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In the solution of the Klein-Gordon equation for the shutter problem, we prove that, at internuclear distances, a relativistic beam of Pi-mesons has a probability density which oscillates in time in a similar way to the spatial dependence in optical Fresnel diffraction from a straight edge. However, for an extreme-relativistic beam, the Fresnel oscillations turn into quantum damped beat oscillations. We prove that quantum beat oscillations are the consequence, at extreme-relativistic velocities, of the interference between the initial incident wave function, and the Green’s function in the relativistic shutter problem. This is a pure quantum relativistic phenomenon.

Quantum beat oscillations are a common subject in Atomic and Molecular Spectroscopy. In Atomic physics, the term quantum beat refers to a superposed oscillatory behavior in the light intensity emitted by some suddenly excited atomic systems in their subsequent decay [

In this same context is the work of Villavicencio et al. [

Diffraction in time oscillations is a pure quantum phenomenon, and similar oscillations arise at the moment of closing and opening gates in nanoscopic circuits [

In this paper we address the following question: are there quantum shutter solutions (not in the Schrödinger’s equation) where quantum beats substitute the usual Fresnel oscillations of diffraction in time? In this paper, we report the positive finding of such result in the diffraction in time of an extreme-relativistic beam of free particles.

The main contribution of this paper is to show the existence of transient quantum beats in the extreme-relativistic quantum shutter problem. After solving the Klein-Gordon equation for the 1D shutter problem, we prove that, at internuclear distances, a relativistic beam of neutral Pi-mesons, has a probability density which oscillates in time in a similar way to the Fresnel oscillations derived in the Schrödinger equation. However, for an extreme-relativistic beam, the Fresnel oscillations turn into quantum damped beats!

In the relativistic shutter problem for spin-0 particles, we assume for all t ≤ 0 , that we have a discontinuous right-moving plane wave in the left side of a perfectly absorbing shutter, and zero to the right:

ψ ( x , t ≤ 0 ) = { e i ( k x − ω t ) x ≤ 0 0 x > 0 = e i ( k x − ω t ) θ ( − x ) , (1)

here θ ( x ) denotes the step function, and the angular frequency is given by ω = c ( k 2 + μ 2 ) 1 / 2 and μ ≡ m c / ℏ . At t = 0 , when the shutter is suddenly opened, we have the initial conditions:

ψ ( x , 0 ) = e i k x θ ( − x ) , ∂ ψ ( x , 0 ) ∂ t = − i ω e i k x θ ( − x ) , (2)

which we use to solve the Klein-Gordon (K-G) equation,

∂ 2 ψ ∂ x 2 − ∂ 2 ψ ∂ ( c t ) 2 = μ 2 ψ . (3)

The exact solution of this initial-value problem is given in the appendix A, where we use dimensionless variables for “position” χ and “time” τ defined by

χ ≡ μ x = m c ℏ x , τ ≡ μ c t = m c 2 ℏ t . (4)

For arbitrary ( τ > 0 , χ > 0 ), on the right-hand side of the shutter, we have the exact K-G wave function solution:

2 e i Ω τ θ ( τ − χ ) ψ ( χ , τ ; Ω ) = e i Ω χ − ∫ χ τ d u e i Ω u { χ J 1 [ ξ ( χ , u ) ] ξ ( χ , u ) + i κ J 0 [ ξ ( χ , u ) ] } ≡ e i Ω χ + i κ ( C 0 + i S 0 ) − χ ( C 1 + i S 1 ) . (5)

Here, we have defined both dimensionless “wave-number” κ and “angular-frequency” Ω as:

ξ ( χ , τ ) ≡ τ 2 − χ 2 , κ ≡ k μ = v / c 1 − ( v / c ) 2 , Ω ≡ ω μ c = 1 1 − ( v / c ) 2 . (6)

We have also defined, the real functions: ( C 0 , S 0 , C 1 , S 1 ) given by:

C 0 ( χ , τ ; Ω ) + i S 0 ( χ , τ ; Ω ) ≡ θ ( τ − χ ) ∫ χ τ e i Ω u J 0 [ ξ ( χ , u ) ] d u , (7a)

C 1 ( χ , τ ; Ω ) + i S 1 ( χ , τ ; Ω ) ≡ θ ( τ − χ ) ∫ χ τ e i Ω u J 1 [ ξ ( χ , u ) ] ξ ( χ , u ) d u . (7b)

Given the K-G wave function ψ ( x , t ) , the probability density is calculated by:

ρ ( x , t ) = i ℏ 2 m c 2 ψ ∗ ∂ ψ ∂ t + c . c . , (8)

which, for dimensionless variables, turn into ρ ( χ , τ ) , we have:

ρ ( χ , τ ) = i 2 ψ ∗ ( χ , τ ) ∂ ψ ( χ , τ ) ∂ τ + c . c . (9)

Notice that the initial relativistic plane-wave, ψ ( χ , τ ≤ 0 ) = exp [ i ( κ χ − Ω τ ) ] , does not have unit density, in fact we have

ρ ( χ , τ ≤ 0 ) = Ω . (10)

