# Higher-Order Terms in BCH Formula?

62 views

### stran...@yahoo.com.au

Jan 16, 2006, 5:16:59 PM1/16/06
to
I'm trying to work out some of the higher-order
terms in the BCH (Baker-Campbell-Hausdorf) formula:

exp(X)exp(Y) = exp(C1 + C2 + C3 + C4 + ...)

Most textbooks only give the C's to 3rd-order, i.e:

where C1 = (X+Y)
C2 = 1/2[X,Y]
C3 = 1/12[X-Y,[X+Y]]

In Varadarajan's "Lie Groups, Lie Algebras & their
Representations", there appears a general recusion
formula for the C_n on p118 (Lemma 2.15.3):

(n+1) C_n+1 = 1/2[X-Y, C_n]
+ SUM1(p) K_2p SUM2(k1..k_2p)
[C_k1,[C_k2,...[c_k2p, X+Y]...]

where SUM1(p) is over p \ge 1 and 2p \le n
SUM2(k's) is over k1,...k_2p > 0
and k1+k2+...k_2p = n

K_2p = B_2p/(2p)!

where B_n is the n'th Bernoulli number.

Using this, Varadarajan's eq(2.15.17) gives C4 as:

C4 = -1/48([Y,[X,[X,Y]]] + [[X,[Y,[X,Y]])

I tried to verify this expression for C4, but I
got something different:

C4 = 1/96[X-Y,[X-Y,[X,Y]]]

which, if you expand it, becomes:

C4 = 1/96([X,[X,[X,Y]] + [Y,[Y,[Y,X]]
- [Y,[X,[X,Y]]] - [[X,[Y,[X,Y])

My terms with 3 X's and 3 Y's are absent from
Varadarajan's expression. (According to me, the
SUM term in the general expression is zero in
this case, so perhaps I could be wrong there?)

Has anyone else out there checked Varadarajan's
calculation? Or does anyone know of an alternate
text that gives the general BCH formula against
which I could compare the above?

valid only for finite-dimensional representations,
with X,Y being bounded operators. Is there a
generalized version of the formula which is valid
for unbounded operators on inf-dim spaces? Is it
simply a matter of requiring that X,Y,(X+Y) and
all the commutators share a common domain, or is
there more to it than that?

- MikeM.

### Aaron Bergman

Jan 17, 2006, 7:39:31 AM1/17/06
to
stran...@yahoo.com.au wrote:

> Has anyone else out there checked Varadarajan's
> calculation? Or does anyone know of an alternate
> text that gives the general BCH formula against
> which I could compare the above?

Aaron

### Arnold Neumaier

Jan 17, 2006, 1:11:09 PM1/17/06
to

stran...@yahoo.com.au wrote:
> I'm trying to work out some of the higher-order
> terms in the BCH (Baker-Campbell-Hausdorf) formula:
>
> exp(X)exp(Y) = exp(C1 + C2 + C3 + C4 + ...)
>
> Most textbooks only give the C's to 3rd-order, i.e:
>
> where C1 = (X+Y)
> C2 = 1/2[X,Y]
> C3 = 1/12[X-Y,[X+Y]]
>
> In Varadarajan's "Lie Groups, Lie Algebras & their
> Representations", there appears a general recusion
> formula for the C_n on p118 (Lemma 2.15.3):
>
> (n+1) C_n+1 = 1/2[X-Y, C_n]
> + SUM1(p) K_2p SUM2(k1..k_2p)
> [C_k1,[C_k2,...[c_k2p, X+Y]...]
>
> where SUM1(p) is over p \ge 1 and 2p \le n
> SUM2(k's) is over k1,...k_2p > 0
> and k1+k2+...k_2p = n
>
> K_2p = B_2p/(2p)!
>
> where B_n is the n'th Bernoulli number.
>
> Using this, Varadarajan's eq(2.15.17) gives C4 as:
>
> C4 = -1/48([Y,[X,[X,Y]]] + [[X,[Y,[X,Y]])

In the most recent issue of SIAM Review (Vol. 47, No. 4, December 2005),
epubs.siam.org/sam-bin/getfile/SIREV/articles/41042.pdf
there is an article about a Lie algebra software package LTP,

http://www.cim.mcgill.ca/~migueltt/ltp/ltp.html

The BCH formula is given (in the SIAM article) as an example,
equation (2.3), and the expression given matches V's
expression.

