Operator Identities

This sheet collects exact operator identities that recur in many-body calculations. It is organized for translation and error checking: each formula states the algebra or regularity it needs, and each specialized topic links to its canonical derivation.

The page does not replace the Commutator Identities card, the full Baker–Campbell–Hausdorff treatment, Normal Ordering in Many-Body QM, or Wick’s Theorem Preview. It assembles the identities needed to move among those subjects without duplicating their derivations.

Scope and Safety Conditions

For finite matrices, every algebraic identity below is literal. For unbounded operators, a formal product such as $AB$, $[A,B]$, or $e^A B e^{-A}$ is meaningful only on a suitable common invariant domain or through a justified closure, quadratic form, or spectral construction.

Before manipulating operators, check:

• that every displayed product is defined on the states being used;
• that sums and integrals converge in the required topology;
• that a trace exists and cyclic reordering is allowed;
• that a power series or BCH expansion converges, truncates, or is explicitly formal;
• that field operators at coincident points have been regulated;
• that fermionic operator parity is tracked under reordering;
• that a basis transformation preserves the canonical algebra.

An identity can be algebraically correct and still be invalid on the proposed domain or inside a divergent trace.

Conventions

The ordinary commutator and anticommutator are

$$\begin{aligned} [A,B] &\equiv AB-BA, \\ \{A,B\} &\equiv AB+BA. \end{aligned}$$

For canonical modes, define

$$\eta = \begin{cases} +1, & \text{bosons},\\ -1, & \text{fermions}. \end{cases}$$

Then one compact algebra is

$$\begin{aligned} d_i d_j^\dagger -\eta d_j^\dagger d_i &= \delta_{ij}, \\ d_i d_j -\eta d_j d_i &= 0, \\ d_i^\dagger d_j^\dagger -\eta d_j^\dagger d_i^\dagger &= 0. \end{aligned}$$

Thus $\eta=+1$ gives bosonic commutators and $\eta=-1$ gives fermionic anticommutators. Ordinary commutators remain ordinary commutators even when the operators are fermionic; the graded bracket is introduced separately below.

Adjoint and Hermiticity Rules

The adjoint reverses operator order:

$$(AB)^\dagger = B^\dagger A^\dagger.$$

Consequently,

$$\begin{aligned} [A,B]^\dagger &= -[A^\dagger,B^\dagger], \\ \{A,B\}^\dagger &= \{A^\dagger,B^\dagger\}. \end{aligned}$$

If $A$ and $B$ are Hermitian on the relevant domain, then $[A,B]$ is anti-Hermitian and $i[A,B]$ is Hermitian. A Hamiltonian term written as a bare commutator of Hermitian operators therefore usually needs a factor of $i$ and a real coefficient.

For a unitary $U$,

$$\begin{aligned} (U^\dagger A U)^\dagger &= U^\dagger A^\dagger U, \\ [U^\dagger A U,U^\dagger B U] &= U^\dagger[A,B]U. \end{aligned}$$

Unitary changes of representation preserve spectra, products, commutators, and Hermiticity when all transformed domains are carried along.

Ordinary Commutator Identities

Linearity and antisymmetry give

$$\begin{aligned} [A,\alpha B+\beta C] &= \alpha[A,B]+\beta[A,C], \\ [A,B] &= -[B,A]. \end{aligned}$$

The Leibniz rules are

$$\begin{aligned} [A,BC] &= [A,B]C+B[A,C], \\ [AB,C] &= A[B,C]+[A,C]B. \end{aligned}$$

Two mixed commutator–anticommutator forms are especially useful for fermionic strings:

$$\begin{aligned} [A,BC] &= \{A,B\}C -B\{A,C\}, \\ [AB,C] &= A\{B,C\} -\{A,C\}B. \end{aligned}$$

They are exact associative-algebra identities; $A,B,C$ need not themselves obey canonical anticommutation relations.

