Rotation of singlet state

Question

I have to show that the singlet state $$ |\psi \rangle = \frac {1}{\sqrt2} [\uparrow_z\downarrow_z - \downarrow_z\uparrow_z] $$ is invariant under rotations.

Applying the rotation operator, $$ (R_{1z} \otimes R_{2z})|\psi \rangle = (R_{1z} \otimes R_{2z})\frac {1}{\sqrt2}\uparrow_z\downarrow_z - (R_{1z} \otimes R_{2z})\frac {1}{\sqrt2}\downarrow_z\uparrow_z$$ I am not sure what to do now. We know that $$ \uparrow_z\downarrow_z = \uparrow_z\otimes\downarrow_z$$ Then $$(R_{1z} \otimes R_{2z})\uparrow_z\downarrow_z = (R_{1z} \otimes R_{2z})(\uparrow_z\otimes\downarrow_z)$$

I don't know what to do now. I am not even sure if this is the right way to solve this problem.

(I have been trying for many days but still can't understand it, so please don't close the question.) Edit: Using Jakob's hint, I have done $$(R_{1z} \otimes R_{2z})\uparrow_z\downarrow_z = (R_{1z} \uparrow_z ) \otimes (R_{2z}\downarrow_z) $$ $$(R_{1z} \otimes R_{2z})\downarrow_z\uparrow_z = (R_{1z} \downarrow_z ) \otimes (R_{2z}\uparrow_z) $$ where $$R_{z} = \begin{bmatrix} \cos (\theta/2) -i\sin(\theta/2)& 0 \\ 0 &\cos (\theta/2) +i\sin(\theta/2) \end{bmatrix}$$ and $$\uparrow_z= \begin{bmatrix} 1 \\ 0 \end{bmatrix}$$ $$\downarrow_z= \begin{bmatrix} 0 \\ 1 \end{bmatrix}$$ So,

$$R_{1z} \uparrow_z = \begin{bmatrix} \cos (\theta/2) -i\sin(\theta/2)\\ 0 \end{bmatrix}$$ $$R_{2z} \downarrow_z = \begin{bmatrix} 0 \\ \cos (\theta/2) +i\sin(\theta/2) \end{bmatrix}$$ $$R_{1z} \downarrow_z = \begin{bmatrix} 0 \\ \cos (\theta/2) +i\sin(\theta/2) \end{bmatrix}$$ $$R_{2z} \uparrow_z = \begin{bmatrix} \cos (\theta/2) -i\sin(\theta/2)\\ 0 \end{bmatrix}$$ Then $$(R_{1z} \uparrow_z ) \otimes (R_{2z}\downarrow_z) = \begin{bmatrix} 0\\ \cos^{2} (\theta/2) -i^{2}\sin^{2}(\theta/2)\\ 0\\ 0 \end{bmatrix} $$ $$(R_{1z} \downarrow_z ) \otimes (R_{2z}\uparrow_z) = \begin{bmatrix} 0\\ 0\\ \cos^{2} (\theta/2) -i^{2}\sin^{2}(\theta/2)\\ 0 \end{bmatrix} $$ and $\cos^{2} (\theta/2) -i^{2}\sin^{2}(\theta/2) = 1$ So, $$(R_{1z} \uparrow_z ) \otimes (R_{2z}\downarrow_z) = \begin{bmatrix} 0\\ 1\\ 0\\ 0 \end{bmatrix} = \uparrow_z \otimes \downarrow_z$$ $$(R_{1z} \downarrow_z ) \otimes (R_{2z}\uparrow_z) = \begin{bmatrix} 0\\ 0\\ 1 \\ 0 \end{bmatrix} = \downarrow_z \otimes \uparrow_z$$ Thus, we get back the original state. Is it fine or am I still missing something?

Answer 1

Consider a system of two particles $\alpha$ and $\beta$ each with spin 1/2. They are in the same real three di-mensional space. We rotate the system in this space so each particle by the same rotation, that is $R_{1z}\equiv R_{2z}$.

The rotation in the 2-dimensional Hilbert space $\mathcal H_{\rho}\:(\rho=\alpha,\beta)$ of each particle is represented by a special unitary operator \begin{equation} {{^{\bf 2}}}U_{\alpha}= {^{\bf 2}}U={^{\bf 2}}U_{\beta} \in SU(2) \tag{01}\label{01} \end{equation} so by the special unitary operator \begin{equation} {^{\bf 4}}U={^{\bf 2}} U_{\alpha}\otimes {^{\bf 2}}U_{\beta}= {^{\bf 2}}U\otimes {^{\bf 2}}U \in SU(4) \tag{02}\label{02} \end{equation}
in the product 4-dimensional Hilbert space $\mathcal H=\mathcal H_{\alpha}\otimes \mathcal H_{\beta}$.

For what next see $\:\color{blue}{\textbf{Example A}}\:$ in my answer here What is the symmetry of the pion triplet (π−,π0,π+)?.

