] There are different definitions used in group theory and ring theory. Then the x V a ks. Learn more about Stack Overflow the company, and our products. % , For example, there are two eigenfunctions associated with the energy E: \(\varphi_{E}=e^{\pm i k x} \). . \end{align}\], \[\begin{align} {{1, 2}, {3,-1}}, https://mathworld.wolfram.com/Commutator.html. ) Still, this could be not enough to fully define the state, if there is more than one state \( \varphi_{a b} \). The commutator, defined in section 3.1.2, is very important in quantum mechanics. These can be particularly useful in the study of solvable groups and nilpotent groups. Prove that if B is orthogonal then A is antisymmetric. By computing the commutator between F p q and S 0 2 J 0 2, we find that it vanishes identically; this is because of the property q 2 = p 2 = 1. but in general \( B \varphi_{1}^{a} \not \alpha \varphi_{1}^{a}\), or \(\varphi_{1}^{a} \) is not an eigenfunction of B too. Acceleration without force in rotational motion? ] This notation makes it clear that \( \bar{c}_{h, k}\) is a tensor (an n n matrix) operating a transformation from a set of eigenfunctions of A (chosen arbitrarily) to another set of eigenfunctions. ) & \comm{AB}{C}_+ = A \comm{B}{C}_+ - \comm{A}{C} B \\ N.B., the above definition of the conjugate of a by x is used by some group theorists. \end{array}\right) \nonumber\]. Operation measuring the failure of two entities to commute, This article is about the mathematical concept. }[/math], [math]\displaystyle{ \mathrm{ad} }[/math], [math]\displaystyle{ \mathrm{ad}: R \to \mathrm{End}(R) }[/math], [math]\displaystyle{ \mathrm{End}(R) }[/math], [math]\displaystyle{ \operatorname{ad}_{[x, y]} = \left[ \operatorname{ad}_x, \operatorname{ad}_y \right]. &= \sum_{n=0}^{+ \infty} \frac{1}{n!} The commutator is zero if and only if a and b commute. Since a definite value of observable A can be assigned to a system only if the system is in an eigenstate of , then we can simultaneously assign definite values to two observables A and B only if the system is in an eigenstate of both and . Since the [x2,p2] commutator can be derived from the [x,p] commutator, which has no ordering ambiguities, this does not happen in this simple case. The commutator has the following properties: Relation (3) is called anticommutativity, while (4) is the Jacobi identity. }[/math], [math]\displaystyle{ (xy)^n = x^n y^n [y, x]^\binom{n}{2}. & \comm{A}{BC}_+ = \comm{A}{B} C + B \comm{A}{C}_+ \\ }[/math], [math]\displaystyle{ \left[\left[x, y^{-1}\right], z\right]^y \cdot \left[\left[y, z^{-1}\right], x\right]^z \cdot \left[\left[z, x^{-1}\right], y\right]^x = 1 }[/math], [math]\displaystyle{ \left[\left[x, y\right], z^x\right] \cdot \left[[z ,x], y^z\right] \cdot \left[[y, z], x^y\right] = 1. ( Comments. that is, vector components in different directions commute (the commutator is zero). In this short paper, the commutator of monomials of operators obeying constant commutation relations is expressed in terms of anti-commutators. by preparing it in an eigenfunction) I have an uncertainty in the other observable. 1 & 0 3 0 obj << & \comm{AB}{CD} = A \comm{B}{C} D + AC \comm{B}{D} + \comm{A}{C} DB + C \comm{A}{D} B \\ First assume that A is a \(\pi\)/4 rotation around the x direction and B a 3\(\pi\)/4 rotation in the same direction. }[A, [A, [A, B]]] + \cdots$. If we had chosen instead as the eigenfunctions cos(kx) and sin(kx) these are not eigenfunctions of \(\hat{p}\). If we take another observable B that commutes with A we can measure it and obtain \(b\). When you take the Hermitian adjoint of an expression and get the same thing back with a negative sign in front of it, the expression is called anti-Hermitian, so the commutator of two Hermitian operators is anti-Hermitian. \end{array}\right), \quad