Mixing
bosonic interaction Heisenberg picture
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I am trying to calculate the time evolution of the operator
beginequation
h(k)=sum_k b_k^daggerb_k, .
endequation
Therefore, I go to the Heisenberg picture
$$
h(k ,t) equiv e^fracihbarHt,left( sum_k b_k^daggerb_kright), e^-fracihbarHt , ,
$$
where $H$ is the Bose-Hubbard Hamiltonian in $k-$space
$$
H = sum_k (epsilon_k-mu) b_k^daggerb_k +fracg2Vsum_k,p,qb_k+q^daggerb_p-q^daggerb_k b_p, ,
$$
and the $b-$operators fulfil the bosonic commutation relation $[b_k,b_p^dagger]=delta_k,p$.
Now, I want to evaluate the above time dependent operator $h(k,t)$ by using the BCH formula
$$
e^XYe^-X=sum_m=0^inftyfrac1m![X,Y]_mqquadtextwithquad [X,Y]_0=Y , , textand, , [X,Y]_m=[X,[X,Y]_m-1], .
$$
For $m=1$, I calculated the following commutator relation
$$
[H,h_KE(k)] = \ = fracg2Vsum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-b_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+
b_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-b_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -q, .
$$
As you can see, there are Kronecker deltas in every term. I am not sure how to evaluate them. My first attempt: Separate the terms since the summation $Sigma$ is linear
$$beginalign
&[H,h_KE(k)] = &\ &= fracg2Vleft(sum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-sum_k,p,q,ub_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+ \ +sum_k,p,q,ub_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-sum_k,p,q,ub_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -qright), .
endalign$$
Now, we can consider every Kronecker delta separately and set in the first term $p=u$, in the second $u=k+q$, in the third $k=u$ and in the fourth $u=p-q$. However, this renders $0$. I felt like that my approach is to naive and I actually would not expect $0$ to be the correct answer. So as a second attempt, I tried to get rid of index summations in the Kronecker deltas. So in the 2nd term, I set $k+q=n$ and in the fourth term, I set $p-q=m$ , and in both cases $q$ was replaced by writing it in terms of n and m, respectively. Then I executed the Kronzucker deltas and set $n=q$ in the second term and $m=q$ in the fourth term.
$$beginalign
&[H,h_KE(k)] =\ &= fracg2Vleft(sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_p -q+k^daggerb_k b_p+\+sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_k+p-q^daggerb_k b_pright)
endalign$$
My Question is:
Is this a correct way of calculating this expression? I was hoping that the Kronzucker deltas will simplify my expression a little more, after all I have to calculate higher order terms $m$ and it will get too messy... Thank you in advance
Ilias
summation operator-theory mathematical-physics index-notation
edited Jul 27 at 13:12
joriki
164k10179328
asked Jul 27 at 11:45
Ilias Seifie
332
add a comment |Â
up vote
5
down vote
favorite
I am trying to calculate the time evolution of the operator
beginequation
h(k)=sum_k b_k^daggerb_k, .
endequation
Therefore, I go to the Heisenberg picture
$$
h(k ,t) equiv e^fracihbarHt,left( sum_k b_k^daggerb_kright), e^-fracihbarHt , ,
$$
where $H$ is the Bose-Hubbard Hamiltonian in $k-$space
$$
H = sum_k (epsilon_k-mu) b_k^daggerb_k +fracg2Vsum_k,p,qb_k+q^daggerb_p-q^daggerb_k b_p, ,
$$
and the $b-$operators fulfil the bosonic commutation relation $[b_k,b_p^dagger]=delta_k,p$.
