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Prove the weaker form of the BCH Formula:

$$e^A e^B = e^{A + B + \frac{1}{2}[A,B]} $$

with the assumption $[A, [B, A]] = 0; [B, [B,A]] = 0$

Start with $f(\lambda) = e^{\lambda A} e^{\lambda B} e^{-\lambda (A + B)}$ and establish the differential equation $\frac{df}{d\lambda} = \lambda [A, B] f.$ Then integrate it to obtain the formula.

What I did is I worked backwards.It is very easy to see that $$ \frac{df}{d\lambda} = \lambda[A,B]f$$ $$ \frac{df}{f} = \lambda[A,B] d\lambda $$ $$ \log f = \frac{\lambda^2}{2}[A,B] $$ $$ f = e^{\lambda A} e^{\lambda B} e^{-\lambda (A + B)} = e^{\frac{\lambda^2}{2}[A,B]} $$ $$ e^{\lambda A} e^{\lambda B} = e^{\lambda (A+B)} e^{\frac{\lambda^2}{2} [A,B]}$$ then take $\lambda = 1 $ to obtain

$$e^A e^B = e^{A + B + \frac{1}{2}[A,B]} $$

Now the hard part is obtaining the differential equation. $$f(\lambda) = e^{\lambda A} e^{\lambda B} e^{-\lambda (A + B)}$$ $$ \frac{df}{d\lambda} = Ae^{\lambda A}e^{\lambda B} e^{-\lambda (A+B)} + e^{\lambda A} Be^{\lambda B} e^{-\lambda (A+B)} - e^{\lambda A} e^{\lambda B} e^{-\lambda (A + B)} (A+B) $$

The best I could make from this mess was

$$ \frac{df}{d\lambda} = [A,f] + e^{\lambda A}e^{\lambda B} [B, e^{-\lambda (A+B)}] $$

It's really difficult for me to be able to use the assumptions $[A, [B, A]] = 0; [B, [B,A]] = 0$ nor make $\lambda$ come down the exponent in the differential equation.

Can you please give a hint on how to proceed with this problem? Any helpful hint will be extremely appreciated! Thank you in advance!


marked as duplicate by ACuriousMind, Kyle Kanos, Qmechanic Feb 12 '15 at 16:29

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First, I assume finite dimensional operators: otherwise you need to check certain boundedness conditions on the operators. Because the CBH series is here truncated by the vanishing double commutators, the conditions for linear operators on e.g. $\mathbf{L}^2(\mathbb{R})$ will be mild.

You need to practice operations with $\mathrm{Ad}$. Look up the following. In the Lie group $\mathfrak{G}$ with algebra $\mathfrak{g}$ the tangent vector to the path:

$$\sigma:\mathbb{R}\to\mathfrak{G};\;\sigma(\tau) = e^A\,e^{\tau\,B}\,e^{-A};\;A,\,B\in\mathfrak{g}\tag{1}$$

at the identity is $\mathrm{Ad}(e^A)\,B=\exp(\mathrm{ad}(A))\,B$. Here $\mathrm{Ad}:\mathfrak{G}\to GL(\mathfrak{g})$ is the Adjoint Representation. It is a Lie group homomorphism from the general Lie group $\mathfrak{G}$ to the matrix Lie group $GL(\mathfrak{g})$. Its kernel is the centre of $\mathfrak{G}$. Since it is a homomorphism, we have $\mathrm{Ad}(\gamma\,\zeta) = \mathrm{Ad}(\gamma)\,\mathrm{Ad}(\zeta);\,\forall \gamma,\,\zeta\in\mathfrak{G}$. Another useful identity is:

$$\begin{array}{lcl}\mathrm{Ad}(e^A)\,B &=& \exp(\mathrm{ad}(A))\,B \\&=&B + \mathrm{ad}(A) B + \frac{\mathrm{ad}(A)^2}{2!}\,B +\cdots \\&=& B+ [A,\,B] + \frac{1}{2!}\, [A,\,[A,\,B]] + \cdots\end{array}\tag{2}$$

and this series is universally convergent if the operator $B\mapsto[A,\,B]$ is suitably bounded (e.g. $\left\|[A,\,B]\right\| \leq K(A)\,\left\|B\right\|$ for some $K(A)\in\mathbb{R}$ - this is certainly true in finite dimensions).

Now, by (1) and the homomorphism property ($\mathrm{Ad}(e^{\lambda\,A}\,e^{\lambda\,B}) = \mathrm{Ad}(e^{\lambda\,A})\,\mathrm{Ad}(e^{\lambda\,B})$), you can find that:

$$\begin{array}{lcl}\mathrm{d}_\lambda f &=& A\,e^{\lambda\, A}\,e^{\lambda\,B}\,e^{-\lambda\,(A+B)} + e^{\lambda\, A}\,B\,e^{\lambda\,B}\,e^{-\lambda\,(A+B)} - e^{\lambda\, A}\,e^{\lambda\,B} \,(A+B)\,e^{-\lambda\,A+B)}\\ &=& \left(A + e^{\lambda\,A}\,B\,e^{-\lambda\,A} - e^{\lambda\,A}\,e^{\lambda\,B}\,(A+B)\,e^{-\lambda\,B}\,e^{-\lambda\,A}\right)\,e^{\lambda\, A}\,e^{\lambda\,B}\,e^{-\lambda\,(A+B)} \\ &=&\left(A+\mathrm{Ad}(e^{\lambda\,A})\left(B-\mathrm{Ad}(e^{\lambda\,B})\,(A+B)\right)\right)\,f\end{array}\tag{3}$$

All the above is perfectly general. You need to specialise it to your truncated case. So use the universally convergent (and here truncated to two terms) series (2) to expand $A+\mathrm{Ad}(e^{\lambda\,A})\left(B-\mathrm{Ad}(e^{\lambda\,B})\,(A+B)\right)$ and truncate it for your special case and I think you should make some headway.

A pedantic peeve: although both orders for the name are quite common, the order that accurately reflects the historical precedence is "Campbell-Baker-Hausdorff" as each of the authors made their contributions in 1897/1898 (Campbell), 1905 (Baker) and 1906 (Hausdorff), respectively. Each was aware of their forerunners' work, but, as stated in Fascicule 16 Ch 1 of Bourbaki (1960), "each found the demonstrations of his forerunners unconvincing(!)". That statement always makes me giggle and gives some comfort that I'm not the only one with about a 5% comprehension rate in reading technical literature (I reckon I need to read a paper about 20 times on average to "get" it). An amusing fact is that none of these three actually worked out the series. Instead, they established the theorem that the series was convergent within some neigbourhood of $\mathbf{0}$ in the Lie algebra and comprises linear and Lie bracket operations only. The formula itself is due to Dynkin and was fully worked out in 1947!

  • $\begingroup$ thank you so much for answering! I'll do my best to study your answer, despite my small introductory-level knowledge of lie groups and algebras. $\endgroup$ – quarkleptonboson Feb 12 '15 at 13:49
  • $\begingroup$ @quarkleptonboson I've added another step to Eq. (3) to help you out. Just think of all the operators as square $N\times N$ matrices and all the Lie brackets and multiplications then become concrete matrix multiplications. (2) is always a literal matrix power series, since the group of invertible linear transformations on $\mathfrak{g}$ is always a matrix group. $\endgroup$ – WetSavannaAnimal Feb 12 '15 at 20:42

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