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edited Oct 27 at 12:15

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Suppose we compute an expectation value of $r_{12} r_{13}^{-1}$ over a wave function $\phi_p (1) \otimes \phi_q(2) \otimes \phi_r (3)$, we denote it as $$\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle \tag{1}$$$$\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle. \tag{1}$$

We can insert a completeness relation $\sum_{\kappa \lambda \tau} | \kappa(1) \lambda(2) \tau(3) \rangle \langle \kappa(1) \lambda(2) \tau(3) | = 1 $ on Eqn. (1)

Therefore, $$ \sum_{\kappa \lambda \tau} \langle pqr | r_{12} | \kappa \lambda\tau \rangle \langle \kappa \lambda\tau | r_{13}^{-1} | pqr\rangle = \sum_{\kappa \lambda \tau} \langle pq | r_{12} | \kappa \lambda \rangle \delta_{r\tau} \langle \kappa \tau | r_{13}^{-1} | p r\rangle \delta_{\lambda q} = \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{13}^{-1}| pr \rangle \tag{2} $$ We rename $\langle \kappa(1) r(3) | r_{13}^{-1}| p(1)r(3) \rangle$ as $$\langle\kappa(1) r(2) | r_{12}^{-1}| p (1) r(2) \rangle \equiv \langle \kappa r | r_{12}^{-1} | pr \rangle \tag{3}$$

Eqn. (2) becomes $$ \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{12}^{-1}| pr \rangle \tag{4} $$ If we use $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$ on (4), we find $$ \langle pq| r_{12} |q \rangle \langle r | r_{12}^{-1}| pr \rangle \tag{5} $$

It looks like (1) $\neq $ (5). But where things get wrong? Part of (4) is $|\kappa(1) \rangle \otimes |q(2) \rangle \langle \kappa(1) | \otimes \langle r(2) |$, that justifies the usage of $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$.

Suppose we compute an expectation value of $r_{12} r_{13}^{-1}$ over a wave function $\phi_p (1) \otimes \phi_q(2) \otimes \phi_r (3)$, we denote it as $$\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle \tag{1}$$

We can insert a completeness relation $\sum_{\kappa \lambda \tau} | \kappa(1) \lambda(2) \tau(3) \rangle \langle \kappa(1) \lambda(2) \tau(3) | = 1 $ on Eqn. (1)

Therefore, $$ \sum_{\kappa \lambda \tau} \langle pqr | r_{12} | \kappa \lambda\tau \rangle \langle \kappa \lambda\tau | r_{13}^{-1} | pqr\rangle = \sum_{\kappa \lambda \tau} \langle pq | r_{12} | \kappa \lambda \rangle \delta_{r\tau} \langle \kappa \tau | r_{13}^{-1} | p r\rangle \delta_{\lambda q} = \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{13}^{-1}| pr \rangle \tag{2} $$ We rename $\langle \kappa(1) r(3) | r_{13}^{-1}| p(1)r(3) \rangle$ as $$\langle\kappa(1) r(2) | r_{12}^{-1}| p (1) r(2) \rangle \equiv \langle \kappa r | r_{12}^{-1} | pr \rangle \tag{3}$$

Eqn. (2) becomes $$ \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{12}^{-1}| pr \rangle \tag{4} $$ If we use $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$ on (4), we find $$ \langle pq| r_{12} |q \rangle \langle r | r_{12}^{-1}| pr \rangle \tag{5} $$

It looks like (1) $\neq $ (5). But where things get wrong? Part of (4) is $|\kappa(1) \rangle \otimes |q(2) \rangle \langle \kappa(1) | \otimes \langle r(2) |$, that justifies the usage of $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$.

