How to write magnetic dipole transition hamiltonian using ladder operator?

The magnetic dipole transition Hamiltonian is

$\hat{H}=\frac{e}{2m_ec}\hat{\mathbf{B}}\cdot\hat{\mathbf{L}}$

How do I express it in terms of ladder operator $\hat{L}_+$, $\hat{L}_-$, and the $z$-projecttion angular momentum $L_z$?

I know that $\hat{L}_\pm=\hat{L}_x\pm i\hat{L}_y$. But is it possible to express that Hamiltonian in ladder operator of $\hat{L}_+$, $\hat{L}_-$, and $L_z$ without expressing $\hat{\mathbf{B}}$ in terms of ladder operator?

If the magnetic field is in the $z$-direction, then the operator dot product becomes simply:
$$\hat H = \frac{e}{2m_e c}B_z \hat L_z$$ so there is no need for ladder operators. The $\mathbf{B}$ is not an operator itself, and so if the field is in a different direction, $$\mathbf{B} = (B_x,B_y,B_z)$$ and the Hamiltonian will be
$$\hat H = \frac{e}{2m_e c}(B_x \hat L_x+B_y \hat L_y+B_z \hat L_z)$$ in which you can simply replace the $\hat L_x, \hat L_y$ with their expressions in terms of the ladder operators.