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For free particle with $V=0$ case we get $$<x_n,t_n;x_{n-1},t_{n-1}>=\frac{1}{w(\Delta t)}\exp \left [\frac{im(x_n-x_{n-1})^2}{2\hbar\Delta t}\right ] \tag{6.42}$$ given in eqn. 6.42 of Sakurai (2nd Edition p.127).

Then given $$<x_n,t_n;x_{n-1},t_{n-1}>|_{t_n=t_{n-1}} = \delta(x_n - x_{n-1})\tag{6.43}$$ in equation 6.43.

Then say we obtain $$\frac{1}{w(\Delta t)}=\sqrt{\frac{m}{2\pi i \hbar \Delta t}}.\tag{6.44}$$

How can I find the following constant without using "Propagator concept"?

Qmechanic
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baponkar
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1 Answers1

3

  1. The Feynman fudge factor (6.44) follows from the normalization (6.43) and the heat-kernel representation (6.42) of the Dirac delta distribution (after a Wick rotation).

  2. Alternatively, the Feynman fudge factor (6.44) can be derived from the Hamiltonian phase space path integral, cf. e.g. section V in my Phys.SE answer here.

Qmechanic
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