Now, I want to evaluate the above time dependent operator $h(k,t)$ by using the BCH formula
$$
e^XYe^-X=sum_m=0^inftyfrac1m![X,Y]_mqquadtextwithquad [X,Y]_0=Y , , textand, , [X,Y]_m=[X,[X,Y]_m-1], .
$$
For $m=1$, I calculated the following commutator relation
$$
[H,h_KE(k)] = \ = fracg2Vsum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-b_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+
b_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-b_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -q, .
$$
As you can see, there are Kronecker deltas in every term. I am not sure how to evaluate them. My first attempt: Separate the terms since the summation $Sigma$ is linear
$$beginalign
&[H,h_KE(k)] = &\ &= fracg2Vleft(sum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-sum_k,p,q,ub_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+ \ +sum_k,p,q,ub_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-sum_k,p,q,ub_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -qright), .
endalign$$
Now, we can consider every Kronecker delta separately and set in the first term $p=u$, in the second $u=k+q$, in the third $k=u$ and in the fourth $u=p-q$. However, this renders $0$. I felt like that my approach is to naive and I actually would not expect $0$ to be the correct answer. So as a second attempt, I tried to get rid of index summations in the Kronecker deltas. So in the 2nd term, I set $k+q=n$ and in the fourth term, I set $p-q=m$ , and in both cases $q$ was replaced by writing it in terms of n and m, respectively. Then I executed the Kronzucker deltas and set $n=q$ in the second term and $m=q$ in the fourth term.
$$beginalign
&[H,h_KE(k)] =\ &= fracg2Vleft(sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_p -q+k^daggerb_k b_p+\+sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_k+p-q^daggerb_k b_pright)
endalign$$
My Question is:
Is this a correct way of calculating this expression? I was hoping that the Kronzucker deltas will simplify my expression a little more, after all I have to calculate higher order terms $m$ and it will get too messy... Thank you in advance
Ilias
summation operator-theory mathematical-physics index-notation
edited Jul 27 at 13:12
joriki
164k10179328
asked Jul 27 at 11:45
Ilias Seifie
332
2
Why was this downvoted? It shows clear effort.
– Shaun
Jul 27 at 11:53
There's no dependence on $k$ in $h$; you're summing over $k$; you shouldn't write it as $h(k)$.
– joriki
Jul 27 at 13:09
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I am trying to calculate the time evolution of the operator
beginequation
h(k)=sum_k b_k^daggerb_k, .
endequation
Therefore, I go to the Heisenberg picture
$$
h(k ,t) equiv e^fracihbarHt,left( sum_k b_k^daggerb_kright), e^-fracihbarHt , ,
$$
where $H$ is the Bose-Hubbard Hamiltonian in $k-$space
$$
H = sum_k (epsilon_k-mu) b_k^daggerb_k +fracg2Vsum_k,p,qb_k+q^daggerb_p-q^daggerb_k b_p, ,
$$
and the $b-$operators fulfil the bosonic commutation relation $[b_k,b_p^dagger]=delta_k,p$.
Now, I want to evaluate the above time dependent operator $h(k,t)$ by using the BCH formula
$$
e^XYe^-X=sum_m=0^inftyfrac1m![X,Y]_mqquadtextwithquad [X,Y]_0=Y , , textand, , [X,Y]_m=[X,[X,Y]_m-1], .
$$
For $m=1$, I calculated the following commutator relation
$$
[H,h_KE(k)] = \ = fracg2Vsum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-b_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+
b_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-b_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -q, .
$$
As you can see, there are Kronecker deltas in every term. I am not sure how to evaluate them. My first attempt: Separate the terms since the summation $Sigma$ is linear
$$beginalign
&[H,h_KE(k)] = &\ &= fracg2Vleft(sum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-sum_k,p,q,ub_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+ \ +sum_k,p,q,ub_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-sum_k,p,q,ub_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -qright), .
endalign$$
Now, we can consider every Kronecker delta separately and set in the first term $p=u$, in the second $u=k+q$, in the third $k=u$ and in the fourth $u=p-q$. However, this renders $0$. I felt like that my approach is to naive and I actually would not expect $0$ to be the correct answer. So as a second attempt, I tried to get rid of index summations in the Kronecker deltas. So in the 2nd term, I set $k+q=n$ and in the fourth term, I set $p-q=m$ , and in both cases $q$ was replaced by writing it in terms of n and m, respectively. Then I executed the Kronzucker deltas and set $n=q$ in the second term and $m=q$ in the fourth term.
