Propagator Operator: Clarifying H Acting & Function of (t-t')

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SUMMARY

The discussion clarifies the role of the Hamiltonian in the propagator operator U(t,t') as defined by the equation U(t,t') = exp((-i/h)(t-t')H). The Hamiltonian H acts on the state |psi> without affecting the time variable (t-t'), confirming that time and the Hamiltonian commute. This derivation is specifically applicable to time-independent Hamiltonians, emphasizing the importance of understanding operator ordering in quantum mechanics.

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jewbinson
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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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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.
 

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