The Jacobi identity is

$$\begin{aligned} 0 ={}& [A,[B,C]] \\ &+ [B,[C,A]] \\ &+ [C,[A,B]]. \end{aligned}$$

For a positive integer $n$,

$$[A,B^n] = \sum_{r=0}^{n-1} B^r[A,B]B^{n-1-r}.$$

If $[A,B]$ also commutes with $B$, this reduces to

$$[A,B^n] = n[A,B]B^{n-1}.$$

The reduction is not valid merely because $n$ is large or because $A$ and $B$ have simple spectra.

Tensor-Product Identities

For operators acting on two tensor factors,

$$\begin{aligned} [A\otimes B,C\otimes D] &= \frac12 [A,C]\otimes\{B,D\} \\ &\quad+ \frac12 \{A,C\}\otimes[B,D] \end{aligned}$$

and

$$\begin{aligned} \{A\otimes B,C\otimes D\} &= \frac12 \{A,C\}\otimes\{B,D\} \\ &\quad+ \frac12 [A,C]\otimes[B,D] \end{aligned}$$

If $A$ acts only on subsystem 1 and $B$ only on subsystem 2, then

$$[A\otimes I,I\otimes B] = 0.$$

This commuting-subsystem fact does not apply to spatially separated fermionic field operators represented in a graded tensor product without first declaring the parity convention.

Graded Commutators

For a parity-homogeneous operator $A$, let $|A|=0$ for even parity and $|A|=1$ for odd parity. The graded commutator is

$$[A,B]_{\mathrm g} \equiv AB - (-1)^{|A||B|} BA.$$

It becomes an ordinary commutator if either operator is even and an anticommutator if both are odd. Its graded antisymmetry is

$$[A,B]_{\mathrm g} = - (-1)^{|A||B|} [B,A]_{\mathrm g}.$$

For parity-homogeneous $A,B,C$,

$$\begin{aligned} [A,BC]_{\mathrm g} &= [A,B]_{\mathrm g}C \\ &\quad+ (-1)^{|A||B|} B[A,C]_{\mathrm g}. \end{aligned}$$

The graded bracket is the natural algebraic tool for canonical fermion generators. Physical Hamiltonians and observables are usually even, so their dynamics is still written with the ordinary commutator.

Similarity Transformations

Define

$$\operatorname{ad}_A(B) \equiv [A,B].$$

The Hadamard lemma is

$$\begin{aligned} e^A B e^{-A} &= e^{\operatorname{ad}_A}B \\ &= \sum_{n=0}^{\infty} \frac{1}{n!} \operatorname{ad}_A^n(B). \end{aligned}$$

If

$$[A,B] = \lambda B$$

with scalar $\lambda$, then

$$e^A B e^{-A} = e^\lambda B.$$

If nested commutators terminate, the series is finite. Otherwise, convergence or a formal asymptotic interpretation must be supplied.

The BCH series starts as

$$\begin{aligned} \log(e^Ae^B) &= A+B+\frac12[A,B] \\ &\quad+ \frac1{12}[A,[A,B]] \\ &\quad+ \frac1{12}[B,[B,A]] +\cdots . \end{aligned}$$

If $[A,B]$ commutes with both $A$ and $B$, all higher nested commutators vanish and

$$e^Ae^B = e^{A+B+\frac12[A,B]}.$$

Use the canonical Baker–Campbell–Hausdorff page for convergence, product formulas, displacement operators, and time-ordering boundaries.

Derivative of an Operator Exponential

For a differentiable operator family $A(\lambda)$,

$$\begin{aligned} \frac{d}{d\lambda}e^{A(\lambda)} &= \int_0^1 ds\, e^{(1-s)A} \\ &\qquad\times \frac{dA}{d\lambda} e^{sA}. \end{aligned}$$

If $[A,dA/d\lambda]=0$, the scalar-looking rule is recovered:

$$\frac{d}{d\lambda}e^A = \frac{dA}{d\lambda}e^A.$$

Without that commutator condition, pulling $dA/d\lambda$ through the exponential is generally wrong.

Under trace-class conditions, cyclicity simplifies the derivative:

$$\frac{d}{d\lambda} \operatorname{Tr}e^{A(\lambda)} = \operatorname{Tr} \left[ \frac{dA}{d\lambda} e^{A(\lambda)} \right].$$

This trace identity remains valid even when $A$ and $dA/d\lambda$ do not commute.