$$beginalign
&[H,h_KE(k)] =\ &= fracg2Vleft(sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_p -q+k^daggerb_k b_p+\+sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_k+p-q^daggerb_k b_pright)
endalign$$
My Question is:
Is this a correct way of calculating this expression? I was hoping that the Kronzucker deltas will simplify my expression a little more, after all I have to calculate higher order terms $m$ and it will get too messy... Thank you in advance
Ilias
summation operator-theory mathematical-physics index-notation
edited Jul 27 at 13:12
joriki
164k10179328
asked Jul 27 at 11:45
Ilias Seifie
332
I am trying to calculate the time evolution of the operator
beginequation
h(k)=sum_k b_k^daggerb_k, .
endequation
Therefore, I go to the Heisenberg picture
$$
h(k ,t) equiv e^fracihbarHt,left( sum_k b_k^daggerb_kright), e^-fracihbarHt , ,
$$
where $H$ is the Bose-Hubbard Hamiltonian in $k-$space
$$
H = sum_k (epsilon_k-mu) b_k^daggerb_k +fracg2Vsum_k,p,qb_k+q^daggerb_p-q^daggerb_k b_p, ,
$$
and the $b-$operators fulfil the bosonic commutation relation $[b_k,b_p^dagger]=delta_k,p$.
Now, I want to evaluate the above time dependent operator $h(k,t)$ by using the BCH formula
$$
e^XYe^-X=sum_m=0^inftyfrac1m![X,Y]_mqquadtextwithquad [X,Y]_0=Y , , textand, , [X,Y]_m=[X,[X,Y]_m-1], .
$$
For $m=1$, I calculated the following commutator relation
$$
[H,h_KE(k)] = \ = fracg2Vsum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-b_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+
b_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-b_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -q, .
$$
As you can see, there are Kronecker deltas in every term. I am not sure how to evaluate them. My first attempt: Separate the terms since the summation $Sigma$ is linear
$$beginalign
&[H,h_KE(k)] = &\ &= fracg2Vleft(sum_k,p,q,u
b_k +q^daggerb_p -q^daggerb_k b_udelta_p ,u-sum_k,p,q,ub_u^daggerb_p -q^daggerb_k b_pdelta_u ,k +q+ \ +sum_k,p,q,ub_k +q^daggerb_p -q^daggerb_p b_udelta_k ,u-sum_k,p,q,ub_u^daggerb_k +q^daggerb_k b_pdelta_u ,p -qright), .
endalign$$
Now, we can consider every Kronecker delta separately and set in the first term $p=u$, in the second $u=k+q$, in the third $k=u$ and in the fourth $u=p-q$. However, this renders $0$. I felt like that my approach is to naive and I actually would not expect $0$ to be the correct answer. So as a second attempt, I tried to get rid of index summations in the Kronecker deltas. So in the 2nd term, I set $k+q=n$ and in the fourth term, I set $p-q=m$ , and in both cases $q$ was replaced by writing it in terms of n and m, respectively. Then I executed the Kronzucker deltas and set $n=q$ in the second term and $m=q$ in the fourth term.
$$beginalign
&[H,h_KE(k)] =\ &= fracg2Vleft(sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_p -q+k^daggerb_k b_p+\+sum_k,p,q
b_k +q^daggerb_p -q^daggerb_k b_p-sum_k,p,qb_q^daggerb_k+p-q^daggerb_k b_pright)
endalign$$
My Question is:
Is this a correct way of calculating this expression? I was hoping that the Kronzucker deltas will simplify my expression a little more, after all I have to calculate higher order terms $m$ and it will get too messy... Thank you in advance
Ilias
summation operator-theory mathematical-physics index-notation
edited Jul 27 at 13:12
joriki
164k10179328
asked Jul 27 at 11:45
Ilias Seifie
332
edited Jul 27 at 13:12
joriki
164k10179328
edited Jul 27 at 13:12
joriki
164k10179328
edited Jul 27 at 13:12
joriki
164k10179328
164k10179328
asked Jul 27 at 11:45
Ilias Seifie
332
asked Jul 27 at 11:45
Ilias Seifie
332
asked Jul 27 at 11:45
Ilias Seifie
332
332
2
Why was this downvoted? It shows clear effort.
– Shaun
Jul 27 at 11:53
There's no dependence on $k$ in $h$; you're summing over $k$; you shouldn't write it as $h(k)$.
