Double slit experiment - Detection probability using Path integral formulation

Question

I'm really interested in the formalism of path integrals, so I am wanting to better my understanding conceptually and computationally.

Consider we have a free particle of mass $m$ an $x=L$ distance away from a detector at $t=0$ and a barrier of with two slits at $y=\pm a$. Both the barrier and detector extended from $-\infty$ to $\infty$ in the y-direction. The distance of the barrier slit to the particle is $x=L/2$.

What I am attempting to determine is the probability for the particle to be measured in the detector $\mathbb{P}(y,T)$ at position $(L,y)$. Where $T$ is the time that it arrives at the detector.

---

The probability to detect this particle is determined by the possible amplitude of its trajectory (assuming it follows the classical path):

$$ \mathbb{P}(y,T) = |A'_{1} + A'_{2}|^{2} $$ where $A'$ is the amplitude from the propagator.

the particle can follow only two paths, primarily the path of the first slit to the detector I will call $\mathcal{O}_{1}$ and through the second slit to the detector $\mathcal{O}_{2}$. Each path has independent time trajectories.

The overlap of having the particle travel to the detector can be defined as $$ \begin{align} \langle B,t=T |A,t=0\rangle \end{align}$$ where $B$ is the point source on the screen and $A$ is the particles source of origin.

The total probability for the particle to take either path over time $T$ to the detector is 1, so the above can be written as a propogator:

$$\begin{align} \langle B,T | A,0 \rangle = \int_{0}^{T} dt_{1} \langle B,T |\mathcal{O}_{1},t_{1} \rangle \langle \mathcal{O}_{1},t_{1}| A,0 \rangle + \int_{0}^{T} dt_{2} \langle B,T |\mathcal{O}_{2},t_{2} \rangle \langle \mathcal{O}_{2},t_{2}| A,0 \rangle \end{align}$$

The propagator for free particle in general is defined as: $$ \begin{align} \mathcal{U}_{free}(q',t' ; q,t_{0}) = \sqrt{\frac{m}{2 \pi i \hbar (t' -t_{0})}} \exp( - \frac{i}{\hbar} \left( \frac{m}{2} \frac{ \Delta q^{2} }{\Delta t} \right) ) \end{align} $$

So for one of the portions in propagation: $$ \begin{align} \langle \mathcal{O}_{1},t_{1} | A,0\rangle &= \sqrt{\frac{m}{2 \pi i \hbar t_{1}}} \exp( - \frac{i}{\hbar} \left( \frac{m}{2} \frac{ x_{1}^{2} }{t_{1}} \right) ) \end{align} $$ where $x_{1}$ is the position at the slit $(\Delta x = x_1 - 0)$

Applying this to all four and combining it gives me then:

$$ \begin{align} \langle B,T | A,0 \rangle = \frac{m}{2 \pi i \hbar}\left( \int_{0}^{T} dt_{1} \frac{ \exp( -\frac{i m}{2\pi} \left( \frac{x_{1}^{2}}{t_{1}} - \frac{x_{1}'^{2}}{T-t_{1}} \right) ) }{t_{1}(T - t_{1})} + \int_{0}^{T} dt_{2} \frac{ \exp( -\frac{i m}{2\pi} \left( \frac{x_{2}^{2}}{t_{2}} - \frac{x_{2}'^{2}}{T-t_{2}} \right) ) }{t_{2}(T - t_{2})} \right) \end{align} $$

So now, this is the part I am stuck with.

1. How do properly account both x and y dimensions in this integral I have above?
2. How do I then propely account for the slit widths?

---

Another approach is to simply look at the action of the particle since we know its lagrangian, and work from there and take into consideration that the particles approach to the slit will be symmetric to each of its paths.

$$ \begin{align} \mathcal{U} = \int_{x=L/2,t=t'}^{L,T} \int_{y=0,t=0}^{y,T} \mathcal{D} e^{i\int dt \mathcal{L}} \end{align}$$

where the lagrangian (not density) of our free particle is $$ \begin{align} \mathcal{L} = \frac{m}{2}( \dot{x}^{2} + \dot{y}^{2} ) \end{align}$$

Answer 1

As I understand it, your main challenge is to add a second dimension to the formulation of the problem you already have. Below I will give a proposed sketch for how to do this.

First, we can generalize the propagator for a free particle to $\mathbb{R}^n$. The computation should be similar to the one in 1D. Write in terms of $\exp^{-iH(t'-t_0)}$, insert $\int dp_i |p_i \rangle\langle p_i|$ a couple of times, apply the Hamiltonian to $\langle p|$, simplify the expression and do the integral. You should check this result, but I believe it is $$ \begin{align} \mathcal{U}_{free}(\vec{q}',t' ; \vec{q},t_{0}) = (\frac{m}{2 \pi i \hbar (t' -t_{0})})^{\frac{n}{2}} \exp( - \frac{i}{\hbar} \left( \frac{m}{2} \frac{ (\Delta q)^{2} }{\Delta t} \right) ) \end{align} $$ where $(\Delta q)^2 = \sum\limits_{i=1}^{n}(q_i)^2$ now.

Next, suppose the slit has a width $\epsilon$, so that it is located between $y=\pm a-\epsilon$ and $y=\pm a+\epsilon$. This means that the particle has to pass through $(x,y)$ where $y\in [-a-\epsilon, -a+\epsilon]\cup[a-\epsilon, a+\epsilon]$. We take this into account by adding an additional integral over $y$. Writing $(x_1,y_1)$ for the location of the first slit it passes through and likewise for the second slit, we get

$$\begin{align} \langle B,T | A,0 \rangle &= \int_{0}^{T} dt_{1} \int\limits_{-a-\epsilon}^{-a+\epsilon} dy_1 \langle B,T |(x_1,y_1),t_{1} \rangle \langle (x_1,y_1),t_{1}| A,0 \rangle \\ &+ \int_{0}^{T} dt_{1} \int\limits_{a-\epsilon}^{a+\epsilon} dy_2 \langle B,T |(x_2,y_2),t_{2} \rangle \langle (x_2,y_2),t_{2}| A,0 \rangle \end{align}$$

Inserting the expression for the free propagator and performing the integral should then give you the final expression. I haven't done this calculation and it is not certain there is an expression in terms of elementary functions (i.e. you need to check whether the integral can be computed explicitly).