Normalization of Slater Determinant and Antisymmetry

Question

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?

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