I Commutator identity

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I'm having a little trouble proving the following identity that is used in the derivation of the Baker-Campbell-Hausdorff Formula: $$[e^{tT},S] = -t[S,T]e^{tT}$$ It is assumed that [S,T] commutes with S and T, these being linear operators. I tried opening both sides and comparing terms to no avail, I feel like I'm missing something really dumb here. Can someone give me a hand?
 

fresh_42

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##\left(\exp(\operatorname{ad}(tT))- 1\right)(S)= \operatorname{ad}(tT)(S)## which is true, if ##\left( \operatorname{ad}(T) \right)^n(S)=0\, , \, n>1\,.##
This follows from ##[[S,T],X]=0\, , \,X\in \{\,S,T\,\}##. Hence we have
$$
\left(\exp(\operatorname{ad}(tT))- 1\right)(S)=\operatorname{Ad}(\exp(tT))(S) -S = \exp(tT)S\exp(-tT) - S = \operatorname{ad}(tT)(S) = tTS-tST
$$
and thus
$$
\exp(tT)S -S \exp(tT) = [\exp(tT),S] =[e^{tT},S] = [tT,S]\exp(tT)=-t[S,T]\exp(tT) = -t[S,T]e^{tT}
$$
 
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Sorry, I'm not that comfortable with the adjoint representation yet. Let me start by seeing if I understand the first condition, if we set n=2 for instance:
$$\left( \operatorname{ad}(T) \right)^2(S) = \operatorname{ad}(T)(\operatorname{ad}(T)(S) ) = \operatorname{ad}(T)([T,S]) = [T, [T,S]] = 0$$ It's zero by the assumption, and I'm guessing the more general condition follows from induction? I don't see how to use it to prove the identity though, in particular I don't get this step: $$ \exp(tT)S\exp(-tT) - S = \operatorname{ad}(tT)(S) $$ Could you clarify things a bit? Thanks in advance.
 

fresh_42

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Sorry, I'm not that comfortable with the adjoint representation yet. Let me start by seeing if I understand the first condition, if we set n=2 for instance:
$$\left( \operatorname{ad}(T) \right)^2(S) = \operatorname{ad}(T)(\operatorname{ad}(T)(S) ) = \operatorname{ad}(T)([T,S]) = [T, [T,S]] = 0$$ It's zero by the assumption, and I'm guessing the more general condition follows from induction?
In this case we don't need induction, it is clear right away. ##(\operatorname{ad}T)^n(S)=\underbrace{[T,[T,[\ldots,[T}_{n\text{ times }},S]\ldots ]=0## since already the most inner expression ##[T,[T,S]]=0## and applying more linear transformations cannot change this. We only need at least two applications, i.e. ##n\geq 2\,.##
I don't see how to use it to prove the identity though, in particular I don't get this step: $$ \exp(tT)S\exp(-tT) - S = \operatorname{ad}(tT)(S) $$ Could you clarify things a bit? Thanks in advance.
Prior to BCH it is usually proven, that ##\operatorname{Ad}\circ \exp = \exp\circ \operatorname{ad}##, i.e. ##((\operatorname{Ad}\circ \exp)(tT))(S) = ((\exp\circ \operatorname{ad})(tT))(S)\,.## The exponential function plays the role of integration in the sense, that it maps tangent vectors ##tT \in \mathfrak{g}## of the Lie algebra into the Lie group ##g:=\exp(tT)=e^{tT} \in G##. Moreover the adjoint representation ##\operatorname{Ad}## of the Lie group on its Lie algebra acts by conjugation, as it comes from the inner automorphisms of the group. This means ##(\operatorname{Ad}(g))(S)=gSg^{-1}\,.## Hence

\begin{align*}
\exp(tT) S \exp(-tT)&= gSg^{-1}\\ &=(\operatorname{Ad}(g))(S)\\ &=(\operatorname{Ad}(e^{tT}))(S)\\ &= ((\exp\circ \operatorname{ad})(tT))(S)\\
&= \exp(\operatorname{ad}(tT))(S)\\
&= \left( \sum_{n=0}^\infty \dfrac{1}{n!}(\operatorname{ad}(tT))^n \right)(S)\\
&= \sum_{n=0}^\infty \dfrac{1}{n!}\left( (\operatorname{ad}(tT))^n (S) \right)\\
&= 1(S) + (\operatorname{ad}(tT))(S) + \sum_{n=2}^\infty \dfrac{1}{n!}\underbrace{\left( (\operatorname{ad}(tT))^n (S) \right)}_{= 0}\\
&= S + (\operatorname{ad}(tT))(S)\\
&= S + [tT,S]
\end{align*}

If you want to read it a bit more detailed, however, still rough due to format, you might want to have a look on:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-representations/
 
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I've read your series of articles, really well written I must say. It amazes me how vast this area is, I thought I knew enough Lie theory after reading Tapp, Stillwell and a few mathematical physics books here and there, but your articles are full of theorems and formalism I never heard of. That being said, you didn't need to go into such detail in your reply, it just didn't occur to me to do an expansion right there. Silly me, thanks a bunch for your help!
 

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