– joriki
Jul 27 at 13:09
add a comment |Â
2
Why was this downvoted? It shows clear effort.
– Shaun
Jul 27 at 11:53
There's no dependence on $k$ in $h$; you're summing over $k$; you shouldn't write it as $h(k)$.
– joriki
Jul 27 at 13:09
2
2
Why was this downvoted? It shows clear effort.
– Shaun
Jul 27 at 11:53
Why was this downvoted? It shows clear effort.
– Shaun
Jul 27 at 11:53
There's no dependence on $k$ in $h$; you're summing over $k$; you shouldn't write it as $h(k)$.
– joriki
Jul 27 at 13:09
There's no dependence on $k$ in $h$; you're summing over $k$; you shouldn't write it as $h(k)$.
– joriki
Jul 27 at 13:09
add a comment |Â
1 Answer
1
active
oldest
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up vote
3
down vote
accepted
Your zero result is correct. There's nothing naive about the way you evaluated the Kronecker deltas; that's exactly the way to evaluate them. Your second attempt is just unnecessarily complicated and will lead to the same zero result if you shift the indices back.
The commutator is zero because your operator $h$ is the number operator and the Bose–Hubbard Hamiltonian $H$ leaves the number of quanta invariant (since each term in the second sum annihilates two quanta with momentum $k$ and $p$ and creates two quanta with momentum $k+q$ and $p-q$), so it commutes with the number operator.
answered Jul 27 at 13:06
joriki
164k10179328
Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$
– Ilias Seifie
Jul 27 at 13:33
sorry, here is the rest: $$ sum_kip,qb_p^daggerb_qdelta_p+k,q-b_q^daggerb_p+kdelta_q,p neq 0 $$ According to the Kronecker deltas, this should also be zero
– Ilias Seifie
Jul 27 at 13:36
@IliasSeifie: Assuming that the sum is over both terms (to clarify that, I'd put parentheses around them), you're right, this should be zero. I can think of two explanations for this: a) a typo/error by your professor or b) something related to the domain of the momenta. I've been assuming that they have either infinite or periodic domain and you can shift them arbitrarily; if you're summing them over some other domain, that might not be the case; e.g. $delta_p+k,q$ might never be $1$ because $p+k$ might never be equal to $q$ if the domain is restricted.
– joriki
Jul 27 at 13:50
@IliasSeifie: If you post the part of the lecture notes that provides the context for this statement, I'll be happy to try to say more.
– joriki
Jul 27 at 13:51
I just saw in his lecture notes that every summand has actually a prefactor $V_kV_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time...
– Ilias Seifie
Jul 27 at 13:52
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Your zero result is correct. There's nothing naive about the way you evaluated the Kronecker deltas; that's exactly the way to evaluate them. Your second attempt is just unnecessarily complicated and will lead to the same zero result if you shift the indices back.
The commutator is zero because your operator $h$ is the number operator and the Bose–Hubbard Hamiltonian $H$ leaves the number of quanta invariant (since each term in the second sum annihilates two quanta with momentum $k$ and $p$ and creates two quanta with momentum $k+q$ and $p-q$), so it commutes with the number operator.
answered Jul 27 at 13:06
joriki
164k10179328
Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$
– Ilias Seifie
Jul 27 at 13:33
sorry, here is the rest: $$ sum_kip,qb_p^daggerb_qdelta_p+k,q-b_q^daggerb_p+kdelta_q,p neq 0 $$ According to the Kronecker deltas, this should also be zero
– Ilias Seifie
Jul 27 at 13:36
@IliasSeifie: Assuming that the sum is over both terms (to clarify that, I'd put parentheses around them), you're right, this should be zero. I can think of two explanations for this: a) a typo/error by your professor or b) something related to the domain of the momenta. I've been assuming that they have either infinite or periodic domain and you can shift them arbitrarily; if you're summing them over some other domain, that might not be the case; e.g. $delta_p+k,q$ might never be $1$ because $p+k$ might never be equal to $q$ if the domain is restricted.
– joriki
Jul 27 at 13:50
@IliasSeifie: If you post the part of the lecture notes that provides the context for this statement, I'll be happy to try to say more.
