Anticommutators

The anticommutator of two operators is the symmetric product

$$\{A,B\}=AB+BA.$$

It complements the antisymmetric commutator

$$[A,B]=AB-BA.$$

Together, the two brackets recover either ordering:

$$AB = \frac12\{A,B\} + \frac12[A,B],$$

and

$$BA = \frac12\{A,B\} - \frac12[A,B].$$

For bounded self-adjoint observables, this split separates two physically different pieces of an ordered product:

• $\{A,B\}/2$ is Hermitian and carries the real symmetric correlation.
• $[A,B]/2$ is anti-Hermitian and carries order sensitivity.

Anticommutators enter symmetrized covariance, the Robertson–Schrödinger uncertainty relation, Pauli and Clifford-type algebras, open-system generators, and fermionic canonical relations. Those appearances share notation but not always the same physical role.

The longer reusable identity list remains in Commutators and Anticommutators. Here the emphasis is on the physical meaning of symmetrized products.

Helpful background. Commutators explains order reversal and adjoints; correlations and covariance explains centered fluctuations.

Anticommutators and Operator Order

The algebra below uses bounded operators, including finite matrices. The final domain section explains the extra care required for unbounded observables.

Acting on a state,

$$\{A,B\}|\psi\rangle = A(B|\psi\rangle) + B(A|\psi\rangle).$$

The anticommutator adds the two orderings. It does not measure their difference and is not an alternative compatibility test.

If

$$\{A,B\}=0,$$

then

$$AB=-BA,$$

and $A$ and $B$ are said to anticommute.

This differs sharply from commuting:

$$[A,B]=0 \quad\Longleftrightarrow\quad AB=BA.$$

If two operators anticommute, then

$$[A,B]=2AB.$$

Thus nonzero anticommuting operators are generally noncommuting. The logical exception is worth remembering: if $AB=BA=0$, then the operators both commute and anticommute. Orthogonal-support projectors provide such examples.

Basic Algebra

The anticommutator is bilinear:

$$\{A,\alpha B+\beta C\} = \alpha\{A,B\} + \beta\{A,C\},$$

and similarly in the first slot. It is symmetric:

$$\{A,B\}=\{B,A\}.$$

Special cases include

$$\{A,A\}=2A^2,$$

and

$$\{A,I\}=2A.$$

Unlike the commutator, the anticommutator does not obey an ordinary Leibniz rule. A useful mixed identity is

$$[A,\{B,C\}] = \{[A,B],C\} + \{B,[A,C]\}.$$

For finite matrices, cyclicity gives

$$\operatorname{Tr}\{A,B\} = 2\operatorname{Tr}(AB),$$

which need not vanish. This contrasts with

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

Hermiticity and the Symmetric Observable Product

If $A$ and $B$ are Hermitian matrices, then

$$\begin{aligned} \{A,B\}^{\dagger} &=(AB+BA)^{\dagger} \\ &=BA+AB \\ &=\{A,B\}. \end{aligned}$$

Therefore

$$A\circ B \equiv \frac12\{A,B\}$$

is Hermitian and can itself represent an observable in the finite-dimensional setting. The operation $A\circ B$ is called the Jordan product.

The ordinary product $AB$ need not be Hermitian, because

$$(AB)^{\dagger}=BA.$$

The Jordan product removes the anti-Hermitian ordering part and keeps the observable-valued symmetric part.

For a density operator $\rho$,

$$\operatorname{Tr} \left( \rho\,\{A,B\} \right)$$

is real whenever the relevant products and traces are well-defined.

The Jordan Product Is Not Associative

The Jordan product is commutative:

$$A\circ B=B\circ A,$$

but it is generally not associative. Direct expansion gives

$$(A\circ B)\circ C -A\circ(B\circ C) = \frac14[B,[A,C]].$$

Nested commutators therefore measure the failure of the symmetric product to associate. Parentheses cannot be dropped from repeated Jordan products.

For example, with Pauli matrices,

$$(\sigma_x\circ\sigma_x)\circ\sigma_y = I\circ\sigma_y = \sigma_y,$$

whereas

$$\sigma_x\circ (\sigma_x\circ\sigma_y) = \sigma_x\circ0 =0.$$

The product is still highly structured; its nonassociativity is not a defect. It encodes a different algebraic aspect of observables from the Lie bracket defined by the commutator.

