Mixing
Boson operator algebra - unitary transformation
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
What is the simple way to evaluate
$$ e^i alpha n(n-1)a^dagger e^-i alpha n(n-1)$$
and
$$ e^i alpha n(n-1)a e^-i alpha n(n-1)$$
where $n=a^dagger a$ and $a$ and $a^dagger$ are the boson annihilation/creation operators for bosons and $alpha$ is a constant?
matrix-calculus noncommutative-algebra adjoint-operators matrix-exponential
asked Aug 1 at 16:53
Galuoises
105
add a comment |Â
up vote
1
down vote
favorite
What is the simple way to evaluate
$$ e^i alpha n(n-1)a^dagger e^-i alpha n(n-1)$$
and
$$ e^i alpha n(n-1)a e^-i alpha n(n-1)$$
where $n=a^dagger a$ and $a$ and $a^dagger$ are the boson annihilation/creation operators for bosons and $alpha$ is a constant?
matrix-calculus noncommutative-algebra adjoint-operators matrix-exponential
asked Aug 1 at 16:53
Galuoises
105
Out of curiosity, is there a physical interpretation of these operators? (I don't mean $n$, $a$, and $a^dagger$. I am curious about the two operators you are asking about.)
– Batominovski
Aug 1 at 16:59
Basically I am trying to rewrite the Hamiltonian of the Bose-Hubbard model in the rotating frame with the onsite interaction....so I have to transform the operators $a$ and $a^dagger$
– Galuoises
Aug 1 at 17:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is the simple way to evaluate
$$ e^i alpha n(n-1)a^dagger e^-i alpha n(n-1)$$
and
$$ e^i alpha n(n-1)a e^-i alpha n(n-1)$$
where $n=a^dagger a$ and $a$ and $a^dagger$ are the boson annihilation/creation operators for bosons and $alpha$ is a constant?
matrix-calculus noncommutative-algebra adjoint-operators matrix-exponential
asked Aug 1 at 16:53
Galuoises
105
What is the simple way to evaluate
$$ e^i alpha n(n-1)a^dagger e^-i alpha n(n-1)$$
and
$$ e^i alpha n(n-1)a e^-i alpha n(n-1)$$
where $n=a^dagger a$ and $a$ and $a^dagger$ are the boson annihilation/creation operators for bosons and $alpha$ is a constant?
matrix-calculus noncommutative-algebra adjoint-operators matrix-exponential
asked Aug 1 at 16:53
Galuoises
105
asked Aug 1 at 16:53
Galuoises
105
asked Aug 1 at 16:53
Galuoises
105
asked Aug 1 at 16:53
Galuoises
105
105
Out of curiosity, is there a physical interpretation of these operators? (I don't mean $n$, $a$, and $a^dagger$. I am curious about the two operators you are asking about.)
– Batominovski
Aug 1 at 16:59
Basically I am trying to rewrite the Hamiltonian of the Bose-Hubbard model in the rotating frame with the onsite interaction....so I have to transform the operators $a$ and $a^dagger$
– Galuoises
Aug 1 at 17:07
add a comment |Â
Out of curiosity, is there a physical interpretation of these operators? (I don't mean $n$, $a$, and $a^dagger$. I am curious about the two operators you are asking about.)
– Batominovski
Aug 1 at 16:59
Basically I am trying to rewrite the Hamiltonian of the Bose-Hubbard model in the rotating frame with the onsite interaction....so I have to transform the operators $a$ and $a^dagger$
– Galuoises
Aug 1 at 17:07
Out of curiosity, is there a physical interpretation of these operators? (I don't mean $n$, $a$, and $a^dagger$. I am curious about the two operators you are asking about.)
– Batominovski
Aug 1 at 16:59
Out of curiosity, is there a physical interpretation of these operators? (I don't mean $n$, $a$, and $a^dagger$. I am curious about the two operators you are asking about.)
– Batominovski
Aug 1 at 16:59
Basically I am trying to rewrite the Hamiltonian of the Bose-Hubbard model in the rotating frame with the onsite interaction....so I have to transform the operators $a$ and $a^dagger$
– Galuoises
Aug 1 at 17:07
Basically I am trying to rewrite the Hamiltonian of the Bose-Hubbard model in the rotating frame with the onsite interaction....so I have to transform the operators $a$ and $a^dagger$
– Galuoises
Aug 1 at 17:07
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
You can evaluate them on states with definite particle number. Let $|krangle$ be a state such that $n|krangle=k|krangle$. Then
begineqnarray*
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)|krangle
&=&
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^mathrm i alpha (k+1)ka^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^2mathrm i alpha ka^dagger|krangle
\
&=&
a^daggermathrm e^2mathrm i alpha n|krangle;.
endeqnarray*
Since $|krangle$ was arbitrary and the states with definite particle number span the state space, this implies
$$
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)=a^daggermathrm e^2mathrm i alpha n;.
$$
Analogously, you can obtain
$$
mathrm e^mathrm i alpha n(n-1)amathrm e^-mathrm i alpha n(n-1)=amathrm e^-2mathrm i alpha(n-1);.
