\overleftarrow{{{\mathfrak{T}}}_1}^{+} {|\overleftarrow{\psi_m}\rangle} &=& \frac{1}{\sqrt{2}}\sum_{k=0}^{m-1} \sqrt{\frac{m+1}{(k+1)(m-k)}} {|\overleftarrow{\psi_k}\overleftarrow{\psi_m}_{-1-k}\rangle}, \notag \\ 
\overleftarrow{{{\mathfrak{T}}}_1}^{+}{|\overrightarrow{\psi_m}\rangle} &=& \frac{1}{\sqrt{2}}\sum_{k=0}^{m-1} \sqrt{\frac{m-k}{(k+1)(m+1)}}\left({|\overleftarrow{\psi_k}\overrightarrow{\psi_m}_{-1-k}\rangle}+{|\overrightarrow{\psi_m}_{-1-k}\overleftarrow{\psi_k}\rangle}\right) \notag \\
&+& \frac{1}{\sqrt{2(m+1)}}\sum_{k=0}^{m} \left({|\phi^\downarrow_k\phi^\uparrow_{m-k}\rangle}-{|\phi^\uparrow_k\phi^\downarrow_{m-k}\rangle}\right). \label{eq:leftT1+} \end{aligned}$$ The other two ${{\mathfrak{psu}}}(1|1)^2$ generators, the ${{\mathfrak{T}}}^-$, can then be computed at this order via hermitian conjugation. Up to gauge transformations, and a rescaling of $g$, this solution is completely fixed by the symmetry constraints.

Order $g^2$ \[sec:g2\]
======================

At $\mathcal{O}(g^2)$, ${\delta{\mathfrak{D}}}$ and the ${{\mathfrak{psu}}}(1,1|2)$ generators receive quantum corrections. From (\[eq:psU11\^2Alg\]), it is straightforward to compute $${\delta{\mathfrak{D}}}_2= 2 {\left\{\overrightarrow{{{\mathfrak{T}}}_1}^+,\overleftarrow{{{\mathfrak{T}}}_1}^-\right\}}=2 {\left\{\overrightarrow{{{\mathfrak{T}}}_1}^-,\overleftarrow{{{\mathfrak{T}}}_1}^+\right\}}.$$ As first shown in [@Beisert:2003jj], ${\delta{\mathfrak{D}}}_2$ acts by projecting two adjacent sites onto modules of definite ${{\mathfrak{psu}}}(2,2|4)$ “spin” $j$ with coefficient $h(j)$. $h$ gives the harmonic numbers $$h(k) = \sum_{k'=1}^{k}\frac{1}{k'}=\psi(k+1)-\psi(1).$$ The harmonic numbers will play an essential role in ${\delta{\mathfrak{D}}}_4$ as well.

${\delta{\mathfrak{D}}}$ is the only generator we need to compute because, as we now explain, once we know ${\delta{\mathfrak{D}}}$ the full ${{\mathfrak{psu}}}(1,1|2) \times {{\mathfrak{psu}}}(1|1)^2$ algebra’s action is fixed by group theory. Knowing ${\delta{\mathfrak{D}}}$ means knowing its eigenstates and eigenvalues. Multiplets are then formed by states of equal eigenvalues, and the generators of ${{\mathfrak{psu}}}(1,1|2) \times {{\mathfrak{psu}}}(1|1)^2$ must connect the states of a multiplet with factors determined by group theory.

However, for our method of computing ${\delta{\mathfrak{D}}}$, it is essential to compute 