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Let's say we have the two many-body states $$ |\psi_k\rangle=\left(\prod_{k=1}^n c_k^\dagger\right)|0\rangle ,\qquad |\psi_\lambda\rangle=\left(\prod_{\lambda=1}^n c_\lambda^\dagger\right)|0\rangle \tag{1}$$ where $k$ and $\lambda$ label the states in different basis. These are related by $$ c_k=\sum_\lambda \langle k| \lambda \rangle c_\lambda ,\qquad c_k^\dagger=\sum_\lambda \langle \lambda |k\rangle c_\lambda^\dagger\tag{2}$$ Are there techniques to evaluate $\langle\psi_k|\psi_\lambda\rangle$? My first approach is to write

$$ |\psi_\lambda \rangle = \left(\prod_{\lambda'} \sum_k \langle \lambda'|k\rangle c_k^\dagger\right)|0\rangle\tag{3}$$ and then try to expand the sums and match operators in the same basis. This, however, seems to be a nightmare! Is there a different approach?

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    $\begingroup$ Now expand out (3) and collect all the coefficients corresponding to the state $|\psi_\lambda\rangle$. The result is a Slater determinant. $\endgroup$
    – Meng Cheng
    Commented Mar 17, 2022 at 23:44
  • $\begingroup$ It looks like following this approach I will get that the answer is the determinant of the matrix $\langle k | \lambda \rangle$. Is this correct? $\endgroup$
    – Ivan
    Commented Mar 17, 2022 at 23:49
  • $\begingroup$ that is correct. $\endgroup$
    – Meng Cheng
    Commented Mar 18, 2022 at 0:01

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Computing the overlap of many-body wavefunctions is in general a fairly difficult task. There are some clever algorithms that reduce the nightmarish-ness - you may find e.g. the one presented in this paper to be of interest.

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