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I've been studying the double slit experiment, particularly focusing on the approach that involves solving the Schrödinger equation for a free particle in two spatial dimensions. Specifically, I'm using the time-dependent Schrödinger equation:

$$ i\hbar \frac{\partial \Psi(x,y,t)}{\partial t} = -\frac{\hbar^2}{2m} \left( \frac{\partial^2 \Psi(x,y,t)}{\partial x^2} + \frac{\partial^2 \Psi(x,y,t)}{\partial y^2} \right) $$

This approach considers the wave function $\Psi$ of a free particle in a two-dimensional space, with specific conditions to represent two slits. The problem is solved numerically to analyze the probability distribution $|\Psi(x,y,t)|^2$ and observe the interference pattern. I understand this approach in terms of Hilbert space, the preferred orthonormal basis $\{|x,y\rangle\}$, the Hamiltonian, and the position as an observable Hermitean operator.

However, I'm struggling to relate this approach to the one discussed in Feynman's lectures, the paper "Space-time approach to Non-relativistic QM", and in "QFT in a nutshell" by A. Zee. These sources discuss the probability amplitude for a particle "to propagate from $A$ to $B$", but I'm unable to see the connection with the Schrödinger equation approach.

My questions are:

  1. How do these two approaches to the double slit experiment relate to each other?
  2. In the context of the probability amplitude "for propagations" approach, what is the Hilbert space, and how does it compare to the Hilbert space in the Schrödinger equation approach?
  3. Is this second approach formulated with the same postulates of QM?

Any insights or explanations to bridge my understanding of these two methodologies would be greatly appreciated.

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  • $\begingroup$ Link to abstract pages? $\endgroup$
    – Qmechanic
    Commented Jan 8 at 19:10
  • $\begingroup$ What is it about the double slit experiment that warrants the invocation of path integrals? What do you hope to gain from this very advanced approach? $\endgroup$
    – my2cts
    Commented Jan 8 at 23:41
  • $\begingroup$ @my2cts Being not a physicist, I didn't think that the second approach was a very advanced approach. On the contrary, that was the first explanation I found to the double slit experiment, so I thought that was the simplest explanation! I thought I was missing a very elemental link to QM postulates $\endgroup$ Commented Jan 9 at 6:15
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    $\begingroup$ @A.J.Pan-Collantes The basic idea is simple enough but from the answer you can see that the practice is rather involved. $\endgroup$
    – my2cts
    Commented Jan 9 at 9:48

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The time evolution of a quantum state can be viewed in terms of the propagator $\hat U(t)$: $$|\psi(t)\rangle = \hat U(t)|\psi(0)\rangle$$ Expressed in the position basis this is $$\psi(\mathbf x,t) = \langle\mathbf x|\int \hat U(t)|\mathbf x'\rangle\langle\mathbf x'|\psi(0)\rangle\mathrm d^3\mathbf x' = \int U(\mathbf x,\mathbf x',t)\psi(\mathbf x',0)\mathrm d^3\mathbf x'$$ Where $U(\mathbf x,\mathbf x',t) \equiv \langle\mathbf x|\hat U(t)|\mathbf x'\rangle$. In this form the propagator $U(\mathbf x,\mathbf x',t)$ can be interpreted as taking the probability amplitude of the particle being at the event $(\mathbf x',0)$ and "propagating" it to the event $(\mathbf x,t)$. Both approaches you mentioned are just different ways of computing the propagator.

The Schrodinger equation is a differential equation for the propagator: \begin{align} i\hbar\frac{\mathrm d}{\mathrm dt}|\psi(t)\rangle = \hat H(t)|\psi(t)\rangle &\implies i\hbar\frac{\mathrm d}{\mathrm dt}\left(\hat U(t)|\psi(0)\rangle\right) = \hat H(t)\left(\hat U(t)|\psi(0)\rangle\right) \\ &\implies i\hbar\frac{\mathrm d}{\mathrm dt}\hat U(t) = \hat H(t)\hat U(t) \end{align}

The path integral computes the propagator by integrating the action functional $\mathcal S[\mathbf r] = \int_0^t\mathcal L(\mathbf r,\dot{\mathbf r},t')\mathrm dt'$ over all possible paths $\mathbf r$ between the events $(\mathbf x',0)$ and $(\mathbf x,t)$: $$\psi(\mathbf x,t) = \mathcal N(t)\iint\psi(\mathbf x',0)e^{i\mathcal S[\mathbf r]\over\hbar}\mathcal D[\mathbf r]\mathrm d^3\mathbf x' \implies U(\mathbf x,\mathbf x',t) = \mathcal N(t)\int e^{i\mathcal S[\mathbf r]\over\hbar}\mathcal D[\mathbf r]$$

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  • $\begingroup$ The propagator is (I guess) a key concept I was missing. Thank you very much. Can you clarify what is $\mathcal{N}$? $\endgroup$ Commented Jan 9 at 7:30
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    $\begingroup$ @A.J.Pan-Collantes $\mathcal N$ is the normalization factor which is determined by setting $\int |\psi(\mathbf x, t)|^2\mathrm d^3\mathbf x = 1$ $\endgroup$
    – Er Jio
    Commented Jan 9 at 15:33

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