How does Fermi's golden rule yield a specific rate?
I'm fairly sure the states $\require{\begingroup} \begingroup \newcommand{\ket}[1]{|#1\rangle} \newcommand{\braket}[1]{\langle #1 \rangle} \ket{i}$ and $\ket{f}$ are eigenstates of the unperturbed Hamiltonian $\hat{H}_0$. Therefore, by virtue of the Hermiticity of $\hat{H}_0$, they are necessarily orthonormal

$$\braket{i|f} = \delta_{ij}$$

therefore if we define $\hat{V'} = \hat{V} + k$ where $\hat{V}$ is the perturbation Hamiltonian and  $k$ is some arbitrary constant

$$\begin{align}
\braket{i|\hat{V'}|f} &amp;= \braket{i|\hat{V}|f} + \braket{i|k|f} \\
&amp;= \braket{i|\hat{V}|f} + k\braket{i|f} \\
&amp;= \braket{i|\hat{V}|f}
\end{align}$$

if $\ket{i}$ and $\ket{f} \endgroup$ are different states. If they are the same state, then you are calculating the transition probability from one state to itself - not a very useful concept.