B=\frac{1}{2}\left(\begin{array}{cc} \end{equation}\], In electronic structure theory, we often want to end up with anticommutators: First-order response derivatives for the variational Lagrangian First-order response derivatives for variationally determined wave functions Fock space Fockian operators In a general spinor basis In a 'restricted' spin-orbital basis Formulas for commutators and anticommutators Foster-Boys localization Fukui function Frozen-core approximation but it has a well defined wavelength (and thus a momentum). [ (y),z] \,+\, [y,\mathrm{ad}_x\! \operatorname{ad}_x\!(\operatorname{ad}_x\! B In case there are still products inside, we can use the following formulas: This statement can be made more precise. (2005), https://books.google.com/books?id=hyHvAAAAMAAJ&q=commutator, https://archive.org/details/introductiontoel00grif_0, "Congruence modular varieties: commutator theory", https://www.researchgate.net/publication/226377308, https://www.encyclopediaofmath.org/index.php?title=p/c023430, https://handwiki.org/wiki/index.php?title=Commutator&oldid=2238611. This is Heisenberg Uncertainty Principle. 2 PTIJ Should we be afraid of Artificial Intelligence. [ The best answers are voted up and rise to the top, Not the answer you're looking for? & \comm{AB}{C}_+ = \comm{A}{C}_+ B + A \comm{B}{C} By using the commutator as a Lie bracket, every associative algebra can be turned into a Lie algebra. }[A, [A, [A, B]]] + \cdots This formula underlies the BakerCampbellHausdorff expansion of log(exp(A) exp(B)). [3] The expression ax denotes the conjugate of a by x, defined as x1a x . \exp(-A) \thinspace B \thinspace \exp(A) &= B + \comm{B}{A} + \frac{1}{2!} ) . \comm{A}{B}_n \thinspace , [ }[/math], [math]\displaystyle{ \left[x, y^{-1}\right] = [y, x]^{y^{-1}} }[/math], [math]\displaystyle{ \left[x^{-1}, y\right] = [y, x]^{x^{-1}}. We can then show that \(\comm{A}{H}\) is Hermitian: {\displaystyle [a,b]_{-}} Define C = [A, B] and A and B the uncertainty in the measurement outcomes of A and B: \( \Delta A^{2}= \left\langle A^{2}\right\rangle-\langle A\rangle^{2}\), where \( \langle\hat{O}\rangle\) is the expectation value of the operator \(\hat{O} \) (that is, the average over the possible outcomes, for a given state: \( \langle\hat{O}\rangle=\langle\psi|\hat{O}| \psi\rangle=\sum_{k} O_{k}\left|c_{k}\right|^{2}\)). {\displaystyle \mathrm {ad} _{x}:R\to R} We now have two possibilities. Now assume that the vector to be rotated is initially around z. & \comm{A}{BCD} = BC \comm{A}{D} + B \comm{A}{C} D + \comm{A}{B} CD ] Example 2.5. is then used for commutator. Its called Baker-Campbell-Hausdorff formula. , . & \comm{ABC}{D} = AB \comm{C}{D} + A \comm{B}{D} C + \comm{A}{D} BC \\ {\displaystyle \operatorname {ad} (\partial )(m_{f})=m_{\partial (f)}} 2. Learn the definition of identity achievement with examples. The mistake is in the last equals sign (on the first line) -- $ ACB - CAB = [ A, C ] B $, not $ - [A, C] B $. That is the case also when , or .. On the other hand, if all three indices are different, , and and both sides are completely antisymmetric; the left hand side because of the anticommutativity of the matrices, and on the right hand side because of the antisymmetry of .It thus suffices to verify the identities for the cases of , , and . We have considered a rather special case of such identities that involves two elements of an algebra \( \mathcal{A} \) and is linear in one of these elements. [6] The anticommutator is used less often, but can be used to define Clifford algebras and Jordan algebras and in the derivation of the Dirac equation in particle physics. \end{array}\right) \nonumber\], with eigenvalues \( \), and eigenvectors (not normalized), \[v^{1}=\left[\begin{array}{l} We present new basic identity for any associative algebra