Superposition of states, spin 1/2

bznm

1. 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: Related Advanced Physics Homework Help News on Phys.org

blue_leaf77

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

Last edited:

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

Homework Helper
Did you post the complete question?

yes..

Orodruin

Staff Emeritus
Homework Helper
Gold Member
2018 Award
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!

"Superposition of states, spin 1/2"

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