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To simplify things a bit, let's take a single qubit and a single qutrit for comparison.

First, the amplitude damping channel (giving e.g. emission of a photon) for a qubit is $\mathcal E\left(\rho\right) = E_0\rho E_0^\dagger + E_1\rho E_1^\dagger$, where $$E_0 = \begin{pmatrix}1 && 0 \\ 0 &&\sqrt{1-\gamma}\end{pmatrix}, \quad E_1 = \begin{pmatrix} 0 && \sqrt \gamma \\ 0 &&0\end{pmatrix},$$ with $E_0$ being required for normalisation. This can also be written as $E_0 = \left|0\rangle\langle 0\right| + \sqrt{1-\gamma}\left|1\rangle\langle 1\right|$ and $E_1 = \sqrt{\gamma}\left|0\rangle\langle 1\right|$. When $\gamma = 1$, the amplitude damping channel gives the state $$\rho = \begin{pmatrix} 1 && 0 \\ 0 && 0\end{pmatrix} = \left|0\rangle\langle 0\right|$$ with certainty. However, this only applies at $0$ temperature, so to represent what happens at finite temperature, the generalised amplitude damping channel needs to be used. This has $\mathcal E\left(\rho\right) = \sum_kE_k\rho E_k^\dagger$, where $$E_0 = \sqrt p\begin{pmatrix}1 && 0 \\ 0 &&\sqrt{1-\gamma}\end{pmatrix}, \quad E_1 = \sqrt p\begin{pmatrix} 0 && \sqrt \gamma \\ 0 &&0\end{pmatrix},$$ $$E_2 = \sqrt{1-p}\begin{pmatrix}\sqrt{1-\gamma} && 0 \\ 0 &&1\end{pmatrix}, \quad E_3 = \sqrt{1-p}\begin{pmatrix} 0 && 0 \\ \sqrt \gamma &&0\end{pmatrix}.$$ Each of these can be split into four sections - an element in the upper left (bottom right) corner causes the amplitude of the upper left (bottom right) corner to decrease (here, this is really just a normalisation factor) while having a single element in a matrix, which is in the top right (bottom left) causes loss (gain) from the bottom right (top left) to the top left (bottom right) element of the density matrix. Now when $\gamma = 1$, this gives the state $$\rho = \begin{pmatrix}p && 0\\ 0 && 1-p\end{pmatrix} = p\left|0\rangle\langle 0\right| + \left(1-p\right)\left|1\rangle\langle 1\right|.$$

This shows that after a long time, the loss and gain cancel (i.e. no more loss or gain occurs), although this state is not very useful for computation, so lets try extending this a bit and adding another qubit. Lets go a bit further than that and take a pair of coupled spin-half fermions (such as a pair of electrons). This gives 4 states - the singlet state $S = \frac{1}{\sqrt 2}\left(\left|\uparrow\downarrow\right> - \left|\downarrow\uparrow\right>\right)$. It also gives a subspace of triplet states $T = \left\lbrace \left|\uparrow\uparrow\right>, \frac{1}{\sqrt 2}\left(\left|\uparrow\downarrow\right> + \left|\downarrow\uparrow\right>\right), \left|\downarrow\downarrow\right>\right\rbrace = \left\lbrace T_0, T_1, T_2\right\rbrace$.

Describing this as a qudit with $d=4$ and using an equivalent 'qudit generalised amplitude damping channel' gives a few possible interaction as in the qubit case:

loss from the triplet subspace to the singlet state

gain from the singlet state to the triplet subspace

amplitude damping within the triplet subspace.

By itself, this still doesn't help very much, so lets place this pair of spin-half particles in the centre of a larger system (a 'spin bath', here used as the environment, mediating the interaction) and allow it to interact. As the states in the triplet subspace are symmetric and it is in the centre of a bath, the probability rate of amplitude damping on the first qubit will, on average, equal the rate of amplitude damping on the second qubit. This means that, instead of having a single qudit amplitude damping channel, there are two copies of the same generalised amplitude damping channel, which reduces the number of possible interactions. In the limit of long time and taking $p=1/2$ (this is just setting the system to a certain non-zero temperature), these are, ignoring normalisation:

gain on the $\left|\downarrow\downarrow\right> = T_2$ state: $$T_2\rightarrow T_2' \propto T_2 + \beta T_1 + \beta^2T_0$$

loss on the $\left|\uparrow\uparrow\right> = T_0$ state: $$T_0\rightarrow T_0' \propto T_0 + \beta T_1 + \beta^2T_2$$

gain and loss on the $\frac{1}{\sqrt 2}\left(\left|\uparrow\downarrow\right>\pm \left|\downarrow\uparrow\right>\right)$ states ($S$ and $T_1$ states): $$T_1\rightarrow T_1' \propto \left(1+\beta^2\right)T_1 + \beta \left(T_0+T_2\right)$$ $$S\rightarrow S' \propto\left(1-\beta^2\right) S.$$

This shows that oscillations in the triplet subspace occur instead of decay, meaning that the triplet subspace is a decoherence free subsystem and can be used as a qutrit. In reality, interactions are more complicated and there will be other types of noise, so there will still be some decoherence, but the reasoning is still the same in that pairing two spin-half particles allows for the triplet state to be used as a decoherence free subsystem (or at least have less decoherence than a qubit) to mitigate the effects of some types of noise.