Show that the many-body wavefunction for bosons is symmetric


 * 1) Consider without loss of generality two bodies.
 * 2) Find \(\psi(x_1,x_2)\).
 * 3) Set up the wavefunction in terms of creation operators: \(\psi(x_1,x_2)= \frac{1}{\sqrt{2}}\left(\langle x_1 | \hat{a}_{s_1}^{\dagger} |0\rangle\langle x_2 | \hat{a}_{s_2}^{\dagger} |0\rangle + \langle x_2 | \hat{a}_{s_2}^{\dagger} |0\rangle\langle x_1 | \hat{a}_{s_1}^{\dagger} |0\rangle\right)\)
 * 4) Expand: \(\psi(x_1,x_2)= \frac{1}{\sqrt{2}}\left(\langle x_1 x_2 | \hat{a}_{s_1}^{\dagger}\hat{a}_{s_2}^{\dagger} + \hat{a}_{s_2}^{\dagger}\hat{a}_{s_1}^{\dagger} |00\rangle\right)\)
 * 5) Use the commutation relation \([\hat{a},\hat{a}^{\dagger}]=0\) to simplify: \(\psi(x_1,x_2)= \frac{2}{\sqrt{2}}\left(\langle x_1 x_2 | \hat{a}_{s_1}^{\dagger}\hat{a}_{s_2}^{\dagger}|00\rangle\right)\)
 * 6) Swap the labels of the particles: \(\psi(x_1,x_2)= \frac{2}{\sqrt{2}}\left(\langle x_2 x_1 | \hat{a}_{s_2}^{\dagger}\hat{a}_{s_1}^{\dagger}|00\rangle\right)\)
 * 7) Find \(\psi(x_2,x_1)\).
 * 8) Set up the wavefunction in terms of creation operators: \(\psi(x_2,x_1)= \frac{1}{\sqrt{2}}\left(\langle x_2 | \hat{a}_{s_1}^{\dagger} |0\rangle\langle x_1 | \hat{a}_{s_2}^{\dagger} |0\rangle + \langle x_1 | \hat{a}_{s_2}^{\dagger} |0\rangle\langle x_2 | \hat{a}_{s_1}^{\dagger} |0\rangle\right)\)
 * 9) Expand: \(\psi(x_2,x_1)= \frac{1}{\sqrt{2}}\left(\langle x_2 x_1 | \hat{a}_{s_2}^{\dagger}\hat{a}_{s_1}^{\dagger} + \hat{a}_{s_1}^{\dagger}\hat{a}_{s_2}^{\dagger} |00\rangle\right)\)
 * 10) Use the commutation relation \([\hat{a},\hat{a}^{\dagger}]=0\) to simplify: \(\psi(x_2,x_1)= \frac{2}{\sqrt{2}}\left(\langle x_2 x_1 | \hat{a}_{s_2}^{\dagger}\hat{a}_{s_1}^{\dagger}|00\rangle\right)\)
 * 11) Substitute the expression for \(\psi(x_2,x_1)\) into the one for \(\psi(x_1,x_2)\): \(\psi(x_1,x_2) = \psi(x_2,x_1)\)