I am learning about the angular momentum addition rule. we have two particles, both with s=1/2 and L=0, and we need to find the total J of the system. We know that J = J1+J2, J1-J2, so in our case J=1 or 0. For case 1, $$m_j = 1,0,-1$$ and for case 0 $$m_j = 0$$
The eigenstates are
$$|1,1 \rangle = |1/2,1/2\rangle_1 \otimes |1/2,1/2\rangle_2 $$
$$|1,0 \rangle = 1/\sqrt(2) *[ |1/2,-1/2\rangle_1 \otimes|1/2,1/2\rangle_2 + |1/2,1/2\rangle_1 \otimes|1/2,-1/2\rangle_2]$$
$$|1,-1 \rangle = |1/2,-1/2\rangle_1 \otimes |1/2,-1/2\rangle_2 $$
and finally
$$|0,0 \rangle = 1/\sqrt(2) *[ |1/2,-1/2\rangle_1 \otimes|1/2,1/2\rangle_2 – |1/2,1/2\rangle_1 \otimes|1/2,-1/2\rangle_2]$$
What I don't get is the physical difference between $$|1,0 \rangle $$ and $$|0,0 \rangle $$
I mean, both particles have opposite spin in both cases.
My question is ,what experiment could help us tell these two states apart ?
What came to mind is that |0,0> means that when we measure the spin in ANY axis, we get 0, while |1,0> means that it is 0 in the z axis and none 0 in the others. But how could this be true, for case |0,0> if i am certain about the measurement in one axis, lets say x, i an uncertain about the others. So how can i be sure that i wont get the same spin for both particles in the y axis thus getting a state
$$ |0,m_y =1\rangle $$
I feel that maybe |0,0> is entangled while |1,0> is not, but can only provide hand waving arguments for it.
I go the info from here, as well as some handwritten notes from my professor(talking about a neutron and an electron, but give the same solution).
Thank you for your time.
Best Answer
The two states are:
$$|0,0\rangle = \frac 1 {\sqrt 2}\big( |+-\rangle - |-+\rangle \big) $$
$$|1,0\rangle = \frac 1 {\sqrt 2}\big( |+-\rangle + |-+\rangle \big) $$
which are both combinations of simultaneous eigenstates of $\hat S_z^{(1)}$ and $\hat S_z^{(2)}$ with different eigenvalues.
The only difference between the total spin eigenstates is the relative phase of single particle eigenstates. That's it.
Since relative phase is entirely quantum mechanical, it's difficult to give "physical" reason that will satisfy classical intuition.
From images available on the internet, angular momentum addition is shown as:
the spin-1/2 is:
and spin-1
So for $|1,0\rangle$, the individual eigenstates are in-phase and add up to $S=1$, $S_z=0$, while out-of-phase is a complete cancelation. The latter case is straight forward, while in the former case, it's also straight forward to get $S_z=0$. To get the $S=1$, you just have to imagine the little cones tilted,. well.. in-phase, so they wind up in the disk representing $|1, 0\rangle$. Not very satisfying, but it's pretty much all we got other than math.