This means, as we will show, that the time-evolved density ρ ( χ > 0 , τ > 0 ) will oscillate in time around the initial value: ρ = Ω . Using Equation (5) we get for arbitrary ( τ > 0 , χ > 0 ) the exact K-G probability density:

8 ρ ( χ , τ ) Ω θ ( τ − χ ) = [ e i Ω χ − χ ( C 1 + i S 1 ) − i κ ( C 0 + i S 0 ) − i χ Ω J 1 ( ξ ) ξ e i Ω τ + κ Ω J 0 ( ξ ) e i Ω τ ] × [ e − i Ω χ − χ ( C 1 − i S 1 ) + i κ ( C 0 − i S 0 ) ] + c . c . (11)

The presence of the function θ ( τ − χ ) implies, as expected, that for times 0 ≤ t ≤ x / c the density vanishes, in full agreement with the relativity theory.

At this point, we apply our present results to a relativistic beam of neutral Pi-mesons ( π 0 ) which are 0-spin particles. We know that Pi-mesons are the carriers of nuclear forces and for this, and no other reason, we assume a particle detector fixed at a distance x from the shutter of about, x = 1.7 × 10 − 15 m , where nuclear forces have a maximum range. All mesons are unstable and decay in various ways. Neutral Pi-mesons have a mean-life given by: t 0 ≈ 10 − 17 sec , which for a mass of, m = 135 MeV / c 2 , we have, τ 0 ≈ 10 + 7 . Consequently, our theoretical predictions for diffraction in time for neutral Pi-mesons will make sense only if the predicted transient behavior happens in a range of times Δ τ having orders of magnitude less than τ 0 . Next, we will show that for all assumed conditions of position and energy, this condition is well satisfied.

Let us assume a relativistic beam of monochromatic, ω = c ( k 2 + μ 2 ) 1 / 2 , μ ≡ m c / ℏ , neutral Pi-mesons ( m = 135 MeV / c 2 ) with “low” velocities, v / c = 0.5 , and a particle detector at, x = 1.7 × 10 − 15 m . Using the exact result in Equation (12) we show in

The main features of this plot are: 1) As expected, for times 0 ≤ t ≤ x / c the density vanishes. 2) In spite of the fact that the exact relativistic density given in Equation (12) looks different from the Schrödinger one, however, for t > x / c the damped oscillations we see in

The next obvious question is: how does diffraction in time look like for extreme-

relativistic values of v/c? In this section, we show properties of extreme relativistic diffraction in time which, as far as we know, have never been reported before.

As we gradually increase the relativistic velocities v/c, the diffraction in time pattern gradually changes from Fresnel into damped beats! To illustrate this, in

Finally in

Two simultaneous properties are needed for the existence of quantum beats. The

first is that in the probability density, which is a quadratic expression of the wave function, we have products of two different oscillatory functions. In our case in Equation (12) we have the most important contribution of these products:

[ sin ( Ω τ ) + cos ( Ω τ ) ] J 0 ( τ 2 − χ 2 ) ( τ ≥ χ ) (12)

where the agular frequency Ω is given by Ω ≡ 1 1 − ( v / c ) 2 .

The second property is that only for extreme relativistic velocities the two oscillatory functions must have similar angular frequencies. Here, for v / c = 0.998 , ( Ω = 15 ), the Bessel J 0 becomes a slow oscillating function of time and each trigonometric is a fast one. See

In conclusion, quantum beat oscillations are the consequence, at extreme relativistic velocities, of the interference between the initial incident wave function: e i Ω τ , and the Green’s function of the relativistic shutter problem: J 0 ( τ 2 − χ 2 ) . This a pure quantum relativistic phenomenon. We claim that our result is original and has never been reported before.

Let us consider the Klein-Gordon (K-G) equation for ψ ( χ , τ ) defined in the infinite range − ∞ < χ < ∞ ,

∂ 2 ψ ∂ χ 2 − ∂ 2 ψ ∂ τ 2 = ψ . (13)

We demand bounded boundary conditions: lim | x | → ∞ | ψ ( χ , τ ) | < ∞ , and the initial conditions, corresponding to the shutter problem:

ψ ( χ , 0 ) = e i κ χ θ ( − χ ) , ∂ ψ ( χ , 0 ) ∂ τ = − i Ω e i κ χ θ ( − χ ) (14)

where Ω ≡ 1 + κ 2 .

To solve the K-G equation, we begin taking the Laplace transform of Equation (14). Denoting,

ϕ ( χ , s ) ≡ L [ ψ ( χ , τ ) ] ≡ ∫ 0 ∞ e − s τ ψ ( χ , τ ) d τ ,

we find the differential equation

d 2 ϕ ( χ , s ) d χ 2 − ( s 2 + 1 ) ϕ ( χ , s ) = − ( s − i Ω ) e i κ χ θ ( − χ ) (15)

which holds in the range − ∞ < χ < ∞ .

