal one-particle basis ${\mathcal E=(|1\rangle,\ldots,|M\rangle)}$ , the normalized vacuum ${|0\rangle}$ , and the creation maps in Definition 7.6. For ${\mathbf n=(n_1,\ldots,n_M)\in{\{0,1\}}^M}$ , define

${|\mathbf n\rangle}_{\mathrm F} :=\prod_{i=1}^{M}{(\hat c_i^\dagger)}^{n_i}|0\rangle ={(\hat c_1^\dagger)}^{n_1}\cdots{(\hat c_M^\dagger)}^{n_M}|0\rangle.$

Smaller mode indices are written to the left, so the rightmost creator acts first. The all-zero tuple gives the vacuum. The vector lies in ${\mathcal H_-^{(N)}}$ , where ${N=\sum_i n_i}$ ; ${i}$ labels a mode and ${n_i}$ its occupation, not a persistent particle. The family ${\{{|\mathbf n\rangle}_{\mathrm F}:\mathbf n\in{\{0,1\}}^M\}}$ is the fermionic occupation-number basis, with normalization, orthogonality, and completeness established next.

For normalization, write ${|I\rangle:=\hat c_{i_1}^\dagger\cdots \hat c_{i_N}^\dagger|0\rangle}$ for the increasing list ${i_1<\cdots<i_N}$ , and remove its least index ${i_1}$ to obtain ${|I'\rangle}$ . Because ${i_1}$ is absent from ${I'}$ , CAR moves ${\hat c_{i_1}}$ through all its creators to annihilate the vacuum. Consequently,

${\|\hat c_{i_1}^\dagger|I'\rangle\|}^2 =\langle I'|(\hat I-\hat c_{i_1}^\dagger\hat c_{i_1})|I'\rangle =\langle I'|I'\rangle.$

Induction from the unit vacuum proves that every ordered state has unit norm. For different occupied sets ${I\ne J}$ of the same size, choose ${i\in I\setminus J}$ . Creator anticommutation exposes ${\hat c_i^\dagger}$ in ${|I\rangle}$ up to a sign, while ${\hat c_i|J\rangle=0}$ by the same commute-to-vacuum argument. Thus ${\langle J|I\rangle=0}$ . States with different particle numbers are orthogonal by the direct-sum construction. This proves orthonormality using CAR, before deriving any coordinate-action formula.

For completeness, repeated use of the creator’s prepend rule gives the projector characterization of the newly defined state. For ${N\ge1}$ ,

${|\mathbf n\rangle}_{\mathrm F} =|i_1\rangle\wedge\cdots\wedge|i_N\rangle =\sqrt{N!}\,\hat P_-^{(N)}\Phi_{\mathbf n}, \qquad i_1<\cdots<i_N.$

Every product-basis tensor either contains a repeated mode, whose antisymmetrization vanishes, or sorts to one of these ordered tensors up to permutation parity. Applying the projector to a product-basis expansion of an arbitrary sector vector therefore expands it in the ordered creator states. The vacuum supplies ${N=0}$ , so the ${2^M}$ states span the finite-mode fermionic Fock space. CAR established their orthonormality; the projector construction establishes that no sector vectors are missing.

Fermionic occupation actions from CAR

Before reading Figure 7.2, fix the order of three modes and insert a fermion in mode 3 of ${|100\rangle}$ . Count occupied modes to the left before deciding the sign.

Figure 7.2. The fermionic sign can be read either from exchanging occupied creators or from reordering one-particle factors in the corresponding antisymmetric state.

Recall the two-factor definition: ${|3\rangle\wedge|1\rangle =(|3\rangle\otimes|1\rangle-|1\rangle\otimes|3\rangle)/\sqrt2 =-|1\rangle\wedge|3\rangle}$ . Here the prepended one-particle factor crosses exactly one occupied earlier factor, so ${\hat c_3^\dagger|100\rangle=-|101\rangle}$ . The diagram makes the crossing count explicit for this state. We now obtain the coordinate actions from CAR, keeping the declared mode order fixed.

For arbitrary ${\mathbf n\in{\{0,1\}}^M}$ , define the number of occupied earlier modes by ${s_i(\mathbf n):=\sum_{j<i}n_j}$ . Move a creator ${\hat c_i^\dagger}$ from the left into its position in the defining ordered product. Each occupied earlier mode contributes one minus sign by CAR. An already occupied mode leaves ${{(\hat c_i^\dagger)}^2=0}$ ; an empty mode gives the ordered state with that mode filled.

For annihilation, move ${\hat c_i}$ through the same earlier creators. If mode ${i}$ is empty, the annihilator passes through the remaining creators and annihilates the vacuum. If that mode is occupied, use ${\hat c_i\hat c_i^\dagger= \hat I-\hat c_i^\dagger\hat c_i}$ . The second term vanishes on the remaining creator word because mode ${i}$ is absent there. Both operations therefore carry the same factor ${{(-1)}^{s_i(\mathbf n)}}$ .

On ${\mathcal D_-}$ , the coordinate actions derived from CAR are therefore

$\begin{aligned} \hat c_i^\dagger|\mathbf n\rangle &={(-1)}^{s_i(\mathbf n)}(1-n_i)|n_1\ldots1_i\ldots n_M\rangle,\\[0.4em] \hat c_i|\mathbf n\rangle &={(-1)}^{s_i(\mathbf n)}n_i|n_1\ldots0_i\ldots n_M\rangle. \end{aligned}$

Projector check of both occupation bases

The construction above defines the fermionic family through ordered creators. Its projector characterization also permits a direct check of normalization, orthogonality, and completeness, alongside the bosonic family.

The bosonic normalization counts repetitions: among the ${N!}$ permutations, exactly ${\prod_i n_i!}$ leave the product tensor unchanged. Because the symmetrizer is an orthogonal projector,

${\|\hat P_+^{(N)}\Phi_{\mathbf n}\|}^2 =\langle\Phi_{\mathbf n},\hat P_+^{(N)}\Phi_{\mathbf n}\rangle =\frac{\prod_i n_i!}{N!}.$

For distinct fermion modes only the identity permutation leaves the tensor unchanged, giving projected squared norm ${1/N!}$ ; a repeated mode cancels when its two slots are exchanged.

To check orthogonality, take distinct occupation patterns ${\mathbf m\ne\mathbf n}$ with the same total particle number ${N}$ . Permutations preserve occupation counts, so their projected tensors have disjoint supports in the orthonormal product basis. Every cross term in the inner product vanishes:

$\bigl\langle\hat P_\pm^{(N)}\Phi_{\mathbf m}, \hat P_\pm^{(N)}\Phi_{\mathbf n}\bigr\rangle=0 \qquad(\mathbf m\ne\mathbf n).$

To check spanning, expand any vector in ${\mathcal H_1^{\otimes N}}$ in the product basis, and apply ${\hat P_\pm^{(N)}}$ term by term. Every nonzero projected product tensor is proportional to the occupation state with its occupation counts; antisymmetrization kills tensors with a repeated mode. For a vector already in ${\mathcal H_\pm^{(N)}}$ , the projector returns that vector. The occupation states therefore span the whole sector, including the vacuum when ${N=0}$ .

Different particle-number sectors are orthogonal by the Fock direct-sum construction. For a general Fock-space vector, its finite sector sums converge in norm. Combining sector spanning with the established normalization and orthogonality gives

$\begin{aligned} \mathcal F_+(\mathcal H_1) &=\overline{\operatorname{span}}\left\{ {|\mathbf n\rangle}_{\mathrm B}:\mathbf n\in\mathbb N_0^M\right\},\\ \mathcal F_-(\mathcal H_1) &=\operatorname{span}\left\{ {|\mathbf n\rangle}_{\mathrm F}:\mathbf n\in{\{0,1\}}^M\right\}. \end{aligned}$

The overline denotes closure in the Hilbert-space norm: general bosonic vectors need not be finite linear combinations of occupation states. With the declared finite number of modes, the fermionic family has ${2^M}$ members and needs no closure. Thus each normalized family is an orthonormal basis of its Fock space, not only of a fixed-number sector.

Changing the declared fermion-mode order changes the coordinate phase of some basis vectors because a wedge changes sign under an odd permutation. It does not change physical predictions when all states and operators are transformed consistently. Omitting the order while quoting a sign makes the calculation irreproducible.

Direct calculation of orthogonality and completeness

The following calculation expands the preceding basis argument using only the ordered one-particle basis and the exchange projectors. Fix ${N\ge1}$ and ${M<\infty}$ . For a word ${\mathbf i=(i_1,\ldots,i_N)}$ , write ${|\mathbf i\rangle:=|i_1\rangle\otimes\cdots\otimes|i_N\rangle}$ . The product-basis inner product is

$\langle\mathbf j|\mathbf i\rangle =\prod_{r=1}^N\delta_{j_r,i_r}, \qquad \mathbf i,\mathbf j\in{\{1,\ldots,M\}}^N.$

Let ${q_r(\mathbf n)}$ be the mode label in slot ${r}$ of the ordered tensor ${\Phi_{\mathbf n}}$ : mode ${i}$ occurs ${n_i}$ times, in increasing mode order. Write ${g_{\mathbf n}:=\prod_i n_i!}$ . In each inner-product calculation below, ${\mathbf m}$ and ${\mathbf n}$ have total occupation ${N}$ . Self-adjointness and idempotence give ${\langle\hat P_\pm\Phi_{\mathbf m},\hat P_\pm\Phi_{\mathbf n}\rangle =\langle\Phi_{\mathbf m},\hat P_\pm\Phi_{\mathbf n}\rangle}$ , with the same fixed ${N}$ understood.