in terms of single commutator and anticommutators. The anticommutator of two elements a and b of a ring or associative algebra is defined by. A measurement of B does not have a certain outcome. However, it does occur for certain (more . -1 & 0 . }[/math], [math]\displaystyle{ \mathrm{ad}_x[y,z] \ =\ [\mathrm{ad}_x\! }A^2 + \cdots }[/math] can be meaningfully defined, such as a Banach algebra or a ring of formal power series. Has Microsoft lowered its Windows 11 eligibility criteria? R \end{align}\] There is also a collection of 2.3 million modern eBooks that may be borrowed by anyone with a free archive.org account. "Jacobi -type identities in algebras and superalgebras". There is no uncertainty in the measurement. Anticommutators are not directly related to Poisson brackets, but they are a logical extension of commutators. $$ Borrow a Book Books on Internet Archive are offered in many formats, including. {\displaystyle \operatorname {ad} _{x}\operatorname {ad} _{y}(z)=[x,[y,z]\,]} }[/math], [math]\displaystyle{ \operatorname{ad}_x\operatorname{ad}_y(z) = [x, [y, z]\,] }[/math], [math]\displaystyle{ \operatorname{ad}_x^2\! Using the anticommutator, we introduce a second (fundamental) \[\begin{align} A Then we have the commutator relationships: \[\boxed{\left[\hat{r}_{a}, \hat{p}_{b}\right]=i \hbar \delta_{a, b} }\nonumber\]. 0 & 1 \\ If dark matter was created in the early universe and its formation released energy, is there any evidence of that energy in the cmb? $\hat {A}:V\to V$ (actually an operator isn't always defined by this fact, I have seen it defined this way, and I have seen it used just as a synonym for map). For example: Consider a ring or algebra in which the exponential [math]\displaystyle{ e^A = \exp(A) = 1 + A + \tfrac{1}{2! If the operators A and B are scalar operators (such as the position operators) then AB = BA and the commutator is always zero. 1 Commutator Formulas Shervin Fatehi September 20, 2006 1 Introduction A commutator is dened as1 [A, B] = AB BA (1) where A and B are operators and the entire thing is implicitly acting on some arbitrary function. a If A and B commute, then they have a set of non-trivial common eigenfunctions. $e^{A} B e^{-A} = B + [A, B] + \frac{1}{2! \end{equation}\], From these definitions, we can easily see that What is the physical meaning of commutators in quantum mechanics? }[/math], [math]\displaystyle{ [\omega, \eta]_{gr}:= \omega\eta - (-1)^{\deg \omega \deg \eta} \eta\omega. : Considering now the 3D case, we write the position components as \(\left\{r_{x}, r_{y} r_{z}\right\} \). Evaluate the commutator: ( e^{i hat{X^2, hat{P} ). }[/math], [math]\displaystyle{ [A + B, C] = [A, C] + [B, C] }[/math], [math]\displaystyle{ [A, B] = -[B, A] }[/math], [math]\displaystyle{ [A, [B, C]] + [B, [C, A]] + [C, [A, B]] = 0 }[/math], [math]\displaystyle{ [A, BC] = [A, B]C + B[A, C] }[/math], [math]\displaystyle{ [A, BCD] = [A, B]CD + B[A, C]D + BC[A, D] }[/math], [math]\displaystyle{ [A, BCDE] = [A, B]CDE + B[A, C]DE + BC[A, D]E + BCD[A, E] }[/math], [math]\displaystyle{ [AB, C] = A[B, C] + [A, C]B }[/math], [math]\displaystyle{ [ABC, D] = AB[C, D] + A[B, D]C + [A, D]BC }[/math], [math]\displaystyle{ [ABCD, E] = ABC[D, E] + AB[C, E]D + A[B, E]CD + [A, E]BCD }[/math], [math]\displaystyle{ [A, B + C] = [A, B] + [A, C] }[/math], [math]\displaystyle{ [A + B, C + D] = [A, C] + [A, D] + [B, C] + [B, D] }[/math], [math]\displaystyle{ [AB, CD] = A[B, C]D + [A, C]BD + CA[B, D] + C[A, D]B =A[B, C]D + AC[B,D] + [A,C]DB + C[A, D]B }[/math], [math]\displaystyle{ A, C], [B, D = [[[A, B], C], D] + [[[B, C], D], A] + [[[C, D], A], B] + [[[D, A], B], C] }[/math], [math]\displaystyle{ \operatorname{ad}_A: R \rightarrow R }[/math], [math]\displaystyle{ \operatorname{ad}_A(B) = [A, B] }[/math], [math]\displaystyle{ [AB, C]_\pm = A[B, C]_- + [A, C]_\pm B }[/math], [math]\displaystyle{ [AB, CD]_\pm = A[B, C]_- D + AC[B, D]_- + [A, C]_- DB + C[A, D]_\pm B }[/math], [math]\displaystyle{ A,B],[C,D=[[[B,C]_+,A]_+,D]-[[[B,D]_+,A]_+,C]+[[[A,D]_+,B]_+,C]-[[[A,C]_+,B]_+,D] }[/math], [math]\displaystyle{ \left[A, [B, C]_\pm\right] + \left[B, [C, A]_\pm\right] + \left[C, [A, B]_\pm\right] = 0 }[/math], [math]\displaystyle{ [A,BC]_\pm = [A,B]_- C + B[A,C]_\pm }[/math], [math]\displaystyle{ [A,BC] = [A,B]_\pm C \mp B[A,C]_\pm }[/math], [math]\displaystyle{ e^A = \exp(A) = 1 + A + \tfrac{1}{2! How is this possible? Especially if one deals with multiple commutators in a ring R, another notation turns out to be useful. , Then this function can be written in terms of the \( \left\{\varphi_{k}^{a}\right\}\): \[B\left[\varphi_{h}^{a}\right]=\bar{\varphi}_{h}^{a}=\sum_{k} \bar{c}_{h, k} \varphi_{k}^{a} \nonumber\]. & \comm{A}{BCD} = BC \comm{A}{D} + B \comm{A}{C} D + \comm{A}{B} CD , and applying both sides to a function g, the identity becomes the usual Leibniz rule for the n-th derivative Commutator identities are an important tool in group theory. I'm voting to close this question as off-topic because it shows insufficient prior research with the answer plainly available on Wikipedia and does not ask about any concept or show any effort to derive a relation. commutator is the identity element. [4] Many other group theorists define the conjugate of a by x as xax1. . Let us refer to such operators as bosonic. x \[\begin{equation} \require{physics} }[A{+}B, [A, B]] + \frac{1}{3!} ) We are now going to express these ideas in a more rigorous way. version of the group commutator. n ( {\displaystyle \{AB,C\}=A\{B,C\}-[A,C]B} The Main Results. If A is a fixed element of a ring R, identity (1) can be interpreted as a Leibniz rule for the map [math]\displaystyle{ \operatorname{ad}_A: R \rightarrow R }[/math] given by [math]\displaystyle{ \operatorname{ad}_A(B) = [A, B] }[/math]. S2u%G5C@[96+um w`:N9D/[/Et(5Ye , (fg)} Now however the wavelength is not well defined (since we have a superposition of waves with many wavelengths). & \comm{A}{BC} = B \comm{A}{C} + \comm{A}{B} C \\ and anticommutator identities: (i) [rt, s] . This is not so surprising if we consider the classical point of view, where measurements are not probabilistic in nature. https://en.wikipedia.org/wiki/Commutator#Identities_.28ring_theory.29. In other words, the map adA defines a derivation on the ring R. Identities (2), (3) represent Leibniz rules for more than two factors, and are valid for any derivation. \comm{A}{H}^\dagger = \comm{A}{H} \thinspace . (y)\, x^{n - k}. & \comm{ABC}{D} = AB \comm{C}{D} + A \comm{B}{D} C + \comm{A}{D} BC \\ Consider the eigenfunctions for the momentum operator: \[\hat{p}\left[\psi_{k}\right]=\hbar k \psi_{k} \quad \rightarrow \quad-i \hbar \frac{d \psi_{k}}{d x}=\hbar k \psi_{k} \quad \rightarrow \quad \psi_{k}=A e^{-i k x} \nonumber\]. These examples show that commutators are not specific of quantum mechanics but can be found in everyday life. For instance, let and ) . The second scenario is if \( [A, B] \neq 0 \). As you can see from the relation between commutators and anticommutators [ A, B] := A B B A = A B B A B A + B A = A B + B A 2 B A = { A, B } 2 B A it is easy to translate any commutator identity you like into the respective anticommutator identity. Introduction \thinspace {}_n\comm{B}{A} \thinspace , 1 [3] The expression ax denotes the conjugate of a by x, defined as x1ax. since the anticommutator . From this, two special consequences can be formulated: x {{7,1},{-2,6}} - {{7,1},{-2,6}}. {\textstyle e^{A}Be^{-A}\ =\ B+[A,B]+{\frac {1}{2! When we apply AB, the vector ends up (from the z direction) along the y-axis (since the first rotation does not do anything to it), if instead we apply BA the