Due to the presence of the step function θ ( − χ ) , the origin χ = 0 is a singular point where the function and its first derivative must be continuous. This fact suggests to break the infinite range into the left ( χ ≤ 0 ) and right ( χ ≥ 0 ) ranges, having different differential equation for each one. For the left-hand side of the shutter, χ ≤ 0 , we define ϕ < ( χ , s ) as the solution of the differential equation:

d 2 ϕ < d χ 2 − ( s 2 + 1 ) ϕ < = − ( s − i Ω ) e i κ χ , (16)

and for the right-hand side, χ ≥ 0 , we define ϕ > ( χ , s ) as the solution of

d 2 ϕ > d χ 2 − ( s 2 + 1 ) ϕ > = 0. (17)

Here both functions ϕ < and ϕ > must be bounded: ( ϕ < at χ → − ∞ ) and ( ϕ > at χ → + ∞ ). The important boundary condition is that the two functions and their corresponding first derivatives must be continuous at the interface, χ = 0 .

Equations (17) and (18) are ordinary second order differential equations and their solution is readily obtained. Taking into account the boundary conditions at ± ∞ we have:

ϕ < ( χ , s ) = A e χ s 2 + 1 + 1 s + i Ω e i κ χ ( χ ≤ 0 ) (18)

and

ϕ > ( χ , s ) = B e − χ s 2 + 1 ( χ ≥ 0 ) (19)

The constants A and B are fixed from the continuity conditions at the interface χ = 0 :

ϕ < ( 0 , s ) = ϕ > ( 0 , s ) , d ϕ < ( 0 , s ) / d χ = d ϕ > ( 0 , s ) / d χ

We have a set of coupled algebraic equations for the constants A and B:

A + 1 s + i Ω = B (20)

A s 2 + 1 + i κ s + i Ω = − B s 2 + 1 (21)

with solutions given by:

2 A = − 1 s + i Ω ( 1 + i κ s 2 + 1 ) (22)

2 B = 1 s + i Ω ( 1 − i κ s 2 + 1 ) (23)

Substituting Equation (23) into Equation (19) and Equation (24) into Equation (20) we get, in space ( χ , s ), the exact solution for relativistic diffraction in time.

For ( χ ≤ 0 ) we have the solution:

ϕ < ( χ , s ) = − 1 2 1 s + i Ω ( 1 + i κ s 2 + 1 ) e χ s 2 + 1 + 1 s + i Ω e i κ χ , (24)

and for ( χ ≥ 0 ) we have:

ϕ > ( χ , s ) = 1 2 1 s + i Ω ( 1 − i κ s 2 + 1 ) e − χ s 2 + 1 . (25)

To get the time solution in the variable τ , we must invert the Laplace transforms. We readily find in Laplace Transforms Tables [

L − 1 [ 1 s + i Ω ] = e − i Ω τ (26)

L − 1 [ e − | χ | s 2 + 1 ] = δ ( τ − | χ | ) − | χ | J 1 ( τ 2 − χ 2 ) τ 2 − χ 2 θ ( τ − | χ | ) (27)

L − 1 [ e − | χ | s 2 + 1 s 2 + 1 ] = J 0 ( τ 2 − χ 2 ) θ ( τ − | χ | ) (28)

where J 0 ( x ) and J 1 ( x ) denote Bessel functions. Next, using the convolution theorem, and after some simplifications we have:

L − 1 [ e − | χ | s 2 + 1 ( s + i Ω ) s 2 + 1 ] = θ ( τ − | χ | ) ∫ | χ | τ d u e − i Ω ( τ − u ) J 0 ( u 2 − χ 2 ) (29)

and

L − 1 [ e − | χ | s 2 + 1 s + i Ω ] = θ ( τ − | χ | ) [ e − i Ω ( τ − | χ | ) − | χ | ∫ | χ | τ d u e − i Ω ( τ − u ) J 1 ( u 2 − χ 2 ) u 2 − χ 2 ] (30)

With the help of Equations (30) and (31) we have the final solutions:

For χ ≤ 0 we get, as expected, the incident and reflected wave:

ψ < ( χ , τ ) = e i ( κ χ − Ω τ ) − 1 2 θ ( τ + χ ) [ e − i Ω ( χ + τ ) + χ ∫ − χ τ d u e i Ω ( u − τ ) J 1 ( u 2 − χ 2 ) u 2 − χ 2 + i κ ∫ − χ τ d u e i Ω ( u − τ ) J 0 ( u 2 − χ 2 ) ] , (31)

and for χ ≥ 0 we get the transmitted wave:

ψ > ( χ , τ ) = 1 2 θ ( τ − χ ) [ e i Ω ( χ − τ ) − χ ∫ χ τ d u e − i Ω ( τ − u ) J 1 ( u 2 − χ 2 ) u 2 − χ 2 − i κ ∫ χ τ d u e − i Ω ( τ − u ) J 0 ( u 2 − χ 2 ) ] , (32)

We see that in fact, the 1D shutter problem is nothing but a particular time-dependent scattering problem, and here we have the exact relativistic solution.

The author declares no conflicts of interest regarding the publication of this paper.

Godoy, S. (2021) Transient Quantum Beat Oscillations in Extreme-Relativistic Diffraction in Time. Journal of Modern Physics, 12, 1-9. https://doi.org/10.4236/jmp.2021.121001