For bosons, substitute the normalization factors and then the permutation average. The inverse slot index follows the convention in Definition 7.1:

$\begin{aligned} \bigl\langle{|\mathbf m\rangle}_{\mathrm B}, {|\mathbf n\rangle}_{\mathrm B}\bigr\rangle &=\frac{N!}{\sqrt{g_{\mathbf m}g_{\mathbf n}}} \langle\Phi_{\mathbf m},\hat P_+^{(N)}\Phi_{\mathbf n}\rangle\\ &=\frac{1}{\sqrt{g_{\mathbf m}g_{\mathbf n}}} \sum_{\pi\in\mathfrak S_N} \prod_{r=1}^N\delta_{q_r(\mathbf m),q_{\pi^{-1}(r)}(\mathbf n)}\\ &=\frac{g_{\mathbf n}}{\sqrt{g_{\mathbf m}g_{\mathbf n}}} \delta_{\mathbf m,\mathbf n} =\delta_{\mathbf m,\mathbf n}. \end{aligned}$

If the occupations differ, no permutation matches all slots, so every product of deltas is zero. If they agree, precisely ${g_{\mathbf n}}$ permutations rearrange only identical labels and contribute one. This calculates both orthogonality and unit normalization.

For fermions, restrict both occupation tuples to ${{\{0,1\}}^M}$ , so each ordered word has distinct labels. Substitution gives

$\begin{aligned} \bigl\langle{|\mathbf m\rangle}_{\mathrm F}, {|\mathbf n\rangle}_{\mathrm F}\bigr\rangle &=N!\langle\Phi_{\mathbf m},\hat P_-^{(N)}\Phi_{\mathbf n}\rangle\\ &=\sum_{\pi\in\mathfrak S_N}\operatorname{sgn}(\pi) \prod_{r=1}^N\delta_{q_r(\mathbf m),q_{\pi^{-1}(r)}(\mathbf n)}\\ &=\delta_{\mathbf m,\mathbf n}. \end{aligned}$

For equal ordered words, only the identity permutation matches every slot, and its sign is positive. For different occupation patterns, none does. If a word instead repeats a mode, let ${\tau}$ exchange two equal slots. Then ${\hat U_\tau\Phi_{\mathbf n}=\Phi_{\mathbf n}}$ ; pairing ${\pi}$ with ${\pi\tau}$ in the antisymmetrizer gives

$\operatorname{sgn}(\pi\tau)\hat U_{\pi\tau}\Phi_{\mathbf n} =-\operatorname{sgn}(\pi)\hat U_\pi\Phi_{\mathbf n}.$

The paired terms cancel, so the repeated-mode projection is zero. It is excluded from the normalized fermionic family. This calculation uses permutation parity, not anticommuting creation operators.

For a concrete bosonic check, take two modes and two particles. The three occupation states are

${|20\rangle}_{\mathrm B}=|1,1\rangle,\qquad {|11\rangle}_{\mathrm B}=\frac{|1,2\rangle+|2,1\rangle}{\sqrt2},\qquad {|02\rangle}_{\mathrm B}=|2,2\rangle.$

Direct multiplication gives

$\begin{aligned} \bigl\langle{|20\rangle}_{\mathrm B},{|11\rangle}_{\mathrm B}\bigr\rangle &=\frac{\delta_{1,1}\delta_{1,2}+\delta_{1,2}\delta_{1,1}}{\sqrt2}=0,\\ {\bigl\|{|11\rangle}_{\mathrm B}\bigr\|}^2 &=\frac{1+0+0+1}{2}=1. \end{aligned}$

Every symmetric two-particle vector on these two modes has equal coefficients on ${|1,2\rangle}$ and ${|2,1\rangle}$ . For arbitrary complex ${\alpha,\beta,\gamma}$ , its expansion is therefore

$\begin{aligned} \Psi^{(2)} &=\alpha|1,1\rangle+\beta(|1,2\rangle+|2,1\rangle)+\gamma|2,2\rangle\\ &=\alpha{|20\rangle}_{\mathrm B}+\sqrt2\beta{|11\rangle}_{\mathrm B} +\gamma{|02\rangle}_{\mathrm B},\\ {\|\Psi^{(2)}\|}^2&=|\alpha|^2+2|\beta|^2+|\gamma|^2. \end{aligned}$

For a fermionic example, use three modes and two particles. Define ${{|i,j\rangle}_-:=(|i,j\rangle-|j,i\rangle)/\sqrt2}$ for ${i<j}$ . Then

${|110\rangle}_{\mathrm F}={|1,2\rangle}_-,\qquad {|101\rangle}_{\mathrm F}={|1,3\rangle}_-,\qquad {|011\rangle}_{\mathrm F}={|2,3\rangle}_-.$

For example, the product-basis rule gives

$\begin{aligned} \bigl\langle{|110\rangle}_{\mathrm F},{|101\rangle}_{\mathrm F}\bigr\rangle &=\frac12\bigl(\delta_{1,1}\delta_{2,3}-\delta_{1,3}\delta_{2,1}\\ &\qquad-\delta_{2,1}\delta_{1,3}+\delta_{2,3}\delta_{1,1}\bigr)=0. \end{aligned}$

Antisymmetry sets diagonal coefficients to zero and pairs each coefficient with its negative under slot exchange. Thus the general vector is

$\begin{aligned} \Psi^{(2)}&=\sum_{i<j}A_{ij}(|i,j\rangle-|j,i\rangle)\\ &=\sqrt2\left(A_{12}{|110\rangle}_{\mathrm F} +A_{13}{|101\rangle}_{\mathrm F}+A_{23}{|011\rangle}_{\mathrm F}\right),\\ {\|\Psi^{(2)}\|}^2&=2\left(|A_{12}|^2+|A_{13}|^2+|A_{23}|^2\right). \end{aligned}$

For the general spanning calculation, let ${\mathbf n(\mathbf i)}$ count the occurrences of each mode in a word ${\mathbf i}$ . Choose a permutation ${\pi_{\mathbf i}}$ such that ${|\mathbf i\rangle=\hat U_{\pi_{\mathbf i}}\Phi_{\mathbf n(\mathbf i)}}$ . The exchange law of the projector and the occupation-state normalization give

$\begin{aligned} \hat P_+^{(N)}|\mathbf i\rangle &=\sqrt{\frac{g_{\mathbf n(\mathbf i)}}{N!}} {|\mathbf n(\mathbf i)\rangle}_{\mathrm B},\\ \hat P_-^{(N)}|\mathbf i\rangle &=\begin{cases} \dfrac{\operatorname{sgn}(\pi_{\mathbf i})}{\sqrt{N!}} {|\mathbf n(\mathbf i)\rangle}_{\mathrm F},&\text{all labels distinct},\\ 0,&\text{a label repeats}. \end{cases} \end{aligned}$

In the fermionic nonzero case the labels are distinct, so the sorting permutation is unique. For bosons its nonuniqueness does not matter because the symmetric projector removes the permutation. For any sector vector ${\Psi^{(N)}}$ , expand in the full product basis and project:

$\Psi^{(N)}=\sum_{\mathbf i}A_{\mathbf i}|\mathbf i\rangle =\hat P_\pm^{(N)}\Psi^{(N)} =\sum_{\mathbf i}A_{\mathbf i}\hat P_\pm^{(N)}|\mathbf i\rangle.$

Grouping these finitely many terms by their occupation counts constructs the expansion coefficients explicitly. For each allowed pattern of total occupation ${N}$ , they are

$\begin{aligned} C^{\mathrm B}_{\mathbf n} &=\sqrt{\frac{g_{\mathbf n}}{N!}} \sum_{\mathbf i:\,\mathbf n(\mathbf i)=\mathbf n}A_{\mathbf i},\\ C^{\mathrm F}_{\mathbf n} &=\frac{1}{\sqrt{N!}} \sum_{\mathbf i:\,\mathbf n(\mathbf i)=\mathbf n} \operatorname{sgn}(\pi_{\mathbf i})A_{\mathbf i}. \end{aligned}$

Substituting these coefficients into the preceding projected sum gives a linear combination of occupation states equal to the original sector vector. This proves spanning without assuming it. The inner products already calculated show that the coefficients are its occupation-state overlaps and are unique. For ${N=0}$ , the only coefficient multiplies the normalized vacuum.

To pass from sectors to Fock space, use ${\mathcal I_+=\mathbb N_0^M}$ and ${\mathcal I_-={\{0,1\}}^M}$ . In the next two formulas ${|\mathbf n\rangle}$ and ${C_{\mathbf n}}$ refer to the appropriate bosonic or fermionic family, and ${|\mathbf n|:=\sum_i n_i}$ . For ${\boldsymbol\psi=(\psi^{(0)},\psi^{(1)},\ldots)}$ , the finite-sector expansions give

$\begin{aligned} \psi^{(N)}&=\sum_{\substack{\mathbf n\in\mathcal I_\pm\\|\mathbf n|=N}} C_{\mathbf n}|\mathbf n\rangle,\\ {\|\boldsymbol\psi\|}^2 &=\sum_{N=0}^{\infty}{\|\psi^{(N)}\|}^2 =\sum_{\mathbf n\in\mathcal I_\pm}|C_{\mathbf n}|^2<\infty. \end{aligned}$

Subtract the expansion truncated at particle number ${L\in\mathbb N_0}$ . Orthogonality of distinct sectors and the finite-sector norm identities give the exact squared error:

${\left\|\boldsymbol\psi- \sum_{\substack{\mathbf n\in\mathcal I_\pm\\|\mathbf n|\le L}} C_{\mathbf n}|\mathbf n\rangle\right\|}^2 =\sum_{N>L}{\|\psi^{(N)}\|}^2 =\sum_{\substack{\mathbf n\in\mathcal I_\pm\\|\mathbf n|>L}} |C_{\mathbf n}|^2\longrightarrow0.$

The limit follows from the convergent nonnegative series in the preceding norm identity. It proves that finite occupation-state sums are dense in the bosonic Fock space. For the finite-mode fermionic space, the tail is exactly zero once ${L\ge M}$ , and the ${2^M}$ occupation states already span it by finite sums. The direct inner-product and expansion calculations thus justify the basis claim used in the main argument.