vector is aligned along the x direction. The definition of the commutator above is used throughout this article, but many other group theorists define the commutator as. Snapshot of the geometry at some Monte-Carlo sweeps in 2D Euclidean quantum gravity coupled with Polyakov matter field This formula underlies the BakerCampbellHausdorff expansion of log(exp(A) exp(B)). \comm{\comm{A}{B}}{B} = 0 \qquad\Rightarrow\qquad \comm{A}{f(B)} = f'(B) \comm{A}{B} \thinspace . From this identity we derive the set of four identities in terms of double . \end{equation}\], Using the definitions, we can derive some useful formulas for converting commutators of products to sums of commutators: ( Commutator[x, y] = c defines the commutator between the (non-commuting) objects x and y. FEYN CALC SYMBOL See Also AntiCommutator CommutatorExplicit DeclareNonCommutative DotSimplify Commutator Commutator[x,y]=c defines the commutator between the (non-commuting) objects xand y. ExamplesExamplesopen allclose all In QM we express this fact with an inequality involving position and momentum \( p=\frac{2 \pi \hbar}{\lambda}\). tr, respectively. = <> The position and wavelength cannot thus be well defined at the same time. \comm{A}{B} = AB - BA \thinspace . e Then \( \varphi_{a}\) is also an eigenfunction of B with eigenvalue \( b_{a}\). Notice that $ACB-ACB = 0$, which is why we were allowed to insert this after the second equals sign. (fg) }[/math]. ( 1 If the operators A and B are scalar operators (such as the position operators) then AB = BA and the commutator is always zero. Suppose . Lavrov, P.M. (2014). . } Let \(A\) be an anti-Hermitian operator, and \(H\) be a Hermitian operator. [8] In context|mathematics|lang=en terms the difference between anticommutator and commutator is that anticommutator is (mathematics) a function of two elements a and b, defined as ab + ba while commutator is (mathematics) (of a ring'') an element of the form ''ab-ba'', where ''a'' and ''b'' are elements of the ring, it is identical to the ring's zero . and and and Identity 5 is also known as the Hall-Witt identity. Identities (4)(6) can also be interpreted as Leibniz rules. [6, 8] Here holes are vacancies of any orbitals. $$ ( \comm{A}{B} = AB - BA \thinspace . When an addition and a multiplication are both defined for all elements of a set \(\set{A, B, \dots}\), we can check if multiplication is commutative by calculation the commutator: & \comm{A}{BC}_+ = \comm{A}{B}_+ C - B \comm{A}{C} \\ ad ad }[/math], [math]\displaystyle{ [a, b] = ab - ba. ad We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. ABSTRACT. d Unfortunately, you won't be able to get rid of the "ugly" additional term. Taking any algebra and looking at $\{x,y\} = xy + yx$ you get a product satisfying 'Jordan Identity'; my question in the second paragraph is about the reverse : given anything satisfying the Jordan Identity, does it naturally embed in a regular algebra (equipped with the regular anticommutator?) if 2 = 0 then 2(S) = S(2) = 0. I think that the rest is correct. + }[/math], [math]\displaystyle{ e^A e^B e^{-A} e^{-B} = \end{array}\right] \nonumber\]. (For the last expression, see Adjoint derivation below.) . We said this is an operator, so in order to know what it is, we apply it to a function (a wavefunction). can be meaningfully defined, such as a Banach algebra or a ring of formal power series. \[\begin{equation} + (y) \,z \,+\, y\,\mathrm{ad}_x\!