Mixing
Taylor series for $e^i(H+varepsilon A)$
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$$e^i(H+varepsilon A) = e^iH(I+i varepsilon A + o(varepsilon))$$
Is it correct?
$H$ and $A$ are hermitian matrix and $[H,A]neq0$. If the one I gave is not the exact solution how do I expand it in series?
matrices
edited Aug 6 at 15:56
Davide Morgante
1,942220
asked Aug 6 at 15:07
user582108
85
 |Â
show 1 more comment
up vote
1
down vote
favorite
$$e^i(H+varepsilon A) = e^iH(I+i varepsilon A + o(varepsilon))$$
Is it correct?
$H$ and $A$ are hermitian matrix and $[H,A]neq0$. If the one I gave is not the exact solution how do I expand it in series?
matrices
edited Aug 6 at 15:56
Davide Morgante
1,942220
asked Aug 6 at 15:07
user582108
85
1
Do $H$ & $A$ commute? If not then no...
– copper.hat
Aug 6 at 15:10
$H$ & $A$ don't commute
– user582108
Aug 6 at 15:15
1
This might involve the Baker-Campbell-Hausdorff formula.
– Luke
Aug 6 at 15:26
1
Use the power-series representation of $e^X$ with $X = A + epsilon B$ and use that $(A+epsilon B)^n = A^n + epsilon (A^n-1B + ldots + B A^n-1) + mathcalo(epsilon)$ to get a series which will be something like $e^A+epsilon B = e^A(1 + epsilon B) + epsilon (ldots)$ (with the last term being compilated and containing all sorts of commutators). Will not be pretty.
– Winther
Aug 6 at 15:32
2
We can check that $$ e^X+epsilon Y = e^X left( 1 + epsilon int_0^1 e^-tX Y e^tX , dt + mathcalO(epsilon^2) right). $$
– Sangchul Lee
Aug 6 at 15:41
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$e^i(H+varepsilon A) = e^iH(I+i varepsilon A + o(varepsilon))$$
Is it correct?
$H$ and $A$ are hermitian matrix and $[H,A]neq0$. If the one I gave is not the exact solution how do I expand it in series?
matrices
edited Aug 6 at 15:56
Davide Morgante
1,942220
asked Aug 6 at 15:07
user582108
85
$$e^i(H+varepsilon A) = e^iH(I+i varepsilon A + o(varepsilon))$$
Is it correct?
$H$ and $A$ are hermitian matrix and $[H,A]neq0$. If the one I gave is not the exact solution how do I expand it in series?
matrices
edited Aug 6 at 15:56
Davide Morgante
1,942220
asked Aug 6 at 15:07
user582108
85
edited Aug 6 at 15:56
Davide Morgante
1,942220
edited Aug 6 at 15:56
Davide Morgante
1,942220
edited Aug 6 at 15:56
Davide Morgante
1,942220
1,942220
asked Aug 6 at 15:07
user582108
85
asked Aug 6 at 15:07
user582108
85
asked Aug 6 at 15:07
user582108
85
85
1
Do $H$ & $A$ commute? If not then no...
– copper.hat
Aug 6 at 15:10
$H$ & $A$ don't commute
– user582108
Aug 6 at 15:15
1
This might involve the Baker-Campbell-Hausdorff formula.
– Luke
Aug 6 at 15:26
1
Use the power-series representation of $e^X$ with $X = A + epsilon B$ and use that $(A+epsilon B)^n = A^n + epsilon (A^n-1B + ldots + B A^n-1) + mathcalo(epsilon)$ to get a series which will be something like $e^A+epsilon B = e^A(1 + epsilon B) + epsilon (ldots)$ (with the last term being compilated and containing all sorts of commutators). Will not be pretty.
– Winther
Aug 6 at 15:32
2
We can check that $$ e^X+epsilon Y = e^X left( 1 + epsilon int_0^1 e^-tX Y e^tX , dt + mathcalO(epsilon^2) right). $$
– Sangchul Lee
Aug 6 at 15:41
 |Â
show 1 more comment
1
Do $H$ & $A$ commute? If not then no...
– copper.hat
Aug 6 at 15:10
$H$ & $A$ don't commute
– user582108
Aug 6 at 15:15
1
This might involve the Baker-Campbell-Hausdorff formula.
– Luke
Aug 6 at 15:26
1
Use the power-series representation of $e^X$ with $X = A + epsilon B$ and use that $(A+epsilon B)^n = A^n + epsilon (A^n-1B + ldots + B A^n-1) + mathcalo(epsilon)$ to get a series which will be something like $e^A+epsilon B = e^A(1 + epsilon B) + epsilon (ldots)$ (with the last term being compilated and containing all sorts of commutators). Will not be pretty.
