commutator anticommutator identities

e If [A, B] = 0 (the two operator commute, and again for simplicity we assume no degeneracy) then \(\varphi_{k} \) is also an eigenfunction of B. \end{equation}\], \[\begin{equation} \end{align}\], \[\begin{align} x e A ad }[/math], [math]\displaystyle{ \operatorname{ad}_{xy} \,\neq\, \operatorname{ad}_x\operatorname{ad}_y }[/math], [math]\displaystyle{ x^n y = \sum_{k = 0}^n \binom{n}{k} \operatorname{ad}_x^k\! Consider first the 1D case. . The most important 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). The following identity follows from anticommutativity and Jacobi identity and holds in arbitrary Lie algebra: [2] See also Structure constants Super Jacobi identity Three subgroups lemma (Hall-Witt identity) References ^ Hall 2015 Example 3.3 We can write an eigenvalue equation also for this tensor, \[\bar{c} v^{j}=b^{j} v^{j} \quad \rightarrow \quad \sum_{h} \bar{c}_{h, k} v_{h}^{j}=b^{j} v^{j} \nonumber\]. $$ (49) This operator adds a particle in a superpositon of momentum states with &= \sum_{n=0}^{+ \infty} \frac{1}{n!} The general Leibniz rule, expanding repeated derivatives of a product, can be written abstractly using the adjoint representation: Replacing x by the differentiation operator [math]\displaystyle{ \partial }[/math], and y by the multiplication operator [math]\displaystyle{ m_f: g \mapsto fg }[/math], we get [math]\displaystyle{ \operatorname{ad}(\partial)(m_f) = m_{\partial(f)} }[/math], and applying both sides to a function g, the identity becomes the usual Leibniz rule for the n-th derivative [math]\displaystyle{ \partial^{n}\! 1 ) The set of all commutators of a group is not in general closed under the group operation, but the subgroup of G generated by all commutators is closed and is called the derived group or the commutator subgroup of G. Commutators are used to define nilpotent and solvable groups and the largest abelian quotient group. Example 2.5. N n = n n (17) then n is also an eigenfunction of H 1 with eigenvalue n+1/2 as well as . [A,BC] = [A,B]C +B[A,C]. Hr (1) there are operators aj and a j acting on H j, and extended to the entire Hilbert space H in the usual way }[/math], [math]\displaystyle{ [x, zy] = [x, y]\cdot [x, z]^y }[/math], [math]\displaystyle{ [x z, y] = [x, y]^z \cdot [z, y]. Many identities are used that are true modulo certain subgroups. /Filter /FlateDecode xZn}'q8/q+~"Ysze9sk9uzf~EoO>y7/7/~>7Fm`dl7/|rW^1W?n6a5Vk7 =;%]B0+ZfQir?c a:J>S\{Mn^N',hkyk] If the operators A and B are scalar operators (such as the position operators) then AB = BA and the commutator is always zero. ] A and \( \hat{p} \varphi_{2}=i \hbar k \varphi_{1}\). (And by the way, the expectation value of an anti-Hermitian operator is guaranteed to be purely imaginary.) Consider the eigenfunctions for the momentum operator: \[\hat{p}\left[\psi_{k}\right]=\hbar k \psi_{k} \quad \rightarrow \quad-i \hbar \frac{d \psi_{k}}{d x}=\hbar k \psi_{k} \quad \rightarrow \quad \psi_{k}=A e^{-i k x} \nonumber\]. \end{align}\] Recall that for such operators we have identities which are essentially Leibniz's' rule. These examples show that commutators are not specific of quantum mechanics but can be found in everyday life. In the proof of the theorem about commuting observables and common eigenfunctions we took a special case, in which we assume that the eigenvalue \(a\) was non-degenerate. Browse other questions tagged, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site. A From this identity we derive the set of four identities in terms of double . -1 & 0 A B %PDF-1.4 1 Rename .gz files according to names in separate txt-file, Ackermann Function without Recursion or Stack. R \comm{A}{B}_n \thinspace , From MathWorld--A Wolfram : ) [4] Many other group theorists define the conjugate of a by x as xax1. & \comm{AB}{C} = A \comm{B}{C} + \comm{A}{C}B \\ The position and wavelength cannot thus be well defined at the same time. First assume that A is a \(\pi\)/4 rotation around the x direction and B a 3\(\pi\)/4 rotation in the same direction. \end{equation}\], \[\begin{equation} Especially if one deals with multiple commutators in a ring R, another notation turns out to be useful. It is easy (though tedious) to check that this implies a commutation relation for . [ The anticommutator of two elements a and b of a ring or associative algebra is defined by {,} = +. We now want to find with this method the common eigenfunctions of \(\hat{p} \). For example: Consider a ring or algebra in which the exponential [math]\displaystyle{ e^A = \exp(A) = 1 + A + \tfrac{1}{2! We can then look for another observable C, that commutes with both A and B and so on, until we find a set of observables such that upon measuring them and obtaining the eigenvalues a, b, c, d, . Let \(\varphi_{a}\) be an eigenfunction of A with eigenvalue a: \[A \varphi_{a}=a \varphi_{a} \nonumber\], \[B A \varphi_{a}=a B \varphi_{a} \nonumber\]. Consider for example: \[A=\frac{1}{2}\left(\begin{array}{ll} We said this is an operator, so in order to know what it is, we apply it to a function (a wavefunction). 