Past exam of the theoretical physics II course of the University of Cambridge / 2012 / Paper 1 / 5 / c

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- Solution c

Solution

The Fresnel convolution of free-particle time slices follows by completing the square:

\frac{( x _{k} - x _{0} ) ^{2}}{k} + (x_{k + 1} - x_{k})^{2} = \frac{k + 1}{k} (x_{k} - \frac{x _{0} + k x _{k + 1}}{k + 1})^{2} + \frac{( x _{k + 1} - x _{0} ) ^{2}}{k + 1} .

The second term is independent of

x_{k}

. The shifted Fresnel integral in the first term gives

\frac{2 πi ℏ ϵ}{m} \frac{k}{k + 1} exp [\frac{im ( x _{k + 1} - x _{0} ) ^{2}}{2ℏ ϵ ( k + 1 )}] .

Here

k

is the positive slice index, not a wave number. The phase is fixed by the same convergence prescription as the short-time kernel.

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