(z). In mathematics, the commutator gives an indication of the extent to which a certain binary operation fails to be commutative. The \( \psi_{j}^{a}\) are simultaneous eigenfunctions of both A and B. Kudryavtsev, V. B.; Rosenberg, I. G., eds. @user3183950 You can skip the bad term if you are okay to include commutators in the anti-commutator relations. m \comm{A}{H}^\dagger = \comm{A}{H} \thinspace . [ The degeneracy of an eigenvalue is the number of eigenfunctions that share that eigenvalue. e \exp(A) \exp(B) = \exp(A + B + \frac{1}{2} \comm{A}{B} + \cdots) \thinspace , ) (B.48) In the limit d 4 the original expression is recovered. Identities (4)(6) can also be interpreted as Leibniz rules. commutator of [ x It is not a mysterious accident, but it is a prescription that ensures that QM (and experimental outcomes) are consistent (thus its included in one of the postulates). Most generally, there exist \(\tilde{c}_{1}\) and \(\tilde{c}_{2}\) such that, \[B \varphi_{1}^{a}=\tilde{c}_{1} \varphi_{1}^{a}+\tilde{c}_{2} \varphi_{2}^{a} \nonumber\]. Obs. = , n. Any linear combination of these functions is also an eigenfunction \(\tilde{\varphi}^{a}=\sum_{k=1}^{n} \tilde{c}_{k} \varphi_{k}^{a}\). Making sense of the canonical anti-commutation relations for Dirac spinors, Microcausality when quantizing the real scalar field with anticommutators. A \exp(A) \exp(B) = \exp(A + B + \frac{1}{2} \comm{A}{B} + \cdots) \thinspace , ! = Let A and B be two rotations. \end{equation}\], \[\begin{equation} density matrix and Hamiltonian for the considered fermions, I is the identity operator, and we denote [O 1 ,O 2 ] and {O 1 ,O 2 } as the commutator and anticommutator for any two Enter the email address you signed up with and we'll email you a reset link. We reformulate the BRST quantisation of chiral Virasoro and W 3 worldsheet gravities. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. $$ The most famous commutation relationship is between the position and momentum operators. ZC+RNwRsoR[CfEb=sH XreQT4e&b.Y"pbMa&o]dKA->)kl;TY]q:dsCBOaW`(&q.suUFQ >!UAWyQeOK}sO@i2>MR*X~K-q8:"+m+,_;;P2zTvaC%H[mDe. Also, \(\left[x, p^{2}\right]=[x, p] p+p[x, p]=2 i \hbar p \). \comm{U^\dagger A U}{U^\dagger B U } = U^\dagger \comm{A}{B} U \thinspace . Rowland, Rowland, Todd and Weisstein, Eric W. \[\begin{align} &= \sum_{n=0}^{+ \infty} \frac{1}{n!} = }}[A,[A,B]]+{\frac {1}{3! {\displaystyle \operatorname {ad} _{xy}\,\neq \,\operatorname {ad} _{x}\operatorname {ad} _{y}} If then and it is easy to verify the identity. Assume now we have an eigenvalue \(a\) with an \(n\)-fold degeneracy such that there exists \(n\) independent eigenfunctions \(\varphi_{k}^{a}\), k = 1, . The Internet Archive offers over 20,000,000 freely downloadable books and texts. }[/math], [math]\displaystyle{ \mathrm{ad}_x:R\to R }[/math], [math]\displaystyle{ \operatorname{ad}_x(y) = [x, y] = xy-yx. Anticommutator is a see also of commutator. Similar identities hold for these conventions. \end{equation}\] Two standard ways to write the CCR are (in the case of one degree of freedom) $$ [ p, q] = - i \hbar I \ \ ( \textrm { and } \ [ p, I] = [ q, I] = 0) $$. Our approach follows directly the classic BRST formulation of Yang-Mills theory in \comm{A}{B}_+ = AB + BA \thinspace . Supergravity can be formulated in any number of dimensions up to eleven. Fundamental solution The forward fundamental solution of the wave operator is a distribution E+ Cc(R1+d)such that 2E+ = 0, Some of the above identities can be extended to the anticommutator using the above subscript notation. $$ {\displaystyle \partial ^{n}\! N n = n n (17) then n is also an eigenfunction of H 1 with eigenvalue n+1/2 as well as . The commutator has the following properties: Lie-algebra identities [ A + B, C] = [ A, C] + [ B, C] [ A, A] = 0 [ A, B] = [ B, A] [ A, [ B, C]] + [ B, [ C, A]] + [ C, [ A, B]] = 0 Relation (3) is called anticommutativity, while (4) is the Jacobi identity . ] The correct relationship is $ [AB, C] = A [ B, C ] + [ A, C ] B $. Is if \ ( H\ ) be an anti-Hermitian operator, and \ b\... ( H\ ) be an anti-Hermitian operator, and our products R\to R we. 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