Where did this term in the path integral come from?

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SUMMARY

The discussion centers on the derivation of path integrals, specifically the propagator defined as $$K_{q_{j+1},q_j}=\langle q_{j+1}|e^{-iH\delta t}|q_j\rangle$$. The confusion arises from the expansion of the propagator, particularly regarding the term $$-\frac{1}{2} H² \delta t²$$ and its classification as $$o(\delta t²)$$. This notation indicates that the term is of a smaller order than $$\delta t²$$ and is thus negligible for small $$\delta t$$. The key takeaway is the understanding of asymptotic notation in the context of perturbative expansions in quantum mechanics.

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PORFIRIO I
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I'm trying to derive the path integral, but got stuck on this part.
I'm trying to derive the path integrals, but this step got me confused:

Consider the propagator

$$K_{q_{j+1},q_j}=\langle q_{j+1}|e^{-iH\delta t}|q_j\rangle $$

Knowing that ##\delta t## is small, we can expand it as

$$K_{q_{j+1},q_j}=\langle q_{j+1}|(1-iH\delta t-\frac 1 2 H² \delta t²+...)|q_j\rangle = \langle q_{j+1}|q_j\rangle -i\delta t\langle q_{j+1}|H|q_j\rangle +o(\delta t²)$$

I don't understand why the third term in the expansion ##-\frac 1 2 H²\delta t²## became this ##o(\delta t²)## and why it isn't inside the brackets like the other terms. what am I missing? Thanks in advance.
 
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That means a term of the order of ##(\delta t)^2##, which for small ##\delta t## is taken to be negligible.
 
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