Propogation Amplitude

1. Oct 4, 2005

ghotra

Consider the amplitude for a free particle to propogate from $\vec{x}_0$ to $$\vec{x}$$:

$$U(t) = \left\langle \vec{x} | e^{-i H t} | \vec{x}_0 \right\rangle$$

I guess I don't understand what it means for a particle to propogate from one "position" to the next. If the particle is not in a position eigenstate, then it does not have a well-defined position. If so, then how can we even talk about propogating from one position to the next.

Bottom line: $| \vec{x}_0 \rangle$ is a wave and does not have a well-defined position. So how can we talk about positions?

I am seeking clarification on this.

2. Oct 4, 2005

DavidK

To begin with, $|\vec{x}_0\rangle$ normally denote an eigenstate of the position operator. Hence, it has a well defined position.

The state $e^{-iHt} |\vec{x}_0\rangle$ is in general not an eigenstate of the position operator, and that is why we can only talk about transition probabilities for the particle, i.e. we can only know the probability of measuring the particle in position $\vec{x}$.

3. Oct 4, 2005

vanesch

Staff Emeritus
You are right of course: if the particle was not in a position eigenstate, then this is not the answer. It was the answer to the question: IF THE PARTICLE WERE IN A POSITION EIGENSTATE then what's the amplitude for it to be in ANOTHER POSITION EIGENSTATE after some time t has elapsed.

So you can now say: big deal, I wasn't in an eigenstate in the first place, so what does this help me ? The answer is: the superposition principle !
EVERY state can be written as a superposition of position eigenstates. For instance, if psi(x) is the "wave function" then the state is:

$$|\psi \rangle = \int \psi(x) |x\rangle dx$$

And applying the time evolution operator to this one, we can "shift it through the integral" to make it act only on |x>. So with the knowledge of $$\langle x| U(t,t_0) | x_0 \rangle$$ we have in fact solved all quantum mechanical questions we could ask about the system. It's sometimes called a "Green's function" of the problem.

cheers,
Patrick.