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Can the path integral be derived in the following way?

$$ \left< \psi \right| \hat{U} \left| \psi \right>=\left< \psi \right| e^{-i t (\hat{T}+\hat{V})/\hbar} \left| \psi \right> $$

Since $\hat{T}$ and $\hat{V}$ do not commute we use the Trotter product formula:

$$ \left< \psi \right| \hat{U} \left| \psi \right> = \lim _{N \to \infty} \left< \psi \right| \left( e^{-i t \hat{T}/(N\hbar)} e^{-i t \hat{V}/(N\hbar)} \right)^N \left| \psi \right> $$

I can then inject the normalization condition $1=\int_{-\infty}^\infty \left| \psi \right> \left< \psi \right| dx$:

$$ \left< \psi \right| \hat{U} \left| \psi \right> = \lim _{N \to \infty} \left< \psi \right| \left( e^{-i t \hat{T}/(N\hbar)} e^{-i t \hat{V}/(N\hbar)} \int_{-\infty}^\infty \left| \psi \right> \left< \psi \right| dx \right)^N \left| \psi \right> $$

Question :

Is the role of $1=\int_{-\infty}^\infty \left| \psi \right> \left< \psi \right| dx$ to "simulate a measurement" of $\psi$ over all steps in time (so that the particle is localized as it travels) and the tradeoff is that the expression contains infinitely many such localized paths. Is this the origin of the interpretation the path integral as a "sum over classical paths".

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  • $\begingroup$ Due diligence? $\endgroup$ – Cosmas Zachos Mar 7 at 18:44
  • $\begingroup$ Are you completely at peace with the WP procedure? The you should delete your question. $\endgroup$ – Cosmas Zachos Mar 9 at 18:43

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