Superposition of states, spin 1/2

[bznm]

Homework Statement

Two particles, their spin are 1/2.

The hamiltonian is $H=\gamma s_1 \cdot s_2$

At t=0, the state $|\alpha(0)>$ is such as $s_{1z}|\alpha(0)>=\hbar/2 |\alpha(0)>$ and $s_{2z}|\alpha(0)>=\hbar/2 |\alpha(0)>$. Find the state $|\alpha(0)>$.2. The attempt at a solution
I think that $|\alpha(0)>=|1,1>$ in the base of total angular momentum, i.e. |1/2, 1/2> in the other one. But the correct result seems to be $|\alpha(0)>=1/ \sqrt2 [|1,1; 1/2, 1/2> + 1/ \sqrt 2 (|1,0; 1/2, 1/2>+|0,0; 1/2, 1/2>)]$.. what's wrong?

I used the following table:

 

[blue_leaf77]

$|\alpha(0)>=1/ \sqrt2 [|1,1; 1/2, 1/2> + 1/ \sqrt 2 (|1,0; 1/2, 1/2>+|0,0; 1/2, 1/2>)]$.

I don't see why an unnormalized state should be the correct answer, besides which bases does it use on the RHS?

EDIT: No, I didn't pay a closer attention on the parentheses, yes this state is normalized.

 

[bznm]

I haven't undestood the result in the book. Can you explain me how you would solve this point?

 

[bznm]

I haven't understood why in the result appear the states |1,0;1/2,1/2> and |0,0;1/2,1/2>. They don't have s1z=1/2 AND s2z=1/2. They are linear combination of $s_{1z}=\pm 1/2$ AND $s_{2z}=\pm 1/2$! :(

 

[blue_leaf77]

Did you post the complete question?

 

[bznm]

yes..

 

[Orodruin]

You are correct, the only state which is an eigenstate of both $s_{1z}$ and $s_{2z}$ with eigenvalues $\hbar/2$ is the state $|\uparrow\rangle \otimes |\uparrow\rangle$.

 

[bznm]

thanks to all of you!