Adding a particle in a chosen mode

Take orthonormal modes ${|1\rangle,|2\rangle}$ , fix their order, and define ${|\pm\rangle=(|1\rangle\pm|2\rangle)/\sqrt2}$ . Keep the input state fixed at ${|+\rangle}$ . Which two-particle vectors result from adding ${|1\rangle}$ , ${|2\rangle}$ , ${|+\rangle}$ , or ${|-\rangle}$ ?

Write ${{|11\rangle}_{\mathrm F}:=|1\rangle\wedge|2\rangle =(|1\rangle\otimes|2\rangle-|2\rangle\otimes|1\rangle)/\sqrt2}$ . The two-slot antisymmetrizer gives $\begin{aligned} \hat c_1^\dagger|+\rangle &=\sqrt2\,\hat P_-^{(2)}(|1\rangle\otimes|+\rangle) =\frac{{|11\rangle}_{\mathrm F}}{\sqrt2},\\[0.4em] \hat c_2^\dagger|+\rangle &=\sqrt2\,\hat P_-^{(2)}(|2\rangle\otimes|+\rangle) =-\frac{{|11\rangle}_{\mathrm F}}{\sqrt2}. \end{aligned}$ The minus sign follows from swapping one-particle factors in an antisymmetric tensor; no operator anticommutator has been assumed. Linearity in the added state now gives ${\hat c_\pm^\dagger=(\hat c_1^\dagger\pm\hat c_2^\dagger)/\sqrt2}$ , so $\hat c_+^\dagger|+\rangle=0, \qquad \hat c_-^\dagger|+\rangle={|11\rangle}_{\mathrm F}.$ Adding a particle in the already occupied mode gives zero, whereas adding one in its orthogonal partner gives a normalized two-particle state. Exclusion therefore holds in the rotated mode basis as well as in the original one. The two initial outputs have squared norm ${1/2}$ ; they are not separately renormalized. This example illustrates particle creation in a chosen mode. Its projector calculation independently checks the canonical relations already established without occupation coordinates.

Bosonic basis actions from normalization

For the bosonic actions, let ${\Phi_{\mathbf n}}$ be the ordered tensor in the occupation-basis definition. Symmetrizing ${|i\rangle\otimes\hat P_+^{(N)}\Phi_{\mathbf n}}$ gives the same result as symmetrizing ${\Phi_{\mathbf n+\mathbf e_i}}$ , where ${\mathbf e_i}$ adds one occupation in mode ${i}$ . Comparing the two basis normalizations gives

$\sqrt{N+1}\, \frac{\sqrt{N!/\prod_jn_j!}} {\sqrt{(N+1)!/[(n_i+1)!\prod_{j\ne i}n_j!]}} =\sqrt{n_i+1}.$

Taking the adjoint matrix elements in the orthonormal basis gives the annihilation amplitude. The resulting bosonic actions on the common core are

$\begin{aligned} \hat a_i^\dagger|\mathbf n\rangle &=\sqrt{n_i+1}|\mathbf n+\mathbf e_i\rangle,\\[0.4em] \hat a_i|\mathbf n\rangle &=\begin{cases}\sqrt{n_i}|\mathbf n-\mathbf e_i\rangle,&n_i\geq1,\\0,&n_i=0,\end{cases} \end{aligned}$

Projector check of the fermionic coordinate actions

The operator-algebra derivation can be checked directly against the projector definition. This also displays explicitly where its normalization and each coordinate sign enter.

For fermions, let ${k_1<\cdots<k_N}$ be the occupied modes and ${\Phi_{\mathbf n}:=|k_1\rangle\otimes\cdots\otimes|k_N\rangle}$ , with ${N\geq1}$ . The normalized occupation state is ${|\mathbf n\rangle=\sqrt{N!}\,\hat P_-^{(N)}\Phi_{\mathbf n}}$ . We apply Definition 7.6 directly to this state as an independent check of the coordinate actions.

Full antisymmetrization absorbs the antisymmetrizer on the last ${N}$ slots:

$\hat P_-^{(N+1)}\bigl(\hat I\otimes\hat P_-^{(N)}\bigr) =\hat P_-^{(N+1)}.$

Indeed, a permutation of those slots contributes its sign once in the inner antisymmetrizer and once when it acts on the full antisymmetrizer. The two signs multiply to one, so every term in the average gives the same projector. Substituting the normalized state into the creation definition therefore gives

$\begin{aligned} \hat c_i^\dagger|\mathbf n\rangle &=\sqrt{N+1}\,\hat P_-^{(N+1)} \bigl(|i\rangle\otimes|\mathbf n\rangle\bigr)\\ &=\sqrt{N+1}\sqrt{N!}\,\hat P_-^{(N+1)} \bigl(|i\rangle\otimes\hat P_-^{(N)}\Phi_{\mathbf n}\bigr)\\ &=\sqrt{(N+1)!}\,\hat P_-^{(N+1)} \bigl(|i\rangle\otimes\Phi_{\mathbf n}\bigr). \end{aligned}$

The coefficient is exactly the normalization of an ${N+1}$ -particle fermionic occupation state. For the vacuum, the defining rule already gives ${\hat c_i^\dagger|0\rangle=|i\rangle}$ ; no zero-slot projector is needed.

If ${n_i=0}$ , sorting the prepended factor into the declared mode order crosses ${s_i(\mathbf n):=\sum_{j<i}n_j}$ occupied factors. Each interchange reverses the sign; empty modes contribute no factor. With ${\mathbf e_i}$ adding one occupation in mode ${i}$ , this gives

$\hat c_i^\dagger|\mathbf n\rangle ={(-1)}^{s_i(\mathbf n)}\sqrt{(N+1)!}\, \hat P_-^{(N+1)}\Phi_{\mathbf n+\mathbf e_i} ={(-1)}^{s_i(\mathbf n)}|\mathbf n+\mathbf e_i\rangle \qquad(n_i=0).$

If ${n_i=1}$ , two factors are equal. Exchanging them fixes the tensor but reverses its antisymmetric projection, so that projection is zero. Thus both the unit magnitude and the blocking factor follow from the projector definition without using any operator anticommutator.

For annihilation with ${n_i=1}$ , put ${\mathbf m=\mathbf n-\mathbf e_i}$ . The defining adjoint relation in the orthonormal occupation basis gives

$\langle\mathbf m|\hat c_i|\mathbf n\rangle ={\langle\mathbf n|\hat c_i^\dagger|\mathbf m\rangle}^* ={(-1)}^{s_i(\mathbf n)}.$

Here ${s_i(\mathbf m)=s_i(\mathbf n)}$ , because removing mode ${i}$ does not change occupations to its left. Every other matrix element vanishes: creation in mode ${i}$ cannot produce ${|\mathbf n\rangle}$ from a different basis vector. If ${n_i=0}$ , no input can reach ${|\mathbf n\rangle}$ by creation in that mode, so annihilation gives zero, including on the vacuum.

Number operators and checks of the basis actions

All products below act on the finite-particle domain ${\mathcal D_\pm=\mathcal F_{\pm,\mathrm{alg}}}$ : states with nonzero components in only finitely many particle-number sectors. The number operator ${\hat N}$ and every creation or annihilation map preserve this domain, so their finite products are well defined there. The bosonic maps are unbounded and are not defined on the entire bosonic Fock space. The number operators read the occupation labels; their diagonal definitions also specify the full self-adjoint domains.

Let ${\mathcal I_+=\mathbb N_0^M}$ for bosons and ${\mathcal I_-={\{0,1\}}^M}$ for fermions. For an occupation-basis expansion ${\psi=\sum_{\mathbf n\in\mathcal I_\pm} \psi_{\mathbf n}|\mathbf n\rangle}$ , define the mode number operator by

$\mathcal D(\hat n_i) :=\left\{\psi:\sum_{\mathbf n\in\mathcal I_\pm} n_i^2{|\psi_{\mathbf n}|}^2<\infty\right\}, \qquad \hat n_i\psi :=\sum_{\mathbf n\in\mathcal I_\pm} n_i\psi_{\mathbf n}|\mathbf n\rangle.$

The total number operator is the diagonal operator

$\mathcal D(\hat N) :=\left\{\psi:\sum_{\mathbf n\in\mathcal I_\pm} \left(\sum_{i=1}^{M}n_i\right)^2 {|\psi_{\mathbf n}|}^2<\infty\right\}, \qquad \hat N\psi :=\sum_{\mathbf n\in\mathcal I_\pm} \left(\sum_{i=1}^{M}n_i\right) \psi_{\mathbf n}|\mathbf n\rangle.$

These diagonal operators are self-adjoint. Their restrictions to the common core ${\mathcal D_\pm}$ satisfy ${\hat n_i=\hat d_i^\dagger\hat d_i}$ and ${\hat N=\sum_i\hat n_i}$ , where ${\hat d=\hat a}$ or ${\hat c}$ . For bosons, ${\sigma(\hat n_i)=\sigma(\hat N)=\mathbb N_0}$ . For fermions with ${M}$ modes, ${\sigma(\hat n_i)=\{0,1\}}$ and ${\sigma(\hat N)=\{0,1,\ldots,M\}}$ .

Number operators read the occupation labels: ${\hat n_i|\mathbf n\rangle=n_i|\mathbf n\rangle}$ and ${\hat N|\mathbf n\rangle=(\sum_i n_i)|\mathbf n\rangle}$ . On ${\mathcal D_\pm}$ , their actions agree with ${\hat n_i=\hat d_i^\dagger\hat d_i}$ and ${\hat N=\sum_i\hat n_i}$ . Mode occupations are ${n_i\in\mathbb N_0}$ for bosons and ${n_i\in\{0,1\}}$ for fermions; the total is therefore a nonnegative integer, bounded by ${M}$ for fermions. The full self-adjoint operators use the domains in the definition above.

