Propagator operator

1. Jan 21, 2012

jewbinson

So in the attachment, in fact (6), the formula for the propagator rectangled in red...

is the Hamiltonian ACTING on (t-t')?

is the Hamiltonian a function of (t-t')?

or should it be (this is what I think), to be more clear

U(t,t') = exp((-i/h)(t-t')H), so that when acting on a state |psi>, we have

U(t,t')|psi> = exp((-i/h)(t-t')H|psi>)

where H operates on whatever comes next. The last one makes most sense to me because U is overall an operator, so the RHS should also be an operator.

Unless it means like this:

U(t,t')|psi> = exp((-i/h)H[(t-t')|psi>]) ?

I'm really not sure which one...

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Last edited: Jan 21, 2012
2. Jan 22, 2012

xepma

It is just the Hamiltonian multiplied by (t-t'), so it is what you think it is. Here, you don't have to be careful with the ordering of the operator and the time-dependence, because the Hamiltonian never operates on time. Time and the Hamiltonian always commute.

Also keep in mind that this derivation applies only to time-independent Hamiltonians.