# Writing total angular momentum eigenstates in terms of spin and orbital angular momentum eigenstates

I want to understand how we can derive the simultaneous eigenfunctions of the total angular momentum operator and the z component of the total angular momentum operator in terms of the orbital angular momentum and spin operator eigenfunctions. I have found a good resource for this, namely this video. At around 1:14:45, the professor does exactly this for the case of l = 1 and s = 1/2 in a very understandable manner. However, at 1:24:15 he finds the remaining two eigenfunctions (those with j = 1/2) by claiming the states must be orthogonal. I have a few questions regarding this. Firstly, why exactly are these states orthogonal? Is this because the total angular momentum operator must somehow be hermitian? Secondly, in order to calculate the values that the professor claimed the last two eigenfunctions were, I had to use that the functions were unit length. Why is this the case? And lastly: Even with the restriction that the states are orthogonal, there are two possible orientations for the resultant vectors. How do I distinguish between these the obtain the correct result? Thank you.

• it would be better if you could write a self-contained question that does not assume people will have to spend more than one hour watching a video. – ZeroTheHero Sep 29 at 12:12
• relevant time-stamps are in the question. – user208480 Sep 29 at 12:14
• Still... what if the link rots etc? Can you make the effort of properly typesetting your question? – ZeroTheHero Sep 29 at 12:27

The strategy is always the same: to construct states with a given $$j$$, look for the state $$\vert j j\rangle$$ as this state is the only state killed by $$J_+$$ and is an eigenstate of $$J_z$$ with eigenvalue $$j$$.
Hence, write \begin{align} \bigl\vert \textstyle 1/2, 1/2 \rangle = \alpha \vert 1,1\rangle \vert 1/2,-1/2\rangle + \beta \vert 1,0\rangle \vert 1/2,1/2\rangle \, . \end{align} Acting with $$J_+=L_++S_+$$ gives \begin{align} 0=\alpha \vert 1,1\rangle \vert 1/2,1/2\rangle +\sqrt{2}\beta \vert 1,1\rangle \vert 1/2,\vert 1/2\rangle \end{align} from which it follows that $$\alpha=\sqrt{\frac{2}{3}}$$ and $$\beta=-\sqrt{\frac{1}{3}}$$, with the overall sign in agreement with the Condon-Shortley phase convention.
One can also proceed by orthogonality as suggested. States with different $$j$$ but the same $$m_j$$ value must be orthogonal in the same way as angular momentum state with different $$\ell$$ but same $$m_\ell$$ are orthogonal.
• I am not using orthogonality to find $\alpha$ and $\beta$. It follows from $J_+\vert j j\rangle=0$, which forces the relation between $\alpha$ and $\beta$. Since the state is normalized you need $\vert\alpha\vert^2+\vert\beta\vert^2=1$, which completely fixes the values. It will automatically be orthogonal to the $j=3/2,m_j=1/2$ state. – ZeroTheHero Sep 29 at 14:10
• yes. It would still be a valid state of course since any multiple of an eigenstate of $j^2$ and $j_z$ is also an eigenstate: if $\vert\psi_n\rangle$ is a solution to the TISE, so is $-\vert\psi_n\rangle$. That's where Condon-Shortley comes in: it states that the coefficient of $\vert \ell_1\ell_1\rangle\vert s,m_s\rangle$ should be taken as positive. – ZeroTheHero Sep 29 at 14:21