A study on the scattering of matter waves through slits

Abstract

Scattering of matter waves through slits has been explored using the Feynman Path Integral formalism. We explicitly plot the near-zero probability densities to analyse the behaviour near the slit. Upon doing so, intriguing patterns emerge, most notably the braid-like structure in the case of double slits, whose complexity increases as one increases the number of slits. Furthermore, the plot shows the existence of a transition region, where the distribution of near-zero probability points changes from the braided to the fringe-like structure, which has been analysed by explicitly expressing the wavefunction as a hypergeometric function. These patterns are analysed while considering the continuity equation and its consequences for the regions with zero probability density.

Appendix A: Two-dimensional analysis

One can attempt to write the wavefunction in two-dimensional case (2-D) by solving the time-dependent Schrödinger’s equation with an appropriate boundary condition at $y=y_1$ describing the slit and the barrier. Alternatively, one may proceed with a similar set of arguments as discussed in Sect. 2.1 using the FPI and demonstrate how similar it is to the 1-D wavefunction.

For the initial state, consider a collapsed state:

$$\begin{aligned} \psi \left( x,y,t=t_0\right) =\delta \left( x-x_0\right) \delta \left( y-y_0\right) \end{aligned}$$

(67)

The kernel in 2-D is slightly different from that in 1D. By taking Lagrangian of free particle in 2-D, i.e.

$$\begin{aligned} \mathcal {L} = \frac{1}{2} \, m(\dot{x}^2 + \dot{y}^2) \end{aligned}$$

(68)

and following the procedure as given in [8, 18], one obtains the following Kernel for a particle going from position $(x_a, y_a)$ at time $t_a$ to $(x_b, y_b)$ at time $t_b$ :

$$\begin{aligned} K\left( b,a\right) =\frac{m}{2\pi \iota \hbar \left( t_b-t_a\right) } \exp \left\{{\frac{\iota m\left[ \left( x_b-x_a\right) ^2+\left( y_b-y_a\right) ^2\right] }{2\hbar \left( t_b-t_a\right) }}\right \} \end{aligned}$$

(69)

Introducing a slit at a specific value of y, let us choose it to be $y=y_1$, then at the position of the slits, the wavefunction will be

$$\begin{aligned} \psi \left( x_1,y_1,t_1\right) =\frac{m}{2\pi \iota \hbar \left( t_1-t_0\right) } \exp\left\{ {\frac{\iota m\left[ \left( x_1-x_0\right) ^2+\left( y_1-y_0\right) ^2\right] }{2\hbar \left( t_1-t_0\right) }}\right \} \end{aligned}$$

(70)

It is not straightforward to compute the evolution beyond the slit in higher dimensional cases. One reason is that one has to consider the paths contributed from the wavefunction present before the slit, and there is a possibility of backflow of paths as well as looped paths around the slits [10]. In order to understand the salient feature of the scenario, the analysis can be carried out by introducing certain approximations, leading to a considerable simplification of the analysis.

Note that the initial condition (collapsed state) is represented by Dirac delta function, by the uncertainty principle the wavefunction is in the superposition of all the possible momentum states. However, our desired state of analysis in 2-D is a plane-wave having a specific momentum p corresponding to wavelength $\lambda$, given by the de Broglie relation $p\ =\ h/\lambda$. One proposed solution is to slice the wavefunction originating from the collapsed state only to select the part which has the effective wavelength $\lambda$. Consider the kernel of the free particle:

$$\begin{aligned} K=F\left( t\right) \exp\left\{ {\frac{\iota m y^2}{2\hbar t}}\right \} \end{aligned}$$

(71)

F(t) is the normalization function, which depends on the dimensions of the space in the system. To find the effective wavelength, we propose to increment y by $\lambda$. This increment should introduce a phase difference of $2\pi$ in the argument of the complex exponential. Mathematically it is expressed as:

$$\begin{aligned} 2\pi =\frac{m\left( y+\lambda \right) ^2}{2\hbar t}-\frac{my^2}{2\hbar t}=\frac{my\lambda }{\hbar t}+\frac{m\lambda ^2}{2\hbar t} \end{aligned}$$

(72)