– joriki
Jul 27 at 13:51
I just saw in his lecture notes that every summand has actually a prefactor $V_kV_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time...
– Ilias Seifie
Jul 27 at 13:52
 |Â
show 1 more comment
up vote
3
down vote
accepted
Your zero result is correct. There's nothing naive about the way you evaluated the Kronecker deltas; that's exactly the way to evaluate them. Your second attempt is just unnecessarily complicated and will lead to the same zero result if you shift the indices back.
The commutator is zero because your operator $h$ is the number operator and the Bose–Hubbard Hamiltonian $H$ leaves the number of quanta invariant (since each term in the second sum annihilates two quanta with momentum $k$ and $p$ and creates two quanta with momentum $k+q$ and $p-q$), so it commutes with the number operator.
answered Jul 27 at 13:06
joriki
164k10179328
Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$
– Ilias Seifie
Jul 27 at 13:33
sorry, here is the rest: $$ sum_kip,qb_p^daggerb_qdelta_p+k,q-b_q^daggerb_p+kdelta_q,p neq 0 $$ According to the Kronecker deltas, this should also be zero
– Ilias Seifie
Jul 27 at 13:36
@IliasSeifie: Assuming that the sum is over both terms (to clarify that, I'd put parentheses around them), you're right, this should be zero. I can think of two explanations for this: a) a typo/error by your professor or b) something related to the domain of the momenta. I've been assuming that they have either infinite or periodic domain and you can shift them arbitrarily; if you're summing them over some other domain, that might not be the case; e.g. $delta_p+k,q$ might never be $1$ because $p+k$ might never be equal to $q$ if the domain is restricted.
– joriki
Jul 27 at 13:50
@IliasSeifie: If you post the part of the lecture notes that provides the context for this statement, I'll be happy to try to say more.
– joriki
Jul 27 at 13:51
I just saw in his lecture notes that every summand has actually a prefactor $V_kV_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time...
– Ilias Seifie
Jul 27 at 13:52
 |Â
show 1 more comment
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Your zero result is correct. There's nothing naive about the way you evaluated the Kronecker deltas; that's exactly the way to evaluate them. Your second attempt is just unnecessarily complicated and will lead to the same zero result if you shift the indices back.
The commutator is zero because your operator $h$ is the number operator and the Bose–Hubbard Hamiltonian $H$ leaves the number of quanta invariant (since each term in the second sum annihilates two quanta with momentum $k$ and $p$ and creates two quanta with momentum $k+q$ and $p-q$), so it commutes with the number operator.
answered Jul 27 at 13:06
joriki
164k10179328
Your zero result is correct. There's nothing naive about the way you evaluated the Kronecker deltas; that's exactly the way to evaluate them. Your second attempt is just unnecessarily complicated and will lead to the same zero result if you shift the indices back.
The commutator is zero because your operator $h$ is the number operator and the Bose–Hubbard Hamiltonian $H$ leaves the number of quanta invariant (since each term in the second sum annihilates two quanta with momentum $k$ and $p$ and creates two quanta with momentum $k+q$ and $p-q$), so it commutes with the number operator.
answered Jul 27 at 13:06
joriki
164k10179328
answered Jul 27 at 13:06
joriki
164k10179328
answered Jul 27 at 13:06
joriki
164k10179328
answered Jul 27 at 13:06
joriki
164k10179328
164k10179328
Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$
– Ilias Seifie
Jul 27 at 13:33
sorry, here is the rest: $$ sum_kip,qb_p^daggerb_qdelta_p+k,q-b_q^daggerb_p+kdelta_q,p neq 0 $$ According to the Kronecker deltas, this should also be zero
– Ilias Seifie
Jul 27 at 13:36
@IliasSeifie: Assuming that the sum is over both terms (to clarify that, I'd put parentheses around them), you're right, this should be zero. I can think of two explanations for this: a) a typo/error by your professor or b) something related to the domain of the momenta. I've been assuming that they have either infinite or periodic domain and you can shift them arbitrarily; if you're summing them over some other domain, that might not be the case; e.g. $delta_p+k,q$ might never be $1$ because $p+k$ might never be equal to $q$ if the domain is restricted.