Trace Identities

Whenever the products are trace class in the required order,

$$\operatorname{Tr}(AB) = \operatorname{Tr}(BA).$$

Hence

$$\operatorname{Tr}[A,B] = 0.$$

More generally,

$$\operatorname{Tr} \left( U^\dagger A U \right) = \operatorname{Tr}A$$

for unitary $U$ and trace-class $A$. A basis-independent trace can be evaluated in any complete orthonormal basis.

If a density operator is a function of the Hamiltonian,

$$\rho = f(H),$$

then $[\rho,H]=0$ and

$$\operatorname{Tr} \left( \rho[H,A] \right) = 0$$

whenever cyclicity is justified. This is the operator statement that equilibrium expectation values are stationary under the same time-independent Hamiltonian.

In infinite-dimensional systems, neither $\operatorname{Tr}(AB)=\operatorname{Tr}(BA)$ nor $\operatorname{Tr}[A,B]=0$ should be invoked before checking trace-class conditions. Regularized traces can contain anomaly or boundary terms.

Projectors and Block Structure

For an orthogonal projector,

$$P^2 = P = P^\dagger.$$

With $Q=I-P$,

$$\begin{aligned} Q^2 &= Q, \\ PQ &= QP = 0, \\ P+Q &= I. \end{aligned}$$

Every operator decomposes as

$$A = PAP+PAQ+QAP+QAQ.$$

The subspaces are invariant under $A$ precisely when the off-diagonal blocks vanish:

$$[P,A] = 0 \quad\Longleftrightarrow\quad PAQ = QAP = 0.$$

Projection before and after an approximation can give different results. An effective operator $PAP$ does not by itself include virtual excursions through $Q$.

Resolvent Identities

Use the convention

$$R_A(z) \equiv (z-A)^{-1}.$$

For $z$ in the resolvent sets of $A$ and $B$,

$$R_A(z)-R_B(z) = R_A(z)(A-B)R_B(z).$$

For one operator evaluated at two spectral parameters,

$$\begin{aligned} R_A(z)-R_A(w) &= (w-z) \mathcal R_A(z,w), \\ \mathcal R_A(z,w) &\equiv R_A(z)R_A(w). \end{aligned}$$

The derivative is

$$\frac{dR_A(z)}{dz} = -R_A(z)^2.$$

If $A$ is self-adjoint,

$$R_A(z)^\dagger = R_A(z^\ast).$$

Sources that define $(A-z)^{-1}$ instead have corresponding sign changes. The Resolvent Operator owns spectral representations, poles, boundary values, and Green-function conventions.

Canonical Mode Algebra

The unified canonical relation implies

$$d_i d_j^\dagger = \delta_{ij} + \eta d_j^\dagger d_i.$$

Thus

$$\begin{array}{c|c} \text{boson} & a_i a_j^\dagger =\delta_{ij}+a_j^\dagger a_i \\ \text{fermion} & c_i c_j^\dagger =\delta_{ij}-c_j^\dagger c_i \end{array}$$

For one mode,

$$\begin{aligned} aa^\dagger &= 1+a^\dagger a, \\ cc^\dagger &= 1-c^\dagger c. \end{aligned}$$

The sign difference is exact and is the source of Bose enhancement versus Pauli exclusion in many elementary contractions.

Number-Operator Commutators

Define

$$n_i \equiv d_i^\dagger d_i, \qquad \widehat N \equiv \sum_i n_i.$$

For both bosons and fermions,

$$\begin{aligned} [n_i,d_j] &= -\delta_{ij}d_j, \\ [n_i,d_j^\dagger] &= \delta_{ij}d_j^\dagger. \end{aligned}$$

Therefore,

$$\begin{aligned} [\widehat N,d_j] &= -d_j, \\ [\widehat N,d_j^\dagger] &= d_j^\dagger. \end{aligned}$$

If $M_{p,q}$ is a monomial with $p$ creation and $q$ annihilation operators, then

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

This charge-counting identity holds without expanding every canonical swap. The canonical application guide is Number Operators and Conserved Quantities.

