That's no surprise. I don't know of any other example in physics, where a pretty simple thing is made complicated by referring to esoterics rather than maths, than the Stueckelberg trick to interpret the negative-frequency modes in relativistic quantum theory.
The trick consists in recognizing that an interpretation of relativistic wave equations in terms of single-particle wave functions is inconsistent with the physics described when it comes to collisions of particles at relativistic energies, because then you always may create new particles and/or destroy particles in the initial state. That's why it is most convenient to formulate relatistic quantum theory in terms of quantum field theory, i.e., a many-body theory from the very beginning.
Then the Dirac equation for free particles becomes an equation for a fermionic field operator (fermionic, because only with canonical equal-time anticommutation relations you get an energy spectrum that bounded from below and thus only then a stable groundstate exists; this is an example for the famous spin-statistics theorem which says that for relativistic particles you must necessarily quantize the corresponding fields as bosons (fermions) if the fields have integer (half-integer) spin).
Now you can expand the field operator in terms of energy-momentum eigenmodes of the Dirac equation. The only "tricky" thing is to write down annihilation operators in front of the positive-frequency modes and creation operators in front of the negative-frequency modes. Then you have for the field operator
$$\hat{\psi}(t,\vec{x})=\sum_{\sigma \in \{-1/2,1/2\}} \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} \frac{1}{\sqrt{(2 \pi)^3 2 E_{\vec{p}}}} \left [\hat{a}(\vec{p},\sigma) \exp(-\mathrm{i} E_{\vec{p}} t+\mathrm{i}\vec{p} \cdot \vec{x}) + \hat{b}^{\dagger}(\vec{p},\sigma) \exp(+\mathrm{i} E_{\vec{p}}-\mathrm{i} \vec{p} \cdot \vec{x}) \right ].$$
So you have two sorts of particles of the same mass, with annihilation operators ##\hat{a}## and ##\hat{b}##. All energy eigenvalues are positive ##E(\vec{p})=\sqrt{m^2+\vec{p}^2}##. As it turns out, the two sorts of particles have the same mass, positive energy (as it must be for a theory with stable ground state), and opposite charges. After normal ordering, energy, momentum, and charge are given by the operators
$$\hat{H}=\sum_{\sigma} \mathrm{d}^3 \vec{p} E(\vec{p})[\hat{N}_a(\vec{p},\sigma)+\hat{N}_b(\vec{p},\sigma)],$$
$$\hat{\vec{P}} = \sum_{\sigma} \mathrm{d}^3 \vec{p} \vec{p} [\hat{N}_a(\vec{p},\sigma)+\hat{N}_b(\vec{p},\sigma)],$$
$$\hat{Q} = \sum_{\sigma} \mathrm{d}^3 \vec{p} [\hat{N}_a(\vec{p},\sigma)-\hat{N}_b(\vec{p},\sigma)],$$
where the occupation-number operators are given by
$$\hat{N}_a(\vec{p},\sigma)=\hat{a}^{\dagger}(\vec{p},\sigma)\hat{a}(\vec{p},\sigma), \quad \hat{N}_b(\vec{p},\sigma)=\hat{b}^{\dagger}(\vec{p},\sigma) \hat{b}(\vec{p},\sigma).$$
The Dirac field thus always describes particles and their antiparticles.
The Stueckelberg interpretation of the negative-frequency modes is thus precisely to avoid esoteric ideas about things going "backwards in time". This doesn't make sense. The only idea is to introduce an creation instead of an annihilation operator in front of the negative-frequency mode. So the other sign in the exponential is just right to describe a single-particle creation instead of an annihilation process when the field operator is applied to any state in Fock space.
