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Scattering — John S. Townsend
Physics — Quantum Mechanics · identical-particles-antisymmetric-states
Problem Verify for an antisymmetric spatial state under exchange of two particles that
∣ ψ A ⟩ = 1 2 ∫ d 3 r 1 d 3 r 2 ( 1 2 ∣ r 1 , r 2 ⟩ − 1 2 ∣ r 2 , r 1 ⟩ ) × ( 1 2 ⟨ r 1 , r 2 ∣ ψ A ⟩ − 1 2 ⟨ r 2 , r 1 ∣ ψ A ⟩ ) = ∫ d 3 r 1 d 3 r 2 ∣ r 1 , r 2 ⟩ ⟨ r 1 , r 2 ∣ ψ A ⟩ \begin{aligned}
|\psi_A\rangle &= \frac{1}{2} \int d^3r_1\, d^3r_2 \left(\frac{1}{\sqrt{2}}|\mathbf{r}_1, \mathbf{r}_2\rangle - \frac{1}{\sqrt{2}}|\mathbf{r}_2, \mathbf{r}_1\rangle\right) \\
&\quad \times \left(\frac{1}{\sqrt{2}}\langle\mathbf{r}_1, \mathbf{r}_2|\psi_A\rangle - \frac{1}{\sqrt{2}}\langle\mathbf{r}_2, \mathbf{r}_1|\psi_A\rangle\right) \\
&= \int d^3r_1\, d^3r_2\, |\mathbf{r}_1, \mathbf{r}_2\rangle\langle\mathbf{r}_1, \mathbf{r}_2|\psi_A\rangle
\end{aligned} ∣ ψ A ⟩ = 2 1 ∫ d 3 r 1 d 3 r 2 ( 2 1 ∣ r 1 , r 2 ⟩ − 2 1 ∣ r 2 , r 1 ⟩ ) × ( 2 1 ⟨ r 1 , r 2 ∣ ψ A ⟩ − 2 1 ⟨ r 2 , r 1 ∣ ψ A ⟩ ) = ∫ d 3 r 1 d 3 r 2 ∣ r 1 , r 2 ⟩ ⟨ r 1 , r 2 ∣ ψ A ⟩ Browse the problem library