Alright, so I was wondering if anyone could help me figure out from one step to the next...(adsbygoogle = window.adsbygoogle || []).push({});

So we have defined |qt>=exp(iHt/[itex]\hbar[/itex])|q>

and we divide some interval up into pieces of duration τ

Then we consider

<[itex]q_{j+1}[/itex][itex]t_{j+1}[/itex]|[itex]q_{j}[/itex][itex]t_{j}[/itex]>

=<[itex]q_{j+1}[/itex]|e^{-iHτ/[itex]\hbar[/itex]}|[itex]q_{j}[/itex]>

=<[itex]q_{j+1}[/itex]|1-(i/[itex]\hbar[/itex])Hτ+O(τ^{2})|[itex]q_{j}[/itex]>

=[itex]\delta[/itex]([itex]q_{j+1}[/itex]-[itex]q_{j}[/itex])-(iτ/[itex]\hbar[/itex])<[itex]q_{j+1}[/itex]|H|[itex]q_{j}[/itex]>

Dumb question:

Where did the delta function come from? I know where the second term comes from and I'm assuming that the higher order terms are being tossed since the τ^{2}and higher order terms have been deemed sufficiently small...but where does this delta function come from?

Is this something like <[itex]q_{j+1}[/itex]|1|[itex]q_{j}[/itex]> = 1 iff [itex]q_{j+1}[/itex]=[itex]q_{j}[/itex], or something having to do with independence? But...then I don't see why after this we have:

(2[itex]\pi[/itex][itex]\hbar[/itex])^{-1}[itex]\int[/itex]dp e^{(i/[itex]\hbar[/itex])p([itex]q_{j+1}[/itex]-[itex]q_{j}[/itex]})

coming from the delta function term .-. what?

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# Dirac Notation in building Path Integrals

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