How Does Interference Fall?

Abstract

We study how single- and double-slit interference patterns fall in the presence of gravity. First, we demonstrate that universality of free fall still holds in this case, i.e., interference patterns fall just like classical objects. Next, we explore lowest order relativistic effects in the Newtonian regime by employing a recent quantum formalism which treats mass as an operator. This leads to interactions between non-degenerate internal degrees of freedom (like spin in an external magnetic field) and external degrees of freedom (like position). Based on these effects, we present an unusual phenomenon, in which a falling double slit interference pattern periodically decoheres and recoheres. The oscillations in the visibility of this interference occur due to correlations built up between spin and position. Finally, we connect the interference visibility revivals with non-Markovian quantum dynamics.

Appendices

Appendices

A The COW Experiment

The Colella-Overhauser-Werner experiment provided the first evidence of a gravitational effect that is purely quantum mechanical [17]. In this experiment, Colella et al. used a silicon crystal interferometer to split a beam of neutrons, placing one of the beam paths in a higher gravitational potential (see Fig. 5). The difference in the gravitational potential between each arm results in a relative phase shift, which, when recombined, can be measured as modulated intensity. On the length scale of the interferometer, the gravitational field is approximately constant. This allows the relative phase difference between the beams to be calculated using the Wentzel-Kramers-Brillouin (WKB) approximation; that is to integrate the potential difference between the classical trajectories over time [27]. The two vertical paths of the interferometer contribute phases which cancel out, leaving only the horizontal paths. The phase shift is found to be

$$\begin{aligned} \Delta \Phi = \frac{2\pi m_Im_{_G}g A \lambda }{h^2}\sin \phi . \end{aligned}$$

(39)

Fig. 5

COW Interferometer | Left Schematic of the apparatus used in the COW experiment, taken from Ref. [17]. The interferometer is rotated about the axis of the first Bragg angle of diffraction. Right Simplified diagram used to derive the induced phase shift

A phase shift of this form would be predicted for a quantum mechanical particle in the presence of any scalar potential; in this case, it is the Newtonian gravitational potential. A full description of this effect requires only regular quantum mechanics and Newtonian theory, needing no metric description of gravity, but being unexplainable by classical Newtonian gravity alone. It represents the first evidence of gravity interacting in a truly quantum mechanical way. However, from the perspective of quantum theory, this effect is well understood as a scalar Aharanov–Bohm effect and manifests similarly for charged particles in electric potentials [14, 28].

B Single Slit Diffraction in a Newtonian Gravitational Potential

1.1 B.1 Derivation of the Propagator

First consider the Lagrangian for a free particle $L = \frac{1}{2}m\dot{x}^2$. The Feynman propagator is given by

(40)

(41)

This result will be needed when we consider the propagator for a particle in a linear gravitational potential. In this case the Lagrangian is given by $L = \frac{1}{2} m \dot{x}^2 - m g x$, which gives the Feynman propagator

(42)

To simplify this calculation we express the path x(t) in terms of deviations from the classical trajectory $x_c(t)$ which satisfies the Euler-Lagrange equations of motion. The Feynman measure which sums over all possible paths then becomes a sum over all possible deviations from the classical path. The action expressed in terms of this new parametrisation is

(43)

(44)

where , which is evidently the extremised action given by the classical trajectory. The term containing $\dot{x}_c\delta \dot{x}$ can be integrated by parts, realising that the deviations are zero at the endpoints of the path:

$$\begin{aligned} S[x_c(t) + \delta x(t)] = S[x_c(t)] + \int _{t}^{t^{\prime }} \frac{1}{2} m (\delta \dot{x})^2 \, dt + [ \ddot{x} \delta x]_{t}^{t^{\prime }} - \int _{t}^{t^{\prime }}m\delta x \left( \ddot{x}_c + g\right) dt. \end{aligned}$$

(45)

The last two terms vanish ($\ddot{x}_c = -g$) and substituting the rest into Eq. (42) and factoring out the classical action we arrive at

(46)

The remaining Feynman integral over the deviations is recognised as the free particle propagator in Eq. (40), but with the subtle difference being that $x=x^{\prime }=0$. Substituting the integral with the expression from Eq. (41) the propagator becomes

$$\begin{aligned} K_g(x^{\prime },t^{\prime };x,t)= \frac{\exp \left[ \frac{i}{\hbar } S\left[ x_c(t)\right] \right] }{\sqrt{2\pi i \hbar (t^{\prime }- t)/m}}. \end{aligned}$$

(47)

Now, all that remains is to find the explicit form for the classical action. We begin with the classical equation of motion, $\ddot{x}_c = -g$, and solve to find the general solution,

$$\begin{aligned} x_c(t)= -\frac{1}{2} g t^2 + a t + b. \end{aligned}$$

(48)

We now impose the boundary conditions $x(t)= x$ and $x(t^{\prime })= x^{\prime }$ and solve for the constants a and b:

(49)

Solving for a and b gives

(50)

and

(51)

(52)

(53)

(54)

Thus, the action of the path taken from (x, t) to $(x^{\prime }, t^{\prime })$ is

(55)

(56)

(57)

(58)

Finally, substituting this into Eq. (47), we arrive at the complete expression for the propagator for a particle in a gravitational potential,

$$\begin{aligned} K_g(x^{\prime },t^{\prime };x,t)&=\; \frac{\exp \left[ \frac{im}{2\hbar }\left\{ \frac{(x^{\prime }- x)^2}{t^{\prime }-t} - g(x+x^{\prime })(t^{\prime }-t) - \frac{g^2}{12}(t^{\prime }-t)^3\right\} \right] }{\sqrt{2\pi i \hbar (t^{\prime }- t)/m}}\;. \end{aligned}$$

(59)

1.2 B.2 Calculating the Single Slit Wavefunction

We now consider applying this propagator to the problem at hand. Let’s begin by assuming that the slits are long enough to ignore diffraction effects in the y direction. Consider a source of particles at the origin (0, 0) and let a double slit be located at distance $z=D$ metres from the source. Each slit has width 2a with centre located at $x=\pm b$. The screen is then a further L metres away from the slits. The two dimensional propagator required for this problem is given by a free particle propagator in the z-direction multiplied by the gravitational propagator for the x direction as calculated in Eq. (2). This propagator allows us to ask the question of If a particle initially starts at position $\vec {r}= (0,0)$, what is the probability of finding the it at position $\vec {r}\, ^{\prime }=(x,D+L)$ on the screen? This distribution in x will be the the two slit interference pattern that we seek. When computing this amplitude we consider a semi-classical approach. We assume that the ‘trajectory’ of the neutron can be separated into two parts: (a) the path from the source to the slits, followed by (b) the path from the slits to the screen. Quantum mechanically the particles need not pass through the slits and there even exists the possibility of them passing through the slits multiple times before hitting the screen. That being said the probabilities associated with these events are negligible.

The semi-classical approach is valid provided that the majority of the particle’s momentum is in the z direction, such that the wavelength is approximately the z-direction wavelength, $\lambda \approx \frac{2\pi \hbar }{m v_z}$. We assume that this wavelength is much smaller than the z-direction scale lengths D and L in conjunction with the assumption that these are much larger than the x direction scale lengths. This allows us to consider the particles motion in the z-direction as approximately classical and allows for the motion to be partitioned about the slits. The specific propagator $K_g^{(1)}(x,T+\tau ;0,0)$, for the process of starting at point $(x=0,z=0)$ at time $t=0$, passing through position $(\omega \in [b-a,b+a],D)$ at time T and then arriving at position $(x,D+L)$ at time $T+\tau$ will simply be a product of propagator for each independent component of the path, integrated over the slit distribution $\Omega (\omega )$,

$$\begin{aligned} \Omega (\omega )&= \left\{ \begin{array}{cc} 1 &{} b-a< \omega <b+a \\ 0 &{} \text {otherwise} \end{array}\right. , \end{aligned}$$

(60)

$$\begin{aligned} K_g^{(1)}(x,T+\tau ;0,0) =&K_0(D,T;0,0)K_0(D+L,T+\tau ;D,T) \nonumber \\&\quad \quad \times \int _{b-a}^{b+a} K_g(\omega ,T;0,0)K_g(x,T+\tau ;\omega ,T)d\omega . \end{aligned}$$

(61)

Evidently for any particular choice of D and L, the two z propagators will only give global phase which is identical for all x. This global phase has no measurable effects, allowing us to discard the z propagators. This integral is performed in below giving the result

$$\begin{aligned} K_g^{(1)}(x,T+\tau ;0,0) =\,&\frac{e^{i\phi (x)}}{i\sqrt{2 \lambda (D+L)}} \nonumber \\&\quad \times \bigg \{C[\sigma _+(x)] - C[\sigma _-(x)] + iS[\sigma _+(x)]-iS[\sigma _-(x)]\bigg \}, \end{aligned}$$