Number Shifts and Conjugation

For a single mode with $n=d^\dagger d$, polynomial identities give

$$\begin{aligned} f(n)d &= d\,f(n-1), \\ d\,f(n) &= f(n+1)d, \\ f(n)d^\dagger &= d^\dagger f(n+1), \\ d^\dagger f(n) &= f(n-1)d^\dagger. \end{aligned}$$

For bosons these identities act naturally on the finite-particle core. Expressions such as $f(n-1)$ are interpreted through the polynomial or spectral identity, not as a claim that a physical state with occupation $-1$ exists.

The exponential special cases are

$$\begin{aligned} e^{\lambda n}d\,e^{-\lambda n} &= e^{-\lambda}d, \\ e^{\lambda n}d^\dagger e^{-\lambda n} &= e^\lambda d^\dagger. \end{aligned}$$

For $\lambda=-i\theta$, this is the $U(1)$ phase rotation generated by number.

Occupation Projectors and Factorials

For one fermionic mode,

$$\begin{aligned} c^2 &= (c^\dagger)^2 = 0, \\ n^2 &= n. \end{aligned}$$

Thus $n$ is a projector with eigenvalues $0$ and $1$.

For one bosonic mode,

$$(a^\dagger)^m a^m = n(n-1)\cdots(n-m+1).$$

In particular,

$$\begin{aligned} (a^\dagger)^2a^2 &= n(n-1), \\ n^2 &= (a^\dagger)^2a^2+n. \end{aligned}$$

The extra $n$ is the contact term generated by commuting an annihilation operator through a creation operator. It is why $n^2$ and the normal-ordered pair-counting operator are not interchangeable for bosons.

Number-Conserving Bilinears

Let

$$E_{ij} \equiv d_i^\dagger d_j.$$

For either canonical statistics,

$$\begin{aligned} [E_{ij},d_k^\dagger] &= \delta_{jk}d_i^\dagger, \\ [E_{ij},d_k] &= -\delta_{ik}d_j. \end{aligned}$$

The bilinears close under ordinary commutation:

$$[E_{ij},E_{kl}] = \delta_{jk}E_{il} -\delta_{il}E_{kj}.$$

For one-body matrices $A$ and $B$, define their second-quantized lifts

$$\widehat A = \sum_{i,j} d_i^\dagger A_{ij}d_j, \qquad \widehat B = \sum_{i,j} d_i^\dagger B_{ij}d_j.$$

Then

$$[\widehat A,\widehat B] = \widehat{[A,B]}.$$

This identity explains why one-body symmetries lift to Fock space without a statistics-dependent sign. It assumes a common canonical basis and number-conserving bilinears; pairing operators obey a larger algebra.

Fermion Parity

For integer-valued total fermion number,

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

It obeys

$$\mathcal P^2 = I, \qquad \mathcal P^\dagger = \mathcal P.$$

Conjugation gives

$$\begin{aligned} \mathcal P c_i\mathcal P &= -c_i, \\ \mathcal P c_i^\dagger\mathcal P &= -c_i^\dagger. \end{aligned}$$

Every even fermionic monomial commutes with $\mathcal P$ and every odd monomial anticommutes with it. Pairing Hamiltonians can break particle-number conservation while preserving fermion parity.

Canonical Basis Changes

A number-conserving mode transformation

$$\widetilde d_\alpha = \sum_i U_{\alpha i}d_i$$

preserves the canonical algebra when $U$ is unitary on the one-particle space:

$$UU^\dagger = U^\dagger U = I.$$

For a Bogoliubov transformation

$$\gamma = Ud+Vd^\dagger,$$

the compact canonical conditions in the $\eta$ convention are

$$\begin{aligned} UU^\dagger -\eta VV^\dagger &= I, \\ UV^{\mathsf T} -\eta VU^{\mathsf T} &= 0. \end{aligned}$$

For bosons, $\eta=+1$ gives the indefinite-metric minus sign. For fermions, $\eta=-1$ gives plus signs. These equations are necessary algebraic checks; stability, implementability in infinite volume, and vacuum choice require more. The canonical physical derivations live in Bogoliubov Theory and the fermionic quasiparticle pages.