If one goes sufficiently far away from the point of origin such that $y\ \gg \ \lambda$, then one can essentially ignore $\lambda ^2$ contribution. Therefore:

$$\begin{aligned} \lambda =\frac{2\pi \hbar }{m\left( y/t\right) } \end{aligned}$$

(73)

We can interpret this form as following; If one takes the ratio y/t as constant, one is essentially tracing the sliced wavefunction whose effective wavelength is $\lambda$. Therefore, we look only at those paths that conserve this ratio. This idealization allows us to deal with the issue of backflow, as the selected paths encounter the slit only once at specific y/t. Using this ratio, we can make the substitution.

$$\begin{aligned} t=\frac{\lambda m y}{2\pi \hbar } \end{aligned}$$

(74)

The above relation strikingly resembles the dimensional substitution we made in 1-D (Eq. 9). Thus, we can proceed to write the evolution wavefunction beyond the slit at the point $\left( x_2,y_2,t_2\right)$.

$$\begin{aligned} \psi \left( x_2,y_2,t_2\right) =\int _{S^\prime } K\left( b,a\right) \psi \left( x_1,y_1,t_1\right) \, \mathrm{d}x_1 \end{aligned}$$

(75)

The kernel K(b, a) and wave function $\psi (x_1,y_1,t_1)$ appearing here in this expression has been explicitly define in Eqs. (69) and (70) respectively, and $S^{\prime }$ is the set of points lying inside the slit. Upon changing the variables as $x^\prime =x_1-x_0$, $y^\prime =y_1-y_0$, $t^\prime =t_1-t_0$, $x^{\prime \prime }=x_2-x_1$, $y^{\prime \prime }=y_2-y_1$ and $t^{\prime \prime }=t_2-t_1$, Eq. (75) becomes;

$$\begin{aligned}&\psi \left( x_2,y_2,t_2\right) =\int _{S^\prime }F\left( t^\prime \right) \exp \left\{{\frac{\iota m\left[ x^{\prime 2}+y^{\prime 2}\right] }{2\hbar t^\prime }}\right \}\nonumber \\&\quad F\left( t^{\prime \prime }\right) \exp \left\{{\frac{\iota m\left[ x^{\prime \prime 2}+y^{\prime \prime 2}\right] }{2\hbar t^{\prime \prime }}}\right \}\, \mathrm{d}x_1 \end{aligned}$$

(76)

Now, making use of the fact that y/t is a constant (Eq. 73), we get:

$$\begin{aligned}&\psi \left( x_2,y_2,t_2\right) =\tilde{F}\left( y^\prime \right) \tilde{F}\left( y^{\prime \prime }\right) \exp\left\{ {\frac{\iota \pi \left[ y^\prime +y^{\prime \prime }\right] }{\lambda }}\right \}\nonumber \\&\quad \int _{S^\prime } \exp \left\{{\frac{\iota \pi }{\lambda }\left( \frac{x^{\prime 2}}{y^\prime }+\frac{x^{\prime \prime 2}}{y^{\prime \prime }}\right) }\right \}\, \mathrm{d}x_1 \end{aligned}$$

(77)

The factors $\tilde{F}\left( y^\prime \right)$ and $\tilde{F}\left( y^{\prime \prime }\right)$ are obtained by transforming $F\left( t^\prime \right)$ and $F\left( t^{\prime \prime }\right)$, respectively, under the transformation $t\ \mapsto y$. Further substituting

$$\begin{aligned} T\left( y^\prime ,y^{\prime \prime }\right) =\tilde{F}\left( y^\prime \right) \tilde{F}\left( y^{\prime \prime }\right) \exp \left\{{\frac{\iota \pi \left[ y^\prime +y^{\prime \prime }\right] }{\lambda }}\right \}\, , \end{aligned}$$

the wavefunction gets the similar form as was in the case of 1-D.

$$\begin{aligned} \psi \left( x_2,y_2,t_2\right) =T\left( y^\prime ,y^{\prime \prime }\right) \int _{S^\prime } exp{\left\{ \frac{\iota \pi }{\lambda }\left( \frac{x^{\prime 2}}{y^\prime }+\frac{x^{\prime \prime 2}}{y^{\prime \prime }}\right) \right\} }\, \mathrm{d}x_1 \end{aligned}$$

(78)

This suggests that the shape of probability density obtained in the idealized 2-D case is identical to that in 1-D.

