9
$\begingroup$

The Baker-Campbell-Hausdorff problem is to obtain $\log(e^{X}e^{Y})$ where $X,Y$ are appropriate operators. The Dynkin series $$\log(e^{tX}e^{tY})=t(X+Y)+\frac{t^2}{2}[X,Y]+o(t^3)$$ gives an expansion in powers of a perturbation parameter $t$, with higher coefficients expressed in terms of successively more complicated commutators of $X,Y$. The parameter $t$ has been introduced to control the convergence of the series; so long as the operators $X,Y$ are bounded the Dynkin series has a finite radius of convergence (theorem 5 of Suzuki 1997). A more involved approach would be to replace $e^{t X}\to e^{s X}$; the appropriate Dynkin series would then presumably be in powers of $s,t$ and have finite radius of convergence in each variable.

Now suppose we instead consider the product $e^X e^{tY}$ for $t\gg 1$. My intuition had been that there would be an asymptotic series $$\log(e^{X} e^{tY})=tY+X+o(t^{-1}).$$ But this is clearly not the standard Dynkin series: powers of $t$ now only arise from the operator $Y$, and moreover $t=\infty$ will always fall outside the finite radius of convergence. Hence my question:

Is there a known asymptotic series for $\log(e^Xe^{t Y})$ at large $t$?

References: - Suzuki, M. Commun.Math. Phys. (1977) 57: 193. doi:10.1007/BF01614161.

Improve this question
$\endgroup$
8
  • 7
    $\begingroup$ If $X,Y$ generate a nilpotent Lie algebra, then the Dynkin series terminates after finitely many terms and is an exact formula. In particular, in this case $\log(e^X e^{tY})$ can certainly contain terms of order $t^2$ or greater in the large $t$ limit. $\endgroup$
    – Terry Tao
    Commented Mar 29, 2017 at 17:01
  • 1
    $\begingroup$ So the behavior at infinity will be sensitive to the algebraic structure. That's a bit frustrating but in retrospect unsurprising given how BCH simplifies when $[X,Y]$ is central in the subgroup. In my specific case, I actually do know that $[X,[X,Y]]$ is central; beyond that, though, I'd have to specialize further to obtain higher commutators. So I may not yet have enough info to get the asymptotic behavior.. $\endgroup$
    – Semiclassical
    Commented Mar 29, 2017 at 17:32
  • 1
    $\begingroup$ But what can we say if we recall the majorization $\log\sigma(AB)\prec \log\sigma(A) + \log\sigma(B)$, which implies that $\log\sigma(e^Xe^{tY}) \prec \log\sigma(e^X)+|t|\log\sigma(e^Y)$? (here $\sigma$ denotes the singular value map) $\endgroup$
    – Suvrit
    Commented Mar 30, 2017 at 17:14
  • 2
    $\begingroup$ I think the first question you should ask is, what is $\lim_{t\to \infty} e^{t Y}$?. The answer clearly depends strongly on $Y$. For example if all eigenvalues of $Y$ have negative real part the limit is zero. What can you say about $Y$? does it, for example, generate a contraction semigroup? (then answers are known) $\endgroup$
    – lcv
    Commented Mar 30, 2017 at 20:23
  • 1
    $\begingroup$ In other words, not only the leading term may not be $t Y$, as @Carlo Beenakker 's answer shows, but it may actually be $O(1)$ (or not exist). $\endgroup$
    – lcv
    Commented Mar 30, 2017 at 20:33

1 Answer 1

Reset to default
3
$\begingroup$

It may be helpful to consider an example that can be solved exactly; Using the Special-case closed form of the Baker-Campbell-Hausdorff formula one finds that if the commutator $[X,Y]$ evaluates to $$[X,Y]=uX +vY +cI$$ then the desired logarithm of the product of matrix exponentials equals $$\log(e^X e^{tY})=tY+X+f(u,v,t)(uX +vY +cI)$$ $$f(u,v,t)=\frac{(ut-v)e^{ut+v}-ute^{ut}+ve^{v}}{uv(e^{ut}-e^{v})}$$ The large-$t$ limit can now be read off once the sign of $u$ is known: $$\lim_{t\rightarrow\infty}f(u,v,t)=\begin{cases} -1/u&\text{if}\;u<0\\ (t/v)(e^v-1)&\text{if}\;u>0\\ (t/v)[v-1+v(e^v-1)^{-1}]&\text{if}\;u=0 \end{cases}$$ There are no terms greater than order $t$ in the large-$t$ limit for this class of commutators. (I'm actually a bit puzzled how $t^2$ terms and higher might appear at all, an explicit example would help me a lot.)

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ @TerryTao's comment explains how $t^2$ and higher terms can appear, no? Namely, if, say, $X = E_{12} + E_{23}$ and $Y = E_{23} + E_{34}$ are $4 \times 4$ matrices, then, assuming my Mathematica scribbling hasn't gone awry, $e^X e^{t Y}$ is (exactly) the exponential of $X + t Y + \frac1 2[X, tY] + \frac1{12}([X, [X, tY]] + [[X, tY], tY])$. $\endgroup$
    – LSpice
    Commented Mar 30, 2017 at 16:57
  • $\begingroup$ As a physically relevant example, the displacement operator in coherent states satisfies the identity $$\hat{D}(\alpha)=e^{\alpha \hat{a}^\dagger-\alpha^\star \hat{a}}=e^{-|\alpha|^2/2}e^{-\alpha \hat{a}^\dagger}e^{-\alpha^\star \hat{a}}$$ where $[\hat{a},\hat{a}^\dagger]=1$ as typical for ladder operators. Specializing to real $\alpha$ then gives an example with terms of quadratic order. $\endgroup$
    – Semiclassical
    Commented Mar 30, 2017 at 22:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .