How to form the spin triplet - singlet states from two electrons with spin not an eigenstate of $ S_z $ (spins not along z-axis))

Question

For a two electron system, we know that the total $ J^2 $ states (Triplet - Singlet) are related with the $\uparrow \downarrow $ , $\downarrow \uparrow$ , $\uparrow \uparrow$ , $\downarrow \downarrow $ states as: $$ \[ \left(\begin{array}{c} \left|11\right\rangle \\ \left|10\right\rangle \\ \left|1-1\right\rangle \\ \left|00\right\rangle \end{array}\right)=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0\\ 0 & 0 & 0 & 1\\ 0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \end{array}\right)\left(\begin{array}{c} \uparrow\uparrow\\ \uparrow\downarrow\\ \downarrow\uparrow\\ \downarrow\downarrow \end{array}\right) \] $$

But what if we want the singlet-triplet not in terms of $\uparrow = \left(\begin{array}{c}1\\0\end{array}\right)$ , $\downarrow =\left(\begin{array}{c}0\\1\end{array}\right)$ , but in terms of spin states that are at an angle $\theta$ from the $z$ axis?

For example when $\hat\eta $ is at $60^o$ from the $z$ axis the spin up and down states are \begin{eqnarray*} \nearrow & = & \left(\begin{array}{c}\frac{\sqrt{3}}{2}\\ \frac{1}{2}\end{array}\right)=\frac{\sqrt{3}}{2}\uparrow+\frac{1}{2}\downarrow\\ \swarrow & = & \left(\begin{array}{c}-\frac{1}{2}\\\frac{\sqrt{3}}{2} \end{array}\right)=-\frac{1}{2}\uparrow+\frac{\sqrt{3}}{2}\downarrow \end{eqnarray*}

By solving for $\uparrow$ and $\downarrow$ in terms of $\nearrow$ and $\swarrow$ and rewriting the $\uparrow \downarrow $ , $\downarrow \uparrow$ , $\uparrow \uparrow$ , $\downarrow \downarrow $ states in terms of the $\nearrow \nearrow$ ,$\nearrow \swarrow$ , $\swarrow \nearrow$ , $\swarrow \swarrow$

we get (after some algebra): $$\left(\begin{array}{c} \left|11\right\rangle \\ \left|10\right\rangle \\ \left|1-1\right\rangle \\ \left|00\right\rangle \end{array}\right)=\left(\begin{array}{cccc} \frac{3}{4} & -\frac{\sqrt{3}}{4} & -\frac{\sqrt{3}}{4} & \frac{1}{4}\\ \sqrt{\frac{3}{8}} & \sqrt{\frac{1}{8}} & \sqrt{\frac{1}{8}} & -\sqrt{\frac{3}{8}}\\ \frac{1}{4} & \frac{\sqrt{3}}{4} & \frac{\sqrt{3}}{4} & \frac{3}{4}\\ 0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \end{array}\right)\left(\begin{array}{c} \nearrow\nearrow\\ \nearrow\swarrow\\ \swarrow\nearrow\\ \swarrow\swarrow \end{array}\right)$$

What is the general method of writing the Triplet - Singlet in terms of two electron states with spin at an arbitrary axis $\hat\eta (\theta)$ ? Thank you!

Answer 1

You are looking for the Wigner d-functions. They relate angular momentum eigenstates through rotation. As you can see in the link the definition is $$d^{(j)}_{m,m'}(\theta) = \langle jm|e^{-i\theta J_y}|j m' \rangle$$ where $e^{-i\theta J_y}$ is a unitary rotation operator.

We'll have two sets of states: $|jm;0\rangle$ for the original basis and $|jm;\theta\rangle$ for eigenstates of momentum along the axis $\hat{\eta}(\theta)$. Let us assume that we've set up our basis so $\hat{\eta}$ lies in the $x-z$ plane. That way the two bases are related to each other by a rotation along the $y$ axis. The two sets of states are then related by a unitary rotation operator: $|jm;\theta\rangle = e^{-i\theta J_y}|jm;0\rangle $

You can look up the various $d$-functions here: http://pdg.lbl.gov/2002/clebrpp.pdf

Using $|jm;0\rangle$ as a complete set of states we can write $$|jm;\theta\rangle = \sum_{m'} |jm';0\rangle\langle jm';0|jm;\theta\rangle$$ $$= \sum_{m'} |jm';0\rangle\langle jm';0|e^{-i\theta J_y}|jm;0\rangle $$ $$|jm;\theta\rangle= \sum_{m'} d^{(j)}_{m',m}(\theta) |jm';0\rangle$$

So you want to perform two steps. First, rotate your total angular momentum state into the basis you'd like to compare to with the $d$-functions. Then second, use the Clebsch-Gordan coefficients to decompose the total-angular momentum states into tensor-product states. The final result is $$|jm;\theta\rangle= \sum_{j_1,j_2,m_1,m_2} \left(\sum_{m'} d^{(j)}_{m',m}(\theta) \langle j_1 m_1;j_2 m_2|j m'\rangle \right) |j_1 m_1; j_2 m_2;\theta\rangle$$

I'll leave you to compare this expression to the matrix you wrote out in your question, but you'll find that plugging in $\theta = 60^\circ$ gives you exactly what you already have.