Bosonic ladder normalization

For one bosonic mode, determine the amplitudes for adding a quantum to and removing a quantum from ${|2\rangle}$ , then test the difference between the two operator orders. Work on the finite-occupation core, where all products are defined.

Check that the stated ladder coefficients agree with adjointness and the number eigenvalues. Write ${\hat a_i^\dagger|n_i\rangle=\alpha_{n_i}|n_i+1\rangle}$ , choose basis phases so ${\alpha_{n_i}\geq0}$ , and use ${\hat n_i=\hat a_i^\dagger\hat a_i}$ with ${\hat n_i|n_i\rangle=n_i|n_i\rangle}$ . Adjointness gives ${{\lvert\alpha_{n_i-1}\rvert}^2=n_i}$ , hence ${\alpha_{n_i}=\sqrt{n_i+1}}$ . Unit ladder coefficients would give an incorrect norm for ${{(\hat a_i^\dagger)}^2|0\rangle}$ .

Thus ${\hat a^\dagger|2\rangle=\sqrt3|3\rangle}$ and ${\hat a|2\rangle=\sqrt2|1\rangle}$ . Acting again gives ${\hat a\hat a^\dagger|2\rangle=3|2\rangle}$ and ${\hat a^\dagger\hat a|2\rangle=2|2\rangle}$ : their difference returns ${|2\rangle}$ . Independently, ${\|\hat a|2\rangle\|^2=2=\langle2|\hat n|2\rangle}$ , while ${{(\hat a^\dagger)}^2|0\rangle/\sqrt{2!}=|2\rangle}$ is normalized. The vacuum boundary gives ${\hat a|0\rangle=0}$ , not a state with negative occupation. The ladder factors are dimensionless; the supporting displacement application supplies the coefficient with units of length. Next, extend the commutator check to an arbitrary occupation vector.

Extend the ladder check to an arbitrary occupation vector on the common core to prove the bosonic canonical commutation relations (CCR). In mode ${i}$ ,

$\hat a_i\hat a_i^\dagger|\mathbf n\rangle =(n_i+1)|\mathbf n\rangle, \qquad \hat a_i^\dagger\hat a_i|\mathbf n\rangle =n_i|\mathbf n\rangle.$

Subtraction gives the same-mode commutator. Operators on distinct mode entries commute, and equal-type actions commute in the same mode. Thus, on every basis vector and hence on the finite-particle domain ${\mathcal D_+}$ ,

$[\hat a_i,\hat a_j^\dagger]=\delta_{ij}\hat I, \qquad [\hat a_i,\hat a_j]=[\hat a_i^\dagger,\hat a_j^\dagger]=0.$

Fermionic exclusion and exchange sign

Fix the order ${(1,2,3)}$ . Starting from ${|100\rangle}$ , compute the two orders in ${\{\hat c_1,\hat c_2^\dagger\}}$ , and test a forbidden insertion into the occupied first mode. Check the diagonal anticommutator ${\{\hat c_2,\hat c_2^\dagger\}}$ on the same state.

The occupation-basis action derived from Definition 7.6 counts only occupied earlier modes. First, $\hat c_2^\dagger|100\rangle=-|110\rangle, \qquad \hat c_1(-|110\rangle)=-|010\rangle.$ In the reversed order, $\hat c_1|100\rangle=|000\rangle, \qquad \hat c_2^\dagger|000\rangle=|010\rangle.$ The sum vanishes, verifying this off-diagonal anticommutator on the supplied state. A second insertion in its occupied first mode instead gives ${\hat c_1^\dagger|100\rangle=0}$ . For the diagonal check, ${\hat c_2\hat c_2^\dagger|100\rangle=|100\rangle}$ , because the removal sign cancels the insertion sign, whereas ${\hat c_2^\dagger\hat c_2|100\rangle=0}$ . Their sum is the original normalized state. These two checks distinguish cancellation between allowed paths from Pauli blocking of a path. They verify this example; the algebra on every state was established in Theorem 7.1. The supporting coordinate checks recover its same-mode and distinct-mode consequences.

The following statement collects the bosonic CCR proved from the basis actions and the fermionic CAR proved from the projector definition. It also records their common number-operator interpretation.

On the common invariant dense domain ${\mathcal D_+}$ , the bosonic maps of Definition 7.6 satisfy the canonical commutation relations

$[\hat a_i,\hat a_j^\dagger]=\delta_{ij}\hat I, \qquad [\hat a_i,\hat a_j]=[\hat a_i^\dagger,\hat a_j^\dagger]=0.$

On ${\mathcal D_-}$ , the fermionic maps satisfy the canonical anticommutation relations

$\{\hat c_i,\hat c_j^\dagger\}=\delta_{ij}\hat I, \qquad \{\hat c_i,\hat c_j\} =\{\hat c_i^\dagger,\hat c_j^\dagger\}=0.$

On ${\mathcal D_\pm}$ , the products ${\hat d_i^\dagger\hat d_i}$ agree with the restrictions of the self-adjoint number operators in Definition 7.8.

The fermionic basis action gives ${{(\hat c_i^\dagger)}^2=0}$ : after one allowed insertion the mode is occupied, so a second insertion vanishes. This is the algebraic expression of the repeated-factor cancellation in the antisymmetric state.

As a coordinate check of the already established algebra, consider one mode ${j}$ . The two possible occupations give

$\hat c_j\hat c_j^\dagger|\mathbf n\rangle =(1-n_j)|\mathbf n\rangle, \qquad \hat c_j^\dagger\hat c_j|\mathbf n\rangle =n_j|\mathbf n\rangle.$

Their sum is the identity on every occupation state.

The same-mode identity also gives

$\hat n_j^2 =\hat c_j^\dagger\hat c_j\hat c_j^\dagger\hat c_j =\hat c_j^\dagger(\hat I-\hat n_j)\hat c_j =\hat n_j,$

because ${\hat c_j^\dagger\hat n_j\hat c_j=0}$ . This recovers the zero-or-one spectrum already specified by the occupation basis and its diagonal number operator.

The distinct-mode case requires the exchange signs to be counted explicitly. Take ${i<j}$ and first suppose ${n_i=1}$ and ${n_j=0}$ , the only occupations for which both terms in ${\{\hat c_i,\hat c_j^\dagger\}|\mathbf n\rangle}$ can be nonzero.

Acting in the two orders gives the following pair of amplitudes. Filling mode ${j}$ after emptying the earlier mode ${i}$ changes the second crossing count by one. Here ${\mathbf e_i}$ is the unit vector in occupation-coordinate slot ${i}$ :

$\begin{aligned} \hat c_i\hat c_j^\dagger|\mathbf n\rangle &={(-1)}^{s_j(\mathbf n)} {(-1)}^{s_i(\mathbf n)}|\mathbf n-\mathbf e_i+\mathbf e_j\rangle,\\[0.6em] \hat c_j^\dagger\hat c_i|\mathbf n\rangle &={(-1)}^{s_i(\mathbf n)} {(-1)}^{s_j(\mathbf n)-1}|\mathbf n-\mathbf e_i+\mathbf e_j\rangle. \end{aligned}$

The amplitudes are negatives of one another. If the required source occupation is absent or the target is already filled, both terms vanish. Thus ${\{\hat c_i,\hat c_j^\dagger\}=0}$ for ${i\ne j}$ ; the remaining distinct-mode anticommutators follow from the same one-crossing difference.

The subscript on ${\hat c_i}$ names a mode, not a persistent particle. Acting with ${\hat c_i}$ lowers the occupation of mode ${i}$ in each occupied basis component, with its ordering sign, and annihilates components in which that mode is empty; it does not identify “which electron” left. This is why a unitary change of mode basis changes the operator labels while leaving the identical-particle content unchanged.

Restrict the self-adjoint number operators from Definition 7.8 to the finite-particle domain ${\mathcal D_\pm}$ . Its states have nonzero components in only finitely many particle-number sectors. The number operator ${\hat N}$ and every creation or annihilation map preserve this space, so all products below are well defined. Write ${\hat d_i=\hat a_i}$ for bosons and ${\hat d_i=\hat c_i}$ for fermions. The single-mode commutators

$[\hat N,\hat d_i^\dagger]=\hat d_i^\dagger, \qquad [\hat N,\hat d_i]=-\hat d_i$

make the sector arrows quantitative. They follow directly from the canonical algebra; for example,

$[\hat n_\ell,\hat d_i^\dagger] =\delta_{\ell i}\hat d_i^\dagger, \qquad \sum_\ell[\hat n_\ell,\hat d_i^\dagger] =\hat d_i^\dagger.$

Now let ${\hat O=\hat X_1\cdots\hat X_r}$ . Each factor is a creator ${\hat X_m=\hat d_{i_m}^\dagger}$ or an annihilator ${\hat X_m=\hat d_{i_m}}$ ; ${m}$ labels its position in the product and ${i_m}$ its mode. Different factors may use the same mode. On the finite-particle domain, repeated use of ${[\hat N,\hat A\hat B]=[\hat N,\hat A]\hat B+ \hat A[\hat N,\hat B]}$ gives

$[\hat N,\hat O] =\sum_{m=1}^{r}\hat X_1\cdots [\hat N,\hat X_m]\cdots\hat X_r =(n_+-n_-)\hat O.$

Each creator contributes ${+1}$ , and each annihilator contributes ${-1}$ . Hence equal counts are a necessary and sufficient test for a nonzero operator monomial to commute with ${\hat N}$ ; no normal-ordering assumption is needed. The expanded product-rule calculation below shows the intermediate steps.

