Consider a system consisting of two electrons with Slater-Determinant $\vert\chi_1\chi_2\rangle$, where $\chi_1$ and $\chi_2$ are one-electron orbitals (spin-orbitals). The Slater Determinant is normalized, i.e, $\langle\chi_1\chi_2\vert\chi_1\chi_2\rangle=1$. Further, the antisymmetry of $\vert\chi_1\chi_2\rangle$ manifests as


Project now the latter relation on $\langle\chi_1\chi_2\vert$, i.e.,


My Question: What does the relation $\langle\chi_1\chi_2\vert\chi_2\chi_1\rangle=-1$ mean?

  • 1
    $\begingroup$ What do you mean what that relation means? $\endgroup$
    – Physiker
    Commented May 25, 2021 at 9:53
  • $\begingroup$ The states seem not to be orthogonal and therefore I wondered about the minus sign. Is there a physical meaning behind this minus sign or is it related to a certain symmetry? $\endgroup$
    – ewf
    Commented May 25, 2021 at 9:55
  • 2
    $\begingroup$ @ewf Well, it's just the anti-symmetry of the Slater-Determinant, no? $\endgroup$ Commented May 25, 2021 at 9:57

1 Answer 1


Let $|\psi\rangle$ and $|\phi\rangle$ be two arbitrary normalized states. If $\langle\psi|\phi\rangle = 0$, then the states are orthogonal. If $\langle\psi|\phi\rangle = 1$, then the states are the same. If $\langle\psi|\phi\rangle = e^{i\theta}$, for some $\theta\in\mathbb{R}$, then the states are the same up to a global phase. Since the global phase has no bearing on the physics of the system, when $\langle\psi|\phi\rangle = e^{i\theta}$, the states are the same.

So for single Slater determinants, exchange of electrons has no bearing on the physics of the system, meaning that $|\chi_1\chi_2\rangle$ and $|\chi_2\chi_1\rangle$ are the same state. However, the antisymmetry property is important when you consider linear combinations of Slater determinants. For example, the state $$ \frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle + |\chi_1\chi_3\rangle\big)$$ is a different state from $$ \frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle + |\chi_3\chi_1\rangle\big)$$ because the antisymmetry property of Slater determinants now manifests as a change in the relative phase between two Slater determinants in superposition: $$ \frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle +|\chi_3\chi_1\rangle\big) = \frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle - |\chi_1\chi_3\rangle\big) \not=\frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle + |\chi_1\chi_3\rangle\big).$$

  • $\begingroup$ Thanks a lot for your answer! $\endgroup$
    – ewf
    Commented Jun 1, 2021 at 16:19

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