Demonstrating two-particle interference with a one-dimensional delta potential well

Suppose the delta potential well $V(x)=-\alpha\delta(x)$ with depth parameter $\alpha>0$ is located at the coordinate origin, and a particle with mass $m$ incident from the left is described by a normalized Gaussian wave packet at time $t=0$,

where $s<0$ and $1/({2}\sqrt{\Delta})$ represent the initial average value and standard deviation of position, respectively, and $\hbar k_{0}>0$ denotes the average incident momentum. The time-dependent evolution of this wave packet requires solving the Schrödinger equation.

For a particle with definite energy $E$ incident from the left and $k={\sqrt{2mE}/\hbar}>0$, the stationary state solution reads Griffiths ,

where $\beta(k)=m\alpha/\left(\hbar^{2}k\right)$ is related to the well depth parameter $\alpha$ and wave vector $k$. Similarly, the solution for a particle incident from the right with the same definite energy but an opposite momentum $k=-{\sqrt{2mE}/\hbar}<0$, takes the form,

with $\beta(k)=m\alpha/\left(\hbar^{2}k\right)<0$ due to $k<0$. In applying Eqs. (2) and (3), it is convenient to distinguish the incident directions by the sign of $k$ and use a single symbol $\psi_{k}\left({x,t}\right)$ to denote the basis states.

These solutions in Eqs. (2), (3) satisfy ${\int_{-\infty}^{\infty}{\psi_{k}^{*}(x,0)\psi_{k^{{}^{\prime}}}\left({x,0}\right)}}dx=2\pi\delta\left(k-k^{{}^{\prime}}\right)$, and thus form a complete and orthogonal basis. Using linear combinations of these solutions, it is straightforward to derive the wave packet solution. The key is to project the initial state onto the basis state and work out the superposition coefficient,

If the initial state is localized on the left side of the well with $|s|\gg 1/\sqrt{\Delta}$, then

and the wave packet solution can be obtained by substituting Eqs. (2), (3), and (5) into the expression:

In addition, if the average incident momentum is much larger than its uncertainty, i.e., $\ k_{0}\gg\sqrt{\mathrm{\Delta}}$, the solution only has contributions from the left incident basis states, so the wave packet solution simplifies to

with the superposition coefficient approximated as,

This is just the Fourier transform of the initial state. The solution in Eq. (7) is a piecewise function. The incident and reflected components are on one side of the coordinate origin, while the transmitted components are on the other side.

The transmission coefficient and reflection coefficient for monochromatic plane waves are $T_{\text{p}}(k)=1/\left(1+\beta(k)^{2}\right)$ and $R_{\text{p}}(k)=\beta(k)^{2}/\left(1+\beta(k)^{2}\right)$ with $T_{\text{p}}(k)+R_{\text{p}}(k)=1$. After scattering, the incident wave packet is completely decomposed into reflected and transmitted wave packets. The total probability on the right side of the well is

and the transmission coefficient for a wave packet can be defined through its long time limit:

If $\phi(k)$ has negligible contribution from $k<0$, then

The transmission coefficient for a wave packet is the average over all the monochromatic plane wave components.

To visualize the scattering process, either the exact solution Eq. (6) or approximate solution Eq. (7) can be plotted. The approximate solution works well when the superposition coefficient has a complete Gaussian distribution. It is convenient to take $1/\sqrt{\Delta}$ as the unit of length and introduce the following dimensionless quantities: position $X=\sqrt{\Delta}\cdot x$, wave vector $K={k/\sqrt{\Delta}}$, time $\tau=\Delta\hbar t/(2m)$, initial average position $S=\sqrt{\Delta}\cdot s$, central wave vector $K_{0}={k_{0}/\sqrt{\Delta}}$, and well depth parameter $\Lambda=\frac{m\alpha}{\hbar^{2}\sqrt{\Delta}}$. The function $\beta(k)$ can now be expressed as $\beta(k)={\Lambda/K}$.