Fourier Completeness

For a periodic lattice of $L$ sites, one common unitary convention is

$$\begin{aligned} d_j &= \frac1{\sqrt L} \sum_k e^{i\mathbf k\cdot\mathbf r_j} d_{\mathbf k}, \\ d_{\mathbf k} &= \frac1{\sqrt L} \sum_j e^{-i\mathbf k\cdot\mathbf r_j} d_j. \end{aligned}$$

The completeness relations are

$$\begin{aligned} \frac1L\sum_k e^{i\mathbf k\cdot (\mathbf r_i-\mathbf r_j)} &= \delta_{ij}, \\ \frac1L\sum_j e^{i(\mathbf k-\mathbf q)\cdot \mathbf r_j} &= \delta_{\mathbf k,\mathbf q}. \end{aligned}$$

They guarantee that canonical (anti)commutators are preserved. Different sign and normalization conventions are equally valid but must be transformed consistently; see Fourier-Transform Conventions.

Pauli and Fierz Identities

For Pauli matrices,

$$\sigma^a\sigma^b = \delta_{ab}I + i\sum_c \epsilon_{abc}\sigma^c.$$

Therefore,

$$\begin{aligned} [\sigma^a,\sigma^b] &= 2i\sum_c \epsilon_{abc}\sigma^c, \\ \{\sigma^a,\sigma^b\} &= 2\delta_{ab}I. \end{aligned}$$

For ordinary three-vectors $\mathbf u,\mathbf v$,

$$\begin{aligned} (\mathbf u\cdot\boldsymbol\sigma) (\mathbf v\cdot\boldsymbol\sigma) &= (\mathbf u\cdot\mathbf v)I \\ &\quad+ i(\mathbf u\times\mathbf v) \cdot\boldsymbol\sigma. \end{aligned}$$

The two-component completeness relation is

$$\sum_{a=1}^3 (\sigma^a)_{\alpha\beta} (\sigma^a)_{\gamma\delta} = 2\delta_{\alpha\delta} \delta_{\beta\gamma} - \delta_{\alpha\beta} \delta_{\gamma\delta}.$$

This is the basic Pauli Fierz identity. Index order matters: exchanging two indices changes which bilinears are being rearranged. The Pauli Matrix Identity Index owns the broader spin-$1/2$ identity set.

Local Spinful-Fermion Identities

At one site, define

$$\begin{aligned} n &= n_\uparrow+n_\downarrow, \\ D &= n_\uparrow n_\downarrow, \\ \mathbf S &= \frac12 \sum_{\alpha,\beta} c_\alpha^\dagger \boldsymbol\sigma_{\alpha\beta} c_\beta. \end{aligned}$$

Since $n_\sigma^2=n_\sigma$,

$$n^2 = n+2D,$$

so

$$D = \frac12 (n^2-n).$$

The local spin Casimir is

$$\mathbf S^2 = \frac34 (n-2D).$$

These identities can be checked on the empty, spin-up, spin-down, and doubly occupied local states. They are exact for one spinful orbital; multiorbital Hund, orbital, and pair-hopping terms require a larger local algebra.

Gaussian Factorization Reminder

Let $X_i$ be operators linear in canonical creation and annihilation operators, with vanishing one-point functions in a Gaussian state. For a fixed operator ordering, write

$$C_{ij} \equiv \langle X_iX_j\rangle.$$

Then the four-point function has the schematic Wick form

$$\begin{aligned} \langle X_1X_2X_3X_4\rangle &= C_{12}C_{34} \\ &\quad+ \eta C_{13}C_{24} \\ &\quad+ C_{14}C_{23}. \end{aligned}$$

For bosons, $\eta=+1$ and all pairings carry plus signs. For fermions, $\eta=-1$ gives the crossing-pair minus sign. Nonzero means, anomalous contractions, time ordering, contour ordering, and reference-state conventions modify what is called a contraction, not the underlying parity bookkeeping.

This factorization is not valid for a generic interacting non-Gaussian state. Use Wick’s Theorem Preview and Normal Ordering in Many-Body QM for the canonical assumptions and extensions.

