How close is the history state to the ground state in the Kitaev clock construction?

Consider a standard circuit to Hamiltonian reduction in QMA. For example, refer here (Vazirani's lecture notes) or page 235 of here (survey by Gharibian et al).

The history state of the Kitaev clock Hamiltonian $$H$$ is given by:

$$\begin{equation} |\psi_{\text{history}}\rangle = \frac{1}{\sqrt{T + 1}} \sum_{i = 0}^{T} V_{i} V_{i - 1} \cdots V_{1} |00 \cdots 0 \rangle |\psi\rangle |i\rangle_{\text{clock}}, \end{equation}$$ where $$V = V_{T} V_{T-1}\cdots V_{1}$$ is the circuit that the QMA verifier, and $$|\psi\rangle$$ is the state sent by the prover. Note that $$|\psi_{\text{history}}\rangle$$ is not the ground state of $$H$$. But how close is it to the ground state?

In other words, if $$|\psi_{\text{ground}}\rangle$$ is the ground state of the Kitaev clock Hamiltonian $$H$$, I am trying to find

$$\begin{equation} || |\psi_{\text{ground}}\rangle - |\psi_{\text{history}}\rangle ||. \end{equation}$$