Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'd like to write equations for $c_{ij}(t)$,

With a hamiltonian of the form $$H=\sum_{kn}a^{\dagger}_k t_{kn}a_n + \frac{1}{2}\sum_{klmn}a^{\dagger}_k a^{\dagger}_l v_{klmn}a_m a_n$$ with $t_{kn}$ and $v_{klmn}$ as matrix elements of the kinetic energy and potential in the single particle basis. Now I'm operating on a two particle state $$|\psi (t)\rangle =\sum_{ij}a^{\dagger}_i a^{\dagger}_j |0\rangle c_{ij}(t)$$ Now I got the potential part to work, as well as the time derivative part. I'm actually having trouble with the kinetic energy part (hence the title). Here's what I have done. looking just at the first term acting $$\sum_{kn}\sum_{ij}a^{\dagger}_k t_{kn} a_n a^{\dagger}_i a^{\dagger}_j |0\rangle c_{ij}(t)$$ I use the commutation relations to fanangle the set of three operators after $t_{kn}$ into $\delta_{ni}a^{\dagger}_j \pm \delta_{nj}a^{\dagger}_i$ where the $\pm$ is for both fermions and bosons. Inserting this into the above I get $$\sum_{kn}\sum_{ij}t_{kn}a^{\dagger}_k (\delta_{ni}a^{\dagger}_j \pm \delta_{nj}a^{\dagger}_i)|0\rangle c_{ij}(t)$$ Now taking the sum over $n$ I get $$\sum_k \sum_{ij}(t_{ki}a^{\dagger}_k a^{\dagger}_j \pm t_{kj}a^{\dagger}_k a^{\dagger}_i)|0\rangle c_{ij}(t) $$

Now my trouble seems silly to me, but I can't figure out what is the right move here. If I swtich the $i$ and $j$ in the second term after the $\pm$ and $c_{ij}(t)$ is symmetric then it's possible zero, but I don't think that makes sense. My total equation as of now is $$\sum_k \sum_{ij}(t_{ki}a^{\dagger}_k a^{\dagger}_j \pm t_{kj}a^{\dagger}_k a^{\dagger}_i)|0\rangle c_{ij}(t)+\frac{1}{2}\sum_{kl}\sum_{ij}a^{\dagger}_k a^{\dagger}_l v_{klij}|0\rangle c_{ij}(t) \pm \frac{1}{2}\sum_{kl}\sum_{ij}a^{\dagger}_k a^{\dagger}_l v_{klji}|0\rangle c_{ij}(t)=i\hbar \sum_{ij}a^{\dagger}_i a^{\dagger}_j |0\rangle \dot{c}_{ij}(t)$$ I'd like to remove the arbitrary ket vectors in the whole expression to just obtain a simple expression for $c_{ij}(t)$

any help would be great, thanks.

share|cite|improve this question
up vote 2 down vote accepted

I'm not sure if the following is the main issue, but OP writes in the question formulation (v1):

[...] $c_{ij}(t)$ is symmetric [...]

The coefficient $c_{ij}$ is only $i\leftrightarrow j$ symmetric in case of bosons. In case of fermions, the coefficient $c_{ij}$ is $i\leftrightarrow j$ antisymmetric.

share|cite|improve this answer
That's what I needed to know. Thank you. – kηives May 1 '12 at 14:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.