- #1
center o bass
- 560
- 2
Hey I'm tring to grasp the symmetrization principle for identical particles. If the particles are indistinguishble, for example two electrons (not regarding spin), then [tex]\psi(x_1,x_2)[/tex] must correspond to the same physical situation as [tex]\psi(x_2,x_1)[/tex]. It follows that
[tex]|\psi(x_1,x_2)|^2 = |\psi(x_2,x_1)|^2[/tex]
and from this the symmetrization principle follows as
[tex] \psi(x_1,x_2) = \pm \psi(x_2,x_1)[/tex].
This is all okay... However, if we had two electrons 1 and 2 with differents spins along the z-axis m1 and m2. Then how could the physical situation of 1 being at x1 with z-spin m1 and 2 being at x2 with spin m2 be the same as the situation of 2 being at x1 with spin m2 while 1 is at x2 with spin m1?
That is how could the physical situation of
[tex]\psi_{m_1,m_2}(x_1,x_2)[/tex]
be the same as the situation
[tex]\psi_{m_2,m_1}(x_2,x_1)?[/tex]
Could'nt we just mesure the z-spin of the particle at x1 in the two situations and determine the difference?
[tex]|\psi(x_1,x_2)|^2 = |\psi(x_2,x_1)|^2[/tex]
and from this the symmetrization principle follows as
[tex] \psi(x_1,x_2) = \pm \psi(x_2,x_1)[/tex].
This is all okay... However, if we had two electrons 1 and 2 with differents spins along the z-axis m1 and m2. Then how could the physical situation of 1 being at x1 with z-spin m1 and 2 being at x2 with spin m2 be the same as the situation of 2 being at x1 with spin m2 while 1 is at x2 with spin m1?
That is how could the physical situation of
[tex]\psi_{m_1,m_2}(x_1,x_2)[/tex]
be the same as the situation
[tex]\psi_{m_2,m_1}(x_2,x_1)?[/tex]
Could'nt we just mesure the z-spin of the particle at x1 in the two situations and determine the difference?