0 & -1 }[/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! A linear operator $\hat {A}$ is a mapping from a vector space into itself, ie. 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 . We know that if the system is in the state \( \psi=\sum_{k} c_{k} \varphi_{k}\), with \( \varphi_{k}\) the eigenfunction corresponding to the eigenvalue \(a_{k} \) (assume no degeneracy for simplicity), the probability of obtaining \(a_{k} \) is \( \left|c_{k}\right|^{2}\). & \comm{A}{BC} = B \comm{A}{C} + \comm{A}{B} C \\ A and B are real non-zero 3 \times 3 matrices and satisfy the equation (AB) T + B - 1 A = 0. the function \(\varphi_{a b c d \ldots} \) is uniquely defined. \[\begin{align} Enter the email address you signed up with and we'll email you a reset link. \comm{A}{B} = AB - BA \thinspace . [ Moreover, if some identities exist also for anti-commutators . Learn more about Stack Overflow the company, and our products. 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. {\displaystyle [AB,C]=A\{B,C\}-\{A,C\}B} [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. The mistake is in the last equals sign (on the first line) -- $ ACB - CAB = [ A, C ] B $, not $ - [A, C] B $. The commutator of two operators acting on a Hilbert space is a central concept in quantum mechanics, since it quantifies how well the two observables described by these operators can be measured simultaneously. x In mathematics, the commutator gives an indication of the extent to which a certain binary operation fails to be commutative. e If \(\varphi_{a}\) is the only linearly independent eigenfunction of A for the eigenvalue a, then \( B \varphi_{a}\) is equal to \( \varphi_{a}\) at most up to a multiplicative constant: \( B \varphi_{a} \propto \varphi_{a}\). {\displaystyle [a,b]_{-}} }}[A,[A,[A,B]]]+\cdots \ =\ e^{\operatorname {ad} _{A}}(B).} B We now have two possibilities. y \[\begin{align} -i \hbar k & 0 $$ 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. R The elementary BCH (Baker-Campbell-Hausdorff) formula reads }[/math], [math]\displaystyle{ \mathrm{ad}_x:R\to R }[/math], [math]\displaystyle{ \operatorname{ad}_x(y) = [x, y] = xy-yx. [ & \comm{A}{BC}_+ = \comm{A}{B}_+ C - B \comm{A}{C} \\ @user3183950 You can skip the bad term if you are okay to include commutators in the anti-commutator relations. ) Planned Maintenance scheduled March 2nd, 2023 at 01:00 AM UTC (March 1st, We've added a "Necessary cookies only" option to the cookie consent popup, Energy eigenvalues of a Q.H.Oscillator with $[\hat{H},\hat{a}] = -\hbar \omega \hat{a}$ and $[\hat{H},\hat{a}^\dagger] = \hbar \omega \hat{a}^\dagger$. R We would obtain \(b_{h}\) with probability \( \left|c_{h}^{k}\right|^{2}\). . Also, the results of successive measurements of A, B and A again, are different if I change the order B, A and B. B Then, \[\boxed{\Delta \hat{x} \Delta \hat{p} \geq \frac{\hbar}{2} }\nonumber\]. If I want to impose that \( \left|c_{k}\right|^{2}=1\), I must set the wavefunction after the measurement to be \(\psi=\varphi_{k} \) (as all the other \( c_{h}, h \neq k\) are zero). This is indeed the case, as we can verify. A 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: \( \langle\hat{O}\rangle=\langle\psi|\hat{O}| \psi\rangle=\sum_{k} O_{k}\left|c_{k}\right|^{2}\)). a Do anticommutators of operators has simple relations like commutators. wiSflZz%Rk .W `vgo `QH{.;\,5b .YSM$q K*"MiIt dZbbxH Z!koMnvUMiK1W/b=&tM /evkpgAmvI_|E-{FdRjI}j#8pF4S(=7G:\eM/YD]q"*)Q6gf4)gtb n|y vsC=gi I"z.=St-7.$bi|ojf(b1J}=%\*R6I H. Using the commutator Eq. }[/math], [math]\displaystyle{ [a, b] = ab - ba. The most important example is the uncertainty relation between position and momentum. \end{array}\right) \nonumber\]. 2. f The anticommutator of two elements a and b of a ring or associative algebra is defined by. This is the so-called collapse of the wavefunction. This is probably the reason why the identities for the anticommutator aren't listed anywhere - they simply aren't that nice. & \comm{AB}{C} = A \comm{B}{C}_+ - \comm{A}{C}_+ B }[/math], [math]\displaystyle{ \mathrm{ad}_x[y,z] \ =\ [\mathrm{ad}_x\! For 3 particles (1,2,3) there exist 6 = 3! 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]. Guaranteed to be purely imaginary. relation between position and momentum ) then n is also an of. 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