(62)

where $C[u] \,{\equiv } \int _0^u \cos \left( \frac{\pi }{2}x^2\right) dx$ is the Fresnel cosine function, $S[u] \equiv \int _0^u \sin \left( \frac{\pi }{2}x^2\right) dx$ is the Fresnel sine function and $\eta = 1 + \frac{L}{D}$ and

$$\begin{aligned}&\sigma _{\pm }(x) = \sqrt{\frac{2}{\lambda L}\eta }\left\{ (b\pm a) -\frac{x}{\eta } - \frac{1}{2}g \frac{m^2 \lambda ^2}{h^2}DL\right\} ,\end{aligned}$$

(63)

$$\begin{aligned}&\phi (x) = \pi \left\{ \frac{x^2}{\lambda (D+L)} - m g x\frac{\lambda (D+L)}{h^2} - \frac{g^2}{12}\frac{m^4\lambda ^3}{h^4}(D+L)(D-L)^2 \right\} . \end{aligned}$$

(64)

The propagator obtains its name for good reason. An initial wavefunction $\psi _0(x)$ convoluted with the propagator will give the future state of the wavefunction for all time $\psi (x,t) = \int G(x,t;s,0) \psi _0(s)ds$. For the purposes of this calculation we can assume a point source of particles such that the initial spacial distribution of the particle is a $\delta$-function. This however means that the wavefunction is the ‘square root of a $\delta$-function’, which is not guaranteed to be defined. That aside, we can calculate the spatial distribution of the particle at the screen, but in order to have this distribution be normalised, we must account for the fact that a large portion of the wavefunction does not pass through the slit. Thus in actual fact the distribution at the screen is given by the conditional probability to be at position x and time $T+\tau$ given that it was at position $x^{\prime }\in [-a,a]$ at time T. Fortunately as outlined below, the normalised wavefunction is simply the propagator in Eq. (62) multiplied by a factor $\sqrt{\frac{\lambda D}{2a}}$. Finally we arrive at the normalised wavefunction at the screen, for a particle passing through a single slit of width 2a, centred at $x=b$,

$$\begin{aligned} \psi ^\text {(1)}(x)&= \frac{e^{i\phi (x)}}{i2\sqrt{\eta a}} \bigg \{C[\sigma _+(x)] - C[\sigma _-(x)] + iS[\sigma _+(x)]-iS[\sigma _-(x)]\bigg \}, \end{aligned}$$

(65)

with $\phi (x)$ and $\sigma _\pm (x)$ defined in Eqs. (6) and (5), and $\eta = 1+L/D$. The square of this wavefunction will give the probability distribution for the particle at the slit, which is plotted in Fig. 2 for various distances between slit and screen. The pattern clearly appears to shift towards the negative x direction as the screen is moved further from the slit.

Integrating over the Slit Profile

The propagator to arrive at x having passed through a single slit of width 2a with centre at $x=b$ is found by integrating over the slit distribution, $\Omega (\omega )$, which is 1 for $b-a< \omega <b+a$ and 0 otherwise:

$$\begin{aligned} K_g^{(1)}(x;a,b) =&\int _{-\infty }^\infty A(x,\omega )\Omega (\omega )d\omega \end{aligned}$$

(66)

$$\begin{aligned} =&\int _{b-a}^{b+a} K_g(\omega ,T;0,0)K_g(x,T+\tau ;\omega ,T)d\omega \end{aligned}$$

(67)

$$\begin{aligned} =&\int _{-\infty }^\infty d\omega \, \Omega (\omega ) \sqrt{\frac{m}{2\pi i \hbar T}} \sqrt{\frac{m}{2\pi i \hbar \tau }} \exp \left[ \frac{im}{2\hbar }\left\{ \frac{\omega ^2}{T} - g\omega T - \frac{g^2}{12}T^3\right\} \right] \end{aligned}$$

(68)

$$\begin{aligned}&\qquad \times \exp \left[ \frac{im}{2\hbar }\left\{ \frac{(x-\omega )^2}{\tau } - g(x+\omega ) \tau - \frac{g^2}{12}\tau ^3\right\} \right] . \end{aligned}$$

(69)

