# The propagator

1. Nov 23, 2008

### sumeetkd

This might turn out to be a stupid question but ...i couldnt understand

The propagater is defined as exp[-iHt/(h/2pi)]...this would be the matrix
so U(t)|$$\psi$$(x,0)> (matrix . vector) would give |$$\psi$$(x,t)>

What i couldn't understand is that the propagator is a diagonal matrix..but it is obvious that
probability of a particle at x,t will be affected by the entire wave distribution.

2. Nov 23, 2008

### CompuChip

First of all, why do you say that the propagator is a diagonal matrix? In general, in "real" space it is not (while in momentum space, it is, as momentum states have a well-defined energy hence are eigenstates of the Hamiltonian).
Secondly, note that x is just some label here, the dynamics is in the time variable.
Finally, if you consider a string with a wave propagating in it, you will see that it is perfectly possible to describe the motion of each point on the string without referring to other points on it, for example u(t) = cos(t/f - kx), for fixed x.

3. Nov 24, 2008

### sumeetkd

I am a total beginner so there are some leaps in my logic.

In real basis the propagator turns out to be sigma over n of (h^2/2m) d^2n/dx^2n for the diagonal terms rest terms being zero. hence the doubt.

The thing about the string was ok for the classical case, but in the quantum case
the probability at x at a later time would be integral( <x|U|x'><x'|psi(x,0)>).
That is there is a contribution from each point in space to the probability distribution that would exist at (x,t).

Hence i felt there should be off diagonal terms.

The problem is also that if you take it in the p basis and then take x components you get non zero terms but in the real basis i was not able to understand how the off diagonal terms are non zero.