We know that the approach that uses: (i) the so-called Schrödinger ansatz, which, in the case with a spatial variable, is written as:

ψ(x, t) = exp (i/ℏ)S(x, t) , (1.1)

where S(x, t) is the action function for a physical system, (ii) the Hamilton-Jacobi equation, (iii) as well as a certain conceptual context, enabled Prof. Schrödinger to obtain his famous equation. In several articles on the same problem, using the classical Hamilton-Jacobi equation, the approximation that consists of taking the limit ℏ → 0 to eliminate an excess term [13]- [17] is used in the approximate passage to the Schrödinger equation. Only recently, it was shown, for the case of quantum systems in one spatial dimension and when the acting potential does not explicitly depend on the time variable, that one can exactly deduce both the complete Schrödinger equation from the classical Hamilton-Jacobi equation, as well as it’s inverse [18]. In this article, considering the action S(x1, x2, t) for a physical system formed by two free particles of equal mass, with coordinate variables x1 and x2, separately, which are moving in opposite directions, we will determine under which condition, in the form of a differential equation for C(x1 − x2), the following function:

Ψ(x1, x2, t) = C(x1 − x2) exp (i/ℏ)S(x1, x2, t) , (1.2)

will correspond to an analytical function. The expression in (1.2), which besides being an obvi-ous extension of (1.1), is only formal since C(x1 − x2) is not defined. By finding C(x1 − x2), we will determine the corresponding mathematical consequences for Ψ(x1, x2, t); in particular, for “infinite" separation between the particles: (x1 − x2 → ∞). In this context, since we will not use the usual ansatz (1.1), it is obvious that there is no guarantee that function (1.2) verifies the Schrödinger equation.

Finally, in recent literature, we find several tools, approaches, and formal problems in math-ematical physics [19]-[26] and the identification and resolution of new quantum problems [27]- [40] that can be used in new extensions and new contexts for quantum mechanics.

The real and imaginary parts of (2.1), which are directly identifiable:

Ψ(x1, x2, t) = C(x1 − x2) cosΦ(x1, x2, t) + i C(x1 − x2) sinΦ(x1, x2, t),

⇒ u(x1, x2, t) = C(x1 −x2) cosΦ(x1, x2, t), v(x1, x2, t) = C(x1 −x2) sinΦ(x1, x2, t) (2.2)

will not be independent, but will be connected by the Cauchy-Riemann conditions:

∂u ∂v ∂u ∂v = & = − . (2.3) ∂x1 ∂x2 ∂x2 ∂x1

Here it is sufficient to consider only the first of them. Substituting (2.2) in (2.3) we have:

C(x1 − x2)cosΦ = C(x1 − x2)sinΦ . (2.4)

We change the variable η ≡ x1 − x2, and then we derive the expressions in (2.4), obtaining:

∂Φ dC dη ∂Φ dC dη C(η) − sinΦ + cosΦ = C(η) cosΦ + sinΦ. (2.5) ∂x1 dη dx1 ∂x2 dη dx2

Or also,

∂Φ dC ∂Φ dC −C(η) sinΦ + (η) cosΦ = C(η) cosΦ − (η) sinΦ. (2.6) ∂x1 dη ∂x2 dη

Also known as Euler-d’Alembert relations [41].

Grouping terms appropriately we have:

∂Φ dC dC ∂Φ − C(η) (x1, x2) + (η) sinΦ + (η) − C(η) (x1, x2) cosΦ = 0. (2.7) ∂x1 dη dη ∂x2

And being the functions sinΦ and cosΦ linearly independent for all x1, x2, t ∈ ℜ, we obtain, from (2.7), the independent equations:

dC ∂Φ dC ∂Φ = C(η) & = C(η) . (2.8) dη ∂x1 dη ∂x2

After equating the equations in (2.8), we obtain:

∂Φ ∂Φ − = 0, (2.9)

Where it was previously considered that C(x1 − x2) is non-zero, as it should be. Then, from the equation on the left-hand side in (2.8) it follows, by differentiation with respect to the variable x1 (assigned to the first particle), that:

d C dη dC dη ∂Φ ∂ Φ = + C(η) . (2.10) dη dx1 dη dx1 ∂x1 ∂x

Or also, d C dC ∂Φ ∂ Φ = + C(η) . (2.11) dη dη ∂x1 ∂x

And from the equation on the right-hand side in (2.8) it follows, by differentiation with respect to the variable x2:

d C dη dC dη ∂Φ ∂ Φ = + C(η) . (2.12) dη dx2 dη dx2 ∂x2 ∂x

Or also, d C dC ∂Φ ∂ Φ − = − + C(η) . (2.13) dη dη ∂x2 ∂x

Adding equations (2.11) and (2.13) we obtain:

dC ∂Φ ∂Φ ∂ Φ ∂ Φ − + C(η) + = 0. (2.14) dη ∂x1 ∂x2 ∂x ∂x

Considering the result in (2.9) we have that (2.14) simplifies to:

∂ Φ ∂ Φ + = 0, (2.15) ∂x ∂x

Where it has been previously considered that C(x − y) is non-zero. Thus, Φ(x, y) satisfies Laplace’s equation and the equation in (2.9). It can also be said, in the context being considered and according to the notation used, that the action S satisfies Laplace’s equation.