Suppose we compute an expectation value of $r_{12} r_{13}^{-1}$ over a wave function $\phi_p (1) \otimes \phi_q(2) \otimes \phi_r (3)$, we denote it as $$\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle. \tag{1}$$

We can insert a completeness relation $\sum_{\kappa \lambda \tau} | \kappa(1) \lambda(2) \tau(3) \rangle \langle \kappa(1) \lambda(2) \tau(3) | = 1 $ on Eqn. (1)

Therefore, $$ \sum_{\kappa \lambda \tau} \langle pqr | r_{12} | \kappa \lambda\tau \rangle \langle \kappa \lambda\tau | r_{13}^{-1} | pqr\rangle = \sum_{\kappa \lambda \tau} \langle pq | r_{12} | \kappa \lambda \rangle \delta_{r\tau} \langle \kappa \tau | r_{13}^{-1} | p r\rangle \delta_{\lambda q} = \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{13}^{-1}| pr \rangle \tag{2} $$ We rename $\langle \kappa(1) r(3) | r_{13}^{-1}| p(1)r(3) \rangle$ as $$\langle\kappa(1) r(2) | r_{12}^{-1}| p (1) r(2) \rangle \equiv \langle \kappa r | r_{12}^{-1} | pr \rangle \tag{3}$$

Eqn. (2) becomes $$ \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{12}^{-1}| pr \rangle \tag{4} $$ If we use $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$ on (4), we find $$ \langle pq| r_{12} |q \rangle \langle r | r_{12}^{-1}| pr \rangle \tag{5} $$

It looks like (1) $\neq $ (5). But where things get wrong? Part of (4) is $|\kappa(1) \rangle \otimes |q(2) \rangle \langle \kappa(1) | \otimes \langle r(2) |$, that justifies the usage of $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$.

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edited Oct 27 at 11:58

Michael Seifert

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Suppose we compute an expectation value of $r_{12} r_{13}^{-1}$ over a wave function $\phi_p (1) \otimes \phi_q(2) \otimes \phi_r (3)$, we denote it as $\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle $ (1)$$\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle \tag{1}$$

We can insert a completeness relation $\sum_{\kappa \lambda \tau} | \kappa(1) \lambda(2) \tau(3) \rangle \langle \kappa(1) \lambda(2) \tau(3) | = 1 $ on Eqn. (1)

Therefore, $$ \sum_{\kappa \lambda \tau} \langle pqr | r_{12} | \kappa \lambda\tau \rangle \langle \kappa \lambda\tau | r_{13}^{-1} | pqr\rangle = \sum_{\kappa \lambda \tau} \langle pq | r_{12} | \kappa \lambda \rangle \delta_{r\tau} \langle \kappa \tau | r_{13}^{-1} | p r\rangle \delta_{\lambda q} = \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{13}^{-1}| pr \rangle (2) $$$$ \sum_{\kappa \lambda \tau} \langle pqr | r_{12} | \kappa \lambda\tau \rangle \langle \kappa \lambda\tau | r_{13}^{-1} | pqr\rangle = \sum_{\kappa \lambda \tau} \langle pq | r_{12} | \kappa \lambda \rangle \delta_{r\tau} \langle \kappa \tau | r_{13}^{-1} | p r\rangle \delta_{\lambda q} = \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{13}^{-1}| pr \rangle \tag{2} $$ We rename $\langle \kappa(1) r(3) | r_{13}^{-1}| p(1)r(3) \rangle$ as $\langle\kappa(1) r(2) | r_{12}^{-1}| p (1) r(2) \rangle \equiv \langle \kappa r | r_{12}^{-1} | pr \rangle$ (3)$$\langle\kappa(1) r(2) | r_{12}^{-1}| p (1) r(2) \rangle \equiv \langle \kappa r | r_{12}^{-1} | pr \rangle \tag{3}$$

Eqn. (2) becomes $$ \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{12}^{-1}| pr \rangle (4) $$$$ \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{12}^{-1}| pr \rangle \tag{4} $$ If we use $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$ on (4), we find $$ \langle pq| r_{12} |q \rangle \langle r | r_{12}^{-1}| pr \rangle (5) $$$$ \langle pq| r_{12} |q \rangle \langle r | r_{12}^{-1}| pr \rangle \tag{5} $$

It looks like (1) $\neq $ (5). But where things get wrong? Part of (4) is $|\kappa(1) \rangle \otimes |q(2) \rangle \langle \kappa(1) | \otimes \langle r(2) |$, that justifies the usage of $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$.

