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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.

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1 Answer 1

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.

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That's what I needed to know. Thank you. –  kηives May 1 '12 at 14:18

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