Completing the square in $\omega$,

$$\begin{aligned} \frac{\omega ^2}{T} - g\omega T -&\frac{g^2}{12}T^3 + \frac{(x-\omega )^2}{\tau } - g(x+\omega ) \tau - \frac{g^2}{12}\tau ^3\end{aligned}$$

(70)

$$\begin{aligned}&= \frac{\omega ^2}{T} + \frac{\omega ^2}{\tau } - \frac{2 x \omega }{\tau } - g\omega (T + \tau ) +\frac{x^2}{\tau } - gx\tau +\frac{g^2}{4}(T^3 + \tau ^3)\end{aligned}$$

(71)

$$\begin{aligned}&= \omega ^2\frac{T + \tau }{T\tau } - 2\omega \left( \frac{x}{\tau } + \frac{1}{2}g(T+\tau )\right) +\frac{x^2}{\tau } - gx\tau - \frac{g^2}{12}(T^3 + \tau ^3)\end{aligned}$$

(72)

$$\begin{aligned}&=\zeta \left( \omega - \frac{\kappa }{\zeta }\right) ^2 - \frac{\kappa ^2}{\zeta } +\frac{x^2}{\tau } - gx\tau - \frac{g^2}{12}(T^3 + \tau ^3), \end{aligned}$$

(73)

where $\zeta = \frac{T + \tau }{T\tau }$ and $\kappa = \frac{x}{\tau } + \frac{1}{2}g(T+\tau )$. Returning to Eq. (69),

$$\begin{aligned} K_g^{(1)}(x;a,b)&= e^{i\phi (x,T,\tau )}\sqrt{\frac{m}{2\pi i \hbar T}} \sqrt{\frac{m}{2\pi i \hbar \tau }} \int _{-\infty }^\infty d\omega \exp \left[ \frac{im\zeta }{2\hbar }\left( \omega - \frac{\kappa }{\zeta }\right) ^2\right] , \end{aligned}$$

(74)

where $\phi (x,T,\tau ) = \frac{m}{2\hbar }\left( \frac{x^2}{\tau } -\frac{\kappa ^2}{\zeta } - gx\tau - \frac{g^2}{12}(T^3 + \tau ^3)\right)$ is the phase produced by terms not dependent on $\omega$. We make the substitution , and define new limits of integration $\sigma _{\pm }(x) =\sqrt{\frac{m\zeta }{\pi \hbar }}\left( (b\pm a) - x \frac{T}{T+\tau } - \frac{1}{2}gT\tau \right)$:

$$\begin{aligned} K_g^{(1)}(x;a,b)&=e^{i\phi (x,T,\tau )}\sqrt{\frac{2 \pi i \hbar }{m\zeta }}\sqrt{\frac{m^2}{(2\pi i \hbar )^2 T\tau }}\int _{\sigma _-}^{\sigma _+}\exp \left[ \frac{i\pi }{2} v^2\right] dv \end{aligned}$$

(75)

$$\begin{aligned}&= \frac{e^{i\phi (x,T,\tau )}}{\sqrt{(2i)^2 \pi \hbar (T+\tau )/m}} \int _{\sigma _-}^{\sigma _+}\left\{ \cos \left( \frac{i\pi }{2}v^2 \right) + i \sin \left( \frac{i\pi }{2}v^2\right) \right\} \, dv\end{aligned}$$

(76)

$$\begin{aligned}&= \frac{e^{i\phi (x,T,\tau )}}{2i\sqrt{\pi \hbar (T+\tau )/m}}\bigg \{C[\sigma _+(x)] - C[\sigma _-(x)] + iS[\sigma _+(x)]-iS[\sigma _-(x)]\bigg \}, \end{aligned}$$

(77)

where $C[u] {\equiv } \int _0^u \cos \left( \frac{\pi }{2}x^2\right) dx$ is the Fresnel cosine function, $S[u] \equiv \int _0^u \sin \left( \frac{\pi }{2}x^2\right) dx$ is the Fresnel sine function. Now to simplify $\phi (x,T,\tau )$ we first have

$$\begin{aligned} \kappa ^2&= \frac{x^2}{\tau ^2} + \frac{x}{\tau }g(T+\tau ) - \frac{g^2}{12}(T+\tau )^2,\end{aligned}$$

