Density matrix of a mixture of singlets and triplets

Question

I was wondering how to write the density matrix of two spin-$\frac{1}{2}$ in a mixture of singlets and triplets. By definition for a mixture, $$ \hat{\rho}=\sum_{j}w_j\, |\psi_j \rangle\langle \psi_j| . $$ But it makes sense for a mixture, given the singlet $$ |0,0\rangle = \frac{1}{\sqrt{2}}(|\downarrow\uparrow\rangle - |\uparrow\downarrow\rangle) $$ that is a superposition of states, to write $|0,0 \rangle\langle 0,0|$? I think not because it will be coherence.

I was thinking also to do $$ \hat{\rho}=w_1|\uparrow\downarrow \rangle\langle \uparrow\downarrow|+w_2|\downarrow\uparrow \rangle\langle \downarrow\uparrow|+w_3|\uparrow\uparrow \rangle\langle \uparrow\uparrow|+w_4|\downarrow\downarrow \rangle\langle \downarrow\downarrow| $$

and to write it in the basis

$ \left( \begin{matrix} \uparrow\uparrow\\ \uparrow\downarrow\\ \downarrow\uparrow\\ \downarrow\downarrow \end{matrix} \right)$

But is it right?

Answer 1

In principle, if $\vert \psi_i\rangle \in \{\vert\uparrow\uparrow\rangle,\vert\uparrow\downarrow\rangle,\vert\downarrow\uparrow\rangle,\vert\downarrow\downarrow\rangle\}$ then your density matrix will be a sum $$ \rho=\sum_{ij}w_{ij}\vert\psi_j\rangle\langle \psi_i\vert \, . $$ There is no a priori reason to restrict to diagonal terms. If you original Hilbert space is $4\times 4$ (as it is here) your density matrix should have $16$ elements, be hermitian, and satisfy the normalization $\sum_{i} w_{ii}=1$.

Note that it is possible to arrange your spin states so they couple to good angular momentum $j=1$ and $j=0$, i.e use the $\{\vert 1,m\rangle, \vert 00\rangle\}$ basis. In this case, $$ \rho=\sum_{jmm'}w_{jmm'}\vert jm\rangle\langle j'm'\vert \, ,\quad j,j'=\{0,1\}\, ,\quad -j\le m\le j,\quad -j'\le m'\le j', $$