Consider Laplace’s equation for the imaginary part of Ψ(x1, x2, t), that is:

∂ v ∂ v + = 0. (2.16) ∂x ∂x

Then, substituting the expression for v(x1, x2, t), which is deduced from (2.2), in (2.16), we have:

C(η) sinΦ + C(η) sinΦ = 0 (2.17) ∂x ∂x Performing the first derivative on each term in (2.17) we have expressly:

∂ ∂Φ dC ∂ ∂Φ dC C(η) cos Φ + sin Φ + C(η) cos Φ − sin Φ = 0 (2.18) ∂x1 ∂x1 dη ∂x2 ∂x2 dη

In (2.18), after performing the second derivative on each term, we get:

2 2 2

∂Φ ∂ Φ dC ∂Φ d C C(η) − sin Φ + cos Φ + 2 cos Φ + 2 sin Φ + ∂x1 ∂x dη ∂x1 dη

∂ Φ ∂Φ dC ∂Φ + C(η) cos Φ − sin Φ − 2 cos Φ = 0. (2.19) ∂x ∂x2 dη ∂x2

Arranging terms in (2.19) we have:

2 2 2 2

∂Φ ∂Φ ∂ Φ ∂ Φ − C(η) sin Φ + + C(η) cos Φ + + ∂x1 ∂x2 ∂x ∂x

d C dC ∂Φ ∂Φ + 2 sin Φ + 2 cos Φ − = 0. (2.20) dη2 dη ∂x ∂x

Using the results (2.15) and (2.9), equation (2.20) reduces to:

2 2 2

d C 1 ∂Φ ∂Φ − + C(η) sin Φ = 0. (2.21) dη2 2 ∂x ∂x

That is, we have:

2 2 2

d C 1 ∂Φ ∂Φ − + C(η) = 0. (2.22) dη2 2 ∂x ∂x

Now it is convenient to return to the original symbology, Φ(x1, x2, t) ≡ (i/ℏ)S(x1, x2, t), to place the action in (2.22); then, we write:

2 2 2

d C 1 ∂S ∂S − + C(η) = 0. (2.23) dη2 2ℏ2 ∂x ∂x

On the other hand, from the Hamilton-Jacobi equation [14], assuming that the two particles are free and have (the same parameter of) mass m, one can write:

1 ∂S 2 ∂S 2 ∂S + = − m (x1, x2, t). (2.24) 2 ∂x1 ∂x2 ∂t

When the physical system does not have a potential that explicitly depends on time, as in the case with two free particles, one can write the (non-standard) action [17] in an additive and separable manner in its variables:

S(x1, x2, t) = −at + g(x1, x2), a ̸= 0. (2.25)

Where the corresponding equation for the unknown function g(x1, x2) is obtained directly from the Hamilton-Jacobi equation, see (2.31). Thus, we have that (2.24) can be rewritten as:

1 ∂S 2 ∂S 2 + = ma. (2.26) 2 ∂x1 ∂x2

2.3 (Quantum) Equation for C(x1 − x2)

Substituting (2.26) into (2.23) we finally find the equation for C(η):

d C ma − C(η) = 0. (2.27) dη2 ℏ2 Note that the above equation, which is quantum (due to the presence of ℏ), linear and dimen-sionally correct, has as a general solution:

n√ o n √ o ma ma C(x1 − x2) = A exp (x1 − x2) + B exp − (x1 − x2) , (2.28)

but, the physical solution corresponds to taking A = 0, since an infinite growth in the value of a quantum function (in this case, when x1 − x2 → ∞) is not physically acceptable [44]; thus, we have that (2.28) can be reduced to:

( √ ) ma C(x1 − x2) = B exp − (x1 − x2) . (2.29)

Consequently, the function Ψ(x1, x2, t), in (1.2), is formally written as:

( √ ) ( ) ma Ψ(x1, x2, t) = B exp − (x1 − x2) exp (i/ℏ) − at + g(x1, x2) . (2.30)

But, in (2.30), the function g(x1, x2) remains undefined. This function must verify the following equation in partial derivatives, non-linear and non-homogeneous:

