What's the path of least action for fermions off-shell? The Lagrangian of fermions is first order both in space-derivatives and time-derivatives. In the variation of the action one usually fixes both the initial point and end point. I have the following questions:


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*How does the variational principle for fermions formally work so that it's mathematically correct. I do not want to know how to derive the Euler-Lagrange equation from the Dirac Lagrangian. I know how to do that. I don't know how one even gets these equations since a solution to the Euler-Lagrange equations will not generally connect these chosen points. 

*If I have an initial and end point that are not connected by an on-shell path (one that solves the Dirac equation), then how do I calculate the path that minimizes the action? 
 A: The way of doing path-integration in fermions is ussing the grassman numbers. A good link for you will be http://www.int.washington.edu/users/dbkaplan/571_14/Fermion_Path_Integration.pdf
This was what you were asking for? I hope you like it.
The path doesn't really matter. If you are still interested in it, I think that you can calculate it ussing the propagator and putting the restriction of final and initial conditions that you want on the first-quantized wavefunction.
A: We integrate over virtual paths in the fermionic path integral. Without integrating over fermionic field configurations, the stationary and virtual paths (in the fermionic action principle) are rendered ill-defined for various reasons:


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*The fields are Grasmann-odd indeterminates, cf. e.g. this Phys.SE post. 

*The set of boundary conditions (BCs) $$\psi(t_i\!=\!0)~=~0, \qquad \psi^{\dagger}(t_i\!=\!0)~=~0,\qquad\psi(t_f\!=\!0)~=~0, \qquad \psi^{\dagger}(t_f\!=\!0)~=~0,  $$ 
are quantum mechanically incompatible$^1$ with  CARs. [Classically (meaning when $\hbar\to 0$), the first two initial BCs are equivalent. Similarly, the last two final BCs are classically equivalent.] 

*There is a mismatch$^1$ between the number of BCs and the first order nature of the Dirac equation, meaning that the stationary path is overdetermined.
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$^1$ These issues also appear in the Hamiltonian bosonic coherent state path integral, cf. e.g. this Phys.SE post. 