– Winther
Aug 6 at 15:32
2
We can check that $$ e^X+epsilon Y = e^X left( 1 + epsilon int_0^1 e^-tX Y e^tX , dt + mathcalO(epsilon^2) right). $$
– Sangchul Lee
Aug 6 at 15:41
1
1
Do $H$ & $A$ commute? If not then no...
– copper.hat
Aug 6 at 15:10
Do $H$ & $A$ commute? If not then no...
– copper.hat
Aug 6 at 15:10
$H$ & $A$ don't commute
– user582108
Aug 6 at 15:15
$H$ & $A$ don't commute
– user582108
Aug 6 at 15:15
1
1
This might involve the Baker-Campbell-Hausdorff formula.
– Luke
Aug 6 at 15:26
This might involve the Baker-Campbell-Hausdorff formula.
– Luke
Aug 6 at 15:26
1
1
Use the power-series representation of $e^X$ with $X = A + epsilon B$ and use that $(A+epsilon B)^n = A^n + epsilon (A^n-1B + ldots + B A^n-1) + mathcalo(epsilon)$ to get a series which will be something like $e^A+epsilon B = e^A(1 + epsilon B) + epsilon (ldots)$ (with the last term being compilated and containing all sorts of commutators). Will not be pretty.
– Winther
Aug 6 at 15:32
Use the power-series representation of $e^X$ with $X = A + epsilon B$ and use that $(A+epsilon B)^n = A^n + epsilon (A^n-1B + ldots + B A^n-1) + mathcalo(epsilon)$ to get a series which will be something like $e^A+epsilon B = e^A(1 + epsilon B) + epsilon (ldots)$ (with the last term being compilated and containing all sorts of commutators). Will not be pretty.
– Winther
Aug 6 at 15:32
2
2
We can check that $$ e^X+epsilon Y = e^X left( 1 + epsilon int_0^1 e^-tX Y e^tX , dt + mathcalO(epsilon^2) right). $$
– Sangchul Lee
Aug 6 at 15:41
We can check that $$ e^X+epsilon Y = e^X left( 1 + epsilon int_0^1 e^-tX Y e^tX , dt + mathcalO(epsilon^2) right). $$
– Sangchul Lee
Aug 6 at 15:41
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Here is a simple trick to expand OP's matrix exponential. Write $X = iH$ and $Y = iepsilon A$ for brevity. Next, introduce the following function
$$ f(t) = e^-tXe^t(X+Y), qquad a(t) = e^-tXYe^tX. $$
Then it is easy to check that
$$ f'(t) = e^-tXYe^t(X+Y) = a(t)f(t). $$
So it follows that $f$ is the ordered exponential of $a$. In particular,
$$ f(1) = 1 + int_0^1 a(t_1) , dt_1 + int_0^1int_0^t_1 a(t_1)a(t_2) , dt_2 dt_1 + cdots $$
From the estimate $left| int_0^1cdotsint_0^t_k-1 a(t_1)cdots a(t_k) , dt_k cdots dt_1 right| leq frac^kk!$, we realize that the formula above correctly produces the expansion of $e^X+Y$ as perturbation of $e^X$. Plugging the substitution back, we obtain
$$ e^i(H+epsilon A) = e^iH left( 1 + iepsilon int_0^1 e^-itHAe^itH , dt + mathcalO(epsilon^2) right). $$
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Here is a simple trick to expand OP's matrix exponential. Write $X = iH$ and $Y = iepsilon A$ for brevity. Next, introduce the following function
$$ f(t) = e^-tXe^t(X+Y), qquad a(t) = e^-tXYe^tX. $$
Then it is easy to check that
$$ f'(t) = e^-tXYe^t(X+Y) = a(t)f(t). $$
So it follows that $f$ is the ordered exponential of $a$. In particular,
$$ f(1) = 1 + int_0^1 a(t_1) , dt_1 + int_0^1int_0^t_1 a(t_1)a(t_2) , dt_2 dt_1 + cdots $$
From the estimate $left| int_0^1cdotsint_0^t_k-1 a(t_1)cdots a(t_k) , dt_k cdots dt_1 right| leq frac^kk!$, we realize that the formula above correctly produces the expansion of $e^X+Y$ as perturbation of $e^X$. Plugging the substitution back, we obtain
$$ e^i(H+epsilon A) = e^iH left( 1 + iepsilon int_0^1 e^-itHAe^itH , dt + mathcalO(epsilon^2) right). $$
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
add a comment |Â
up vote
1
down vote