Arnold Neumaier

### Arnold Neumaier

Jan 17, 2006, 1:19:40 PM1/17/06
to

stran...@yahoo.com.au wrote:
> I'm trying to work out some of the higher-order
> terms in the BCH (Baker-Campbell-Hausdorf) formula:
>
> exp(X)exp(Y) = exp(C1 + C2 + C3 + C4 + ...)
>
> Most textbooks only give the C's to 3rd-order, i.e:
>
> where C1 = (X+Y)
> C2 = 1/2[X,Y]
> C3 = 1/12[X-Y,[X+Y]]
>
> In Varadarajan's "Lie Groups, Lie Algebras & their
> Representations", there appears a general recusion
> formula for the C_n on p118 (Lemma 2.15.3):
>
> (n+1) C_n+1 = 1/2[X-Y, C_n]
> + SUM1(p) K_2p SUM2(k1..k_2p)
> [C_k1,[C_k2,...[c_k2p, X+Y]...]
>
> where SUM1(p) is over p \ge 1 and 2p \le n
> SUM2(k's) is over k1,...k_2p > 0
> and k1+k2+...k_2p = n
>
> K_2p = B_2p/(2p)!
>
> where B_n is the n'th Bernoulli number.
>
> Using this, Varadarajan's eq(2.15.17) gives C4 as:
>
> C4 = -1/48([Y,[X,[X,Y]]] + [[X,[Y,[X,Y]])

### Arnold Neumaier

Jan 17, 2006, 9:45:03 PM1/17/06
to

stran...@yahoo.com.au wrote:
> I'm trying to work out some of the higher-order
> terms in the BCH (Baker-Campbell-Hausdorf) formula:
>
> exp(X)exp(Y) = exp(C1 + C2 + C3 + C4 + ...)
>
> Most textbooks only give the C's to 3rd-order, i.e:
>
> where C1 = (X+Y)
> C2 = 1/2[X,Y]
> C3 = 1/12[X-Y,[X+Y]]
>
> In Varadarajan's "Lie Groups, Lie Algebras & their
> Representations", there appears a general recusion
> formula for the C_n on p118 (Lemma 2.15.3):
>
> (n+1) C_n+1 = 1/2[X-Y, C_n]
> + SUM1(p) K_2p SUM2(k1..k_2p)
> [C_k1,[C_k2,...[c_k2p, X+Y]...]
>
> where SUM1(p) is over p \ge 1 and 2p \le n
> SUM2(k's) is over k1,...k_2p > 0
> and k1+k2+...k_2p = n
>
> K_2p = B_2p/(2p)!
>
> where B_n is the n'th Bernoulli number.
>
> Using this, Varadarajan's eq(2.15.17) gives C4 as:
>
> C4 = -1/48([Y,[X,[X,Y]]] + [[X,[Y,[X,Y]])

### Eugene Stefanovich

Jan 18, 2006, 1:56:48 PM1/18/06
to
stran...@yahoo.com.au wrote:
> I'm trying to work out some of the higher-order
> terms in the BCH (Baker-Campbell-Hausdorf) formula:
>
> exp(X)exp(Y) = exp(C1 + C2 + C3 + C4 + ...)
>
> Most textbooks only give the C's to 3rd-order, i.e:
>
> where C1 = (X+Y)
> C2 = 1/2[X,Y]
> C3 = 1/12[X-Y,[X+Y]]
[...]

> Has anyone else out there checked Varadarajan's
> calculation? Or does anyone know of an alternate
> text that gives the general BCH formula against
> which I could compare the above?