(78)

$$\begin{aligned} \frac{\kappa ^2}{\zeta }&= \frac{x^2}{\tau ^2}\frac{T \tau }{T+\tau } + \frac{x}{\tau }g(T+\tau )\frac{T \tau }{T+\tau } - \frac{g^2}{12}(T+\tau )^2\frac{T \tau }{T+\tau }\end{aligned}$$

(79)

$$\begin{aligned}&= \frac{x^2 T}{\tau (T+\tau )} + gxT - \frac{g^2}{12}T \tau (T+\tau ) \end{aligned}$$

(80)

to get

$$\begin{aligned} \phi (x,T,\tau )&= \frac{m}{2\hbar }\left\{ \frac{x^2}{\tau } -\frac{\kappa ^2}{\zeta } - gx\tau - \frac{g^2}{12}(T^3 + \tau ^3)\right\} \end{aligned}$$

(81)

$$\begin{aligned}&=\frac{m}{2}\left\{ \frac{x^2(T+\tau ) - x^2 T}{\tau (T+\tau )} - gx(T+\tau ) - \frac{g^2}{12}(T^3 + \tau ^3 - T\tau (T+\tau ))\right\} \end{aligned}$$

(82)

$$\begin{aligned}&= \frac{m}{2}\left\{ \frac{x^2}{T +\tau } - gx(T+\tau ) - \frac{g^2}{12}(T+\tau )(T-\tau )^2\right\} . \end{aligned}$$

(83)

We can make use of the approximation $v_z\gg v_x$ and that $\lambda \approx \frac{h}{mv_z}$ to find expressions $T=\frac{m \lambda D}{h}$ and $\tau =\frac{m\lambda L}{h}$. Thus, we have

$$\begin{aligned} T \pm \tau = \frac{m\lambda (D\pm L)}{h} \quad \text{ and } \quad T\tau = \frac{m^2\lambda ^2}{h^2}DL. \end{aligned}$$

(84)

Using these we get

$$\begin{aligned} \frac{m\zeta }{\pi \hbar }= \frac{2}{\lambda }\left( \frac{1}{D} + \frac{1}{L}\right) \qquad \text{ where } \qquad \zeta = \frac{T+\tau }{T\tau } = \frac{h}{m\lambda }\frac{D+L}{DL} =\frac{h}{m\lambda }\left( \frac{1}{D} + \frac{1}{L}\right) . \end{aligned}$$

(85)

Next, let

$$\begin{aligned} \eta = \frac{T}{T+\tau } = \frac{D}{D+L} = \frac{1}{1+L/D}, \end{aligned}$$

(86)

allowing us to express $\phi$ and $\sigma _{\pm }$ in terms of L, D and $\lambda$:

$$\begin{aligned}&\sigma _{\pm }(x) =\sqrt{\frac{m\zeta }{\pi \hbar }}\left( (b\pm a) - x \frac{T}{T+\tau } - \frac{1}{2}gT\tau \right) \nonumber \\&\qquad \; = \sqrt{\frac{2}{\lambda L}\eta }\left\{ (b\pm a) -\frac{x}{\eta } - \frac{1}{2}g \frac{m^2 \lambda ^2}{h^2}DL\right\} , \end{aligned}$$

(87)

$$\begin{aligned}&\phi (x,T,\tau ) = \frac{m}{2}\left\{ \frac{x^2}{T +\tau } - gx(T+\tau ) - \frac{g^2}{12}(T+\tau )(T-\tau )^2\right\} ,\end{aligned}$$

(88)

$$\begin{aligned}&\phi (x)=\pi \left\{ \frac{x^2}{\lambda (D+L)} - m g x\frac{\lambda (D+L)}{h^2} - \frac{g^2}{12}\frac{m^4\lambda ^3}{h^4}(D+L)(D-L)^2 \right\} . \end{aligned}$$

(89)

This is the form of the propagator given in Eq. (62).

Normalisation of the Distribution at the Screen

As derived in the first section the propagator for the process of starting at position $\vec {r}= (0,0)$, passing through the point $(x^{\prime }\in [b-a,b+a],D)$ and finally being detected at position $\vec {r}\, ^{\prime }=(x,D+L)$ on the screen is not the same as the wave-function at the screen. To obtain this we must first convolve the propagator with a initial wavefunction whose square magnitude is a $\delta$-function giving the wavefunction as seen at the other side of the slit. This wavefunction however will not be normalised due to the fact that only a portion of the initially normalise wavefunction has been propagated beyond the slits. It can be renormalised however by scaling by the probability of passing through the slit. Unfortunately the ‘square root of a $\delta$-function’ is not always well defined as is the case for operators acting on any distribution. We can however attempt to use a Gaussian with variance $\sigma$ as the initial wavefunction, compute the quantity of interest and take the limit $\sigma \rightarrow 0$. Under suitable circumstances the limit will be defined giving the desired result.

