I'm reading the Wikipedia article on DFT, and it says that there is a one-to-one correspondence between the ground state particle density

$$n_0(\vec{r}) = N \int \text{d}^3 r_2 \int \text{d}^3 r_3 \cdots \int \text{d}^3 r_N \Psi_0^{*}(\vec{r}, \vec{r}_2, \ldots, \vec{r}_N) \Psi_0(\vec{r}, \vec{r}_2, \ldots, \vec{r}_N)$$

and the ground state wavefunction $\Psi_0(\vec{r}, \vec{r}_2, \ldots, \vec{r}_N)$ such that $\Psi_0$ is a functional of $n_0$:

$$\Psi_0 = \Psi[n_0]$$

I don't understand how this is possible. It seems to me that you could have an infinite number of wavefunctions, with differing phases, as long as the squared amplitudes end up the same. Could someone explain to me why this isn't possible?

  • 2
    $\begingroup$ Good question! I have a feeling they mean you can determine $\Psi_0$ up to an overall phase, i.e. that there is a one-to-one correspondence between ground state particle densities and $U(1)$ equivalence classes of wavefunctions, but without reading the paper that Wikipedia cites, I'm hesitant to edit the article to reflect that. $\endgroup$
    – David Z
    Commented May 27, 2014 at 0:46
  • 1
    $\begingroup$ Also note that depending who you ask, the wavefunction is defined as an equivalence class of functions, so their statement could be considered correct as written. $\endgroup$
    – d_b
    Commented May 30, 2014 at 16:05

1 Answer 1


The statement is supposed to read "unique up to global phase".

  • 1
    $\begingroup$ What Michael said. Probably in DFT (for simplicity's sake), they choose the phase such that the ground state is real. $\endgroup$
    – Nick
    Commented May 30, 2014 at 15:56

Your Answer

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

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