Sum and Difference Squares

Anticommutators are the cross terms in operator squares:

$$(A+B)^2 = A^2+B^2+\{A,B\},$$

and

$$(A-B)^2 = A^2+B^2-\{A,B\}.$$

Consequently,

$$\{A,B\} = \frac12 \left[ (A+B)^2-(A-B)^2 \right].$$

This polarization identity shows how the symmetric product can be recovered from observable squares.

If $A$ and $B$ anticommute, the cross term vanishes:

$$(A+B)^2=A^2+B^2.$$

Centered Observables and Covariance

For observables $A$ and $B$ in a state $\rho$, define centered operators

$$\delta A = A-\langle A\rangle I,$$

and

$$\delta B = B-\langle B\rangle I.$$

Here $\delta A$ is an operator, while $\Delta A=\sqrt{\langle(\delta A)^2\rangle}$ will denote its standard deviation. The mean is

$$\langle A\rangle=\operatorname{Tr}(\rho A).$$

The symmetrized covariance is

$$\operatorname{Cov}_{\rho}(A,B) = \frac12 \left\langle \{\delta A,\delta B\} \right\rangle.$$

Expanding the centered operators gives

$$\operatorname{Cov}_{\rho}(A,B) = \frac12 \langle AB+BA\rangle - \langle A\rangle\langle B\rangle.$$

For Hermitian $A$ and $B$, this is real and symmetric under exchange:

$$\operatorname{Cov}_{\rho}(A,B) = \operatorname{Cov}_{\rho}(B,A).$$

The ordered centered moment splits as

$$\langle\delta A\,\delta B\rangle = \operatorname{Cov}_{\rho}(A,B) + \frac12\langle[A,B]\rangle.$$

Thus

$$\operatorname{Cov}_{\rho}(A,B) = \operatorname{Re} \langle\delta A\,\delta B\rangle,$$

while the commutator supplies its imaginary part. The statistical interpretation, compatible-observable joint distributions, and connected correlations are developed in Correlations and Covariance.

Covariance Matrices Are Positive Semidefinite

For a family of Hermitian observables $A_1,\ldots,A_r$, define

$$\Gamma_{jk} = \frac12 \left\langle \{\delta A_j,\delta A_k\} \right\rangle.$$

For any real vector $\mathbf u$, let

$$O_{\mathbf u} = \sum_j u_j\delta A_j.$$

Then

$$\begin{aligned} \mathbf u^{\mathsf T}\Gamma\mathbf u &= \left\langle O_{\mathbf u}^2 \right\rangle \\ &\ge0. \end{aligned}$$

Therefore $\Gamma$ is a real positive-semidefinite matrix. The anticommutator is what makes the matrix symmetric while retaining the physical quadratic fluctuation of every real linear combination of observables.

Robertson–Schrödinger Uncertainty

The stronger two-observable uncertainty relation is

$$(\Delta A)^2(\Delta B)^2 \ge \left| \frac12 \left\langle \{\delta A,\delta B\} \right\rangle \right|^2 + \left| \frac1{2i} \langle[A,B]\rangle \right|^2.$$

Equivalently,

$$(\Delta A)^2(\Delta B)^2 \ge \operatorname{Cov}_{\rho}(A,B)^2 + \left| \frac1{2i} \langle[A,B]\rangle \right|^2.$$

The commutator term measures the antisymmetric obstruction, while the anticommutator term records aligned fluctuations. A state can have $\langle[A,B]\rangle=0$ and still obey a nontrivial bound because its symmetrized covariance is nonzero.

The Cauchy–Schwarz derivation, equality condition, and domain assumptions belong to General Uncertainty Relations.

Pauli Matrices: Dot and Cross Products

For real vectors $\mathbf a$ and $\mathbf b$, the Pauli identity is

$$(\mathbf a\cdot\boldsymbol\sigma) (\mathbf b\cdot\boldsymbol\sigma) = (\mathbf a\cdot\mathbf b)I + i(\mathbf a\times\mathbf b) \cdot\boldsymbol\sigma.$$

Taking symmetric and antisymmetric parts gives

$$\frac12 \left\lbrace \mathbf a\cdot\boldsymbol\sigma, \mathbf b\cdot\boldsymbol\sigma \right\rbrace = (\mathbf a\cdot\mathbf b)I,$$

and

$$\frac12 \left[ \mathbf a\cdot\boldsymbol\sigma, \mathbf b\cdot\boldsymbol\sigma \right] = i(\mathbf a\times\mathbf b) \cdot\boldsymbol\sigma.$$

In component form,

$$\{\sigma_i,\sigma_j\} = 2\delta_{ij}I,$$

and

$$[\sigma_i,\sigma_j] = 2i \sum_k\epsilon_{ijk}\sigma_k.$$

Orthogonal directions satisfy

$$\mathbf a\cdot\mathbf b=0 \quad\Longrightarrow\quad \left\lbrace \mathbf a\cdot\boldsymbol\sigma, \mathbf b\cdot\boldsymbol\sigma \right\rbrace =0.$$

They anticommute but do not commute unless one operator vanishes. The spin and rotation interpretation belongs to Pauli Matrices.