Suppose we compute an expectation value of $r_{12} r_{13}^{-1}$ over a wave function $\phi_p (1) \otimes \phi_q(2) \otimes \phi_r (3)$, we denote it as $\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle $ (1)

We can insert a completeness relation $\sum_{\kappa \lambda \tau} | \kappa(1) \lambda(2) \tau(3) \rangle \langle \kappa(1) \lambda(2) \tau(3) | = 1 $ on Eqn. (1)

Therefore, $$ \sum_{\kappa \lambda \tau} \langle pqr | r_{12} | \kappa \lambda\tau \rangle \langle \kappa \lambda\tau | r_{13}^{-1} | pqr\rangle = \sum_{\kappa \lambda \tau} \langle pq | r_{12} | \kappa \lambda \rangle \delta_{r\tau} \langle \kappa \tau | r_{13}^{-1} | p r\rangle \delta_{\lambda q} = \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{13}^{-1}| pr \rangle (2) $$ We rename $\langle \kappa(1) r(3) | r_{13}^{-1}| p(1)r(3) \rangle$ as $\langle\kappa(1) r(2) | r_{12}^{-1}| p (1) r(2) \rangle \equiv \langle \kappa r | r_{12}^{-1} | pr \rangle$ (3)

Eqn. (2) becomes $$ \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{12}^{-1}| pr \rangle (4) $$ If we use $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$ on (4), we find $$ \langle pq| r_{12} |q \rangle \langle r | r_{12}^{-1}| pr \rangle (5) $$

It looks like (1) $\neq $ (5). But where things get wrong? Part of (4) is $|\kappa(1) \rangle \otimes |q(2) \rangle \langle \kappa(1) | \otimes \langle r(2) |$, that justifies the usage of $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$.

Suppose we compute an expectation value of $r_{12} r_{13}^{-1}$ over a wave function $\phi_p (1) \otimes \phi_q(2) \otimes \phi_r (3)$, we denote it as $$\langle pqr | r_{12} r_{13}^{-1} |pqr \rangle \tag{1}$$

We can insert a completeness relation $\sum_{\kappa \lambda \tau} | \kappa(1) \lambda(2) \tau(3) \rangle \langle \kappa(1) \lambda(2) \tau(3) | = 1 $ on Eqn. (1)

Therefore, $$ \sum_{\kappa \lambda \tau} \langle pqr | r_{12} | \kappa \lambda\tau \rangle \langle \kappa \lambda\tau | r_{13}^{-1} | pqr\rangle = \sum_{\kappa \lambda \tau} \langle pq | r_{12} | \kappa \lambda \rangle \delta_{r\tau} \langle \kappa \tau | r_{13}^{-1} | p r\rangle \delta_{\lambda q} = \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{13}^{-1}| pr \rangle \tag{2} $$ We rename $\langle \kappa(1) r(3) | r_{13}^{-1}| p(1)r(3) \rangle$ as $$\langle\kappa(1) r(2) | r_{12}^{-1}| p (1) r(2) \rangle \equiv \langle \kappa r | r_{12}^{-1} | pr \rangle \tag{3}$$

Eqn. (2) becomes $$ \sum_\kappa \langle pq| r_{12} |\kappa q \rangle \langle \kappa r | r_{12}^{-1}| pr \rangle \tag{4} $$ If we use $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$ on (4), we find $$ \langle pq| r_{12} |q \rangle \langle r | r_{12}^{-1}| pr \rangle \tag{5} $$

It looks like (1) $\neq $ (5). But where things get wrong? Part of (4) is $|\kappa(1) \rangle \otimes |q(2) \rangle \langle \kappa(1) | \otimes \langle r(2) |$, that justifies the usage of $\sum_{\kappa} |\kappa \rangle \langle \kappa | = 1$.