I am trying to calculate the time evolution of the operator \begin{equation} h(k)=\sum_k b_k^{\dagger}b_k\, . \end{equation} Therefore, I go to the Heisenberg picture $$ h(k ,t) \equiv e^{\frac{i}{\hbar}Ht}\,\left( \sum_k b_k^{\dagger}b_k\right)\, e^{-\frac{i}{\hbar}Ht} \, , $$ where $H$ is the Bose-Hubbard Hamiltonian in $k-$space $$ H = \sum_k (\epsilon_k-\mu) b_k^{\dagger}b_k +\frac{g}{2V}\sum_{k,p,q}b_{k+q}^{\dagger}b_{p-q}^{\dagger}b_k b_p\, , $$ 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 $$ e^XYe^{-X}=\sum_{m=0}^{\infty}\frac{1}{m!}[X,Y]_m\qquad\text{with}\quad [X,Y]_0=Y \, \, \text{and}\, \, [X,Y]_m=[X,[X,Y]_{m-1}]\, . $$ For $m=1$, I calculated the following commutator relation $$ [H,h_{KE}(k)] = \\ = \frac{g}{2V}\sum_{k,p,q,u} b_{k +q}^{\dagger}b_{p -q}^{\dagger}b_k b_u\delta_{p ,u}-b_{u}^{\dagger}b_{p -q}^{\dagger}b_k b_p\delta_{u ,k +q}+ b_{k +q}^{\dagger}b_{p -q}^{\dagger}b_p b_u\delta_{k ,u}-b_{u}^{\dagger}b_{k +q}^{\dagger}b_k b_p\delta_{u ,p -q}\, . $$ 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 $$\begin{align} &[H,h_{KE}(k)] = &\\ &= \frac{g}{2V}\left(\sum_{k,p,q,u} b_{k +q}^{\dagger}b_{p -q}^{\dagger}b_k b_u\delta_{p ,u}-\sum_{k,p,q,u}b_{u}^{\dagger}b_{p -q}^{\dagger}b_k b_p\delta_{u ,k +q}+ \\ +\sum_{k,p,q,u}b_{k +q}^{\dagger}b_{p -q}^{\dagger}b_p b_u\delta_{k ,u}-\sum_{k,p,q,u}b_{u}^{\dagger}b_{k +q}^{\dagger}b_k b_p\delta_{u ,p -q}\right)\, . \end{align}$$ 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. $$\begin{align} &[H,h_{KE}(k)] =\\ &= \frac{g}{2V}\left(\sum_{k,p,q} b_{k +q}^{\dagger}b_{p -q}^{\dagger}b_k b_p-\sum_{k,p,q}b_{q}^{\dagger}b_{p -q+k}^{\dagger}b_k b_p+\\+\sum_{k,p,q} b_{k +q}^{\dagger}b_{p -q}^{\dagger}b_k b_p-\sum_{k,p,q}b_{q}^{\dagger}b_{k+p-q}^{\dagger}b_k b_p\right) \end{align}$$

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


  • 3
    $\begingroup$ Why was this downvoted? It shows clear effort. $\endgroup$
    – Shaun
    Commented Jul 27, 2018 at 11:53
  • $\begingroup$ There's no dependence on $k$ in $h$; you're summing over $k$; you shouldn't write it as $h(k)$. $\endgroup$
    – joriki
    Commented Jul 27, 2018 at 13:09

1 Answer 1


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.

  • $\begingroup$ Thank you. The reason for, why I thought this is naive is because of my Professors lecture notes. There, he states that $$ $\endgroup$ Commented Jul 27, 2018 at 13:33
  • $\begingroup$ sorry, here is the rest: $$ \sum_{kip,q}b_p^{\dagger}b_q\delta_{p+k,q}-b_q^{\dagger}b_{p+k}\delta_{q,p} \neq 0 $$ According to the Kronecker deltas, this should also be zero $\endgroup$ Commented Jul 27, 2018 at 13:36
  • $\begingroup$ @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. $\endgroup$
    – joriki
    Commented Jul 27, 2018 at 13:50
  • $\begingroup$ @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. $\endgroup$
    – joriki
    Commented Jul 27, 2018 at 13:51
  • $\begingroup$ I just saw in his lecture notes that every summand has actually a prefactor $V_{k}V_pV_q$. So it is basically a weighted sum and hence does not give zero. Thank you very much for your time... $\endgroup$ Commented Jul 27, 2018 at 13:52

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