Can the ’combined limit’ be performed for all potentials ?

7 Can the ’combined limit’ be performed for all potentials ?

We know now that three potentials exist which, for properly chosen initial wave-packets, lead to deterministic equations of motion in the limit $ℏ → 0$ . We shall refer to such potentials for brevity as ’deterministic potentials’. A (complete) reconstruction of NM from QT requires that all (or almost all) potentials are deterministic. In this section we ask if this can be true.

As a first point we note that the probability density $ρ (x, t)$ of all deterministic wave packets takes, by definition, a very specific functional form, namely one that reduces, like Eq. (26), in the limit $ℏ → 0$ to a delta function. This fixes essentially one of our two dynamic variables; we have two differential equations for a single unknown variable $S (x, t )$ . It seems unlikely that this overdetermined system of equations admits solutions for $S (x, t)$ for arbitrary potentials $V$ .

The second point to note is, that the existence of a deterministic limit does not only fix the functional form of $ρ$ but also its argument. Let us assume, that a deterministic solution for $ρ$ and $S$ , with $ρ (x, t )$ taking the form (26) with unspecified $ϵ (t )$ , exists. The probability density $ρ (x, t)$ depends necessarily on $⃗x - ⃗r (t )$ , where the position vector $⃗r(t )$ is a solution of Newton’s equation for the same potential $V (x )$ that occurs in the Schrödinger equation. The crucial point is that this dependence is not created by the limiting process $ℏ → 0$ but is already present for finite $ℏ$ , in the exact quantum-mechanical solution. For given initial conditions it has been created, so to say, by the quantum-theoretical formalism. This implies that $⃗r (t )$ describes for finite $ℏ$ not the time-dependence of a particle trajectory but of a position expectation value. Since $⃗r(t )$ is (again for finite $ℏ$ ) the solution of Newton’s equations, such equations for expectation values must already be present in the quantum-theoretical formalism. Of course, in the deterministic limit $ℏ → 0$ the difference between particle trajectories and expectation values vanishes, but the important point is that Newton’s equations must hold already for finite $ℏ$ . We conclude that the existence of the equations of motion of NM for position expectation values is a necessary condition for the existence of deterministic potentials.

This line of reasoning leads to a mathematical condition for deterministic potentials. Let us assume that we have a deterministic potential $(det ) V (x )$ in our quantum-theoretical ( $ℏ$ finite) problem. We calculate the expectation value $----- xk (t )$ as defined by (17) using the deterministic probability density (26). We obtain $----- xk (t ) = rk (t )$ , i.e. the expectation value follows the time-dependence of the trajectory [the peak of $ρ (x, t )$ ]. The latter must fulfill Newton’s equation with the force derived from $(det ) V (x )$ , otherwise the deterministic limit could not exist. Using these facts in Ehrenfest’s theorems (15) and (16) we obtain immediately the following integral equation for deterministic potentials
$∫ (det ) 3 (3) (det ) F k (r ) = d x δϵ (x - r )F k (x ),$ (30)

where $(det ) (det ) F (x ) = - ∂ V (x ) ∕ ∂ xk k$ . The quantity $ρ (x, t )$ has been renamed $(3 ) δ ϵ (x - r (t ))$ in order to show the convolution-type structure of the equation. Note that (30) is only for $ϵ > 0$ a constraint for $det V (x )$ . It is easy to see that a particular solution is given by $V (det ) (x ) = a + b x + c x x k k i,k i k$ , where the coefficients may depend on time. This is essentially a linear combination of the three deterministic potentials $xn, n = 0, 1, 2$ found already in the last section. According to a theorem by Titchmarsh [24] (a simple proof may be obtained with the help of the theory of generalized functions [15], see section 10) other solutions of (30) do not exist. This theorem shows that the ’combined limit’ cannot be performed for all potentials. Although the present treatment does not cover all conceivable physical situations, the results obtained so far imply already definitively that the limit $ℏ → 0$ of QT does not agree with NM.

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