1.1 A1: Steady-state analysis of the 2-D case

For a steady-state solution, the probability density must have ‘settled’ in the region and will not change with time. This will be equivalent to solving the time-independent Schrödinger’s equation with appropriate boundary conditions at $y=y_1$, which can further be written in the form of the Helmholtz equation

$$\begin{aligned} \frac{\mathrm{d}^2\psi }{\mathrm{d}z^2}\ \ =\ \ -\frac{\mathrm{d}^2\psi }{\mathrm{d}x^2}\ \ -\ k^2\psi \end{aligned}$$

(79)

and the relevant boundary condition would be a Sommerfeld condition (see, for example, [34] for details).

For the case of FPI, in order to introduce a steady-state, one needs to have a continuously emitting source which can be implemented through a boundary condition as follows

$$\begin{aligned} \psi _0\left( x,y_0,t\right) =\delta \left( x-x_0\right) \exp \left\{{\frac{\iota E t}{\hbar }}\right \} \end{aligned}$$

(80)

Here, E is the energy of the non-interacting particles that are introduced into the system.

With this modification, the source is continuously injecting particles in the system. Therefore, to compute the probability density at a point in space, it is required to consider the $\psi$ introduced at previous times as well. To simplify the analysis we employ the time slicing method, where the entire time-interval $[0,\infty )$ has been sliced into countable number of bins, each with a fixed size $\epsilon$.

Let $l\epsilon$ be the time elapsed since a wavefunction has been introduced to the system, where l is a positive integer. For the sake of brevity, we use $D=x^{\prime 2}+y^{\prime 2}$, here the symbols have the usual meaning as defined before. Hence we can write the steady-state wavefunction as:

$$\begin{aligned} \psi \left( D\right) =\sum _{l=1}^{\infty }\psi \left( D,l\epsilon \right) \end{aligned}$$

(81)

where for a free particle,

$$\begin{aligned} \psi \left( D,l\epsilon \right) =F\left( l\epsilon \right) \exp \left\{ {\frac{\iota m D}{2\hbar l\epsilon }}\right \} \exp \left \{{\frac{\iota E l\epsilon }{\hbar }}\right \}\end{aligned}$$

(82)

Noting that $F\left( l\epsilon \right) \propto 1 / l \epsilon$, one can effectively ignore wavefunctions for which time elapsed is large such that $F(l\epsilon ) \approx 0$. Also, the wavefunctions corresponding to the very small time elapses will have very rapid variations of phase. Thus, their overall contribution will be effectively negated.

Consider a wavefunction for which time elapsed is $t^\prime$ and look at the wavefunction from neighbouring times, i.e. $t^\prime \pm \Delta t$, where $\Delta t\ll t^\prime$. These are given by:

$$\begin{aligned} \psi \left( D,t^\prime \right)= & {} F\left( t^\prime \right) \exp \left \{{\frac{\iota m D}{2\hbar \left( t^\prime \right) }} \right \}\exp \left \{ {\frac{\iota E\left( t^\prime \right) }{\hbar }} \right \}\end{aligned}$$

(83)

$$\begin{aligned} \psi \left( D,t^\prime \pm \Delta t\right)= & {} F\left( t^\prime \pm \Delta t\right) \exp\left \{ {\frac{\iota m D}{2\hbar \left( t^\prime \pm \Delta t\right) \ }} \right \} \exp \left \{{\frac{\iota E\left( t^\prime \pm \Delta t\right) }{\hbar }} \right \}\end{aligned}$$

(84)

Since $\Delta t\ll t^\prime$, we can write

$$\begin{aligned} \frac{1}{t^\prime \pm \Delta t}\simeq \frac{1\mp \left( \Delta t/t^\prime \right) }{t^\prime }\, . \end{aligned}$$

