# I Two Identical non-entangled Particle System

1. Jul 26, 2016

### TimeRip496

$$|\psi(x_1,x_2)|^2=|\psi(x_2,x_1)|^2$$
$$\psi(x_1,x_2)=+/-\psi(x_2,x_1)$$
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?

2. Jul 26, 2016

### Truecrimson

In three dimensions, the phase $e^{i \alpha}$ reduces to $\pm 1$ by asserting that swapping two particles twice should be the same as doing nothing. So the swapping operator must have eigenvalues $\pm 1$. In two dimensions, you can get anyons besides bosons and fermions. For more than two particles, you can get states that are not symmetric nor antisymmetric.

If there is no external label such as location (e.g. one particle is in D.C. while another is in Moscow) you can't distinguish between the first and the second terms.

Last edited: Jul 26, 2016
3. Jul 27, 2016

### TimeRip496

But how do you get $$\psi(x_1,x_2)=+/-\psi(x_2,x_1)$$ from
$$|\psi(x_1,x_2)|^2=|\psi(x_2,x_1)|^2$$?

4. Jul 27, 2016

### Truecrimson

Without simply asserting it, I don't think there is a satisfactory justification for the first line other than the spin-statistics theorem in 3+1 (or higher)-dimensional quantum field theories. That's why anyons are possible in 2+1 dimensions.

Edit: Source