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Jan 16, 2006, 5:16:59 PM1/16/06

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I'm trying to work out some of the higher-order

terms in the BCH (Baker-Campbell-Hausdorf) formula:

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?

Another question: Varadarajan's exposition appears

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?

Thanks in advance.

- MikeM.

Jan 17, 2006, 7:39:31 AM1/17/06

to

In article <1137389208....@g49g2000cwa.googlegroups.com>,

stran...@yahoo.com.au wrote:

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?

<http://www.math.niu.edu/~rusin/known-math/00_incoming/BCH>

Aaron

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,

The package runs under Maple and can be downloaded from

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

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]])

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]])

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]]

[...]> 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.

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

about CBH for operators.

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

about CBH for operators.

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

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

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".

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

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

Formulaes, which treats "time-dependent" operators as well, you can

find at

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

Yu.N. Kosovtsov

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