We will begin with an initial wavefunction that is the square root of a Gaussian

$$\begin{aligned} \psi _\sigma (x) = g_\sigma (x)&= \frac{1}{(2\pi \sigma ^2)^{\frac{1}{4}}} e^{-\frac{x^2}{4\sigma ^2}}. \end{aligned}$$

(90)

However we notice that the square root of a Gaussian is simply another Gaussian of variance $\rho =\sigma \sqrt{2}$ multiplied by the factor $(8\pi \sigma ^2)^\frac{1}{4}$. So the initial function can represented as

$$\begin{aligned} \psi _\rho (x)&= (4\pi \rho ^2)^\frac{1}{4} \frac{e^{-\frac{x^2}{2\rho ^2}}}{\sqrt{2\pi \rho ^2}} = (4\pi \rho ^2)^\frac{1}{4} g_\rho (x). \end{aligned}$$

(91)

Convolving this with the propagator $K_g^{(1)}(x,T+\tau ;x_0,0)$ will give the un-normalised wavefunction at the screen:

$$\begin{aligned} \psi _\rho (x,T+\tau ) = \int _{\infty }^\infty dx_0 K_g^{(1)}(x,T+\tau ;x_0,0)\psi _\rho (x_0). \end{aligned}$$

(92)

To renormalise this, we scale by the probability of the particle passing through the slit. The probability of the particle being in $x\in [b-a,b+a]$ at time T is

$$\begin{aligned} P(x\in [b-a,b+a];T) = \int _{b-a}^{b+a} \, \left| \int _{-\infty }^{\infty }K_g(x^{\prime },T;x_0,0)\psi _\rho (x_0)dx_0\right| ^2dx^{\prime }, \end{aligned}$$

(93)

which gives that the renormalised wavefunction at the screen is

$$\begin{aligned} \psi _\rho ^{\prime }(x,T+\tau )&= \frac{\psi (x,T+\tau )}{\sqrt{\int _{-a}^a \, |\int _{-\infty }^{\infty }K_g(x^{\prime },T;x_0,0)\psi _\rho (x_0)dx_0|^2dx^{\prime }}}\end{aligned}$$

(94)

$$\begin{aligned}&=\frac{\int _{\infty }^\infty K_g^{(1)}(x,T+\tau ;x_0,0)\psi _\rho (x_0)dx_0}{\sqrt{\int _{b-a}^{b+a} \, |\int _{-\infty }^{\infty }K_g(x^{\prime },T;x_0,0)\psi _\rho (x_0)dx_0|^2dx^{\prime }}}\end{aligned}$$

(95)

$$\begin{aligned}&= \frac{(4\pi \rho ^2)^\frac{1}{4}\int _{\infty }^\infty K_g^{(1)}(x,T+\tau ;x_0,0)g_\rho (x_0)dx_0}{(4\pi \rho ^2)^\frac{1}{4}\sqrt{\int _{b-a}^{b+a} \, |\int _{-\infty }^{\infty }K_g(x^{\prime },T;x_0,0)g_\rho (x_0)dx_0|^2dx^{\prime }}}. \end{aligned}$$

(96)

Now the limit $\sigma \rightarrow 0$ can equivalently be taken as $\rho \rightarrow 0$. The Gaussians $g_\rho (x)$ then become delta functions $\delta (x)$:

$$\begin{aligned} \psi ^{\prime }(x,T+\tau )&= \lim _{\rho -\rightarrow 0}\frac{\int _{\infty }^\infty K_g^{(1)}(x,T+\tau ;x_0,0)g_\rho (x_0)dx_0}{\sqrt{\int _{b-a}^{b+a} \, |\int _{-\infty }^{\infty }K_g(x^{\prime },T;x_0,0)g_\rho (x_0)dx_0|^2dx^{\prime }}}\qquad \end{aligned}$$

(97)

$$\begin{aligned}&= \frac{K_g^{(1)}(x,T+\tau ;0,0)}{\sqrt{\int _{b-a}^{b+a} \, |K_g(x^{\prime },T;0,0)|^2dx^{\prime }}} = \frac{K_g^{(1)}(x,T+\tau ;0,0)}{\sqrt{\int _{b-a}^{b+a} \,(2\pi \hbar T/m)^{-1}dx^{\prime }}}\qquad \end{aligned}$$