Anticommuting Hermitian Involutions

Suppose $A$ and $B$ are Hermitian and satisfy

$$A^2=B^2=I, \qquad \{A,B\}=0.$$

For real $\alpha$ and $\beta$,

$$\begin{aligned} (\alpha A+\beta B)^2 &=\alpha^2A^2+\beta^2B^2 +\alpha\beta\{A,B\} \\ &=(\alpha^2+\beta^2)I. \end{aligned}$$

Hence every eigenvalue $\lambda$ of $\alpha A+\beta B$ obeys

$$\lambda^2=\alpha^2+\beta^2.$$

The possible eigenvalues are therefore

$$\lambda=\pm\sqrt{\alpha^2+\beta^2}.$$

Another useful consequence is

$$ABA=-B,$$

because $AB=-BA$ and $A^2=I$. Conjugation by $A$ reverses the sign of $B$.

No nonzero vector can be a simultaneous eigenvector of both involutions. If

$$A|\psi\rangle=a|\psi\rangle, \qquad B|\psi\rangle=b|\psi\rangle,$$

with $a,b\in\{+1,-1\}$, then

$$\{A,B\}|\psi\rangle = 2ab|\psi\rangle \ne0,$$

contradicting anticommutation.

Spectral Pairing

More generally, if

$$\{A,B\}=0$$

and

$$A|a\rangle=a|a\rangle,$$

then

$$\begin{aligned} A(B|a\rangle) &=-BA|a\rangle \\ &=-a(B|a\rangle). \end{aligned}$$

Whenever $B|a\rangle\ne0$, it is an $A$ eigenvector with eigenvalue $-a$. Thus an invertible anticommuting operator pairs eigenspaces at $a$ and $-a$; the inverse provides a bijection, so their multiplicities agree.

Zero modes require separate care. If $B|a\rangle=0$, no partner is produced, and if $a=0$, the sign-reversed eigenvalue is the same eigenvalue.

The Anticommutator of Positive Operators Need Not Be Positive

Hermiticity does not imply positivity. Let

$$P= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad Q= \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix}.$$

Both $P$ and $Q$ are rank-one projectors and therefore positive. Yet

$$\{P,Q\} = \begin{pmatrix} 1&1/2\\ 1/2&0 \end{pmatrix}.$$

Its eigenvalues are

$$\lambda_{\pm} = \frac{1\pm\sqrt2}{2}.$$

Because $\lambda_-<0$, the anticommutator is not positive. Symmetrizing a product restores Hermiticity, not positivity.

Open-System Preview: Lindblad Dissipation

For a bounded operator $L$, a single-channel Lindblad dissipator has the form

$$\mathcal D_L(\rho) = L\rho L^{\dagger} - \frac12 \left\lbrace L^{\dagger}L, \rho \right\rbrace.$$

The jump term alone changes the trace. The anticommutator term supplies the matching symmetric subtraction:

$$\operatorname{Tr}(L\rho L^{\dagger}) = \operatorname{Tr}(L^{\dagger}L\rho),$$

while

$$\frac12 \operatorname{Tr} \left\lbrace L^{\dagger}L, \rho \right\rbrace = \operatorname{Tr}(L^{\dagger}L\rho).$$

Therefore

$$\operatorname{Tr}\mathcal D_L(\rho)=0.$$

The full completely positive dynamics, Hamiltonian term, and many-channel form belong to the Lindblad–GKSL Equation.

Fermionic Preview

Fermionic creation and annihilation operators satisfy canonical anticommutation relations

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

Setting $i=j$ in the second relation gives

$$2c_i^2=0,$$

so

$$c_i^2=0.$$

No mode can be created or annihilated twice in succession. The number operator

$$n_i=c_i^{\dagger}c_i$$

obeys

$$n_i^2=n_i,$$

so its eigenvalues are $0$ and $1$.