$$
answered Aug 1 at 17:41
joriki
164k10179328
1
Thank you so much! I was trying much more harder paths without any reasons
– Galuoises
Aug 2 at 12:34
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
You can evaluate them on states with definite particle number. Let $|krangle$ be a state such that $n|krangle=k|krangle$. Then
begineqnarray*
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)|krangle
&=&
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^mathrm i alpha (k+1)ka^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^2mathrm i alpha ka^dagger|krangle
\
&=&
a^daggermathrm e^2mathrm i alpha n|krangle;.
endeqnarray*
Since $|krangle$ was arbitrary and the states with definite particle number span the state space, this implies
$$
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)=a^daggermathrm e^2mathrm i alpha n;.
$$
Analogously, you can obtain
$$
mathrm e^mathrm i alpha n(n-1)amathrm e^-mathrm i alpha n(n-1)=amathrm e^-2mathrm i alpha(n-1);.
$$
answered Aug 1 at 17:41
joriki
164k10179328
1
Thank you so much! I was trying much more harder paths without any reasons
– Galuoises
Aug 2 at 12:34
add a comment |Â
up vote
1
down vote
accepted
You can evaluate them on states with definite particle number. Let $|krangle$ be a state such that $n|krangle=k|krangle$. Then
begineqnarray*
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)|krangle
&=&
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^mathrm i alpha (k+1)ka^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^2mathrm i alpha ka^dagger|krangle
\
&=&
a^daggermathrm e^2mathrm i alpha n|krangle;.
endeqnarray*
Since $|krangle$ was arbitrary and the states with definite particle number span the state space, this implies
$$
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)=a^daggermathrm e^2mathrm i alpha n;.
$$
Analogously, you can obtain
$$
mathrm e^mathrm i alpha n(n-1)amathrm e^-mathrm i alpha n(n-1)=amathrm e^-2mathrm i alpha(n-1);.
$$
answered Aug 1 at 17:41
joriki
164k10179328
1
Thank you so much! I was trying much more harder paths without any reasons
– Galuoises
Aug 2 at 12:34
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You can evaluate them on states with definite particle number. Let $|krangle$ be a state such that $n|krangle=k|krangle$. Then
begineqnarray*
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)|krangle
&=&
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^mathrm i alpha (k+1)ka^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^2mathrm i alpha ka^dagger|krangle
\
&=&
a^daggermathrm e^2mathrm i alpha n|krangle;.
endeqnarray*
Since $|krangle$ was arbitrary and the states with definite particle number span the state space, this implies
$$
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)=a^daggermathrm e^2mathrm i alpha n;.
$$
Analogously, you can obtain
$$
mathrm e^mathrm i alpha n(n-1)amathrm e^-mathrm i alpha n(n-1)=amathrm e^-2mathrm i alpha(n-1);.
$$
answered Aug 1 at 17:41
joriki
164k10179328
You can evaluate them on states with definite particle number. Let $|krangle$ be a state such that $n|krangle=k|krangle$. Then
begineqnarray*
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)|krangle
&=&
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^mathrm i alpha (k+1)ka^daggermathrm e^-mathrm i alpha k(k-1)|krangle
\
&=&
mathrm e^2mathrm i alpha ka^dagger|krangle
\
&=&
a^daggermathrm e^2mathrm i alpha n|krangle;.
endeqnarray*
Since $|krangle$ was arbitrary and the states with definite particle number span the state space, this implies
$$
mathrm e^mathrm i alpha n(n-1)a^daggermathrm e^-mathrm i alpha n(n-1)=a^daggermathrm e^2mathrm i alpha n;.
$$
Analogously, you can obtain
$$
mathrm e^mathrm i alpha n(n-1)amathrm e^-mathrm i alpha n(n-1)=amathrm e^-2mathrm i alpha(n-1);.
$$
answered Aug 1 at 17:41
joriki
164k10179328
edited Aug 1 at 18:11
answered Aug 1 at 17:41
joriki
164k10179328
answered Aug 1 at 17:41
joriki
164k10179328
answered Aug 1 at 17:41
joriki
164k10179328
164k10179328
1
Thank you so much! I was trying much more harder paths without any reasons
– Galuoises
Aug 2 at 12:34
add a comment |Â
1
Thank you so much! I was trying much more harder paths without any reasons
– Galuoises
Aug 2 at 12:34
1
1
Thank you so much! I was trying much more harder paths without any reasons
– Galuoises
Aug 2 at 12:34
Thank you so much! I was trying much more harder paths without any reasons
– Galuoises
Aug 2 at 12:34
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2869272%2fboson-operator-algebra-unitary-transformation%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Out of curiosity, is there a physical interpretation of these operators? (I don't mean $n$, $a$, and $a^dagger$. I am curious about the two operators you are asking about.)
– Batominovski
Aug 1 at 16:59
Basically I am trying to rewrite the Hamiltonian of the Bose-Hubbard model in the rotating frame with the onsite interaction....so I have to transform the operators $a$ and $a^dagger$
– Galuoises
Aug 1 at 17:07