C = X + Y + 1/2[X,Y] + 1/12[[X,Y],Y] + 1/12[[Y,X],X]
+ 1/24 [[[Y,X],X],Y] - 1/720 [[[[X,Y],Y],Y],Y]
+ 1/360 [[[[X,Y],Y],Y],X] + 1/360 [[[[Y,X],X],X],Y]
- 1/120 [[[[X,Y],Y],X],Y] - 1/120 [[[[Y,X],X],Y],X]...

G. H. Weiss and A. A. Maradudin, The Baker-Hausdorff
formula and a problem in crystal physics, J. Math. Phys. 3 (1962),
771.

Eugene.

### Dave Elliott

Jan 18, 2006, 1:57:32 PM1/18/06
to
The first appendix of "Lie Groups, Convex Cones and
Semigroups" by Hilgert, Hofmann and Lawson
QA 387 .H535 1989
has Dynkin's formulas.
CBH converges for completely normable real (or complex) Lie algebras
for which the bracket is
continuous.
There are also papers in the math--physics literature

### José Carlos Santos

Jan 21, 2006, 5:52:33 AM1/21/06
to
On 16-01-2006 22:16, stran...@yahoo.com.au wrote:

> I'm trying to work out some of the higher-order
> terms in the BCH (Baker-Campbell-Hausdorf) formula:
>
> exp(X)exp(Y) = exp(C1 + C2 + C3 + C4 + ...)
>
> Most textbooks only give the C's to 3rd-order, i.e:
>
> where C1 = (X+Y)
> C2 = 1/2[X,Y]
> C3 = 1/12[X-Y,[X+Y]]

Indeed. One exception is Roger Godement's "Introduction à la théorie des
groupes de Lie" (Springer-Verlag, 2004) and he provides an interesting
explanation of why is it true that that "[m]ost textbooks only give the
C's to 3rd-order".

Best regards,

Jose Carlos Santos

### Arnold Neumaier

Jan 22, 2006, 1:30:10 PM1/22/06
to

Why don't you give this explanation for those who don't have the book?

Arnold Neumaier

### José Carlos Santos

Jan 22, 2006, 5:00:14 PM1/22/06
to
Arnold Neumaier wrote:

>>> I'm trying to work out some of the higher-order
>>> terms in the BCH (Baker-Campbell-Hausdorf) formula:
>>>
>>> exp(X)exp(Y) = exp(C1 + C2 + C3 + C4 + ...)
>>>
>>> Most textbooks only give the C's to 3rd-order, i.e:
>>>
>>> where C1 = (X+Y)
>>> C2 = 1/2[X,Y]
>>> C3 = 1/12[X-Y,[X+Y]]
>>
>> Indeed. One exception is Roger Godement's "Introduction à la théorie des
>> groupes de Lie" (Springer-Verlag, 2004) and he provides an interesting
>> explanation of why is it true that that "[m]ost textbooks only give the
>> C's to 3rd-order".
>
> Why don't you give this explanation for those who don't have the book?

First of all, those who have the book will find it at section 6.6.

Godement explicitly performs the calculation of those terms whose total
degree does no exceed 4. Before he starts simplifying, this takes nine
lines, with seven or eight terms in each. He also writes (I'm
translating) that someone at the beginning of the [twentieth] century
had the charity of performing these calculations in full, thereby
allowing other authors to leave it as "notes" or "exercises".

### Arnold Neumaier

Jan 22, 2006, 5:01:50 PM1/22/06
to

Why don't you give this explanation for those who don't have the book?

Arnold Neumaier

### koso...@escort.lviv.net

Feb 2, 2006, 1:33:18 PM2/2/06
to
Elementary explanation and more general approach to BCH and related
Formulaes, which treats "time-dependent" operators as well, you can
find at

http://arxiv.org/abs/math-ph/0409035

Yu.N. Kosovtsov