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I am given the Hamiltonian, in an exercise called plasmons, and where $\langle, \rangle $ denotes the expectation value.

$$ H = \sum_{k} \varepsilon_k a_k^{\dagger} a_k + \sum_{k_1,k_2,q} V_q a_{k_1+q}^{\dagger} a_{k_2+q}^\dagger a_{k_2} a_{k_1}$$

and I am supposed to write down the time-dependence equation for $\langle a_{k-Q}^{\dagger} a_k \rangle$.

Now, I know that this is the Heisenberg equation of motion which is

$$-i \hbar \partial_t \langle a_{k-Q}^{\dagger} a_k \rangle = \langle [H, a_{k-Q}^{\dagger} a_k ] \rangle $$ in that case.

I am supposed to end up with

$$- i \hbar \partial_t \langle a_{k-Q}^{\dagger} a_k \rangle = (\varepsilon_{k-Q} - \varepsilon_{k} ) \langle a_{k-Q}^{\dagger} a_k \rangle +V_Q(\langle a_{k}^{\dagger} a_k \rangle - \langle a_{k-Q}^{\dagger} a_{k-Q} \rangle ) \sum_{k_2} \langle a_{k_2-Q}^{\dagger} a_{k_2} \rangle + \sum_{q} V_q (\langle a_{k-q}^{\dagger} a_{k-q} \rangle - \langle a_{k-Q+q}^{\dagger} a_{k-Q+q} \rangle)\langle a_{k-Q}^{\dagger} a_k \rangle$$

Now I actually managed to get the first two terms, but I don't see how to the potential:

$$+V_Q(\langle a_{k}^{\dagger} a_k \rangle - \langle a_{k-Q}^{\dagger} a_{k-Q} \rangle ) \sum_{k_2} \langle a_{k_2-Q}^{\dagger} a_{k_2} \rangle + \sum_{q} V_q (\langle a_{k-q}^{\dagger} a_{k-q} \rangle - \langle a_{k-Q+q}^{\dagger} a_{k-Q+q} \rangle)\langle a_{k-Q}^{\dagger} a_k \rangle$$

We are also allowed to use Hartree Fock factorizations in the potential in order to avoid a coupling of the "calculated expectation values to higher expectation values", but I am not sure what this actually means.

From the lecture I would guess that this means something like $$\langle a_1^{\dagger} a_2^{\dagger} a_3 a_4 \rangle \sim \langle a_1^{\dagger}a_4 \rangle \langle a_2^{\dagger} a_3 \rangle - \langle a_1^{\dagger} a_3 \rangle \langle a_2^{\dagger} a_4 \rangle.$$

By the way: I have one rather simples questions about this: Does anybody know if this model assumes that $k$ is discrete or continuous?

-(Maybe you could post this in the comments).

If anything is unclear please let me know.

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  • $\begingroup$ As regarding to $\langle a^\dagger_{k-Q}a_l\rangle$ - I guess you problem is that you mess vacuum of the free Hamiltonian with the vacuum of the full system. The first one is being annihilated by $a_k$, while the vacuum state for the full Hamiltonian is not. The expression you are calcualting is in the vacuum of the full system. $\endgroup$
    – Fedxa
    Commented Nov 22, 2014 at 19:49
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    $\begingroup$ No. Example-state $|\psi\rangle=(a^\dagger_1+a^\dagger_2)|0\rangle$. Then $\langle\psi|a^\dagger_2 a_1|\psi\rangle=1$ $\endgroup$
    – Fedxa
    Commented Nov 22, 2014 at 20:16
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    $\begingroup$ The problem is not horrifying, by the way. I think you are asked actually to perform a Hartree-Fock mean-field on the interacting electronic Hamiltonian. A detailed explanation can be found in H. Bruus and K. Flensberg, “Many-Body Quantum Theory in Condensed Matter Physics”, chp. 4 (“Mean field theory”). $\endgroup$
    – AlQuemist
    Commented Nov 14, 2015 at 14:08
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    $\begingroup$ The first equation looks wrong. How can the interaction increase the momentum of both interacting particles? $\endgroup$
    – DanielSank
    Commented Sep 25, 2016 at 8:20
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    $\begingroup$ The title of this post is not helpful. Please use a more descriptive title. $\endgroup$
    – DanielSank
    Commented Jan 3, 2017 at 18:57

