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Display information for equation id:math.219095.28 on revision:219095

* Page found: Commutator (eq math.219095.28)

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TeX (original user input): 

 e^{A}Be^{-A}=B+[A,B]+\frac{1}{2!}[A,[A,B]]+\frac{1}{3!}[A,[A,[A,B]]]+\cdots \equiv e^{\operatorname{ad}(A)} B.

TeX (checked): 

e^{A}Be^{-A}=B+[A,B]+{\frac {1}{2!}}[A,[A,B]]+{\frac {1}{3!}}[A,[A,[A,B]]]+\cdots \equiv e^{\operatorname {ad} (A)}B.

LaTeXML (experimental; uses MathML) rendering

MathML (12.88 KB / 1.854 KB) :

e A ⁢ B ⁢ e - A = B + [ A , B ] + 1 2 ! ⁢ [ A , [ A , B ] ] + 1 3 ! ⁢ [ A , [ A , [ A , B ] ] ] + ⋯ ≡ e ad ⁡ ( A ) ⁢ B . superscript 𝑒 𝐴 𝐵 superscript 𝑒 𝐴 𝐵 𝐴 𝐵 1 2 𝐴 𝐴 𝐵 1 3 𝐴 𝐴 𝐴 𝐵 ⋯ superscript 𝑒 ad 𝐴 𝐵 {\displaystyle e^{{A}}Be^{{-A}}=B+[A,B]+{\frac{1}{2!}}[A,[A,B]]+{\frac{1}{3!}}% [A,[A,[A,B]]]+\cdots\equiv e^{{\operatorname{ad}(A)}}B.} 
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SVG (16.785 KB / 4.558 KB) :

e Superscript upper A Baseline times upper B times e Superscript negative upper A Baseline equals upper B plus left-bracket upper A comma upper B right-bracket plus StartFraction 1 Over 2 factorial EndFraction times left-bracket upper A comma left-bracket upper A comma upper B right-bracket right-bracket plus StartFraction 1 Over 3 factorial EndFraction times left-bracket upper A comma left-bracket upper A comma left-bracket upper A comma upper B right-bracket right-bracket right-bracket plus midline-horizontal-ellipsis identical-to e Superscript a d left-parenthesis upper A right-parenthesis Baseline times upper B period

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MathML (2.455 KB / 529 B) :

e A B e − A = B + [ A , B ] + 1 2 ! [ A , [ A , B ] ] + 1 3 ! [ A , [ A , [ A , B ] ] ] + ⋯ ≡ e ad ⁡ ( A ) B . {\displaystyle e^{A}Be^{-A}=B+[A,B]+{\frac {1}{2!}}[A,[A,B]]+{\frac {1}{3!}}[A,[A,[A,B]]]+\cdots \equiv e^{\operatorname {ad} (A)}B.} 
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    <annotation encoding="application/x-tex">{\displaystyle e^{A}Be^{-A}=B+[A,B]+{\frac {1}{2!}}[A,[A,B]]+{\frac {1}{3!}}[A,[A,[A,B]]]+\cdots \equiv e^{\operatorname {ad} (A)}B.}</annotation>
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SVG (11.091 KB / 3.965 KB) :

{\displaystyle e^{A}Be^{-A}=B+[A,B]+{\frac {1}{2!}}[A,[A,B]]+{\frac {1}{3!}}[A,[A,[A,B]]]+\cdots \equiv e^{\operatorname {ad} (A)}B.}

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