Normalization of Slater Determinant and Antisymmetry

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

$$\vert\chi_2\chi_1\rangle=-\vert\chi_1\chi_2\rangle.$$

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

$$\langle\chi_1\chi_2\vert\chi_2\chi_1\rangle=-\langle\chi_1\chi_2\vert\chi_1\chi_2\rangle=-1.$$

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

• What do you mean what that relation means? Commented May 25, 2021 at 9:53
• 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?
– ewf
Commented May 25, 2021 at 9:55
• @ewf Well, it's just the anti-symmetry of the Slater-Determinant, no? Commented May 25, 2021 at 9:57

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: $$\frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle +|\chi_3\chi_1\rangle\big) = \frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle - |\chi_1\chi_3\rangle\big) \not=\frac{1}{\sqrt{2}}\big(|\chi_1\chi_2\rangle + |\chi_1\chi_3\rangle\big).$$