Dirac Notation in building Path Integrals

[Elwin.Martin]

Alright, so I was wondering if anyone could help me figure out from one step to the next...
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(τ²)|[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 τ² 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?

 

[Bill_K]

Your last question... because that's what the Fourier transform of a delta function is:

δ(x) ≡ (1/2π) ∫e^ixy dy

and put x = q_j+1 - q_j, y = p/ħ

 

[Morgoth]

your 1st Q;
it is the normalization of q_i, q_j