∂g ∂g + = 2ma , (2.31)

a solution to this is found directly:

g(x1, x2) = ϵ ma (x1 + x2) , ϵ = ±1. (2.32)

Then, the total function, describing the state for two particles, is written as:

n √ o n o ma √ Ψ(x1, x2, t) = B exp − (x1 − x2) exp (i/ℏ) ϵ ma (x1 + x2) − at . (2.33)

The amplitude of Ψ(x1, x2, t), that is, the non-oscillating factor, progressively attenuates as the physical system moves. As a result, function (2.33) vanishes for infinite separation between the particles:

When: (x1 − x2) → ∞ then: Ψ(x1, x2, t) → 0. (2.34) That is, a manifestation of the quantum description, the constant ℏ, previously present in this function, disappears. In this way, the analytical function in (2.33), under the mathematical con-dition expressed by the limit (x1 − x2) → ∞, which should correspond to the classical limit of quantum mechanics, vanishes at the quantum-classical boundary . Which equation is satisfied by function (2.33)? This will be considered in section 2.5. Let us first look at an aspect of the mathematical compatibility of function (2.33), which should correspond, by hypothesis, to the total state of two free particles.

For free particles that move independently of each other, their total function should correspond to the product of the functions of each particle. We will see below that function (2.33) can be decomposed as a product of functions:

Ψ(x1, x2, t) = ψ1(x1, t) ψ2(x2, t). (2.35)

(Here we are mainly interested in compatibility and not in univocal identification) being the expressions of the partial functions:

n √ o n o ma √ a ψ1(x1, t) = B exp − x1 exp (i/ℏ) ϵ ma x1 − t , (2.36)

Of which, according to the formalism of quantum mechanics, would not be a collapse of the function.

Assuming that Ψ(x1, x2, t) in (2.33) verifies a linear equation, we also have the possibility of writing: Ψ(x1, x2, t) = ψ1(x1, t) ψ2(x2, t)|ϵ=+1 + ψ1(x1, t) ψ2(x2, t)|ϵ=−1 = 2ψ1(x1, t) ψ2(x2, t).

n √ o n o ma √ a ψ2(x2, t) = B exp x2 exp (i/ℏ) ϵ ma x2 − t , (2.37)

Where we correctly note, in the phase angle in (2.36) and (2.37), the coefficient (a/2) in the time-dependent term, which corresponds to half the energy of the complete system, a. Assuming that the particles move away from each other from the common origin (x1 = x2 = 0) of the reference direction, we would have to take as a consequence x2 = −x1. Therefore, considering x1 > 0 and choosing ϵ = 1, we have:

n √ o n √ o 1/2 ma ma a ψ1(x1, t) = B exp − x1 exp i x1 − t , (2.38) ℏ ℏ 2ℏ n √ o n √ o 1/2 ma ma a ψ2(−x1, t) = B exp − x1 exp − i x1 + t . (2.39) ℏ ℏ 2ℏ Note that the two functions have the same amplitude and that this attenuates with the increase of x1. For the functions described above, we have that the following definitions:

k1 = k2 ≡ ma/ℏ & ω1 = ω2 ≡ a/2ℏ, (2.40)

are dimensionally compatible with the wavenumber (k) and the frequency (ω). Thus, the ex-pressions (2.38) and (2.39) represent, separately, two plane waves that, moving away from each other in opposite directions, attenuate as they propagate and whose product coincides with the function (2.33). We thus verify that the function Ψ(x1, x2, t) in (2.33) is compatible with the initial assumption that this function corresponds to two free particles.

We can directly obtain the equation for the function Ψ by taking advantage of the fact that its real and imaginary parts, in (2.2), must verify Laplace’s equation, separately. However, to highlight a specific relationship between the equation for Ψ(x1, x2, t) and the equation for C(x1 − x2), we will go another way. In particular, it is interesting to verify whether the equation for Ψ is related to the Schrödinger equation.

From the formal expression in (1.2) we have, by derivation in relation to the variable x1 that: n o n o ∂Ψ i ∂S dC dη = C(η) exp iS(x1, x2, t)/ℏ + exp iS(x1, x2, t)/ℏ , (2.41) ∂x1 ℏ ∂x1 dη dx1

A second derivation, in relation to the same variable, generates the expression:

2 2 n o ∂ Ψ 1 ∂S = C(η) − exp iS(x1, x2, t)/ℏ + ∂x ℏ ∂x1

2 n o n o i ∂ S dC dη i ∂S + C(η) exp iS(x1, x2, t)/ℏ + exp iS(x1, x2, t)/h + ℏ ∂x dη dx1 ℏ ∂x1 n o 2 n o dC i ∂S d C + exp iS(x1, x2, t)/ℏ + exp iS(x1, x2, t)/ℏ . (2.42) dη ℏ ∂x dη2

