n*s
发帖数: 752 | 1 is there a way to approximate
e^(A+B)
where A and B do not commute and B is much smaller than A
thx very much |
v**g
发帖数: 20 | 2 According to the Zassenhaus formula,
e^(A+B)=e^A e^B e^{-1/2[A,B]} ...
are you looking for something like this?
actually this formula can be found in almost any book on advanced quantum
mechanics, I think.
see also:
http://en.wikipedia.org/wiki/Baker-Campbell-Hausdorff_formula |
h****r
发帖数: 391 | 3 Matrix Mathematics
by Dennis S. Bernstein
Princeton Univ. Press
2005
这本书里有好几种计算matrix exponential的方法
【在 n*s 的大作中提到】
: is there a way to approximate
: e^(A+B)
: where A and B do not commute and B is much smaller than A
: thx very much
|
g****t
发帖数: 31659 | 4 这本书好象几百页的索引，晕死。
Matrix Mathematics
by Dennis S. Bernstein
Princeton Univ. Press
2005
这本书里有好几种计算matrix exponential的方法
【在 h****r 的大作中提到】
: Matrix Mathematics
: by Dennis S. Bernstein
: Princeton Univ. Press
: 2005
: 这本书里有好几种计算matrix exponential的方法
|
c*******h
发帖数: 1096 | 5 If A is diagonalizable, let A=UDU^{-1} where U and D contain eigenvectors
and eigenvalues. Then exp(A)=exp(UDU^{-1})=U*exp(D)*U^{-1}.
If B is very small w.r.t. A, consider it a perturbation to A. Then
sensitivity
theorems on eigenvectors and eigenvalues of A can be applied under some
conditions. All what says is that differences between eigenvectors/
eigenvalues
of A and those of A+B can be bounded. Hence the difference between exp(A)
and exp(A+B) can be bounded.
In other words, using exp(A) to
【在 n*s 的大作中提到】
: is there a way to approximate
: e^(A+B)
: where A and B do not commute and B is much smaller than A
: thx very much
|
D*******a
发帖数: 3688 | 6 我记得有个矩阵同时对角化的定理，可能可以用
【在 c*******h 的大作中提到】
: If A is diagonalizable, let A=UDU^{-1} where U and D contain eigenvectors
: and eigenvalues. Then exp(A)=exp(UDU^{-1})=U*exp(D)*U^{-1}.
: If B is very small w.r.t. A, consider it a perturbation to A. Then
: sensitivity
: theorems on eigenvectors and eigenvalues of A can be applied under some
: conditions. All what says is that differences between eigenvectors/
: eigenvalues
: of A and those of A+B can be bounded. Hence the difference between exp(A)
: and exp(A+B) can be bounded.
: In other words, using exp(A) to
|
c*******h
发帖数: 1096 | 7 that requires A and B commute
【在 D*******a 的大作中提到】
: 我记得有个矩阵同时对角化的定理，可能可以用
|
n*s
发帖数: 752 | 8 i know this formula
the problem is that it seems to me that
e^A e^B e^{-1/2[A,B]} is not close to e^(A+B)
for me, (A+B) = (P+Q)t
and t could be big, i only have Q << P
【在 v**g 的大作中提到】
: According to the Zassenhaus formula,
: e^(A+B)=e^A e^B e^{-1/2[A,B]} ...
: are you looking for something like this?
: actually this formula can be found in almost any book on advanced quantum
: mechanics, I think.
: see also:
: http://en.wikipedia.org/wiki/Baker-Campbell-Hausdorff_formula
|
n*s
发帖数: 752 | 9 exp(A) is my first trial
a first order contribution is missing in this approximation
【在 c*******h 的大作中提到】
: If A is diagonalizable, let A=UDU^{-1} where U and D contain eigenvectors
: and eigenvalues. Then exp(A)=exp(UDU^{-1})=U*exp(D)*U^{-1}.
: If B is very small w.r.t. A, consider it a perturbation to A. Then
: sensitivity
: theorems on eigenvectors and eigenvalues of A can be applied under some
: conditions. All what says is that differences between eigenvectors/
: eigenvalues
: of A and those of A+B can be bounded. Hence the difference between exp(A)
: and exp(A+B) can be bounded.
: In other words, using exp(A) to
|
c*******h
发帖数: 1096 | 10 if you want the first order term, a natural way is using (I+B)*exp(A)
or exp(A)*(I+B), or even exp(B)*exp(A) or exp(A)*exp(B).
you still can derive some bounds by help of some eigenvalue sensitivity
results.
【在 n*s 的大作中提到】
: exp(A) is my first trial
: a first order contribution is missing in this approximation
|
n*s
发帖数: 752 | 11 i don't find it natual though,
exp(A+B)=exp(A)exp(B)exp(-[A,B]/2)...
all those exponentials have 1st order term in B
even keeping quite a few those expoentials, the RHS deviate the LHS quite
violently as I turn the t parameter big
【在 c*******h 的大作中提到】
: if you want the first order term, a natural way is using (I+B)*exp(A)
: or exp(A)*(I+B), or even exp(B)*exp(A) or exp(A)*exp(B).
: you still can derive some bounds by help of some eigenvalue sensitivity
: results.
|
c*******h
发帖数: 1096 | 12 that may be possible. consider the case when A and B are scalars.
the difference between exp(A+B) and (1+B)exp(A) w.r.t. exp(A)
grows exponentially as the magnitude of B grows, even if the ratio
of A and B is kept fixed.
the fact is that first order approximation to exponentials is far
below satisfatory unless the magnitude of the variable is small.
【在 n*s 的大作中提到】
: i don't find it natual though,
: exp(A+B)=exp(A)exp(B)exp(-[A,B]/2)...
: all those exponentials have 1st order term in B
: even keeping quite a few those expoentials, the RHS deviate the LHS quite
: violently as I turn the t parameter big
|
