Example of "motion" in Quantum Mechanics

Statement
Suppose we have a (2+1)-dimensional spacetime. I have just detected a quantum particle at the position $A = (x, y) = (0, 0)$ with time $t = 0$ . Obviously, at time $t = 0$ , the probability of finding it at a different position, such as $(2, 6)$ , is 0. But what is the probability of detecting the particle at $B = (2, 6)$ at $t = 2$ ? Assume $m = 1$ .

Development
To calculate the probability of detecting a quantum particle at a different position at a later time, you would generally use the principles of quantum mechanics, specifically the wave function evolution. Since you're dealing with a quantum particle, its behavior is governed by the Schrodinger equation:

i ℏ \frac{\partial}{\partial t} Ψ (x, y, t) = \hat{H} Ψ (x, y, t) .

To solve this problem, you would:

Define the Initial State: Determine the wave function $Ψ (x, y, 0)$ at $t = 0$ . Since you detected the particle at $A = (0, 0)$ at $t = 0$ , the wave function would be sharply peaked around this point. A Dirac delta, or something like that.
Set Up the Hamiltonian: For a particle with mass $m = 1$ , in a potential $V (x, y)$ , the Hamiltonian is typically $\hat{H} = - \frac{ℏ^{2}}{2 m} \nabla^{2} + V (x, y)$ . You would need to know the potential energy function $V (x, y)$ to proceed.
Evolve the Wave Function: Use the Schrödinger equation to evolve the wave function from $t = 0$ to $t = 2$ . This involves solving partial differential equations, which can be complex and may require numerical methods.
Calculate the Probability: The probability of finding the particle at position $B = (2, 6)$ at $t = 2$ is given by the square of the magnitude of the wave function at that point and time: $P (2, 6, 2) = | Ψ (2, 6, 2) |^{2}$ .

Aside question
What if I want to compute the probability of being at $B = (2, 6)$ but going first through $C = (1, 1)$ at time $t = 1$ ? How should I proceed?

Answer

Compute the Amplitude for $(1, 1)$ at $t = 1$ : First, you would calculate the probability amplitude for the particle to move from $(0, 0)$ at $t = 0$ to $(1, 1)$ at $t = 1$ . This involves solving the time-dependent Schrödinger equation as before.
Compute the Amplitude for $(2, 6)$ at $t = 2$ : Next, calculate the probability amplitude for moving from $(1, 1)$ at $t = 1$ to $(2, 6)$ at $t = 2$ . Again, this involves the Schrödinger equation for this specific leg of the journey.
Multiply the Amplitudes: The total amplitude for the path going through $(1, 1)$ at $t = 1$ is the product of the amplitudes of the two legs of the journey. Why? I don't understand this yet. See section 2 in Feynman "Space-time approach to non-relativistic quantum mechanics".
Calculate the Probability: The probability of the particle being at $(2, 6)$ at $t = 2$ , given it passed through $(1, 1)$ at $t = 1$ , is the square of the magnitude of the total amplitude calculated in the previous step.

I think these two questions are fundamental to understand Feynman's path integral formulation.