These commutators explain the sector connections in Figure 7.3: ${[\hat N,\hat d_i^\dagger]=\hat d_i^\dagger}$ sends a vector from sector ${N}$ to sector ${N+1}$ , whereas ${[\hat N,\hat d_i^\dagger\hat d_j]=0}$ keeps every fixed-number sector invariant. The first identity describes an algebraic map. Only the terms actually present in a Hamiltonian determine whether the physical dynamics changes particle number.

Let ${\hat O}$ be a nonzero operator monomial containing ${n_+}$ creation maps and ${n_-}$ annihilation maps, all defined on a common domain invariant under ${\hat N}$ and the factors of ${\hat O}$ . Then

$[\hat N,\hat O]=(n_+-n_-)\hat O.$

Thus ${\hat O}$ preserves every particle-number sector exactly when ${n_+=n_-}$ . Hermiticity is a different property: a Hermitian sum such as ${\hat c_i^\dagger\hat c_j^\dagger+\hat c_j\hat c_i}$ , with ${i\ne j}$ , contains opposite sector changes and generally does not commute with ${\hat N}$ .

Particle-number sectors and mode maps

Which arrows in Figure 7.3 must connect neighboring particle numbers, and where may a number-conserving Hamiltonian have nonzero matrix blocks? State whether an arrow by itself proves contact with a particle reservoir.

Figure 7.3. Bosonic (top) and fermionic (bottom) particle-number sectors for four one-particle modes. The fermion arrows show available creation and annihilation maps; successive steps may use different modes. The example below adds particles in modes 4, 3, 2, then 1, with positive amplitudes in basis order 1, 2, 3, 4. Repeated creation in the same fermion mode gives zero. Annihilation sends the vacuum to zero, and creation sends the fully occupied fermion sector to zero; the vacuum sector and zero vector are distinct. Bosonic sectors continue beyond four particles. Blue loops show a number-conserving Hamiltonian acting within each sector.

The arrows show algebraic domains and codomains. A physical Hamiltonian changes particle number only if it contains terms whose net number of creation and annihilation operators differs. Merely writing ${\hat c_i^\dagger}$ as a useful map does not assert that such a term occurs in ${\hat H}$ .

The number-operator commutator now explains the sector map. Creation and annihilation connect neighboring sectors, whereas a Hamiltonian commuting with the total number operator acts within each sector, as the blue loops show. The arrows describe available maps; the Hamiltonian selects the physical dynamics.

From mode products to hopping

The simplest products already generate useful operators. For fermions and distinct modes ${i\ne j}$ , read the factors from right to left:

$\hat n_i=\hat c_i^\dagger\hat c_i, \qquad \hat T_{ij}:=\hat c_i^\dagger\hat c_j, \qquad \hat H_{ij}:=h_{ij}\hat T_{ij}+h_{ij}^*\hat T_{ji}.$

The first product counts an occupation; the second removes one from mode ${j}$ and adds one in mode ${i}$ , with the exchange sign already fixed. The transfer ${\hat T_{ij}}$ is not generally Hermitian, whereas ${\hat T_{ij}^\dagger=\hat T_{ji}}$ makes ${\hat H_{ij}}$ Hermitian. The coefficient ${h_{ij}}$ has units of energy and describes the physical coupling; statistics alone does not determine it. All three products preserve total particle number. Factor order specifies algebraic composition; it does not assert that a tunneling process consists of two separate physical removal and injection events. The double-dot and two-site calculations below evaluate these products explicitly. The supporting comparison with unsigned occupation flips translates the same physical operators into an equivalent finite-system representation.

One hopping action in a spinful double dot

Return to Example 7.3 with a spinful double dot. The task is to find the signed hopping amplitude and check it against reversed order and particle-number conservation. Order spin orbitals as ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ . Apply ${\hat c_{2\uparrow}^\dagger\hat c_{1\uparrow}}$ to ${|1100\rangle}$ .

Act from right to left, counting occupied earlier modes at each step. The first annihilation produces ${|0100\rangle}$ . The creation operator for the third mode passes one occupied earlier mode, so

$\hat c_{2\uparrow}^\dagger\hat c_{1\uparrow}|1100\rangle =-|0110\rangle.$

Reverse the order of the two operators. The rightmost creation crosses two occupied earlier modes and therefore has positive sign; the following annihilation of the first mode also has positive sign:

$|1100\rangle \xrightarrow{\ \hat c_{2\uparrow}^\dagger\ } |1110\rangle \xrightarrow{\ \hat c_{1\uparrow}\ } |0110\rangle.$

Thus ${\hat c_{1\uparrow}\hat c_{2\uparrow}^\dagger|1100\rangle =+|0110\rangle}$ , and the two orders cancel in ${\{\hat c_{1\uparrow},\hat c_{2\uparrow}^\dagger\}|1100\rangle=0}$ . The anticommutation relation determines the relative sign of the two operator orders.

As a check, the hopping monomial has one creation and one annihilation operator and preserves total ${N=2}$ , consistent with both initial and final states lying in the same sector.

In the double-dot realization, ${|1100\rangle}$ means that both spin modes on the first dot are occupied in the declared ordering. The monomial transfers the spin-up occupation to the second dot. A charge sensor distinguishes the initial charge pattern ${(2,0)}$ from the final pattern ${(1,1)}$ , but it does not measure the minus sign of one amplitude in isolation. That sign becomes observable only through interference with another path to the same final state. The ordering convention fixes how these amplitudes are represented: changing conventions must transform all amplitudes consistently, whereas dropping a relative sign changes an interference prediction.

Using the four-mode order ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ , calculate ${\hat c_{1\downarrow}^\dagger\hat c_{2\downarrow}|1011\rangle}$ . Write the intermediate state and each sign. Then decide whether the monomial commutes with ${\hat N}$ without evaluating a full commutator.

The two-site hopping problem in site and energy bases

For ${\hat H=-t(\hat c_1^\dagger\hat c_2+\hat c_2^\dagger\hat c_1)}$ , with real hopping energy ${t}$ , find the one-particle eigenstates and express the same operator in their mode basis. Suppress spin and declare the ordered one-particle site basis ${\mathcal B_{\mathrm{site}}=(|1\rangle,|2\rangle)}$ , where ${|1\rangle=\hat c_1^\dagger|0\rangle}$ , ${|2\rangle=\hat c_2^\dagger|0\rangle}$ .

Direct action fixes the matrix before diagonalization; the signed-action method is the one just checked in Example 7.4:

$\hat H|1\rangle=-t|2\rangle, \qquad \hat H|2\rangle=-t|1\rangle, \qquad {[\hat H]}_{\mathcal B_{\mathrm{site}}} =\begin{pmatrix}0&-t\\-t&0\end{pmatrix}.$

The normalized eigenvectors and energies are

$|+\rangle=\frac{|1\rangle+|2\rangle}{\sqrt2}, \quad E_+=-t, \qquad |-\rangle=\frac{|1\rangle-|2\rangle}{\sqrt2}, \quad E_-=+t.$

Define mode maps that are even and odd under site interchange, a spatial symmetry of the one-particle modes distinct from particle-slot exchange: ${\hat c_\pm=(\hat c_1\pm\hat c_2)/\sqrt2}$ . The inverse relations are ${\hat c_1=(\hat c_++\hat c_-)/\sqrt2}$ and ${\hat c_2=(\hat c_+-\hat c_-)/\sqrt2}$ . Therefore

$\hat c_1^\dagger\hat c_2+\hat c_2^\dagger\hat c_1 =\frac12\left[(\hat c_+^\dagger+\hat c_-^\dagger) (\hat c_+-\hat c_-) +(\hat c_+^\dagger-\hat c_-^\dagger) (\hat c_++\hat c_-)\right] =\hat n_+-\hat n_-.$

The cross terms cancel exactly. Substitution into the same operator yields

$\hat H=-t\,\hat c_+^\dagger\hat c_+ +t\,\hat c_-^\dagger\hat c_-.$

The trace and determinant of the one-particle block, ${0}$ and ${-t^2}$ , agree in both bases. At fixed site-basis phases, changing the sign of ${t}$ exchanges which combination is lower in energy but preserves the level separation ${2|t|}$ . For an isolated bond, one can also remove that sign by rephasing one site basis state and transforming all readouts consistently. Its spectrum alone cannot measure the sign. This basis rephasing is distinct from changing the declared fermionic mode order. The supporting periodic-chain calculation applies the same test: transform both coefficients and mode operators, leaving the physical operator unchanged. The isolated two-site spectrum alone does not specify reservoir occupations or a readout protocol.

Basis changes transform mode operators

Double-dot readout

The two-site calculation supplies symmetric and antisymmetric energy modes. Their creators and adjoints are the even and odd maps already used in Example 7.1. Linearity in the added one-particle state makes the same transformation valid on every sector; the hopping energy coefficients multiply the number operators of those energy modes. Rewriting a state in another basis leaves it unchanged. The selected observable determines which occupation the apparatus measures [PPLG10b].

Before reading Figure 7.4, take the symmetric state ${|+\rangle}$ . Predict its two site-occupation probabilities, its energy-mode occupations, and which of those two descriptions a local charge sensor reads without a preceding basis rotation.

Figure 7.4. The same one-particle ray has different occupation coordinates in the site and energy-mode bases. Local charge sensing reads site occupation; energy spectroscopy resolves the level splitting. Inferring mode populations also requires a preparation and readout protocol.

The diagram represents the same state in two bases. In the site basis, ${|+\rangle=(|1\rangle+|2\rangle)/\sqrt2}$ gives ${\langle\hat n_1\rangle=\langle\hat n_2\rangle=1/2}$ . In the energy basis it gives ${\langle\hat n_+\rangle=1}$ and ${\langle\hat n_-\rangle=0}$ . The two descriptions are consistent because both the state coordinates and the operators transform with the chosen basis. A local sensor measures ${\hat n_1}$ or ${\hat n_2}$ , not ${\hat n_+}$ , unless an additional operation first rotates the measurement basis.