These consequences follow directly from the algebra. An explicit occupation-number basis additionally requires a chosen order of modes to track fermionic signs; see Fermionic Anticommutation Relations.

For graded operator algebras, the commutator and anticommutator are unified by

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

Here $|A|,|B|\in\{0,1\}$ denote even or odd parity for homogeneous operators. When both operators are odd, the graded commutator becomes an ordinary anticommutator. This grading is a structural statement, not a license to replace commutators by anticommutators arbitrarily.

Domain Caveats

For unbounded operators, both products must exist. The natural domain is

$$\mathcal D(\{A,B\}) = \mathcal D(AB) \cap \mathcal D(BA).$$

Hermiticity calculations assume that adjoints and products are defined on appropriate domains. Even if $A$ and $B$ are self-adjoint separately, the sum $AB+BA$ on a naive intersection need not automatically be self-adjoint.

Covariance and uncertainty formulas require finite second moments. Their quadratic-form versions use inner products such as $\operatorname{Re}\langle\delta A\psi,\delta B\psi\rangle$ and need only $\psi\in\mathcal D(A)\cap\mathcal D(B)$. Writing an actual product expectation $\langle\psi|AB|\psi\rangle$ requires the stronger product-domain condition. In infinite-dimensional examples, always distinguish a formal algebraic expression from a closed or self-adjoint operator with a declared domain.

Common Pitfalls

Anticommuting does not mean compatible. For example, $\sigma_x$ and $\sigma_z$ anticommute and have no common eigenvector. Orthogonal projectors are a special exception: both their product orderings vanish.

Centering is essential for covariance. Subtract $\langle A\rangle I$ and $\langle B\rangle I$ before taking the symmetrized product. Otherwise the result also contains the product of means.

Symmetrizing does not preserve positivity. The positive-projector example above has a negative anticommutator eigenvalue. A Hermitian operator can have either sign.

Exercises

Exercise 1: Ordered-product decomposition

Starting from the definitions of the commutator and anticommutator, derive

$$AB = \frac12\{A,B\} + \frac12[A,B].$$

If $A$ and $B$ are Hermitian, identify the Hermitian and anti-Hermitian pieces.

Solution

Adding the two brackets gives

$$\begin{aligned} \{A,B\}+[A,B] &=(AB+BA) +(AB-BA) \\ &=2AB. \end{aligned}$$

Dividing by $2$ yields the formula. For Hermitian $A$ and $B$, $\{A,B\}/2$ is Hermitian and $[A,B]/2$ is anti-Hermitian.

Exercise 2: Jordan associator

Let $A\circ B=\{A,B\}/2$. Prove

$$(A\circ B)\circ C -A\circ(B\circ C) = \frac14[B,[A,C]].$$

Solution

Expanding the first association gives

$$(A\circ B)\circ C = \frac14 \left( ABC+BAC +CAB+CBA \right).$$

The second gives

$$A\circ(B\circ C) = \frac14 \left( ABC+ACB +BCA+CBA \right).$$

Subtracting,

$$\begin{aligned} (A\circ B)\circ C -A\circ(B\circ C) &= \frac14 \left( BAC-BCA -ACB+CAB \right) \\ &=\frac14[B,[A,C]]. \end{aligned}$$

Exercise 3: Pauli-vector decomposition

Use

$$\sigma_i\sigma_j = \delta_{ij}I + i\sum_k\epsilon_{ijk}\sigma_k$$

to derive the anticommutator and commutator of $\mathbf a\cdot\boldsymbol\sigma$ and $\mathbf b\cdot\boldsymbol\sigma$.

Solution

Expand both vector operators in components:

$$\begin{aligned} (\mathbf a\cdot\boldsymbol\sigma) (\mathbf b\cdot\boldsymbol\sigma) &= \sum_{i,j}a_ib_j\sigma_i\sigma_j \\ &=(\mathbf a\cdot\mathbf b)I +i(\mathbf a\times\mathbf b) \cdot\boldsymbol\sigma. \end{aligned}$$

Reversing $\mathbf a$ and $\mathbf b$ keeps the dot product and reverses the cross product. Therefore

$$\left\lbrace \mathbf a\cdot\boldsymbol\sigma, \mathbf b\cdot\boldsymbol\sigma \right\rbrace = 2(\mathbf a\cdot\mathbf b)I,$$

and

$$\left[ \mathbf a\cdot\boldsymbol\sigma, \mathbf b\cdot\boldsymbol\sigma \right] = 2i(\mathbf a\times\mathbf b) \cdot\boldsymbol\sigma.$$

Exercise 4: Anticommuting involutions

Suppose $A,B$ are Hermitian matrices on a nonzero space, $A^2=B^2=I$, and $\{A,B\}=0$. For real $\alpha,\beta$ not both zero, show that

$$C=\alpha A+\beta B$$

has both eigenvalues $\pm\sqrt{\alpha^2+\beta^2}$ with equal multiplicities.

