1.5.4 The Baker-Campbell-Hausdorff Formula

When work­ing with ma­tri­ces, we cannot assume matrix prod­ucts com­mute, as is the case for many of the el­e­ments in the Lie al­ge­bra $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$. Re­gard­ing the matrix ex­po­nen­tial of el­e­ments in the set­ting of non-com­mu­ta­tive Lie al­ge­bras, we can state that the fol­low­ing,

rep­re­sent­ing com­mu­ta­tive ma­tri­ces, does not hold for non-com­mut­ing matrix groups

and in­stead, ex­press the group prod­uct, as

The Baker-Camp­bell-Haus­dorff the­o­rem states just this, for $Z$, as the sum of the Lie al­ge­bra el­e­ments $X,Y$, in ad­di­tion to a series of Lie brack­et terms de­fined by $X,Y$. So, for our Lie groups, the BCH for­mu­la proves that the Lie brack­ets of $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$ de­ter­mine the prod­uct op­er­a­tions for the Lie group. We define the ad­joint op­er­a­tor $ad_{X}(Y)$, as the Lie brack­et op­er­a­tion

and for a pos­i­tive in­te­ger $n$, nested com­mu­ta­tors can be de­scribed, as

The­o­rem (1.5.37) – We can now state and prove that¹

Proof – We define

and com­put­ing the Taylor series of $Y(t)$, around the point, $t = 0$, shows that $Y(0) = Y$, and

Higher de­riv­a­tives are simple to cal­cu­late and relate to nested ad­joint op­er­a­tors, by

By sub­sti­tut­ing $t = 1$ into the Taylor series, we have sat­is­fied the ini­tial state­ment $e^{X}Ye^{- X} = \exp\left( ad_{X} \right)(Y)$.

We define the fol­low­ing aux­il­iary func­tions and their power series

that sat­is­fy

to for­mal­ly state the Baker-Camp­bell-Haus­dorff for­mu­la

for $g(z)$, the aux­il­iary func­tion we just de­fined, and the BCH for­mu­la for $\ln\left( e^{X}e^{Y} \right)$ will con­verge pro­vid­ed that $\left\| e^{X}e^{Y} - \epsilon \right\| < 1$, for $\epsilon$, the iden­ti­ty matrix and $\left\| \ldots \right\|$, the matrix norm. We can now expand the BCH for­mu­la using the Taylor series de­f­i­n­i­tion of $g(z)$, to show that

holds for con­ver­gent series. To demon­strate this con­cep­tu­al­ly, we use the fol­low­ing two simple el­e­ments $X,Y$

and com­pute the fol­low­ing matrix ex­po­nen­tials $\in G$

For $e^{X + Y}$, the matrix log­a­rithm simply equals the sum of the orig­i­nal el­e­ments $X,Y$

and we in­ci­den­tal­ly see that $e^{Z}$ equals the Lie al­ge­bra $- Y_{3} \in \mathfrak{s}_{3\mathbb{C}}^{\Delta}.$To find the tan­gent space for $e^{Z}$, we take ad­van­tage of the fact that these ma­tri­ces are di­ag­o­nal­iz­able, ex­press­ing $\ln\left( e^{Z} \right)$, as

for $V$, a uni­tary square matrix of the eigen­vec­tors of $e^{Z}$, and $D$, the nat­ur­al log of the di­ag­o­nal matrix of eigen­val­ues of $e^{Z}$. Cal­cu­lat­ing $D,V$, and $V^{- 1}$, for $e^{Z}$, gives us the fol­low­ing ma­tri­ces

and the matrix log of $D$, equals

The matrix log­a­rithm cal­cu­la­tion be­comes

and to verify that $e^{\ln\left( e^{Z} \right)}$ does indeed equal our ini­tial result, we find

As we have shown, the Baker-Camp­bell-Haus­dorff for­mu­la is the series of Lie brack­et op­er­a­tions of the form

in which the por­tion of the series $X + Y$, is the com­mu­ni­tive al­ge­bra part of $Z$, while the re­main­der is its non-com­mu­ni­tive part. From what we cal­cu­lat­ed in our ex­am­ple, we can see that the non-com­mu­ni­tive por­tion of $Z$ is

and this can be ex­pressed, as

With a basic un­der­stand­ing of the BCH for­mu­la, we can apply it to 2 el­e­ments in the $\mathfrak{s}_{3\mathbb{C}}^{\Delta}$ Lie al­ge­bra for $X_{1}{,X}_{2} \in U(3)$, de­fined