(98)

$$\begin{aligned}&= K_g^{(1)}(x,T+\tau ;0,0) \sqrt{\frac{\frac{h}{m} T}{2a}} = K_g^{(1)}(x,T+\tau ;0,0) \sqrt{\frac{\frac{m\lambda D}{h} \frac{h}{m}}{2a}}\qquad \end{aligned}$$

(99)

$$\begin{aligned}&= K_g^{(1)}(x,T+\tau ;0,0) \sqrt{\frac{\lambda D}{2a}},\qquad \end{aligned}$$

(100)

where $|K_g(x^{\prime },T;0,0)|^2$ was taken from Eq. (59). Thus we see that the normalised wavefunction at the screen is given by multiplying the propagator by the factor $\sqrt{\frac{\lambda D}{2a}}$.

C Tracing Out Spin from a Matrix Propagator

This is best achieved using the density operator prescription. The pure density operator $\rho$ for a quantum state $\left| \psi \right\rangle$ is $\rho = \left| \psi \right\rangle \left\langle \psi \right|$. For a state comprised of two subsystems, we can ignore the state of a subsystem by tracing it out. This is given by the operation ${{\mathrm{tr}}}_B[X_{AB}] = \sum _k \left\langle k \right| _B X_{AB} \left| k \right\rangle _B$, where $X_{AB}$ is an operator on the composite system AB and $\{\left| k \right\rangle _B\}$ forms a complete basis for subsystem B.

Here, we would like to trace out the spin state. The initial density operator is $\rho _0 = \left| {\chi _0}\big \rangle \big \langle {\chi _0}\right| \otimes \left| {\psi _0}\big \rangle \big \langle {\psi _0}\right|$, where $\langle {x}|{\psi _0}\rangle = \psi _0(x)$, is the spatial distribution of the particle, and it is assumed that, initially, the spatial location of the particle is uncorrelated with the spin state. The state of the system at later time t is given by $\rho (t) = U(t)\rho _0U^\dagger (t)$. We can represent the time evolution operator in terms of the propagator by making use of the resolution of the identity $\sum _{\{\chi ,\chi ^{\prime }\}\in \{\downarrow ,\uparrow \}}\int dx \,dx^{\prime }\left| {x^{\prime },\chi ^{\prime }}\big \rangle \big \langle {x,\chi }\right| = \mathbbm {1}$:

$$\begin{aligned} U(t)=\sum _{\{\chi ,\chi ^{\prime }\}\in \{\downarrow ,\uparrow \}}\int dx \,dx ^{\prime }K_g^{\chi ^{\prime }\!,\chi }(x^{\prime },t;x,0) \left| {x^{\prime },\chi ^{\prime }}\big \rangle \big \langle {x,\chi }\right| , \end{aligned}$$

(101)

which gives that,

$$\begin{aligned} \rho (t) =&\sum \int dx \, dx^{\prime }dy\, dy^{\prime }K_g^{\chi ^{\prime }\!,\chi }(x^{\prime },t;x,0)K_g^{^*\phi ^{\prime }\!,\phi }(y^{\prime },t;y,0)\left| x^{\prime },\chi ^{\prime } \right\rangle \nonumber \\&\qquad \times \left\langle { {\chi ^{\prime }}x,\chi }\big |{\rho _0}\big |{ {\chi ^{\prime }}y,\phi }\right\rangle \left\langle y^{\prime },\phi ^{\prime } \right| , \end{aligned}$$

(102)

where the sum is over all spin variables. We can then take the trace over the spin subspace to give,

$$\begin{aligned} \tilde{\rho }(t) = {{\mathrm{tr}}}_\text {spin}[\rho (t)] =&\sum \int dx \, dx^{\prime }dy\, dy^{\prime }K_g^{\chi ^{\prime }\!,\chi }(x^{\prime },t;x,0)K_g^{^*\phi ^{\prime }\!,\phi }(y^{\prime },t;y,0) \nonumber \\&\qquad \times \left\langle { {\chi ^{\prime }}x,\chi }\big |{\rho _0}\big |{ {\chi ^{\prime }}y,\phi }\right\rangle \langle {\chi ^{\prime }}|{\phi ^{\prime }}\rangle \left| {x^{\prime }}\big \rangle \big \langle {y^{\prime }}\right| . \end{aligned}$$

