6 field, ¯ F in Eq. (9), to be zero. This means we are now solving with the following choice of initial conditions In what follows, we shall set M = 1 and so all quantities will be in units of M . Before solving for the time evolution, we consider the asymptotic state, which will be a stationary solution of Eq. (49). That is, we think of the double well equation (49) as a Schrodinger equation with Hamiltonian The corresponding Hamiltonian for a single potential well centered at x = 0 is This has exactly one bound state with eigenvalue ω 2 0 = m 2 − 1 [6]. (Recall that we are working in units with M = 1.) So for the double well potential, at least when the two wells are well separated, there must be exactly two bound states, which can be approximated as The energies of these two bound states are nearly iden- tical, ω 2 ± = m 2 − 1 ± e − $^{Γ}$, Γ( m, L ) ≫ 1, split by ex- ponentially small corrections due to tunneling between the double wells. These bound states are the stationary states that the system can evolve into. We also observe that there is an apparent instability in the current model when m 2 < 1, since then ω 2 0 < 0 and the bound state solution can grow exponentially. To understand this instability, we examine the double well action in (45). If the potential V ( x ) is deep enough, there will be a region where m 2 + V ( x ) is sufficiently negative, that it becomes favorable for φ to grow without bound in this region. In terms of the fermionic model, the well is so deep that it is favorable to pull fermion pairs out of the vacuum and put them at the bottom of the potential. The evolution of the initial data, φ$_{0}$ , in this setup can be evolved formally by writing where the summation is over both bound and continuum states of H . We can check that the initial conditions are satisfied by setting t = 0 in the factor cos( ω$_{n}$t ) and in its time derivative. We expect that, as time progresses, the continuum states will disperse to infinity, leaving behind only the initial overlap with the bound state. While for- mally correct, the expansion in Eq. (56) is only useful if we know the full eigenspectrum of the double well poten- tial. In the absence of the eigenspectrum, it is easier to numerically evolve the equation of motion. We have evolved Eq. (49) using the explicit Crank- Nicholson algorithm with two iterations with first-order absorbing boundary conditions. The runs were done on very large lattices so that boundary effects are minimal. In Fig. 3 we plot the current envelopes at x = 0 versus time for several different parameters, disregarding the rapid oscillations between the envelopes. Similarly, in Fig. 4 we show the behavior of the electric field at x = 0. On the log-log plot in Fig. 5, it is clear that the envelopes decay as a power law. A fit gives where N ≈ 14 is a factor which could depend on the dimensionless product ML , where M − 1 is the width of the wells (see Eq. (46)). The electrical conductivity is therefore again given by Eq. (41), as for setup 1. φ ( t = 0 , x ) = tanh( M ( x + L )) − tanh( M ( x − L )) ≡ φ$_{0}$ ( x ) ˙ φ ( t = 0 , x ) = 0 (50) Hψ$_{n}$ = ω 2 $_{n}$ψ$_{n}$ (51) H ≡ − ∂ 2 x + m 2 + V ( x + L ) + V ( x − L ) (52) H$_{1}$ = − ∂ 2 x + m 2 + V ( x ) (53) ψ ∝ sech( x ) (54) ψ ∼ sech( x − L ) ± sech( x + L ) (55) E$_{rms}$ = N √ g t , j x rms = gN √ g t (57) φ ( t, x ) = ∑ n cos ( ω$_{n}$t ) ψ$_{n}$ ( x ) 〈 ψ$_{n}$ | φ$_{0}$ 〉 , (56) A. Asymptotic State B. Time Evolution ( ∂ 2 + m 2 + V ( x + L ) + V ( x − L )) φ = 0 (49) FIG. 3: Envelopes of the plots of the current at the center of the capacitor, j $^{x}$( t, x = 0), versus time. From black to light grey, the curves represent, respectively, the evolution for m = 1 . 5 , 2 and 2.5. The rapid oscillations between the envelopes are not shown. 0 200 400 600 800 1000 Time GLYPH<LParen1> t GLYPH<RParen1> GLYPH<Minus> 15 GLYPH<Minus> 10 GLYPH<Minus> 5 0 5 10 15 j x GLYPH<LParen1> t,x GLYPH<Equal> 0 GLYPH<RParen1>