Fast Validation Checks

1. Adjoint: reverse the order of every factor before simplifying.
2. Parity: count fermionic swaps or use the graded bracket.
3. Charge: verify $[\widehat N,O]$ from creation-minus-annihilation count.
4. Small matrices: test spin or finite-mode identities in the smallest nontrivial representation.
5. Vacuum action: apply both sides to $\lvert0\rangle$ and low-occupation states.
6. Trace: check trace-class conditions before cyclic reordering.
7. Domains: identify a common invariant core for unbounded products.
8. Basis change: verify the canonical matrix conditions before using transformed modes.
9. Normal ordering: retain every contraction term unless an approximation is declared.
10. Limits: check bosonic and fermionic specializations separately when using $\eta$ notation.

Common Mistakes

• Using an anticommutator where the Heisenberg equation requires an ordinary commutator.
• Applying an ungraded Leibniz rule to a graded bracket.
• Replacing $[A,B^n]$ by $n[A,B]B^{n-1}$ without the additional commutation condition.
• Writing $d e^A/d\lambda=A'e^A$ when $[A,A']\ne0$.
• Assuming $\operatorname{Tr}[A,B]=0$ for products that are not trace class.
• Forgetting the minus sign in $c_ic_j^\dagger=\delta_{ij}-c_j^\dagger c_i$.
• Treating a fermion occupation $n_i$ as an unrestricted integer rather than a projector.
• Replacing bosonic $n^2$ by $(a^\dagger)^2a^2$ and dropping the contact term.
• Assuming particle-number-breaking pairing also breaks fermion parity.
• Applying Wick factorization to a non-Gaussian state.
• Using a resolvent identity with the signs from the opposite $(A-z)^{-1}$ convention.
• Comparing Pauli Fierz formulas without matching index order.

Cross-Links

• Symbols and Conventions
• Commutator Identities
• Commutators and Anticommutators
• Baker–Campbell–Hausdorff Formula
• Number Operators and Conserved Quantities
• Normal Ordering in Many-Body QM
• Wick’s Theorem Preview
• Bosonic Operators in Many-Body Models
• Fermionic Operators in Many-Body Models
• Topological Superconductors applies the Majorana bilinear and fermion-parity identities to boundary modes, finite-size splitting, and parity-sensitive operations.
• Resolvent Operator
• Pauli Matrix Identity Index
• Fourier-Transform Conventions

References

• M. Reed and B. Simon, Methods of Modern Mathematical Physics, Vol. I: Functional Analysis, revised and enlarged ed., Academic Press (1980) — domains, adjoints, traces, spectral theory, and resolvents.
• B. C. Hall, Lie Groups, Lie Algebras, and Representations, 2nd ed., Springer (2015) — exponentials, adjoint actions, BCH, and Lie-algebra identities.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover (2003) — canonical many-body operators, Green functions, and diagrammatic algebra.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Westview Press (1998) — second quantization, coherent states, Wick expansion, and thermal methods.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press (2010) — fermionic and bosonic Gaussian algebras, symmetries, and field-theory conventions.
• H. Bruus and K. Flensberg, Many-Body Quantum Theory in Condensed Matter Physics, Oxford University Press (2004) — operator methods, Fock-space identities, and response calculations.

Exercises

1. Derive a mixed product identity

Prove

$$[A,BC] = \{A,B\}C -B\{A,C\}$$

directly from the definitions.

Solution

Expand the right-hand side:

$$\begin{aligned} X &\equiv \{A,B\}C-B\{A,C\}, \\ X &= (AB+BA)C \\ &\quad- B(AC+CA) \\ &= ABC+BAC \\ &\quad- BAC-BCA \\ &= ABC-BCA \\ &= [A,BC]. \end{aligned}$$

No statistics assumption was used. The identity follows only from associativity and the definitions of the two brackets.

2. Conjugate a charged operator

Suppose $[\widehat N,O]=qO$. Show that

$$e^{\lambda\widehat N} O e^{-\lambda\widehat N} = e^{\lambda q}O.$$

Apply the result to $d_j$ and $d_j^\dagger$.

