Setup
Suppose we have some Hamiltonian $H$ which is known to exhibit spontaneous symmetry breaking (SSB), at least in some parameter regime. For simplicity, we might consider the 2D Ising model below its critical temperature.
Now suppose we add some weak perturbation such that the new Hamiltonian is
$$ H^{\prime} = H + \lambda W, $$
where we assume that $W$ is bounded above by $H$, $\lVert W \rVert \leq \lVert H \rVert$, and $\lambda \ll 1$ is a dimensionless parameter controlling the strength of the perturbation. Note that I do not assume that $W$ preserves the symmetries of $H$.
Question
Can we say anything about whether or not $H^{\prime}$ will also exhibit SSB? Furthermore, if it does exhibit SSB, how are the new symmetry breaking states related to the old ones? Are they the same, at least up to some $\mathcal{O}(\lambda)$ error? If there is a phase transition, are the critical parameters the same?
I believe the folklore answer to this question is "$H^{\prime}$ will still exhibit SSB for $\lambda \ll 1$, and the new symmetry breaking states are the same as the old ones", at least in the simple case of the Ising model. However, I would be interested to hear if any of this has been rigorously proven, again even if only for some specific models.
Edit
I will attempt to clarify my question. I am mainly interested in the equilibrium states of the perturbed Hamiltonian, and whether they break the symmetry of the original Hamiltonian, while still forming a closed space under the action of the original symmetry group.
Copying from my earlier comments, I believe one example of where this happens is with discrete time crystals. For concreteness let us consider a many-body localized time crystal, following this paper as a reference. There one drives a system of qubits periodically with a two-pulse protocol. This can be described by the Floquet operator
$$ U_{f} = \exp(-i t_{0} H_{\mathrm{MBL}}) \exp\left(i t_{1} \sum_{i} \sigma_{i}^{x}\right), $$
where $H_{\mathrm{MBL}}$ is given by
$$ H_{\mathrm{MBL}} = \sum_{i} (J_{i} \sigma_{i}^{z}\sigma_{i+1}^{z} + h_{i}^{z} \sigma_{i}^{z} + h_{i}^{x} \sigma_{i}^{x}), $$
with the parameters $J_{i}$, $h_{i}^{z}$ and $h_{i}^{x}$ chosen uniformly from $J_{i} \in [(J/2), (3J/2)]$, $h_{i}^{z} \in [0, h^{z}]$ and $h_{i}^{x} \in [0, h]$. If $t_{1} = \pi/2$ then the spins are exactly flipped during the first pulse. In this case, if one takes some observable, say $\sigma^{z}$ for one of the spins, and calculates the Fourier spectrum of its expectation value, one will see a peak at $\nu_{\mathrm{drive}}/2$, where $\nu_{\mathrm{drive}} = (t_{0} + t_{1})^{-1}$ is the drive frequency. In this way, we say that the discrete time translation symmetry is spontaneously broken.
The relevance of this example to my question is that, even if $t_{1} = \pi/2 + \epsilon$ for some small $\epsilon > 0$, this peak in the Fourier spectrum remains pinned at $\nu_{\mathrm{drive}}/2$, even though naively one would expect it to move if the spins aren't perfectly flipped. This is observed in experiments such as this one.
What I am wondering is how common this "robustness to error" is in conventional examples of SSB.
