Assuming that the following manipulations are correct the translational symmetry of your Lagrangian can be gauged by including a scalar gauge field $\phi$ and a one form gauge field $A_{\mu}$.

First of all, assuming that the boundary terms do not contribute we can write the Lagrangian density as $$ \mathscr L=\psi^\dagger i\partial_t\psi-\frac{1}{2m}(\partial^{\mu}\psi)^{\dagger}\partial_{\mu}\psi. $$

Now writing $\psi$ as $\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$ the translation of $\psi$ can be written as $\left[ \begin{array}{cc} 1 & \theta_1+i\theta_2\\ 0 & 1\\ \end{array}\right]\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$

Lie algebra of the group of matrices of the form $\left[ \begin{array}{cc} 1 & \theta_1+i\theta_2\\ 0 & 1\\ \end{array}\right]$ is the set of matrices $\left[ \begin{array}{cc} 0 & a+ib\\ 0 & 0\\ \end{array}\right]$

Now to gauge this symmetry introduce a Lie algebra valued one form $A=A_{\mu}dx^{\mu}$ which under a gauge transformation

$\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]\rightarrow \left[ \begin{array}{cc} 1 & \theta_1(x)+i\theta_2(x)\\ 0 & 1\\ \end{array}\right]\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$

transform as

$A_{\mu}\rightarrow g(x)A_{\mu}g(x)^{-1}+(\partial_{\mu}g(x))g(x)^{-1} $

Where $g(x)=\left[ \begin{array}{cc} 1 & \theta_1(x)+i\theta_2(x)\\ 0 & 1\\ \end{array}\right]$

However we note that the Lagrangian $$ \mathscr L=\psi^\dagger i(\partial_t-A_t)\psi-\frac{1}{2m}((\partial^{\mu}-A^{\mu})\psi)^{\dagger}(\partial_{\mu}-A_{\mu})\psi. $$

is not gauge invariant and neither real.

Obstruction to gauge invariance is the fact that $\psi^{\dagger}$ doesn't transform by right multiplication by $g(x)^{-1}$ but rather by right multiplication by $g(x)^{\dagger}$

To repair gauge invariance one may introduce a matrix valued scalar gauge field $\phi$ which under change of gauge changes as

$\phi(x)\rightarrow (g(x)^{\dagger})^{-1}\phi(x)g(x)^{-1}$

Then we see that the Lagrangian

$$ \mathscr L=\psi^\dagger iexp(\phi(x))(\partial_t-A_t)\psi-\frac{1}{2m}((\partial^{\mu}-A^{\mu})\psi)^{\dagger}exp(\phi(x))(\partial_{\mu}-A_{\mu})\psi. $$

is gauge invariant. However still the Lagrangian is not real. To repair that we can include complex conjugate of each term in it.

Assuming that the following manipulations are correct the translational symmetry of your Lagrangian can be gauged by including a scalar gauge field $\phi$ and a one form gauge field $A_{\mu}$.

First of all, assuming that the boundary terms do not contribute we can write the Lagrangian density as $$ \mathscr L=\psi^\dagger i\partial_t\psi-\frac{1}{2m}(\partial^{\mu}\psi)^{\dagger}\partial_{\mu}\psi. $$

Now writing $\psi$ as $\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$ the translation of $\psi$ can be written as $\left[ \begin{array}{cc} 1 & \theta_1+i\theta_2\\ 0 & 1\\ \end{array}\right]\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$

Lie algebra of the group of matrices of the form $\left[ \begin{array}{cc} 1 & \theta_1+i\theta_2\\ 0 & 1\\ \end{array}\right]$ is the set of matrices $\left[ \begin{array}{cc} 0 & a+ib\\ 0 & 0\\ \end{array}\right]$

Now to gauge this symmetry introduce a Lie algebra valued one form $A=A_{\mu}dx^{\mu}$ which under a gauge transformation

$\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]\rightarrow \left[ \begin{array}{cc} 1 & \theta_1(x)+i\theta_2(x)\\ 0 & 1\\ \end{array}\right]\left[\begin{array}{cc}\psi(x)\\1\end{array}\right]$

transform as

$A_{\mu}\rightarrow g(x)A_{\mu}g(x)^{-1}+(\partial_{\mu}g(x))g(x)^{-1} $

Where $g(x)=\left[ \begin{array}{cc} 1 & \theta_1(x)+i\theta_2(x)\\ 0 & 1\\ \end{array}\right]$

However we note that the Lagrangian $$ \mathscr L=\psi^\dagger i(\partial_t-A_t)\psi-\frac{1}{2m}((\partial^{\mu}-A^{\mu})\psi)^{\dagger}(\partial_{\mu}-A_{\mu})\psi. $$

is not gauge invariant and neither real.

Obstruction to gauge invariance is the fact that $\psi^{\dagger}$ doesn't transform by right multiplication by $g(x)^{-1}$ but rather by right multiplication by $g(x)^{\dagger}$

To repair gauge invariance one may introduce a matrix valued scalar gauge field $\phi$ whose exponential under change of gauge changes as

$exp(\phi(x))\rightarrow (g(x)^{\dagger})^{-1}exp(\phi(x))g(x)^{-1}$

(how does $\phi$ change? I am not sure)

Then we see that the Lagrangian

$$ \mathscr L=\psi^\dagger iexp(\phi(x))(\partial_t-A_t)\psi-\frac{1}{2m}((\partial^{\mu}-A^{\mu})\psi)^{\dagger}exp(\phi(x))(\partial_{\mu}-A_{\mu})\psi. $$

is gauge invariant. However still the Lagrangian is not real. To repair that we can include complex conjugate of each term in it.