The minimal description deliberately omits excited dot orbitals, detailed electrostatics, leads, decoherence, spin–orbit coupling, and detector errors. It can validate sector counting, tunneling signs, and basis-dependent readout. It cannot predict a complete charge-stability diagram or a coherence time [RMHG07b].

Mode covariance under basis change

The two-site readout illustrates a general change of mode coordinates. Let ${(|i\rangle)}$ and ${(|\alpha\rangle)}$ be two orthonormal bases of the same one-particle Hilbert space, related by the unitary matrix ${U_{i\alpha}=\langle i|\alpha\rangle}$ . The mode operators transform as

$\hat d_\alpha^\dagger=\sum_iU_{i\alpha}\hat d_i^\dagger, \qquad \hat d_\alpha=\sum_iU_{i\alpha}^*\hat d_i.$

Unitarity preserves the canonical algebra. For fermions,

$\{\hat c_\alpha,\hat c_\beta^\dagger\} =\sum_{ij}U_{i\alpha}^*U_{j\beta} \{\hat c_i,\hat c_j^\dagger\} ={(U^\dagger U)}_{\alpha\beta}\hat I =\delta_{\alpha\beta}\hat I.$

The calculation states what a unitary change of one-particle basis preserves. It also prevents a common conceptual mistake: site and momentum operators do not create different species of electron. They create the same fermionic degree of freedom in different one-particle modes.

From a populated band to an operator action

The double-dot calculation separated a change of mode coordinates from a change of physical state. We can now make the same distinction in the cosine band of Lecture 4: keep its energy curve fixed and change its occupations. Take independent fermions of one resolved spin species on a six-cell periodic ring, with lattice spacing ${a}$ , zero onsite energy, and hopping scale ${t>0}$ . The energy function is ${\varepsilon(k)=-2t\cos(ka)}$ , corresponding to ${t_{\mathrm{L4}}=-t}$ . The periodic-chain application below derives this dispersion from hopping; here it is the supplied one-particle input.

Use ${\varepsilon(k)=-2t\cos(ka)}$ , ${t>0}$ , for one resolved spin species. Order the six allowed momentum modes by $(k_1a,k_2a,k_3a,k_4a,k_5a,k_6a) =(-\pi,-2\pi/3,-\pi/3,0,\pi/3,2\pi/3), \qquad \hat c_j:=\hat c_{k_j}.$ For this independent-mode model, take the energy operator to be ${\hat H_{\mathrm{band}}=\sum_{j=1}^{6}\varepsilon(k_j)\hat n_j}$ , where ${\hat n_j=\hat c_j^\dagger\hat c_j}$ . Start from the prepared reference ${|\Omega\rangle=|001110\rangle}$ . Evaluate ${\hat c_2^\dagger\hat c_4|\Omega\rangle}$ , find its particle-number and energy changes, and compare with ${\hat c_5^\dagger\hat c_4|\Omega\rangle}$ .

Sampling the supplied dispersion gives $\bigl(\varepsilon(k_1),\ldots,\varepsilon(k_6)\bigr) =t(2,1,-1,-2,-1,1).$ The occupied modes are 3, 4, and 5, so the reference has ${N=3}$ and energy ${E_\Omega=-4t}$ . It fills precisely the modes below zero energy, consistent with Lecture 5’s zero-temperature filling rule at chemical potential ${\mu=0}$ .

Annihilation in mode 4 crosses the occupied mode 3. In the intermediate state, no occupied mode precedes mode 2, so its insertion adds no further sign: $\begin{aligned} \hat c_4|001110\rangle&=-|001010\rangle,\\ \hat c_2^\dagger\hat c_4|001110\rangle&=-|011010\rangle. \end{aligned}$ The normalized final state has occupations in modes 2, 3, and 5. Summing their energies and counting their occupations gives $E_{\mathrm f}=-t,\qquad \Delta E=\varepsilon(k_2)-\varepsilon(k_4)=3t,\qquad \Delta N=0.$ One creator and one annihilator also give ${[\hat N,\hat c_2^\dagger\hat c_4]=0}$ , independently of this state. The energy increase must be supplied by an external operation; the diagonal ${\hat H_{\mathrm{band}}}$ alone does not transfer occupation between modes.

For the alternative target, mode 5 remains occupied after removal from mode 4. Its creation action therefore vanishes: $\hat c_5^\dagger\hat c_4|001110\rangle=0.$ The allowed operation creates an electron above and a missing electron below the reference filling, the number-conserving particle–hole change of Lecture 6. Its minus sign depends on the declared ordering; a dot plot records occupations, and an isolated overall sign does not change their probabilities.

Figure 7.5. A fixed band with two occupation patterns. Filled dots denote occupied modes and open dots empty modes of one resolved spin species; the continuous curve guides the eye between the six allowed momenta. The promoted occupation raises the many-particle energy by 3t while preserving particle number.

Each dot in this example represents a definite ${n_j\in\{0,1\}}$ . Lecture 5’s equilibrium Fermi function instead gives the mean ${\langle\hat n_j\rangle=f(\varepsilon(k_j))}$ , which can lie between zero and one. Both descriptions require a preparation in addition to the band; the energy curve alone specifies neither its filling nor an excitation.

Closing the construction

Products of mode maps generate the observables and Hamiltonians used in condensed matter. A simple density interaction for distinct fermionic modes is

$\hat H_{\mathrm{int}}=U\hat n_i\hat n_j =U\hat c_i^\dagger\hat c_j^\dagger\hat c_j\hat c_i, \qquad i\ne j,\quad U\in\mathbb R.$

It contributes energy ${U}$ when both modes are occupied and zero otherwise. The equality follows by two fermionic interchanges; the real coefficient ${U}$ supplies the interaction energy. Local occupation becomes charge and spin density; bilinears become hopping and current; quartic terms become interactions; and a unitary change of one-particle basis connects sites to bands. Superconducting quasiparticles require a later extension that mixes creation and annihilation maps, beyond the unitary mode rotation derived here. The double dot and optical-lattice double well show the smallest setting in which the algebra, a preparation, and a readout can all be checked against one another.

Fix the order of three fermionic modes and choose one bosonic mode. Without consulting the examples, construct one normalized basis state in each algebra, apply one annihilation and one creation map with every sign or square-root factor shown, and identify the maps’ particle-number domains and codomains. Then perform one unitary change of the fermionic mode basis and state which object changes representation and which physical state does not. If a fermionic sign was assigned without first declaring the mode order, or if the basis change was described as moving persistent particles, the construction is not yet reproducible.

Under a unitary rotation of the one-particle basis, the occupation coordinates and ladder-operator components change together, while the abstract Fock-space vector and operator remain fixed. Their basis coordinates and matrix entries generally change; the canonical algebra, spectra, and consistently evaluated expectation values remain unchanged. Creation and annihilation are maps between declared number sectors; neither the mode label nor the occupation string supplies a persistent identity to an indistinguishable particle.

Next: construct operators from physical matrix elements

Lecture 7 has established sector structure, normalized mode maps, fermionic signs, number counting, basis covariance, and simple operator products. The next task is to construct the corresponding Fock-space operators from supplied one-particle and two-particle matrix elements. The microscopic values of material coupling constants remain model inputs.

For example, an additive one-particle observable with matrix elements ${o_{ij}=\langle i|\hat o|j\rangle}$ is represented by

$\hat O=\sum_{ij}o_{ij}\hat c_i^\dagger\hat c_j.$

This previews the general rule; Lecture 8 derives it and the distinct coefficient conventions for two-particle interactions. Here the counting, hopping, and density-interaction examples establish how the building blocks are used.

Supporting comparison: unsigned occupation flips

Write the same occupied-mode crossing count with insertion index ${j}$ :

$s_j(\mathbf n)=\sum_{i<j}n_i.$

The ordered creator product is the definition of the fermionic state; these signed coordinate actions are its consequences. Their occupation-dependent signs must also be retained when translating physical operators to unsigned flips.

The creation and annihilation maps build exchange signs into products used to construct states and physical operators. Unsigned occupation flips can also represent a finite fermionic system exactly if all operators are translated consistently, with the required occupation-dependent signs. The benefit here is the direct relation between particle creation, basis changes, and compact operator products, rather than a claim of universal computational speed.

On the same finite fermionic occupation basis, define ${\hat b_i^\dagger}$ and ${\hat b_i}$ to change the bit ${n_i}$ without a sign, returning zero when creation is blocked or annihilation has no occupation to remove. They are sufficient to represent the entire finite system. The distinction is which formulas express particle creation or a physical observable.

For the fixed input ${|+\rangle}$ of the particle-creation example, naive combinations ${\hat b_\pm^\dagger=(\hat b_1^\dagger\pm\hat b_2^\dagger)/\sqrt2}$ give

$\hat b_+^\dagger|+\rangle={|11\rangle}_{\mathrm F}, \qquad \hat b_-^\dagger|+\rangle=0.$

Thus these combinations do not create a particle in the rotated modes with the projector definition. This does not make the unsigned representation incomplete: the correct physical creators in that representation include parity strings.