(103)

The spatial probability distribution is then given by the expectation of the position operator $\left\langle {\hat{x}}\right\rangle = {{\mathrm{tr}}}[\tilde{\rho }(t)\hat{x}]$. Making use of the fact that $\langle {\chi ^{\prime }}|{\phi ^{\prime }}\rangle = \delta _{\chi ^{\prime },\phi ^{\prime }}$ and $\langle {y^{\prime }}|{x^{\prime }}\rangle = \delta (x^{\prime }- y^{\prime })$, we find

$$\begin{aligned} \left\langle {\hat{x}}\right\rangle = \sum \int dx \, dy K_g^{\chi ^{\prime }\!,\chi }(x^{\prime },t;x,0)K_g^{^*\chi ^{\prime }\!,\phi }(x^{\prime },t;y,0)\left\langle { {\chi ^{\prime }}x,\chi }\big |{\rho _0}\big |{ {\chi ^{\prime }}y,\phi }\right\rangle . \end{aligned}$$

(104)

At this point, we work with the term $\left\langle { {\chi ^{\prime }}x,\chi }\big |{\rho _0}\big |{ {\chi ^{\prime }}y,\phi }\right\rangle$

$$\begin{aligned} \left\langle { {\chi ^{\prime }}x,\chi }\big |{\rho _0}\big |{ {\chi ^{\prime }}y,\phi }\right\rangle =&\left\langle { {\chi ^{\prime }}x,\chi }\big |{\Big (\left| {\chi _0}\big \rangle \big \langle {\chi _0}\right| \otimes \left| {\psi _0}\big \rangle \big \langle {\psi _0}\right| \Big )}\big |{ {\chi ^{\prime }}y,\phi }\right\rangle \end{aligned}$$

(105)

$$\begin{aligned} =&\left\langle \chi \right| {\Big (\alpha \left| \uparrow \right\rangle + \beta \left| \downarrow \right\rangle \Big )\Big (\alpha ^*\left\langle \uparrow \right| + \beta ^*\left\langle \downarrow \right| \Big )}\left| \phi \right\rangle \langle {x}|{\psi _0}\rangle \langle {\psi _0}|{y}\rangle \end{aligned}$$

(106)

$$\begin{aligned} =&\left\langle \chi \right| {\Big (|\alpha |^2\left| {\uparrow }\big \rangle \big \langle {\uparrow }\right| + |\beta |^2\left| {\downarrow }\big \rangle \big \langle {\downarrow }\right| + \alpha ^*\beta \left| {\downarrow }\big \rangle \big \langle {\uparrow }\right| + \beta *\alpha \left| {\uparrow }\big \rangle \big \langle {\downarrow }\right| \Big )}\left| \phi \right\rangle \nonumber \\&\quad \times \psi _0(x)\psi _0^*(y). \end{aligned}$$

(107)

Since the matrix propagator in Eq. (22) is diagonal, we can immediately discard the $\left| {\downarrow }\big \rangle \big \langle {\uparrow }\right|$ and $\left| {\uparrow }\big \rangle \big \langle {\downarrow }\right|$ terms when substituting into the expression for the spatial distribution in Eq. (104),

$$\begin{aligned} \left\langle {\hat{x}}\right\rangle =&\int dx \, dy |\alpha |^2K_g^{\uparrow ,\uparrow }(x^{\prime },t;x,0)\psi _0(x)\left( K_g^{\uparrow ,\uparrow }(x^{\prime },t;y,0)\psi _0(y)\right) ^*\end{aligned}$$

(108)

$$\begin{aligned}&+ |\beta |^2K_g^{\downarrow ,\downarrow }(x^{\prime },t;x,0)\psi _0(x)\left( K_g^{\downarrow ,\downarrow }(x^{\prime },t;y,0)\psi _0(y)\right) ^* \nonumber \\=&|\alpha |^2\left| \int dx K_g^{\uparrow ,\uparrow }(x^{\prime },t;x,0)\psi _0(x)\right| ^2 +|\beta |^2\left| \int dx K_g^{\downarrow ,\downarrow }(x^{\prime },t;x,0)\psi _0(x)\right| ^2. \end{aligned}$$

(109)

This result is simply a convex sum of the initial spatial distribution evolved by each propagator.