accepted
Here is a simple trick to expand OP's matrix exponential. Write $X = iH$ and $Y = iepsilon A$ for brevity. Next, introduce the following function
$$ f(t) = e^-tXe^t(X+Y), qquad a(t) = e^-tXYe^tX. $$
Then it is easy to check that
$$ f'(t) = e^-tXYe^t(X+Y) = a(t)f(t). $$
So it follows that $f$ is the ordered exponential of $a$. In particular,
$$ f(1) = 1 + int_0^1 a(t_1) , dt_1 + int_0^1int_0^t_1 a(t_1)a(t_2) , dt_2 dt_1 + cdots $$
From the estimate $left| int_0^1cdotsint_0^t_k-1 a(t_1)cdots a(t_k) , dt_k cdots dt_1 right| leq frac^kk!$, we realize that the formula above correctly produces the expansion of $e^X+Y$ as perturbation of $e^X$. Plugging the substitution back, we obtain
$$ e^i(H+epsilon A) = e^iH left( 1 + iepsilon int_0^1 e^-itHAe^itH , dt + mathcalO(epsilon^2) right). $$
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Here is a simple trick to expand OP's matrix exponential. Write $X = iH$ and $Y = iepsilon A$ for brevity. Next, introduce the following function
$$ f(t) = e^-tXe^t(X+Y), qquad a(t) = e^-tXYe^tX. $$
Then it is easy to check that
$$ f'(t) = e^-tXYe^t(X+Y) = a(t)f(t). $$
So it follows that $f$ is the ordered exponential of $a$. In particular,
$$ f(1) = 1 + int_0^1 a(t_1) , dt_1 + int_0^1int_0^t_1 a(t_1)a(t_2) , dt_2 dt_1 + cdots $$
From the estimate $left| int_0^1cdotsint_0^t_k-1 a(t_1)cdots a(t_k) , dt_k cdots dt_1 right| leq frac^kk!$, we realize that the formula above correctly produces the expansion of $e^X+Y$ as perturbation of $e^X$. Plugging the substitution back, we obtain
$$ e^i(H+epsilon A) = e^iH left( 1 + iepsilon int_0^1 e^-itHAe^itH , dt + mathcalO(epsilon^2) right). $$
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
Here is a simple trick to expand OP's matrix exponential. Write $X = iH$ and $Y = iepsilon A$ for brevity. Next, introduce the following function
$$ f(t) = e^-tXe^t(X+Y), qquad a(t) = e^-tXYe^tX. $$
Then it is easy to check that
$$ f'(t) = e^-tXYe^t(X+Y) = a(t)f(t). $$
So it follows that $f$ is the ordered exponential of $a$. In particular,
$$ f(1) = 1 + int_0^1 a(t_1) , dt_1 + int_0^1int_0^t_1 a(t_1)a(t_2) , dt_2 dt_1 + cdots $$
From the estimate $left| int_0^1cdotsint_0^t_k-1 a(t_1)cdots a(t_k) , dt_k cdots dt_1 right| leq frac^kk!$, we realize that the formula above correctly produces the expansion of $e^X+Y$ as perturbation of $e^X$. Plugging the substitution back, we obtain
$$ e^i(H+epsilon A) = e^iH left( 1 + iepsilon int_0^1 e^-itHAe^itH , dt + mathcalO(epsilon^2) right). $$
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
answered Aug 8 at 15:30
Sangchul Lee
85.6k12155253
85.6k12155253
add a comment |Â
add a comment |Â
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1
Do $H$ & $A$ commute? If not then no...
– copper.hat
Aug 6 at 15:10
$H$ & $A$ don't commute
– user582108
Aug 6 at 15:15
1
This might involve the Baker-Campbell-Hausdorff formula.
– Luke
Aug 6 at 15:26
1
Use the power-series representation of $e^X$ with $X = A + epsilon B$ and use that $(A+epsilon B)^n = A^n + epsilon (A^n-1B + ldots + B A^n-1) + mathcalo(epsilon)$ to get a series which will be something like $e^A+epsilon B = e^A(1 + epsilon B) + epsilon (ldots)$ (with the last term being compilated and containing all sorts of commutators). Will not be pretty.
– Winther
Aug 6 at 15:32
2
We can check that $$ e^X+epsilon Y = e^X left( 1 + epsilon int_0^1 e^-tX Y e^tX , dt + mathcalO(epsilon^2) right). $$
– Sangchul Lee
Aug 6 at 15:41