As shown in Fig. 1(a), when the front of the incident wave packet reaches the location of the potential well at $X=0$, the reflected wave component is immediately generated, resulting in interference fringes formed by the superposition of the incident and reflected wave packets in the region $X<0$. After some time, the incident wave packet is fully transformed into two wave packets moving in opposite directions. The time evolution of $P_{\text{R}}$ in Fig. 1(b) shows that the probability saturates to a constant value, which is the transmission coefficient. The higher the incident energy, the greater the transmission coefficient. Under the assumption that $\ k_{0}\gg\sqrt{\mathrm{\Delta}}$, $\left|{\phi(k)}\right|^{2}$ is a complete Gaussian function with mean value $k_{0}$, so the value of $\beta(k)$ at central wave vector $\beta_{0}=\frac{m\alpha}{\hbar^{2}k_{0}}={\Lambda/K_{0}}$ determines the wave packet transmission coefficient $T=1/\left(1+\beta_{0}^{2}\right)$. The 50:50 beam splitter is realized at $\beta_{0}=1$. To change the splitting ratio is to simply change the ratio of $\Lambda/K_{0}$.

If $k_{0}$ is comparable or smaller than $\sqrt{\mathrm{\Delta}}$, then $\left|{\phi(k)}\right|^{2}$ is a truncated Gaussian function in the approximate result of Eq. (8), while it exhibits oscillations in the exact result of Eq. (5), as shown in Fig. 1(d). These two distribution curves give rise to different state evolutions, but make negligible difference in the averaged transmission coefficients. They both lead to transmission coefficients slightly higher than that of a plane wave at the central wave vector, as shown in Fig. 1(c). On the whole, the formula $T=1/\left(1+\beta_{0}^{2}\right)$ is a good estimate for the wave packet transmission coefficient.

Codes for all the subplots are accessible via the online supplementary material SM .

Here, we introduce two tools for analyzing scattering: the joint probability distribution and the probability distribution for separation. the probability that both particles emerge from the same side of the well has also been derived and analyzed, from which the optimal interference conditions can be obtained.

The two-particle interference model is shown in Fig. 2(a), where two particles with mass $m$ are incident from the left and right sides of the delta potential well, respectively. The states of the incident particles at $t=0$ can be written as,

with initial average position $s_{1}<0$, $s_{2}>0$, and average incident momentum $\hbar k_{01}>0$, $\hbar k_{02}<0$. If there is no inter-particle interaction like attraction or repulsion, the two particles will be scattered independently in the potential well. They meet at the well, interfere due to wave packet overlap, convert into reflected and transmitted components, and then leave the scattering region. If the initial wave packets are well-localized on one side of the barrier ($\left|s_{j}\right|\gg 1/\sqrt{\Delta}$), and the average incident momentum is much larger than its uncertainty ($\left|k_{0j}\right|\gg\sqrt{\mathrm{\Delta}}$), then the wave packets evolve as follows,

with superposition coefficients $\phi_{j}(k)=\left(\frac{1}{2\mathrm{\Delta}\pi}\right)^{\frac{1}{4}}e^{i{({k_{0j}-k})}s_{j}}e^{-\frac{{(k-k}_{0j})^{2}}{4\mathrm{\Delta}}},(j=1,2)$. The wave packet transmission coefficients are $T_{j}=1/\left(1+\beta_{0j}^{2}\right)$ with $\beta_{0j}=\frac{m\alpha}{\hbar^{2}k_{0j}}$, $(j=1,2)$.

Since there is no inter-particle interaction, the two-particle wave function for distinguishable particles can be written in the form of a direct product state. However, if the particles are identical bosons or fermions, the wave function has to be symmetrized and anti-symmetrized, respectively.

$\Psi_{+}$, $\Psi_{-}$, $\Psi_{\text{D}}$ represent the two-particle wave functions for bosons, fermions and distinguishable particles, respectively, and the positions of particles 1 and 2 are labeled by $x_{1}$, $x_{2}$. All scenarios can be achieved via cold atoms cold : two sodium-23 or rubidium-87 atoms are identical bosons, two lithium-6 or potassium-40 atoms are identical fermions, and two isotopes with nearly equal mass such as rubidium-87 and rubidium-85 are distinguishable particles.

Each scattered particle has some probability of being on the left or right side of the well, so there are four outgoing channels corresponding to the four quadrants in Fig. 2(b). The quadrants have a clear physical meaning: the first and third quadrants indicate that both particles emerge on the same side of the potential well, while the second and fourth quadrants indicate that they emerge on opposite sides. The symmetry of the wave function under exchange will lead to differences in the joint probability distribution $\left|\Psi_{i}\right|^{2}$ in position space.

