1
$\begingroup$

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:

  1. 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.

  2. 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?

$\endgroup$
  • 2
    $\begingroup$ What is specific here about fermions? Can you answer these questions for the bosonic case? $\endgroup$ – coconut Dec 16 '17 at 9:29
  • $\begingroup$ The equation of motion for fermions is first order. For bosons it's second order. You then have no problem with two boundary conditions - there will always be a solution connecting the initial with the final value. $\endgroup$ – WIMP Dec 17 '17 at 6:26
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I did not ask how to do the path integral for fermions. I asked how to calculate the path itself. Suppose I do the path integral, then all I get is a functional determinante and then what do I do with that? I can estimate it off-shell, ok, but what's the path (or the paths) that belong to that? I suspect they aren't continuous, but there ought to be a way to calculate that, no? $\endgroup$ – WIMP Dec 16 '17 at 8:24
  • $\begingroup$ I don't understand the 2nd paragraph. I have a wave-function, I have a propagator, I have an initial and final condition that are not generally connected by the propagator. Now what? Could you be more specific? $\endgroup$ – WIMP Dec 16 '17 at 8:27
0
$\begingroup$

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:

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

  2. 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.]

  3. 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.

--

$^1$ These issues also appear in the Hamiltonian bosonic coherent state path integral, cf. e.g. this Phys.SE post.

$\endgroup$
  • $\begingroup$ I think there might be a typo in the boundary conditions. Currently, it seems that the third and the fourth are the same as the first and the second. $\endgroup$ – coconut Dec 16 '17 at 20:57
  • $\begingroup$ I don't see how 1) means the paths are ill-defined. As to 2), well, then chose boundary conditions that are compatible with the commutation relations. The stationary path clearly exists if you have boundary conditions that both happen to lie on the same solution, so I don't understand why you think they're ill-defined. See, I can think up many functions that will connect some initial value to an end value. These will generally not solve the Dirac equation. I can then calculate the action, ok. But which path is the one with the biggest weight in the path integral? $\endgroup$ – WIMP Dec 17 '17 at 6:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.