So try this : Represent the particle states by $\:2\times 1\:$ matrices \begin{equation} \uparrow_z = \begin{bmatrix} \:1\:\vphantom{\dfrac12}\\ \:0\:\vphantom{\dfrac12} \end{bmatrix} \qquad \downarrow_z = \begin{bmatrix} \:0\:\vphantom{\dfrac12}\\ \:1\:\vphantom{\dfrac12} \end{bmatrix} \tag{03}\label{03} \end{equation} Express the product singlet state as $\:4\times 1\:$ matrix \begin{equation} |\boldsymbol{1}\rangle =\uparrow_z\downarrow_z-\downarrow_z\uparrow_z = \begin{bmatrix} \:1\:\vphantom{\dfrac12}\\ \:0\:\vphantom{\dfrac12} \end{bmatrix} \otimes \begin{bmatrix} \:0\:\vphantom{\dfrac12}\\ \:1\:\vphantom{\dfrac12} \end{bmatrix} - \begin{bmatrix} \:0\:\vphantom{\dfrac12}\\ \:1\:\vphantom{\dfrac12} \end{bmatrix} \otimes \begin{bmatrix} \:1\:\vphantom{\dfrac12}\\ \:0\:\vphantom{\dfrac12} \end{bmatrix} = \begin{bmatrix} \:0\:\vphantom{\tfrac{a}{b}}\\ \:1\:\vphantom{\tfrac{a}{b}}\\ \:0\:\vphantom{\tfrac{a}{b}}\\ \:0\:\vphantom{\tfrac{a}{b}} \end{bmatrix} - \begin{bmatrix} \:0\:\vphantom{\tfrac{a}{b}}\\ \:0\:\vphantom{\tfrac{a}{b}}\\ \:1\:\vphantom{\tfrac{a}{b}}\\ \:0\:\vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{04}\label{04} \end{equation} that is \begin{equation} |\boldsymbol{1}\rangle = \uparrow_z\downarrow_z-\downarrow_z\uparrow_z = \begin{bmatrix} \hphantom{-}0\hphantom{-}\vphantom{\tfrac{a}{b}}\\ +1\hphantom{-}\vphantom{\tfrac{a}{b}}\\ -1\hphantom{-}\vphantom{\tfrac{a}{b}}\\ \hphantom{-}0\hphantom{-}\vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{05}\label{05} \end{equation} Represent the rotation by a $\:2\times 2\:$ special unitary matrix \begin{equation} {^{\bf 2}}U = \begin{bmatrix} \:\hphantom{-}g & \hphantom{-}h\:\vphantom{\dfrac12}\\ \:-\overline{h} & \hphantom{-}\overline{g}\:\vphantom{\dfrac12} \end{bmatrix} \qquad g\overline{g}+h\overline{h}=1 \tag{06}\label{06} \end{equation} and find the $\:4\times 4\:$ special unitary matrix in the product space, see equation \eqref{02} \begin{equation} {^{\bf 4}}U={^{\bf 2}}U\otimes {^{\bf 2}}U=\left({^{\bf 2}}U\right)^{\otimes 2} \in SU(4) \tag{07}\label{07} \end{equation} Then find what is the result of the application of $\:{^{\bf 4}}U\:$ on the singlet $\:|\boldsymbol{1}\rangle\:$ \begin{equation} {^{\bf 4}}U|\boldsymbol{1}\rangle = ??? \tag{08}\label{08} \end{equation}

Alternatively, due to the fact that the factor spaces $\mathcal H_{\alpha},\mathcal H_{\beta}$ are of the same dimension $\:2$ we could represent the product states as $\:2\times 2\:$ matrices. Especially for the singlet we have ^{\begin{equation} |\psi\rangle =\uparrow_z\downarrow_z-\downarrow_z\uparrow_z = \begin{bmatrix} \:1\:\vphantom{\dfrac12}\\ \:0\:\vphantom{\dfrac12} \end{bmatrix} \begin{bmatrix} \:0\:& 1\vphantom{\dfrac12} \end{bmatrix} - \begin{bmatrix} \:0\:\vphantom{\dfrac12}\\ \:1\:\vphantom{\dfrac12} \end{bmatrix} \begin{bmatrix} \:1\:& 0\vphantom{\dfrac12} \end{bmatrix} = \begin{bmatrix} \:\:0\:&\:1\:\: \vphantom{\dfrac12}\\ \:0\:&\:0\:\: \vphantom{\dfrac12} \end{bmatrix} - \begin{bmatrix} \:\:0\:&\:0\:\: \vphantom{\dfrac12}\\ \:1\:&\:0\:\: \vphantom{\dfrac12} \end{bmatrix} \tag{09}\label{09} \end{equation}} that is \begin{equation} |\psi\rangle =\uparrow_z\downarrow_z-\downarrow_z\uparrow_z = \begin{bmatrix} \hphantom{-}0\:& +1\:\: \vphantom{\dfrac12}\\ -1\:&\hphantom{+}0\:\: \vphantom{\dfrac12} \end{bmatrix} \tag{10}\label{10} \end{equation} Then using the special unitary matrix $\:{^{\bf 2}}U\:$ of \eqref{06} find the product state represented by the following $\:2\times 2\:$ matrix \begin{equation} \left({^{\bf 2}}U\right)|\psi\rangle\left({^{\bf 2}}U\right)^{\boldsymbol{\top}} =\left({^{\bf 2}}U\right)\left(\uparrow_z\downarrow_z-\downarrow_z\uparrow_z\!\!\vphantom{\tfrac12}\right)\left({^{\bf 2}}U\right)^{\boldsymbol{\top}} \tag{11}\label{11} \end{equation} where $\:\left({^{\bf 2}}U\right)^{\boldsymbol{\top}}\:$ the transpose of $\:\left({^{\bf 2}}U\right)$.