On this finite basis, the diagonal counting operator acts as ${\hat n_i|\mathbf n\rangle=n_i|\mathbf n\rangle}$ . Removing and then restoring a bit gives the same action, so ${\hat n_i=\hat b_i^\dagger\hat b_i}$ . Define

$\hat S_i:=\prod_{j<i}(\hat I-2\hat n_j), \qquad \hat c_i=\hat S_i\hat b_i, \qquad \hat c_i^\dagger=\hat b_i^\dagger\hat S_i.$

Each occupied earlier mode contributes minus one, so these identities recover the fermionic actions on every basis vector. Distinct-mode unsigned flips commute; same-mode flips enforce the two-level exclusion constraint. For ${i\ne j}$ , the transfer becomes

$\hat c_i^\dagger\hat c_j =\hat b_i^\dagger \left[\prod_{\min(i,j)<k<\max(i,j)}(\hat I-2\hat n_k)\right]\hat b_j.$

The string is empty for neighboring modes in the declared order, including a two-mode hopping problem. All consistently translated matrix elements agree. For finite ${M}$ , completeness can also be seen from the matrix units

$|\mathbf n\rangle\langle\mathbf m| =\left[\prod_i{(\hat b_i^\dagger)}^{n_i}\right] \left[\prod_i(\hat I-\hat n_i)\right] \left[\prod_i\hat b_i^{m_i}\right].$

Their linear combinations give every operator on this finite Fock space. Anticommuting creators package the same signs into a calculus tied directly to one-particle states; whether that representation is faster for a particular calculation depends on the method and model.

Application: periodic-chain hopping

For a one-dimensional periodic lattice with ${N_{\mathrm c}}$ sites at ${R_j=ja}$ , use the Fourier convention

(7.2)

$\hat c_{k\sigma} =\frac1{\sqrt{N_{\mathrm c}}} \sum_{j=0}^{N_{\mathrm c}-1} \mathrm e^{-\mathrm i kR_j}\hat c_{j\sigma}, \qquad \hat c_{j\sigma} =\frac1{\sqrt{N_{\mathrm c}}} \sum_k\mathrm e^{\mathrm i kR_j}\hat c_{k\sigma}.$

The allowed ${k}$ values are the ${N_{\mathrm c}}$ translation characters from Lectures 3–4. The normalization and inverse signs are fixed together; mixing conventions can introduce a spurious factor of ${N_{\mathrm c}}$ or an inconsistent phase for an oriented hopping term. Reversing both Fourier signs consistently relabels ${k}$ as ${-k}$ ; the even cosine band is unchanged.

Extend Example 7.5 to a periodic chain with ${N_{\mathrm c}\geq3}$ , real ${t,\varepsilon_0}$ , and the Fourier convention in Equation (7.2). Determine the many-particle hopping operator in momentum space and check that the basis change preserves state count and energy. For the spin-independent nearest-neighbor hopping Hamiltonian

$\hat H_t =\varepsilon_0\sum_{j\sigma}\hat n_{j\sigma} -t\sum_{j\sigma} \left(\hat c_{j+1,\sigma}^\dagger\hat c_{j\sigma} +\hat c_{j\sigma}^\dagger\hat c_{j+1,\sigma}\right),$

use ${R_j=ja}$ and the periodic boundary grid.

Substitute the inverse Fourier transform and use the finite site sum to test which momentum pairs survive. One term becomes

$\sum_j\hat c_{j+1,\sigma}^\dagger\hat c_{j\sigma} =\frac1{N_{\mathrm c}} \sum_{j,k,k'} \mathrm e^{-\mathrm i k'R_{j+1}} \mathrm e^{\mathrm i kR_j} \hat c_{k'\sigma}^\dagger\hat c_{k\sigma}.$

Finite Fourier orthogonality evaluates the sum over sites,

$\sum_j\mathrm e^{\mathrm i(k-k')R_j} =N_{\mathrm c}\delta_{kk'},$

and leaves the displacement phase ${\mathrm e^{-\mathrm i ka}}$ . Adding the Hermitian conjugate gives

(7.3)

$\hat H_t =\sum_{k\sigma} \left[\varepsilon_0-2t\cos(ka)\right] \hat c_{k\sigma}^\dagger\hat c_{k\sigma}.$

This is the localized-orbital cosine band expressed as a many-particle operator, with the explicit convention change ${t_{\mathrm{L4}}=-t}$ . The diagonal coefficient is a one-particle energy function; the hatted product is an occupation operator. A filled many-particle state and a one-particle band are not the same mathematical object even when they use the same ${\varepsilon(k)}$ .

This simplification depends on the boundary condition. Periodicity requires ${\mathrm e^{\mathrm i kN_{\mathrm c}a}=1}$ , hence ${k=2\pi n/(N_{\mathrm c}a)}$ . With an antiperiodic boundary condition the allowed values shift by ${\pi/(N_{\mathrm c}a)}$ , but orthogonality and the cosine dispersion retain the same form on the shifted grid. A leftover factor of ${N_{\mathrm c}}$ therefore diagnoses inconsistent Fourier normalization, whereas a shifted grid can be a legitimate boundary convention.

For one spin species, unitarity also gives ${\sum_k\hat c_k^\dagger\hat c_k=\sum_j\hat n_j}$ , so the transform does not create additional one-particle states. At ${t=0}$ , all ${N_{\mathrm c}}$ one-particle energies equal ${\varepsilon_0}$ . For a four-site ring they are ${\varepsilon_0-2t,\varepsilon_0,\varepsilon_0+2t,\varepsilon_0}$ ; their trace ${4\varepsilon_0}$ agrees with the site matrix. Each diagonal coefficient is real and has energy units. Translation-invariant hopping and the specified momentum grid justify this reduction; a site-dependent bond couples momenta and must not be dropped. The supporting displacement application uses the same Fourier construction but changes the operator from fermionic hopping to lattice displacement.

Application: phonon displacement and the zero-mode boundary

The bosonic application extends Lecture 6 using a similar Fourier construction, subject to the reality constraint and treatment of the zero mode in Lecture 6’s acoustic-mode criterion. For a periodic monatomic one-dimensional lattice with atomic mass ${M}$ , the allowed momentum set contains a rigid translation at ${q=0}$ , where ${\omega_0=0}$ . That degree of freedom is a center-of-mass coordinate, not a harmonic oscillator. Define the inversion-invariant positive-frequency set

$\mathcal Q_+:=\{q:\omega_q>0\}, \qquad q\in\mathcal Q_+\Longrightarrow -q\in\mathcal Q_+.$

For a periodic monatomic chain of ${N_{\mathrm c}}$ sites with mass ${M}$ , fix the center of mass and retain the set ${\mathcal Q_+=\{q:\omega_q>0\}}$ , with ${\omega_{-q}=\omega_q}$ . Construct the internal displacement and find ${\langle1_q|\hat u_j^\perp|0\rangle}$ , where ${|1_q\rangle=\hat a_q^\dagger|0\rangle}$ . State how the zero-frequency translation is treated. For each real positive-frequency normal coordinate use the canonical oscillator ${\hat H_Q=\hat P^2/(2M)+M\omega^2\hat Q^2/2}$ , ${[\hat Q,\hat P]=\mathrm i\hbar\hat I}$ , to determine the displacement prefactor.

Use the normalized ladders of Example 7.2 and require a Hermitian real-space observable rather than a separately real coefficient at each complex Fourier label. First set ${\hat Q=\alpha(\hat a+\hat a^\dagger)}$ and ${\hat P=-\mathrm i\beta(\hat a-\hat a^\dagger)}$ , with positive real coefficients. The commutator gives ${2\alpha\beta=\hbar}$ . Inserting these expressions into ${\hat H_Q}$ , cancellation of the ${\hat a^2}$ and ${{(\hat a^\dagger)}^2}$ terms requires ${\beta=M\omega\alpha}$ . Hence ${\alpha=\sqrt{\hbar/(2M\omega)}}$ , and the remaining Hamiltonian is ${\hbar\omega(\hat a^\dagger\hat a+\hat I/2)}$ . Canonical normalization, not Hermiticity alone, fixes the amplitude. The complex traveling-wave mode coordinates retain this normalization and obey ${\hat Q_q=\sqrt{\hbar/(2M\omega_q)} (\hat a_q+\hat a_{-q}^\dagger)}$ . Write the full Hermitian displacement as ${\hat u_j=\hat U_{\mathrm{CM}}+\hat u_j^\perp}$ , where ${\hat U_{\mathrm{CM}}=N_{\mathrm c}^{-1}\sum_j\hat u_j}$ is quantized as a free collective coordinate unless the center of mass is fixed. The internal displacement is

(7.4)

$\hat u_j^\perp =\frac1{\sqrt{N_{\mathrm c}}}\sum_{q\in\mathcal Q_+} \sqrt{\frac{\hbar}{2M\omega_q}} \left(\hat a_q+\hat a_{-q}^\dagger\right) \mathrm e^{\mathrm i qR_j}.$

The coefficient now has a nonzero denominator and units of length. Taking the adjoint and relabeling ${q\to-q}$ returns ${\hat u_j^\perp}$ , provided ${\omega_{-q}=\omega_q}$ . The paired ${q,-q}$ structure makes a real displacement from complex Fourier modes. It is not evidence that phonon number is conserved; ${\hat u_j}$ explicitly connects neighboring occupation sectors. In the fixed-center-of-mass convention used here, ${\hat U_{\mathrm{CM}}=0}$ ; without that convention, its free-particle quantization must be added separately and must never be inserted into the oscillator coefficient as ${1/\sqrt{\omega_0}}$ .

One way to recover the combination is to start from a complex normal coordinate ${\hat Q_q\propto\hat a_q+\hat a_{-q}^\dagger}$ . Hermiticity of the real-space displacement demands ${\hat Q_q^\dagger=\hat Q_{-q}}$ . The inverse Fourier sum then pairs the terms at ${q}$ and ${-q}$ , producing exactly the displayed operator. Using ${\hat a_q+\hat a_q^\dagger}$ for each complex traveling-wave coordinate would generally fail ${\hat Q_q^\dagger=\hat Q_{-q}}$ , so the inverse Fourier sum would not be Hermitian. A real standing-wave basis can use such oscillator coordinates, but its mode labels and reconstruction must be changed consistently.

