First, I will set $e=1$ for simplicity.
Let $\psi_0$ denote the wave function that satisfies the free Schrodinger equation:
\begin{equation}
i \frac{\partial \psi_0}{\partial t} = -\frac{1}{2m}\mathbf{\nabla}^2 \psi_0 + V \psi_0 \tag{1}
\end{equation}
Furthermore, let $\psi$ be the wave function that obeys the Schrodinger equation for a non-vanishing vector potential $\mathbf{A}$:
\begin{equation}
i \frac{\partial \psi}{\partial t} = -\frac{1}{2m}(\mathbf{\nabla}-i\mathbf{A})^2 \psi+ V \psi \tag{2}
\end{equation}
Let us now write:
\begin{equation}
\psi=\exp \left( i \int_{\gamma} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_0
\end{equation}
where $\gamma$ is a path from some arbitrary point $\mathbf{x}_0$ to some other point $\mathbf{x}_1$. We can then write:
\begin{equation}
\left( \mathbf{\nabla} -i \mathbf{A} \right)^2 \psi = \exp \left( i \int_{\gamma} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \mathbf{\nabla}^2 \psi_0
\end{equation}
Substituting this expression into equation $(2)$ gives equation $(1)$. This implies that the wave function of an electrically charged particle travelling through space where $\mathbf{A} \neq 0$ will gain an additional phase.
We know that the wave function at the point $Q$ (see the figure below) is a result of quantum superposition, i.e. we can write:
\begin{equation}
\begin{aligned}
\begin{split}
\psi_{\scriptscriptstyle Q} & = \psi(\mathbf{x},\gamma_1) + \psi(\mathbf{x},\gamma_2) \\&
= \exp \left( i \int_{\gamma_1} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_{0}(\mathbf{x},\gamma_1) + \exp \left( i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_{0}(\mathbf{x},\gamma_2) \\&
= \exp \left( i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \left( \exp \left( i \int_{\gamma_1} \mathbf{A} \cdot \mathrm{d} \mathbf{l} - i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right)\psi_{0}(\mathbf{x},\gamma_1) + \psi_{0}(\mathbf{x},\gamma_2) \right)
\end{split}
\end{aligned}
\end{equation}
We can use Stoke's theorem on the first term inside the brackets, because $\gamma_1-\gamma_2$ is a closed path:
\begin{equation}
\int_{\gamma_1} \mathbf{A} \cdot \mathrm{d} \mathbf{l} - \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} = \int \mathbf{B} \cdot \mathrm{d}\mathbf{S} = F
\end{equation}
where $F$ is the total magnetic flux due to the solenoid through a surface defined by the closed boundary $\gamma_2-\gamma_1$. The wave function at $Q$ can now be written as:
\begin{equation}
\psi_{\scriptscriptstyle Q} = \exp \left( i \int_{\gamma_2} \mathbf{A} \cdot \mathrm{d} \mathbf{l} \right) \left( \exp \left( i F \right)\psi_{0}(\mathbf{x},\gamma_1) + \psi_{0}(\mathbf{x},\gamma_2) \right)
\end{equation}
This shows that the relative phase difference, and thus the interference pattern, is dependent on the magnetic flux due to the solenoid. This is the Aharonov-Bohm effect.
