5
$\begingroup$

The probability amplitude for a particle to travel from $\mathbf{x}_i $ to $\mathbf{x}_f$ in a time $t$ is given by the path integral

$$ \langle \mathbf{x}_f | e^{-iHt} |\mathbf{x}_i \rangle = \int \mathcal{D}\mathbf{x} e^{iS[\mathbf{x}]} $$

where I am integrating over all possible paths $\mathbf{x}: \mathbb{R} \rightarrow \mathbb{R}^3 $ such that $\mathbf{x}(0) = \mathbf{x}_i$ and $\mathbf{x}(t) = \mathbf{x}_f$. Well I was thinking about the double slit experiment and I have heard the idea of a "wavefunction" of a particle that has travelled through a particular slit and I have always wondered what that means exactly. Then I recalled that we define the wavefunction of a system in state $| \psi \rangle$ as $$ \psi(\mathbf{x},t) = \langle \mathbf{x} | \psi(t) \rangle. $$ If, for the double slit experiment, If I could find a Hamiltonian $H$ that describes the effect of the double slit, I could surely chose my state at a time $t$ after releasing the particle at $\mathbf{x}_i$ to be the propagated state $$|\psi(t)\rangle = e^{-iHt}|\mathbf{x}_i\rangle$$ Therefore my wavefunction would be $$ \psi(\mathbf{x},t) = \langle \mathbf{x} | e^{-iHt}|\mathbf{x}_i \rangle $$

which is the path integral as given above. If I could somehow write down the Hamiltonian that describes the behaviour of the double-slit, surely my wavefunction, and hence my path integral, would solve the Schrodinger equation?

$\endgroup$
1
$\begingroup$

If I could somehow write down the Hamiltonian that describes the behavior of the double-slit, surely my wavefunction, and hence my path integral, would solve the Schrodinger equation?

Interesting question, and the answer is yes. I think perhaps the most straight forward reason is that, as you say assuming we know the appropriate Hamiltonian $H$, the state of the system that you've written evolves in time according to the unitary time evolution operator, which operates on solutions of the Schrodinger equation. That is,

$$ \lvert \bf{x}_f \rangle = \mathcal{U}(t)\lvert \bf{x}_i \rangle = e^{-iHt} \lvert \bf{x}_i \rangle$$

for $\frac{\partial H}{\partial t} = 0$.

In this article, they show how the Schrodinger equation is derived from the time evolution of a generic quantum state, which is what you've proposed. There's a lot of formalism here, but what's ultimately the serious task is finding that Hamiltonian which produces sensible predictions for the double slit experiment. Here's a very related question.

$\endgroup$
0
$\begingroup$

I will say something to the general question and then adress the double-slit question more directly. Feel free to jump to part II. I am also ignoring all constants by choosing appropriate units.

I. General Remarks

As another commenter pointed out, indeed the path integral and the Schroedinger formalism are interlinked, in the following sense: Consider a system with a Hamiltonian $H = H(p,x)$, where in the r.h.s. we have expressed the Hamilton operator als an operator function of momentum $p$ and position $x$, e.g.

$H(p,x) = p^2 + V(x). $

Then if we "dequantize" this Hamiltonian to get a classical Hamiltonian function, and if this classical Hamiltonian is quadratic in the momentum variables we can from there calculate the classical action functional

$S[x(t)] = \int ( p(t) \dot{x}(t) - H(p(t),x(t))) dt$

where $p(t)$ is expressed in terms of $x(t)$ and $\dot{x}(t)$.

The statement is now that if $U(x_f,t_f;x_i,t_i)$ is the transition amplitude of a particle propagating from an initial position $x_i$ at time $t_i$ to a final postion $x_f$ at time $t_f$, then on the one hand, it satisfies Schroedingers equation:

$ \left[- i \frac{\partial}{\partial t} + H( - i \partial_x,x) \right] U(x,t;x_i,t_i) = 0$

On the other hand, this transition amplitude is given as a path integral

$U(x_f,t_f;x_i,t_i) = \int_{x(t_i) = x_i}^{x(t_f) = t_f} D[x(t)] e^{ i S[x(t)]} .$

II. Double-slit case

In fact it is not difficult to write down the Schroedinger case for this problem. For this remember that in the position representation, it is just a partial differential equation, and they come with a choice of boundary condition. Namely we can model the double slit by the following PDE

$ i \frac{\partial}{\partial t} \psi_t(x) = - \nabla^2 \psi_t(x) $

subject to the boundary conditions

$ \psi(t,x) = 0 \text{ for } x \in D$

here, $D$ is the region where the double slit is. For example, we can say that

$D = \{(x,y,z) \in \mathbb{R}^3 : x = 0, |y| < 1 OR |y| > 2 \}$

which corresponds of the slits sitting in the $yz$-plane, being infinitely extended in the $z$ direction and extending from $y=-2$ to $y = -1$ and from $y = 1$ to $y = 2$.

Similarily, you may use the path integral with action functional

$S[x(t)] = \int \dot{x}^2(t) d t$

note however that here we do not integrate over all paths, but only over maps $x:\mathbb{R} \mapsto \mathbb{R}^3 - D$; that is, we consider only those paths that go through the slits.

This is however not too helpful, as both descriptions are not very easy to solve. The Schroedinger equation shouldn't be too hard to solve on a computer, and the path integral allows for some semiclassics, which i think is done in Feynman's book on path integration.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.