1 Answer 1

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When you say you "got the first to terms" I guess that means the non-interacting part. So, your trouble seems to be with the interaction terms. I don't want to work the whole thing out since this looks like homework, but maybe this will help:

Probably best to start by working out the commutator: $$ [\sum_{k_1,k_2,q} V_q a_{k_1+q}^{\dagger} a_{k_2+q}^\dagger a_{k_2} a_{k_1},a_{k-Q}^\dagger a_{k}] $$, which it not really so horrible.

You can just repeatedly use the fact that: $$ [AB,C] = A[B,C] + [A,C]B $$

I think it might help to rewrite the commutator of interest like: $$ [a^\dagger_1 a^\dagger_2 a_3 a_4,a_5^\dagger a_6] $$ since it will make it harder to make transcription errors in the writing.

You can do this reduction many ways. For example, you could start off like: $$ [a^\dagger_1 a^\dagger_2 a_3 a_4,a_5^\dagger a_6] = a^\dagger_1 [a^\dagger_2 a_3 a_4,a_5^\dagger a_6] + [a^\dagger_1 ,a_5^\dagger a_6]a^\dagger_2 a_3 a_4 $$ $$ = a^\dagger_1 [a^\dagger_2 a_3 a_4,a_5^\dagger a_6] + a_5^\dagger[a^\dagger_1 , a_6]a^\dagger_2 a_3 a_4 $$ $$ = a^\dagger_1 (a^\dagger_2[ a_3 a_4,a_5^\dagger a_6]+[a^\dagger_2 ,a_5^\dagger a_6]a_3 a_4) + a_5^\dagger[a^\dagger_1 , a_6]a^\dagger_2 a_3 a_4 $$ $$ = a^\dagger_1 (a^\dagger_2[ a_3 a_4,a_5^\dagger ]a_6+a_5^\dagger[a^\dagger_2 , a_6]a_3 a_4) + a_5^\dagger[a^\dagger_1 , a_6]a^\dagger_2 a_3 a_4 $$ $$ = a^\dagger_1 (a^\dagger_2(a_3[ a_4,a_5^\dagger ]+[ a_3 ,a_5^\dagger ]a_4)a_6+a_5^\dagger[a^\dagger_2 , a_6]a_3 a_4) + a_5^\dagger[a^\dagger_1 , a_6]a^\dagger_2 a_3 a_4 $$ $$ = a^\dagger_1 (a^\dagger_2(a_3\delta_{4,5}+\delta_{35}a_4)a_6-a_5^\dagger\delta_{26}a_3 a_4) - a_5^\dagger\delta_{16}a^\dagger_2 a_3 a_4 $$ $$ = a^\dagger_1 a^\dagger_2 a_3 a_6\delta_{4,5}+a^\dagger_1 a^\dagger_2 a_4 a_6\delta_{35}-a^\dagger_1 a_5^\dagger a_3 a_4\delta_{26} - a_5^\dagger a^\dagger_2 a_3 a_4\delta_{16} $$

Ok, so now what? I guess now you can start using the Hartree Fock approximation to figure out the expectation value for: $$ < a^\dagger_1 a^\dagger_2 a_3 a_6\delta_{4,5}+a^\dagger_1 a^\dagger_2 a_4 a_6\delta_{35}-a^\dagger_1 a_5^\dagger a_3 a_4\delta_{26} - a_5^\dagger a^\dagger_2 a_3 a_4\delta_{16} >$$

Then substitute back in the correct index names and do all the summations.

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