Given this and the fact that F is inversely proportional to t we can write

$$\begin{aligned} F\left( t^\prime \pm \Delta t\right) \simeq F\left( t^\prime \right) \left[ 1\mp \left( \Delta t/t^\prime \right) \right] \, . \end{aligned}$$

Using these approximations, one obtains:

$$\begin{aligned}&\psi \left( D,t^\prime \pm \Delta t\right) =F\left( t^\prime \right) \left[ 1\mp \left( \Delta t/t^\prime \right) \right] \nonumber \\&\quad \exp \left\{ \frac{\iota m D}{2 \hbar t^\prime }\left[ 1\mp (\Delta t/ t^\prime )\right] \right\} \exp \left\{ \frac{\iota E t^\prime }{\hbar }\left[ 1\pm (\Delta t/ t^\prime )\right] \right\} \end{aligned}$$

(85)

After a simple rearrangement of terms,we get:

$$\begin{aligned}&\psi \left( D,t^\prime \pm \Delta t\right) =F\left( t^\prime \right) \exp \left\{{\frac{\iota m D}{2\hbar t^\prime }}\right\}\exp \left \{{\frac{\iota E t^\prime }{\hbar }}\right\}\nonumber \\&\quad \left[ 1\mp \left( \Delta t/t^\prime \right) \right] \exp \left \{{\frac{\mp \iota \left( \Delta t/t^\prime \right) }{\hbar }\left[ \frac{m D}{2t^\prime }-Et^\prime \right] }\right\} \end{aligned}$$

(86)

Evidently it can be re-expressed using Eq. (83) as:

$$\begin{aligned} \psi \left( D,t^\prime \pm \Delta t\right) =\psi \left( D,t^\prime \right) \ \left[ 1\mp \left( \Delta t/t^\prime \right) \right] \exp\left\{ {\frac{\mp \iota \left( \Delta t/t^\prime \right) }{\hbar }\left[ \frac{mD}{2t^\prime }-Et^\prime \right] }\right\} \end{aligned}$$

(87)

Recall that E is the energy of the particles emitted by the source and is a constant. In the classical scenario, it is given as $mv^2/2$. Furthermore, v can be taken as the constant ratio of position and time variable. If we take v to be $y^\prime /t^\prime$, then it is evident to write down $Et^\prime =my^{\prime 2}/2t^\prime$. Under these conditions, Eq. (87) can be written as:

$$\begin{aligned} \psi \left( D,t^\prime \pm \Delta t\right) =\psi \left( D,t^\prime \right) \left[ 1\mp \left( \Delta t/t^\prime \right) \right] \exp \left\{ {\frac{\mp \iota \left( \Delta t/t^\prime \right) }{\hbar }\frac{mx^{\prime 2}}{2t^\prime }} \right\}\end{aligned}$$

(88)

If the argument inside the exponent is taken to be small as well, given $x^\prime$ is comparable with $y^\prime$, then it is possible to write:

$$\begin{aligned} \psi \left( D,t^\prime \pm \Delta t\right) =\psi \left( D,t^\prime \right) \left[ 1\mp \left( \Delta t/t^\prime \right) \right] \left[ 1\mp \frac{\iota \left( \Delta t/t^\prime \right) }{\hbar }\frac{mx^{\prime 2}}{2t^\prime }\right] \end{aligned}$$

(89)

Expanding the square brackets and retaining the terms only up to first-order in $\left( \Delta t/t^\prime \right)$, one gets:

$$\begin{aligned} \psi \left( D,t^\prime \pm \Delta t\right) =\psi \left( D,t^\prime \right) \left[ 1\mp \left( 1+\frac{\iota m x^{\prime 2}}{2\hbar t^\prime }\right) \frac{\Delta t}{t^\prime }\right] \end{aligned}$$

(90)

Therefore,

$$\begin{aligned} \psi \left( D,t^\prime +\Delta t\right) +\psi \left( D,t^\prime -\Delta t\right) \approx 2\psi \left( D,t^\prime \right) \end{aligned}$$

(91)

Hence, we can write using Eq. (81):

$$\begin{aligned} \psi \left( D\right) \approx k\psi \left( D,t^\prime \right) \ \end{aligned}$$

(92)

With the condition that $y^\prime /t^\prime$ is constant, such that they are related to energy as, $E=my^{\prime 2}/2t^{\prime 2}$. Therefore, one concludes that the idealization of the 2-D case that has been developed in the previous sub-section is a good approximation to the steady state scenario.