For a distinct pair ${q,-q}$ , put ${A_q=\sqrt{\hbar/(2MN_{\mathrm c}\omega_q)}}$ . Its contribution is $A_q\left[(\hat a_q+\hat a_{-q}^\dagger)\mathrm e^{\mathrm i qR_j} +(\hat a_{-q}+\hat a_q^\dagger)\mathrm e^{-\mathrm i qR_j}\right].$ The two terms are adjoints. With ${|1_q\rangle=\hat a_q^\dagger|0\rangle}$ , only the ${\hat a_q^\dagger}$ term survives in ${\langle1_q|\hat u_j^\perp|0\rangle =A_q\mathrm e^{-\mathrm i qR_j}}$ ; its reverse matrix element is the complex conjugate. Summing over sites gives zero for this nonzero momentum, consistent with fixed center of mass. Also ${A_q}$ has length units and the vacuum variance is the positive sum ${\langle0|{(\hat u_j^\perp)}^2|0\rangle =\sum_{q\in\mathcal Q_+}\hbar/(2MN_{\mathrm c}\omega_q)}$ . At a self-inverse zone-boundary momentum, retain its single term rather than counting a distinct pair twice. No formula here permits inserting ${\omega_0=0}$ into an oscillator denominator. The matrix element supplies the missing bridge from a mode occupation to a displacement probe.

Supporting calculation: particle-number change

The question behind Criterion 7.1 is why the commutator counts creators and annihilators even when the product is not normally ordered. Work on the common finite-particle domain ${\mathcal D_\pm}$ , preserved by ${\hat N}$ and every factor. Begin with two factors ${\hat A,\hat B}$ that preserve this domain. Add and subtract ${\hat A\hat N\hat B}$ :

(7.5)

$\begin{aligned} [\hat N,\hat A\hat B] &=\hat N\hat A\hat B-\hat A\hat B\hat N\\ &=(\hat N\hat A\hat B-\hat A\hat N\hat B) +(\hat A\hat N\hat B-\hat A\hat B\hat N)\\ &=[\hat N,\hat A]\hat B+\hat A[\hat N,\hat B]. \end{aligned}$

The inserted terms cancel exactly. No factors have been exchanged, so this is an ordinary commutator identity for both bosons and fermions; no additional fermionic minus sign belongs in the product rule.

For ${\hat O=\hat X_1\cdots\hat X_r}$ , first apply the two-factor identity to ${\hat X_1}$ and the remaining product. With three factors, repeating it once gives the full expansion:

(7.6)

$\begin{aligned} [\hat N,\hat X_1\hat X_2\hat X_3] &=[\hat N,\hat X_1]\hat X_2\hat X_3 +\hat X_1[\hat N,\hat X_2\hat X_3]\\ &=[\hat N,\hat X_1]\hat X_2\hat X_3 +\hat X_1[\hat N,\hat X_2]\hat X_3 +\hat X_1\hat X_2[\hat N,\hat X_3]. \end{aligned}$

For any number of factors, the same recursion creates one term for each position ${m}$ : replace only ${\hat X_m}$ by its commutator and leave all other factors in their original order. This gives ${[\hat N,\hat O]=\sum_{m=1}^r\hat X_1\cdots [\hat N,\hat X_m]\cdots\hat X_r}$ .

Now take each factor to be a creator or annihilator, with ${\hat d_i=\hat a_i}$ for bosons and ${\hat d_i=\hat c_i}$ for fermions. The earlier single-mode relations read ${[\hat N,\hat X_m]=s_m\hat X_m}$ , where ${s_m=+1}$ for ${\hat X_m=\hat d_{i_m}^\dagger}$ and ${s_m=-1}$ for ${\hat X_m=\hat d_{i_m}}$ . Substitute this into every term:

(7.7)

$\begin{aligned} [\hat N,\hat O] &=\sum_{m=1}^r\hat X_1\cdots(s_m\hat X_m)\cdots\hat X_r\\ &=\sum_{m=1}^r s_m\hat O =\left(\sum_{m=1}^r s_m\right)\hat O\\ &=(n_+-n_-)\hat O. \end{aligned}$

The numbers ${n_+}$ and ${n_-}$ count creation and annihilation factors in the written product. Each scalar ${s_m}$ can be pulled out without reordering operators. Mode labels may repeat; normal ordering is unnecessary. If the monomial is the zero operator, the identity still holds, but its commutation with ${\hat N}$ gives no constraint on the two counts.

Finally let ${|\Psi_N\rangle}$ lie in the ${N}$ -particle sector, so ${\hat N|\Psi_N\rangle=N|\Psi_N\rangle}$ , and write ${q:=n_+-n_-}$ . Rearranging the commutator gives

(7.8)

$\begin{aligned} \hat N\hat O|\Psi_N\rangle &=\bigl(\hat O\hat N+[\hat N,\hat O]\bigr)|\Psi_N\rangle\\ &=N\hat O|\Psi_N\rangle+q\hat O|\Psi_N\rangle\\ &=(N+q)\hat O|\Psi_N\rangle. \end{aligned}$

Thus every nonzero output lies in sector ${N+q}$ ; a blocked or forbidden operation may instead give zero. For a nonzero monomial, preserving every number sector is equivalent to ${q=0}$ , or equal creator and annihilator counts. This establishes the criterion used by the sector map without specifying which monomials occur in a physical Hamiltonian.

Problems

1. Ordered-mode action and sector structure

Use the mode order ${(1\uparrow,1\downarrow,2\uparrow,2\downarrow)}$ .

Evaluate ${\hat c_{2\downarrow}^\dagger\hat c_{1\downarrow}|1100\rangle}$ , showing the intermediate state and both signs.
Evaluate the same monomial on ${|1001\rangle}$ .
Determine its nonzero particle-number blocks and verify its commutator with ${\hat N}$ .
Repeat the block and commutator reasoning for ${\hat c_{1\uparrow}^\dagger\hat c_{2\downarrow}^\dagger+ \hat c_{2\downarrow}\hat c_{1\uparrow}}$ .

2. Bosonic normalization and fermionic exclusion

For one bosonic mode, evaluate ${{(\hat a^\dagger)}^2|0\rangle}$ , its norm, and ${\hat a^2{(\hat a^\dagger)}^2|0\rangle}$ .
Repeat the first operation for one fermionic mode and derive the result from the canonical anticommutation relation rather than quoting Pauli exclusion separately.
For two fermionic modes, define ${|n_1n_2\rangle={(\hat c_1^\dagger)}^{n_1} {(\hat c_2^\dagger)}^{n_2}|0\rangle}$ . Evaluate ${\hat c_2^\dagger\hat c_1^\dagger|0\rangle}$ and explain the sign.
Compare the allowed eigenvalues of the bosonic and fermionic number operators. Explain why an occupation label specifies how many particles occupy a mode rather than identifying a persistent particle.

3. Fourier normalization and a localized particle

On a periodic chain of ${N_{\mathrm c}}$ sites, start from

$\hat c_j=\frac1{\sqrt{N_{\mathrm c}}}\sum_k \mathrm e^{\mathrm i kR_j}\hat c_k.$

Use finite Fourier orthogonality to derive the inverse transform.
Prove that ${\{\hat c_j,\hat c_{j'}^\dagger\}=\delta_{jj'}\hat I}$ if and only if the momentum-space operators obey the corresponding canonical algebra.
Show that the total number operator has the same diagonal form in both bases: ${\sum_j\hat c_j^\dagger\hat c_j= \sum_k\hat c_k^\dagger\hat c_k}$ .
For ${|j_0\rangle=\hat c_{j_0}^\dagger|0\rangle}$ , calculate ${\langle j_0|\hat c_k^\dagger\hat c_k|j_0\rangle}$ and interpret the relation between real-space localization and momentum-space occupation.

References

[Foc32] Fock, V. (1932). “Konfigurationsraum und zweite quantelung.” Zeitschrift für Physik 75 (9–10), 622–647. https://doi.org/10.1007/BF01344458.

[PPLG10b] Petersson, K. D., Petta, J. R., Lu, H., and Gossard, A. C. (2010). “Quantum coherence in a one-electron semiconductor charge qubit.” Physical Review Letters 105 (24), 246804. https://doi.org/10.1103/PhysRevLett.105.246804.

[RMHG07b] Reilly, D. J., Marcus, C. M., Hanson, M. P., and Gossard, A. C. (2007). “Fast single-charge sensing with an RF quantum point contact.” Applied Physics Letters 91 (16), 162101. https://doi.org/10.1063/1.2794995.

Lecture 7: Fock Space and Occupation States

7.1 From exchange symmetry to occupation sectors

7.1.1 Exchange-symmetric sectors

7.1.2 Bosonic occupation states and occupation labels

7.2 From creation maps to canonical algebra

7.2.1 Constructing the mode maps

7.2.2 Annihilation as contraction of an antisymmetric product

7.2.3 Canonical anticommutation from the projector definition

7.2.4 Fermionic occupation states from ordered creation

7.2.5 Fermionic occupation actions from CAR

7.2.6 Projector check of both occupation bases

7.2.7 Direct calculation of orthogonality and completeness

7.2.8 Adding a particle in a chosen mode

7.2.9 Bosonic basis actions from normalization

7.2.10 Projector check of the fermionic coordinate actions

7.2.11 Number operators and checks of the basis actions

7.2.12 Bosonic ladder normalization

7.2.13 Fermionic exclusion and exchange sign

7.2.14 Particle-number sectors and mode maps

7.2.15 From mode products to hopping

7.3 One hopping action in a spinful double dot

7.4 The two-site hopping problem in site and energy bases

7.5 Basis changes transform mode operators

7.5.1 Double-dot readout

7.5.2 Mode covariance under basis change

7.5.3 From a populated band to an operator action

7.5.4 Closing the construction

7.5.5 Next: construct operators from physical matrix elements

7.5.6 Supporting comparison: unsigned occupation flips

7.5.7 Application: periodic-chain hopping

7.5.8 Application: phonon displacement and the zero-mode boundary

7.5.9 Supporting calculation: particle-number change

7.6 Problems