Solution

Repeated commutation gives

$$\operatorname{ad}_{\widehat N}^m(O) = q^mO.$$

The Hadamard series is therefore

$$\begin{aligned} e^{\lambda\widehat N} O e^{-\lambda\widehat N} &= \sum_{m=0}^\infty \frac{\lambda^m}{m!} q^mO \\ &= e^{\lambda q}O. \end{aligned}$$

Since $[\widehat N,d_j]=-d_j$ and $[\widehat N,d_j^\dagger]=d_j^\dagger$,

$$\begin{aligned} e^{\lambda\widehat N} d_j e^{-\lambda\widehat N} &= e^{-\lambda}d_j, \\ e^{\lambda\widehat N} d_j^\dagger e^{-\lambda\widehat N} &= e^\lambda d_j^\dagger. \end{aligned}$$

3. Close the bilinear algebra

For canonical bosons or fermions, derive

$$\begin{aligned} E_{ij} &\equiv d_i^\dagger d_j, \\ [E_{ij},E_{kl}] &= \delta_{jk}E_{il} -\delta_{il}E_{kj}. \end{aligned}$$

Solution

First use the canonical algebra to establish the statistics-independent identities

$$\begin{aligned} [E_{ij},d_k^\dagger] &= \delta_{jk}d_i^\dagger, \\ [E_{ij},d_l] &= -\delta_{il}d_j. \end{aligned}$$

Then apply the ordinary Leibniz rule:

$$\begin{aligned} [E_{ij},E_{kl}] &= [E_{ij},d_k^\dagger]d_l \\ &\quad+ d_k^\dagger[E_{ij},d_l] \\ &= \delta_{jk}d_i^\dagger d_l -\delta_{il}d_k^\dagger d_j \\ &= \delta_{jk}E_{il} -\delta_{il}E_{kj}. \end{aligned}$$

The result is the same for bosons and fermions because each $E_{ij}$ has even fermion parity.

4. Compare bosonic and fermionic occupation squares

Show that $n^2=n$ for one fermionic mode, whereas

$$n^2 = (a^\dagger)^2a^2+n$$

for one bosonic mode.

Solution

For a fermion,

$$\begin{aligned} n^2 &= c^\dagger c c^\dagger c \\ &= c^\dagger(1-c^\dagger c)c \\ &= c^\dagger c - c^\dagger c^\dagger cc. \end{aligned}$$

Since $(c^\dagger)^2=c^2=0$, the last term vanishes and $n^2=n$.

For a boson,

$$\begin{aligned} n^2 &= a^\dagger a a^\dagger a \\ &= a^\dagger(1+a^\dagger a)a \\ &= n+(a^\dagger)^2a^2. \end{aligned}$$

The different sign in the canonical algebra produces the different occupation identity.

5. Reduce the local Hubbard double occupancy

For one spinful fermion orbital, prove

$$\begin{aligned} n &= n_\uparrow+n_\downarrow, \\ D &= n_\uparrow n_\downarrow \\ &= \frac12(n^2-n). \end{aligned}$$

Solution

Expand the square:

$$\begin{aligned} n^2 &= n_\uparrow^2 +n_\downarrow^2 \\ &\quad+ n_\uparrow n_\downarrow +n_\downarrow n_\uparrow. \end{aligned}$$

The two number operators commute, and each is a projector:

$$n_\sigma^2 = n_\sigma.$$

Therefore

$$n^2 = n+2n_\uparrow n_\downarrow,$$

which rearranges to the desired result.

On the four local basis states, both sides have eigenvalues $0,0,0,1$ for empty, spin-up, spin-down, and double occupancy.

6. Prove the first resolvent identity

With $R_A(z)=(z-A)^{-1}$ and $R_B(z)=(z-B)^{-1}$, prove

$$R_A(z)-R_B(z) = R_A(z)(A-B)R_B(z).$$

Solution

Insert the inverse factors on both sides:

$$\begin{aligned} R_A-R_B &= R_A(z-B)R_B \\ &\quad- R_A(z-A)R_B \\ &= R_A(A-B)R_B. \end{aligned}$$

The identity requires $z$ to lie in both resolvent sets. If the convention $(A-z)^{-1}$ is used instead, the corresponding sign arrangement must be rederived rather than copied.