- #1
TimeRip496
- 254
- 5
$$|\psi(x_1,x_2)|^2=|\psi(x_2,x_1)|^2$$
$$\psi(x_1,x_2)=+/-\psi(x_2,x_1)$$
How do they convert they former into the latter one? Is it due to the modulus?
I know the latter can also be written as $$\psi(x_1,x_2)=e^{i\alpha}\psi(x_2,x_1)$$ where the exponential is the phase used to replace +/-.
$$\psi(x_1,x_2)=A[\psi_a(x_1)\psi_b(x_2)\pm\psi_a(x_2)\psi_b(x_1)]$$
As for this, isn't $$\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2)$$? Why do we need to add the additional one?
Is it because, the particles are indistinguishable and thus we can add $$\psi_a(x_2)\psi_b(x_1)$$?
If that is the case, won't $$(A\psi_a(x_1)\psi_b(x_2))^2$$ or $$(A\psi_a(x_2)\psi_b(x_1))^2$$ be 0.5(probability) each?
$$\psi(x_1,x_2)=+/-\psi(x_2,x_1)$$
How do they convert they former into the latter one? Is it due to the modulus?
I know the latter can also be written as $$\psi(x_1,x_2)=e^{i\alpha}\psi(x_2,x_1)$$ where the exponential is the phase used to replace +/-.
$$\psi(x_1,x_2)=A[\psi_a(x_1)\psi_b(x_2)\pm\psi_a(x_2)\psi_b(x_1)]$$
As for this, isn't $$\Psi(x_1,x_2)=\Psi_a(x_1) \Psi_b(x_2)$$? Why do we need to add the additional one?
Is it because, the particles are indistinguishable and thus we can add $$\psi_a(x_2)\psi_b(x_1)$$?
If that is the case, won't $$(A\psi_a(x_1)\psi_b(x_2))^2$$ or $$(A\psi_a(x_2)\psi_b(x_1))^2$$ be 0.5(probability) each?