0
$\begingroup$

I have seen the exchange principle written in two ways, one in terms of coordinates and the other in terms of states:

If $\psi_{AB}(1,2)$ represents particle $A$ in state $1$ and particle $B$ in state $2$ then for bosons: $$\psi_{AB}(1,2)=\psi_{AB}(2,1)$$ and for fermions: $$\psi_{AB}(1,2)=-\psi_{AB}(2,1)$$

and

If $\psi_{AB}(\vec x_1, \vec x_2)$ represents particle $A$ at $\vec x_1$ and particle $B$ at $\vec x_2$ then for bosons: $$\psi_{AB}(\vec x_1,\vec x_2)=\psi_{AB}(\vec x_2,\vec x_1)$$ and for fermions: $$\psi_{AB}(\vec x_1,\vec x_2)=-\psi_{AB}(\vec x_2,\vec x_1)$$

Do both these tell us the exactly same information (i.e. may one hold when the other doesn't) and can this be shown?

$\endgroup$

2 Answers 2

0
$\begingroup$

In the first convention, by state they mean total state (including spatial and spin components) but for the second convention, by state they mean only spatial component. This means that the first convention has more information and hence more restrictive. Particles can be at the same place (following the second convention) yet have different spin states (not following the first convention). So, the second convention is a subset of the first convention by being less restrictive.

By the end of the day, however, fermions should have an antisymmetric total state and bosons should have a symmetric total state.

$\endgroup$
0
$\begingroup$

I am pretty sure that the vector $\vec{x}$ includes spin coordinates. The coordinate itself is an observable which has a continuum spectrum. So a coordinate operator has eigenvalues that characterize a state. So there is no much difference between the first and second definition, since instead of saying an electron has a coordinate $x$, we can also say that the electron is in the state $x$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.