– joriki
Jul 27 at 13:50
@IliasSeifie: If you post the part of the lecture notes that provides the context for this statement, I'll be happy to try to say more.
– joriki
Jul 27 at 13:51
I just saw in his lecture notes that every summand has actually a prefactor $V_kV_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time...
– Ilias Seifie
Jul 27 at 13:52
 |Â
show 1 more comment
Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$
– Ilias Seifie
Jul 27 at 13:33
sorry, here is the rest: $$ sum_kip,qb_p^daggerb_qdelta_p+k,q-b_q^daggerb_p+kdelta_q,p neq 0 $$ According to the Kronecker deltas, this should also be zero
– Ilias Seifie
Jul 27 at 13:36
@IliasSeifie: Assuming that the sum is over both terms (to clarify that, I'd put parentheses around them), you're right, this should be zero. I can think of two explanations for this: a) a typo/error by your professor or b) something related to the domain of the momenta. I've been assuming that they have either infinite or periodic domain and you can shift them arbitrarily; if you're summing them over some other domain, that might not be the case; e.g. $delta_p+k,q$ might never be $1$ because $p+k$ might never be equal to $q$ if the domain is restricted.
– joriki
Jul 27 at 13:50
@IliasSeifie: If you post the part of the lecture notes that provides the context for this statement, I'll be happy to try to say more.
– joriki
Jul 27 at 13:51
I just saw in his lecture notes that every summand has actually a prefactor $V_kV_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time...
– Ilias Seifie
Jul 27 at 13:52
Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$
– Ilias Seifie
Jul 27 at 13:33
Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$
– Ilias Seifie
Jul 27 at 13:33
sorry, here is the rest: $$ sum_kip,qb_p^daggerb_qdelta_p+k,q-b_q^daggerb_p+kdelta_q,p neq 0 $$ According to the Kronecker deltas, this should also be zero
– Ilias Seifie
Jul 27 at 13:36
sorry, here is the rest: $$ sum_kip,qb_p^daggerb_qdelta_p+k,q-b_q^daggerb_p+kdelta_q,p neq 0 $$ According to the Kronecker deltas, this should also be zero
– Ilias Seifie
Jul 27 at 13:36
@IliasSeifie: Assuming that the sum is over both terms (to clarify that, I'd put parentheses around them), you're right, this should be zero. I can think of two explanations for this: a) a typo/error by your professor or b) something related to the domain of the momenta. I've been assuming that they have either infinite or periodic domain and you can shift them arbitrarily; if you're summing them over some other domain, that might not be the case; e.g. $delta_p+k,q$ might never be $1$ because $p+k$ might never be equal to $q$ if the domain is restricted.
– joriki
Jul 27 at 13:50
@IliasSeifie: Assuming that the sum is over both terms (to clarify that, I'd put parentheses around them), you're right, this should be zero. I can think of two explanations for this: a) a typo/error by your professor or b) something related to the domain of the momenta. I've been assuming that they have either infinite or periodic domain and you can shift them arbitrarily; if you're summing them over some other domain, that might not be the case; e.g. $delta_p+k,q$ might never be $1$ because $p+k$ might never be equal to $q$ if the domain is restricted.
– joriki
Jul 27 at 13:50
@IliasSeifie: If you post the part of the lecture notes that provides the context for this statement, I'll be happy to try to say more.
– joriki
Jul 27 at 13:51
@IliasSeifie: If you post the part of the lecture notes that provides the context for this statement, I'll be happy to try to say more.
– joriki
Jul 27 at 13:51
I just saw in his lecture notes that every summand has actually a prefactor $V_kV_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time...
– Ilias Seifie
Jul 27 at 13:52
I just saw in his lecture notes that every summand has actually a prefactor $V_kV_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time...
– Ilias Seifie
Jul 27 at 13:52
 |Â
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2
Why was this downvoted? It shows clear effort.
– Shaun
Jul 27 at 11:53
There's no dependence on $k$ in $h$; you're summing over $k$; you shouldn't write it as $h(k)$.
– joriki
Jul 27 at 13:09