1.2 A2: Validity of the 1-D approximation in the near field region

In the steady state analysis (Appendix A.1), we used the approximation that the $x^\prime$ and $y^\prime$ are comparable. However, at the location of bubbles $x^\prime \gg y^\prime$ (for the description of the bubble refer to Sect. 4). Therefore, one cannot approximate Eqs. (88, 89).

To check the validity of the approximation in the near field region, we have to proceed with the Eq. (88) without the linear approximation.

Consider:

$$\begin{aligned} T= & {} \exp \left\{ \frac{\iota (\Delta t /t^\prime )}{\hbar } \frac{m x^{\prime 2}}{2 t^\prime } \right\} \end{aligned}$$

(93)

$$\begin{aligned} G= & {} \psi (D, t^\prime + \Delta t) + \psi (D, t^\prime - \Delta t) \end{aligned}$$

(94)

Therefore, using the Eq. (88) we can write:

$$\begin{aligned} G= & {} \psi (D, t^\prime ) \left[ \left[ 1-\left( \Delta t/t^\prime \right) \right] T^*+\left[ 1+\left( \Delta t/t^\prime \right) \right] T\right] \end{aligned}$$

(95)

$$\begin{aligned} \therefore G= & {} \ \psi \left( D,\ t^\prime \right) \left[ \left( T+T^*\right) +\frac{\Delta t}{t^\prime }\left( T-T^*\right) \right] \end{aligned}$$

(96)

Assume the argument inside the exponential (Eq. 93) to be $\tilde{T}$, that is:

$$\begin{aligned} \widetilde{T}= & {} \ \frac{\left( \Delta t/t^\prime \right) }{\hbar }\frac{mx^{\prime 2}}{2t^\prime } \end{aligned}$$

(97)

$$\begin{aligned} \Rightarrow G= & {} \psi \left( D,t^\prime \right) \left[ \cos {\left( \widetilde{T}\right) }+\iota \frac{\Delta t}{t^\prime }\sin {\left( \widetilde{T}\right) }\right] \end{aligned}$$

(98)

It is clear that the neighbouring wavefunctions are not simply adding up to the 1-D wavefunction, but have an additional factor to them. To see the cumulative effect of all the neighbouring wavefunctions, we must integrate them over $\frac{\Delta t}{t^\prime }$. Foremost, substituting

$$\begin{aligned} z=\frac{\Delta t}{t^\prime };k=\frac{mx^{\prime 2}}{2\hbar t^\prime }\ \end{aligned}$$

(99)

Implying that the sum described in Eq. (81) can be written as:

$$\begin{aligned} \Rightarrow \psi \left( D\right) =\int _{0}^{a}G\mathrm{d}z \end{aligned}$$

(100)

Here, a is upper limit to the integral, which is less than 1 and greater than zero.

$$\begin{aligned} \therefore \psi \left( D\right) =\psi \left( D,t^\prime \right) \int _{0}^{a}{\cos {\left( kz\right) }+\iota z\sin {\left( kz\right) }}\mathrm{d}z \end{aligned}$$

(101)

Upon evaluating the integral we will get:

$$\begin{aligned} \therefore \psi \left( D\right) =\psi \left( D,t^\prime \right) \frac{1}{k}\ \left[ \sin {\left( kz\right) }+\frac{\iota }{k}\left[ \sin {\left( kz\right) }-kz\cos {\left( kz\right) }\right] \right] _0^{a} \end{aligned}$$

(102)

Observing the above equation we can comment that all the zeroes of the $\psi (D)$ will be identical to that of the zeroes of $\psi (D, t^\prime )$, as the other part of the expression does not have any zeroes. Thus, the Null map for the case of a single source will be considered identical to that obtained in the 1-D formalism. When considering the double slit or dual source, the additional factor will change the Null map in the region. However, near the middle of the slits, the map will nearly be the same as that of 1-D, for similar reasons as discussed in Sect. 5.2.