1
$\begingroup$

In Sakurai's quantum mechanics book 3rd edition page 448, equation 7.88, the book writes

"Kohn and Sham found a way to derive a self-consistent approximation scheme, based on single particle wave functions $\phi_j(\boldsymbol{x})$ that solve a particular, albeit fictitious, one body Schrodinger equation. The $\phi_j(\boldsymbol{x})$ form the multiparticle wave function $\Psi(\boldsymbol{x}_1,\boldsymbol{x}_2,...,\boldsymbol{x}_N)$ trough the $N$-body generalization of 7.60 (the Slater determinant), that is $$\Psi(\boldsymbol{x}_1,\boldsymbol{x}_2,...,\boldsymbol{x}_N)=\phi(\boldsymbol{x}_1)\phi_2(\boldsymbol{x}_2)...\phi_N(\boldsymbol{x}_N)\pm\phi_2(\boldsymbol{x}_1)\phi_1(\boldsymbol{x}_2)...\phi_N(\boldsymbol{x}_N)\pm..."$$

My question is, is there a $\frac{1}{\sqrt{N!}}$ factor missing in the above equation?

$\textbf{Update}$: I suspect the book doesn't bother with normalization, because in the case of symmetric functions and if $\phi_i$'s are not necessairily distinct, the normalization constant would be tedious to write out.

$\endgroup$

1 Answer 1

2
$\begingroup$

Yes, the Slater determinant should have a factor of $1/\sqrt{N!}$ for the state to be properly normalized.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.