Because of the indistinguishable nature of identical particles, it no longer makes sense to number the particles by 1 and 2. Thus, we can introduce the variables $r=x_{1}-x_{2}$ and $R=\left(x_{1}+x_{2}\right)/2$, where $r$ and $R$ are the relative and center of mass coordinates, respectively. The probability distribution for separation of particles in the state of $\Psi_{i}$ ($i=+,-,\text{D}$) is

This definition involves the summation of relative coordinates with opposite signs and the integration over the center of mass coordinates. It gives the statistical description of the inter-particle distance, and is closely linked to the question of whether the two particles stay together after the scattering.

Due to the amplitude interference of two-particle states, bosons tend to come out of the same port, and fermions tend to come out of different ports, while distinguishable particles do not have such a tendency zongshu . After scattering, the single-particle wave functions can be fully decomposed into the superposition of the reflected component $\Phi_{j}^{\text{R}}$ and the transmitted component $\Phi_{j}^{\text{T}}$ as follows ($j=1,2$),

with

These four wave packets will move farther away from the potential well over time, and their shape gets lower and wider due to dispersion. For the case that both particles come out from the right side, the particle incident from the left must be transmitted and the particle incident from the right must be reflected. Thus the probability amplitude of two identical particles appearing on the right side of the potential well is,

where the $+$ and $-$ signs correspond to bosons and fermions, respectively. Combined with Eq. (17), the integration of the probability density $\left|\Psi_{\text{right}}\right|^{2}$ in the entire coordinate space results in the probability that both particles appear on the right side, i.e.,

where the second line adopts the assumption that the wave packet has a narrow range of frequency: i.e., $\left|k_{0j}\right|\gg\sqrt{\mathrm{\Delta}}$. Apparently, the wave packets evolve with time, however, the final scattering probability does not.

Adding the probability of both particles coming out from the left side, the total probability of detecting both particles on the same side is,

where $T_{j}$, $R_{j}=1-T_{j}$ are the transmission and reflection coefficients for the left ($j=1$) and right ($j=2$) incident particles, respectively. The first term $P_{\text{D}}=T_{1}R_{2}+R_{1}T_{2}$ is the result for distinguishable particles, and the second term arises from the exchange symmetry of the identical particles. Due to probability conservation, the probability of two particles emerging from opposite sides equals $(1-P_{+,-,\text{D}})$, which is the coincidence probability measured in HOM experiments zongshu . To make a significant difference, the second term should have a contribution as large as possible. This requires $T_{1}\cong T_{2}\cong 0.5$, $s_{1}+s_{2}\cong 0$, $k_{01}+k_{02}\cong 0$, which means that the two incident particles have the same energy and arrive at a 50:50 beam splitter simultaneously.

We can also analyze spin or polarization degrees of freedom. For simplicity, suppose that the dimension of spin space is 2, and the spin states of the incident particles from the left and right sides are $\left|\left.u\right\rangle\right.$ and $\left.c\middle|\left.u\right\rangle+d\middle|\left.v\right\rangle\right.$, respectively, with $|c|^{2}+|d|^{2}=1$ and $\left|\left.u\right\rangle\right.$, $\left|\left.v\right\rangle\right.$ being orthonormal. Accordingly, the probability amplitude of detecting two particles on the right side becomes,

Analyzing the measurement probability in position space requires summing over the spin degrees of freedom, so there is

The final impact on the total probability is to multiply the overlap coefficient $|c|^{2}$ in the second term, i.e.,

For $c=1$, the spin states are the same, and the interference effect is at its strongest, while for $c=0$, the spin states are orthogonal, and there is no interference effect. In summary, to achieve the best comparison effect, the particles should also have the same spin state. These are the conditions under which the HOM effect occurs zongshu and are here obtained entirely from the wave packet solution. For a simple check, with the optimal interference conditions, i.e., $T_{1}=T_{2}=0.5$, $s_{1}+s_{2}=0$, $k_{01}+k_{02}=0$, $c=1$, then $P_{+}=1$, $P_{-}=0$, and $P_{\text{D}}=0.5$. Deviation from these conditions will weaken, or remove the observation effect. As shown in Eq. (23), the influence from the differences in spin state is quadratic, while that from differences in incident energy and initial distance is exponential decay.