Solution

Squaring gives

$$\begin{aligned} C^2 &=\alpha^2A^2+\beta^2B^2 +\alpha\beta\{A,B\} \\ &=(\alpha^2+\beta^2)I. \end{aligned}$$

If $C|c\rangle=c|c\rangle$, then

$$c^2|c\rangle = (\alpha^2+\beta^2)|c\rangle.$$

Hence $c=\pm\sqrt{\alpha^2+\beta^2}$. Set $D=-\beta A+\alpha B$. Then $\{C,D\}=0$ and $D^2=(\alpha^2+\beta^2)I$, so $D$ is invertible and maps every eigenvector at $c$ to one at $-c$. Both signs occur with equal multiplicities. If $\alpha=\beta=0$, the only eigenvalue is zero instead.

Exercise 5: Qubit covariance

Let

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right),$$

with real $\mathbf r$, $|\mathbf r|\le1$, and real vectors $\mathbf a,\mathbf b$. Let

$$A=\mathbf a\cdot\boldsymbol\sigma, \qquad B=\mathbf b\cdot\boldsymbol\sigma.$$

Show that

$$\operatorname{Cov}_{\rho}(A,B) = \mathbf a\cdot\mathbf b - (\mathbf a\cdot\mathbf r) (\mathbf b\cdot\mathbf r).$$

Solution

The expectation values are

$$\langle A\rangle=\mathbf a\cdot\mathbf r, \qquad \langle B\rangle=\mathbf b\cdot\mathbf r.$$

The Pauli anticommutator gives

$$\frac12\langle\{A,B\}\rangle = \mathbf a\cdot\mathbf b.$$

Subtracting the product of means yields the stated covariance.

Exercise 6: Positivity counterexample

For

$$P= \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix}, \qquad Q= \frac12 \begin{pmatrix} 1&1\\ 1&1 \end{pmatrix},$$

verify that both are positive projectors but $\{P,Q\}$ is not positive.

Solution

Both matrices satisfy $P^2=P=P^{\dagger}$ and $Q^2=Q=Q^{\dagger}$, so their eigenvalues are $0$ and $1$.

Their anticommutator is

$$\{P,Q\} = \begin{pmatrix} 1&1/2\\ 1/2&0 \end{pmatrix}.$$

Its characteristic polynomial is

$$\lambda^2-\lambda-\frac14,$$

with roots

$$\lambda_{\pm} = \frac{1\pm\sqrt2}{2}.$$

The smaller root is negative, so $\{P,Q\}$ is Hermitian but not positive.

Exercise 7: Trace preservation in a dissipator

For

$$\mathcal D_L(\rho) = L\rho L^{\dagger} - \frac12 \{L^{\dagger}L,\rho\},$$

show that $\operatorname{Tr}\mathcal D_L(\rho)=0$.

Solution

Cyclicity gives

$$\operatorname{Tr}(L\rho L^{\dagger}) = \operatorname{Tr}(L^{\dagger}L\rho).$$

For the anticommutator term,

$$\begin{aligned} \frac12\operatorname{Tr} \{L^{\dagger}L,\rho\} &= \frac12 \operatorname{Tr} \left( L^{\dagger}L\rho+\rho L^{\dagger}L \right) \\ &= \operatorname{Tr}(L^{\dagger}L\rho). \end{aligned}$$

The two contributions cancel.

Exercise 8: Fermionic nilpotency and occupation

For one fermionic mode, assume

$$\{c,c^{\dagger}\}=I, \qquad \{c,c\}=0.$$

Show that $c^2=0$ and that $n=c^{\dagger}c$ satisfies $n^2=n$.

Solution

The second relation gives

$$2c^2=0,$$

so $c^2=0$. For the number operator,

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

The last term vanishes because $c^2=0$; equivalently, $(c^{\dagger})^2=0$. Therefore

$$n^2=n.$$

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.
• P. Jordan, J. von Neumann, and E. Wigner, “On an Algebraic Generalization of the Quantum Mechanical Formalism,” Annals of Mathematics 35, 29–64 (1934).
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.