And, after two derivations of (1.2) in relation to the variable x2, we have:

2 2 n o ∂ Ψ 1 ∂S = C(η) − exp iS(x1, x2, t)/h + ∂x ℏ ∂x2

2 n o n o i ∂ S dC i ∂S + C(η) exp iS(x1, x2, t)/ℏ − exp iS(x1, x2, t)/ℏ + ℏ ∂x dη ℏ ∂x2 n o 2 n o dC i ∂S d C − exp iS(x1, x2, t)/ℏ + exp iS(x1, x2, t)/ℏ . (2.43) dη ℏ ∂x dη2

We must not forget that the variables x1 and x2 are independent for Ψ(x1, x2, t), for C(x1 −x2), as well as for g(x1, x2). Adding the expressions (2.42) and (2.43) we have:

∂ Ψ ∂ Ψ ∂x ∂x ( ) ( ) 2 2 n o 1 ∂S ∂S i ∂S ∂S dC = − + Ψ + − exp iS(x1, x2, t)/ℏ + ℏ2 ∂x ∂x ℏ ∂x ∂x dη

1 2 1 2

2 2 2 n o i ∂ S ∂ S d C + + Ψ + 2 exp iS(x1, x2, t)/ℏ . (2.44)

2 2 2

ℏ ∂x ∂x dη

Using the results (2.9) and (2.15), the expression (2.44) is simplified, arriving at:

2 2 2 2 2 n o ∂ Ψ ∂ Ψ 1 ∂S ∂S d C + = − + Ψ + 2 exp iS(x1, x2, t)/ℏ . (2.45)

2 2 2

∂x ∂x ℏ ∂x1 ∂x2 dη

Substituting (2.26) in (2.45), we have,

2 2 2 n o ∂ Ψ ∂ Ψ 2ma d C + = − Ψ + 2 exp iS(x1, x2, t)/ℏ . (2.46)

2 2 2

∂x ∂x ℏ dη

Or also, considering (1.2), we have:

2 2 2 n o ∂ Ψ ∂ Ψ d C ma + = 2 (η) − C(η) exp iS(x1, x2, t)/ℏ . (2.47)

2 2 2 2

∂x ∂x dη ℏ

The term that appears in the center in (2.47) is precisely the one that appears on the left side in equation (2.27); thus, in (2.47), it must be zero. Therefore, we arrive at the expected result:

∂ Ψ ∂ Ψ + = 0 , (2.48) ∂x ∂x

Which is a classical equation (does not contain ℏ). It is a surprise that, since Ψ is a quantum function, it verifies a classical equation. From (2.48), no solution that carries time dependence is expected; however, Ψ, em (2.33), depends, in addition to the coordinate variables x1 and x2, on the time variable t and fulfills equation (2.48). How can this be explained?

The fact that function (2.33), verifying equation (2.48), presents time dependence, can be inter-preted as corresponding to a situation in which this equation comes from a larger equation, with a time derivative term.

In the problem considered here, a special context or aspect of the physical manifestation can be identified, where the temporal dependence is not physically relevant, even though the corresponding “variable" remains in the wave function. Specifically, it is observed, from a phys-ical point of view, that there is no difference between reaching the quantum-classical boundary quickly or slowly. Reaching this boundary determines the “collapse" of a quantum description, which is the same regardless of how quickly this happened. In this scenario, due to the essence of the process, the time variable loses meaning, has no relevance, and only remains as a parameter, so that:

Ψ(x1, x2, t) ≡ Ψ(x1, x2; t), ⇒ = 0. (2.49)

Thus, the complete Schrödinger equation for two free quantum particles (with the same mass parameter m), in the case of 1-dimensional motion:

2 2 2

ℏ ∂ Ψ ∂ Ψ ∂Ψ − + = iℏ . (2.50) 2m ∂x ∂x ∂t is reduced to the classical equation (2.48) in the situation of proximity to the classical-quantum boundary.

Complementarily, and independently of what is developed here, it can be seen in [29] that it was verified that an entangled function with the form:

Ψ(⃗x1, ⃗x2) = Cn |⃗x2 − ⃗x1|/λC ψn(⃗x1) un(⃗x2), (2.51)

which is a generalization of the Einstein-Podolsky-Rosen state [45], where λC is the Compton wavelength of the particles involved, ψn(⃗x1) eigenfunctions of an arbitrary observable for parti-cle 1, etc., satisfies the 3-dimensional version of equation (2.48) under the main requirement that the function Cn |⃗x2 − ⃗x1|/λC , for |⃗x2 − ⃗x1| → ∞, verifies: Cn → 0; that is, that in (2.51) the quantum entanglement that carries this in the context of approximation to the quantum-classical boundary is erased.