- #1
RedX
- 970
- 3
Not every state can be represented by a direct product. Do states that can be written as a direct product have anything special about them?
It seems that states that can be written as a direct product lose correlation between between each individual state. More specifically, stronger than losing correlation, they become independent.
As an example, take two protons with angular momentum zero:
[tex]\psi(r_1,r_2) (|+->-|-+>) [/tex]
which is a direct product of a function of position, and an antisymmetric spin state. By writing the wavefunction as a direct product, you seem to have lost information on whether the spin up proton is near r1 or r2.
Indeed, measurement of position is independent of measurement of spin, and:
[tex]<f(r_1,r_2)g(s_1,s_2)>=<f(r_1,r_2)><g(s_1,s_2)>[/tex]
Going back to the two protons with angular momentum zero, why is it that the ground state is:
[tex]\psi_S(r_1,r_2) (|+->-|-+>) [/tex]
where S stands for symmetric in position. This total expression is definitely antisymmetric which is required for identical fermions. But does the ground state have to be written in a way such that the spin and positions are uncorrelated?
And why is the next excited state:
[tex]\psi_A(r_1,r_2) (spin \;1\; triplet) [/tex]
again uncorrelated?
Does this have to do with the fact that naturally, position doesn't care about spin, that it's not more likely for spin to like the negative x-axis than the positive x axis?
Anyways, are there anymore special properties about states that can be written as direct products?
It seems that states that can be written as a direct product lose correlation between between each individual state. More specifically, stronger than losing correlation, they become independent.
As an example, take two protons with angular momentum zero:
[tex]\psi(r_1,r_2) (|+->-|-+>) [/tex]
which is a direct product of a function of position, and an antisymmetric spin state. By writing the wavefunction as a direct product, you seem to have lost information on whether the spin up proton is near r1 or r2.
Indeed, measurement of position is independent of measurement of spin, and:
[tex]<f(r_1,r_2)g(s_1,s_2)>=<f(r_1,r_2)><g(s_1,s_2)>[/tex]
Going back to the two protons with angular momentum zero, why is it that the ground state is:
[tex]\psi_S(r_1,r_2) (|+->-|-+>) [/tex]
where S stands for symmetric in position. This total expression is definitely antisymmetric which is required for identical fermions. But does the ground state have to be written in a way such that the spin and positions are uncorrelated?
And why is the next excited state:
[tex]\psi_A(r_1,r_2) (spin \;1\; triplet) [/tex]
again uncorrelated?
Does this have to do with the fact that naturally, position doesn't care about spin, that it's not more likely for spin to like the negative x-axis than the positive x axis?
Anyways, are there anymore special properties about states that can be written as direct products?