We can de­ter­mine the matrix ex­po­nen­tial $e^{X_{1} + X_{2}}$ in a sim­i­lar way to what we did for the $U(3)$ group, by using two vari­ables, $x_{1},x_{2}$, in­stead of the three vari­ables. The group form can be trigono­met­ri­cal­ly rep­re­sent­ed as

for $r = \left( \sqrt{x_{1}^{2} - x_{1}x_{2} + x_{2}^{2}} \right),\ \ \gamma = \left( x_{1} + x_{2} \right),\ \ \frac{\partial r}{\partial x_{1}} = \left( \frac{x_{1} - \frac{x_{2}}{2}}{r} \right),\ \ \frac{\partial r}{\partial x_{2}} = \left( \frac{x_{2} - \frac{x_{1}}{2}}{r} \right).$

For the $X_{1},X_{2}$, el­e­ments, we would have that $x_{1} = i$, $x_{2} = i$, and the ex­po­nen­tial map­ping $e^{X_{1} + X_{2}}$ is

For the result of the prod­uct $e^{X_{1}}e^{X_{2}} = e^{Z}$, having the trigono­met­ric form

where once again, $x_{1} = 1$ and $x_{2} = 1$. As $e^{Z}$ is a di­ag­o­nal­iz­able matrix, $e^{Z}$ can be ex­pressed, as

where once again for the $3 \times 3$ square ma­tri­ces, the $V$ matrix rep­re­sents the $e^{Z}$ eigen­vec­tors, and the $D$ matrix, the di­ag­o­nal­ized $e^{Z}$ eigen­val­ues. For the sake of com­plete­ness, any form $e^{X_{1}}e^{X_{2}} = e^{Z},$for $X_{1},X_{2}, \in \mathfrak{u}(3),$where $\mathfrak{u}(3)$ rep­re­sents Lie al­ge­bra el­e­ments of $U(3)$, the eigen­vec­tor and eigen­val­ues ma­tri­ces take the form

for eigen­val­ues $\lambda_{k}$:

and $V$, a matrix com­posed of the eigen­vec­tor columns:

Our ex­am­ple sim­pli­fies since $x_{1} = x_{2} = 1,$and the ma­tri­ces $D$ and $V$ become

Nu­mer­i­cal­ly the ma­tri­ces above ap­prox­i­mate to

and by cal­cu­lat­ing matrix log­a­rithm of the di­ag­o­nal­ized matrix $D$, as simply the log­a­rithms of the in­di­vid­ual en­tries on the main di­ag­o­nal, the result for $\ln(D)$, is

Solv­ing for$\ {\ln(e}^{X_{1}}e^{X_{2}}) = Z$

gives us the $Z$ matrix for $e^{Z}$ $\in U(3) \subset S_{3\mathbb{C}}^{\Delta}$, rep­re­sent­ing the same result the Baker-Camp­bell-Haus­dorff for­mu­la would yield.

To sum­ma­rize, the Baker-Camp­bell-Haus­dorff for­mu­la can be con­sid­ered as fol­lows²: the el­e­ments $X_{1},X_{2} \in {\mathfrak{u}(3)\subset \ \mathfrak{s}}_{3\mathbb{C}}^{\Delta}$ are prim­i­tive; their matrix ex­po­nen­tials $e^{X_{1}},e^{X_{2}}$ are grou­p­like, as is their prod­uct $e^{X_{1}}e^{X_{2}}$; so, the matrix log­a­rithm of $e^{X_{1}}e^{X_{2}}$ is prim­i­tive and can be writ­ten as the in­fi­nite sum of Lie brack­et el­e­ments gen­er­at­ed by $X_{1},X_{2}.$

1. Haber, Howard – "Notes on the Matrix Ex­po­nen­tial and Log­a­rithm", 2019 http://scipp.ucsc.edu/~haber/web­page/Ma­trix­Ex­pLog.pdf
2. Nathan, Ja­cob­son,* "Lie Al­ge­bras", John Wiley & Sons, 1966*